Integrand size = 36, antiderivative size = 393 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((29+i) A+(1+76 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {((28-30 i) A+(75+77 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+29 i) A-(76+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^3 d}+\frac {15 (2 i A-5 B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {7 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{24 a^3 d}+\frac {(i A-B) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{4 a d (a+i a \tan (c+d x))^2}-\frac {3 (2 i A-5 B) \tan ^{\frac {5}{2}}(c+d x)}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
(-1/32-1/32*I)*((29+I)*A+(1+76*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a ^3/d*2^(1/2)-(1/32+1/32*I)*((29+I)*A+(1+76*I)*B)*arctan(1+2^(1/2)*tan(d*x+ c)^(1/2))/a^3/d*2^(1/2)-1/64*((28-30*I)*A+(75+77*I)*B)*ln(1-2^(1/2)*tan(d* x+c)^(1/2)+tan(d*x+c))/a^3/d*2^(1/2)-(1/64+1/64*I)*((1+29*I)*A-(76+I)*B)*l n(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a^3/d*2^(1/2)+15/8*(2*I*A-5*B)*ta n(d*x+c)^(1/2)/a^3/d+7/24*(4*A+11*I*B)*tan(d*x+c)^(3/2)/a^3/d+1/6*(I*A-B)* tan(d*x+c)^(9/2)/d/(a+I*a*tan(d*x+c))^3+1/4*(A+2*I*B)*tan(d*x+c)^(7/2)/a/d /(a+I*a*tan(d*x+c))^2-3/8*(2*I*A-5*B)*tan(d*x+c)^(5/2)/d/(a^3+I*a^3*tan(d* x+c))
Time = 6.17 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {-3 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-3 \sqrt [4]{-1} (29 A+76 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+\sqrt {\tan (c+d x)} \left (-45 (2 A+5 i B)+(-242 i A+598 B) \tan (c+d x)+3 (68 A+163 i B) \tan ^2(c+d x)+48 i (A+2 i B) \tan ^3(c+d x)+16 i B \tan ^4(c+d x)\right )}{24 a^3 d (-i+\tan (c+d x))^3} \]
(-3*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x ]^3*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) - 3*(-1)^(1/4)*(29*A + (76*I)* B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^3*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) + Sqrt[Tan[c + d*x]]*(-45*(2*A + (5*I)*B) + ((-242* I)*A + 598*B)*Tan[c + d*x] + 3*(68*A + (163*I)*B)*Tan[c + d*x]^2 + (48*I)* (A + (2*I)*B)*Tan[c + d*x]^3 + (16*I)*B*Tan[c + d*x]^4))/(24*a^3*d*(-I + T an[c + d*x])^3)
Time = 1.67 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.95, number of steps used = 25, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4078, 27, 3042, 4078, 27, 3042, 4078, 3042, 4011, 3042, 4011, 3042, 4017, 25, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{9/2} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {3 \tan ^{\frac {7}{2}}(c+d x) (3 a (i A-B)+a (A+5 i B) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^2}dx}{6 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {7}{2}}(c+d x) (3 a (i A-B)+a (A+5 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x)^{7/2} (3 a (i A-B)+a (A+5 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{4 a^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {-\frac {\int -\frac {2 \tan ^{\frac {5}{2}}(c+d x) \left (7 a^2 (A+2 i B)-a^2 (5 i A-16 B) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) \left (7 a^2 (A+2 i B)-a^2 (5 i A-16 B) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\int \frac {\tan (c+d x)^{5/2} \left (7 a^2 (A+2 i B)-a^2 (5 i A-16 B) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) \left (15 (2 i A-5 B) a^3+7 (4 A+11 i B) \tan (c+d x) a^3\right )dx}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \tan (c+d x)^{3/2} \left (15 (2 i A-5 B) a^3+7 (4 A+11 i B) \tan (c+d x) a^3\right )dx}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \sqrt {\tan (c+d x)} \left (15 a^3 (2 i A-5 B) \tan (c+d x)-7 a^3 (4 A+11 i B)\right )dx+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \sqrt {\tan (c+d x)} \left (15 a^3 (2 i A-5 B) \tan (c+d x)-7 a^3 (4 A+11 i B)\right )dx+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \frac {-15 (2 i A-5 B) a^3-7 (4 A+11 i B) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)}}dx+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \frac {-15 (2 i A-5 B) a^3-7 (4 A+11 i B) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)}}dx+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\frac {2 \int -\frac {a^3 (15 (2 i A-5 B)+7 (4 A+11 i B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 \int \frac {a^3 (15 (2 i A-5 B)+7 (4 A+11 i B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \int \frac {15 (2 i A-5 B)+7 (4 A+11 i B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {9}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {3 a^2 (-5 B+2 i A) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((29+i) A+(1+76 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) ((1+29 i) A-(76+i) B) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}+\frac {14 a^3 (4 A+11 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {30 a^3 (-5 B+2 i A) \sqrt {\tan (c+d x)}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+2 i B) \tan ^{\frac {7}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{4 a^2}\) |
((I*A - B)*Tan[c + d*x]^(9/2))/(6*d*(a + I*a*Tan[c + d*x])^3) - (-((a*(A + (2*I)*B)*Tan[c + d*x]^(7/2))/(d*(a + I*a*Tan[c + d*x])^2)) + ((3*a^2*((2* I)*A - 5*B)*Tan[c + d*x]^(5/2))/(d*(a + I*a*Tan[c + d*x])) - ((-2*a^3*((1/ 2 + I/2)*((29 + I)*A + (1 + 76*I)*B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d* x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + (1/2 + I /2)*((1 + 29*I)*A - (76 + I)*B)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x] ]/(2*Sqrt[2]))))/d + (30*a^3*((2*I)*A - 5*B)*Sqrt[Tan[c + d*x]])/d + (14*a ^3*(4*A + (11*I)*B)*Tan[c + d*x]^(3/2))/(3*d))/(2*a^2))/(2*a^2))/(4*a^2)
3.2.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Time = 0.07 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.52
| method | result | size |
| derivativedivides | \(\frac {\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-6 B \left (\sqrt {\tan }\left (d x +c \right )\right )+2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (\frac {-5 i \left (7 i B +4 A \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {182 i B}{3}-\frac {98 A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (14 i A -27 B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {2 \left (29 i A -76 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {A}{16}-\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(204\) |
| default | \(\frac {\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-6 B \left (\sqrt {\tan }\left (d x +c \right )\right )+2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {i \left (\frac {-5 i \left (7 i B +4 A \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {182 i B}{3}-\frac {98 A}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (14 i A -27 B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {2 \left (29 i A -76 B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {A}{16}-\frac {i B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\) | \(204\) |
1/d/a^3*(2/3*I*B*tan(d*x+c)^(3/2)-6*B*tan(d*x+c)^(1/2)+2*I*A*tan(d*x+c)^(1 /2)+1/8*I*((-5*I*(7*I*B+4*A)*tan(d*x+c)^(5/2)+(-182/3*I*B-98/3*A)*tan(d*x+ c)^(3/2)+(-27*B+14*I*A)*tan(d*x+c)^(1/2))/(tan(d*x+c)-I)^3+2*(-76*B+29*I*A )/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2))))+4*(1 /16*A-1/16*I*B)/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2 ^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 779 vs. \(2 (294) = 588\).
Time = 0.30 (sec) , antiderivative size = 779, normalized size of antiderivative = 1.98 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \log \left (\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} \log \left (\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} + 29 \, A + 76 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 3 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} \log \left (-\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-841 i \, A^{2} + 4408 \, A B + 5776 i \, B^{2}}{a^{6} d^{2}}} - 29 \, A - 76 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 2 \, {\left (2 \, {\left (-73 i \, A + 174 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - {\left (187 i \, A - 492 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (-11 i \, A + 23 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (-7 i \, A + 10 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]
-1/96*(3*(a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((I*A ^2 + 2*A*B - I*B^2)/(a^6*d^2))*log(2*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)* sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2)) + (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 3*(a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I* c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*log(-2*((a^3*d*e^(2*I*d*x + 2* I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) *sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))* e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 3*(a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^( 6*I*d*x + 6*I*c))*sqrt((-841*I*A^2 + 4408*A*B + 5776*I*B^2)/(a^6*d^2))*log (1/8*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I )/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-841*I*A^2 + 4408*A*B + 5776*I*B^2)/(a^ 6*d^2)) + 29*A + 76*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 3*(a^3*d*e^(8*I*d *x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((-841*I*A^2 + 4408*A*B + 577 6*I*B^2)/(a^6*d^2))*log(-1/8*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I *e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-841*I*A^2 + 44 08*A*B + 5776*I*B^2)/(a^6*d^2)) - 29*A - 76*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3 *d)) + 2*(2*(-73*I*A + 174*B)*e^(8*I*d*x + 8*I*c) - (187*I*A - 492*B)*e^(6 *I*d*x + 6*I*c) + 3*(-11*I*A + 23*B)*e^(4*I*d*x + 4*I*c) - (-7*I*A + 10*B) *e^(2*I*d*x + 2*I*c) - I*A + B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2...
Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 1.02 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.53 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (29 \, A + 76 i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} - \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {60 \, A \tan \left (d x + c\right )^{\frac {5}{2}} + 105 i \, B \tan \left (d x + c\right )^{\frac {5}{2}} - 98 i \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 182 \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 42 \, A \sqrt {\tan \left (d x + c\right )} - 81 i \, B \sqrt {\tan \left (d x + c\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} - \frac {2 \, {\left (-i \, B a^{6} d^{2} \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, A a^{6} d^{2} \sqrt {\tan \left (d x + c\right )} + 9 \, B a^{6} d^{2} \sqrt {\tan \left (d x + c\right )}\right )}}{3 \, a^{9} d^{3}} \]
-(1/16*I + 1/16)*sqrt(2)*(29*A + 76*I*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt (tan(d*x + c)))/(a^3*d) - (1/16*I - 1/16)*sqrt(2)*(A - I*B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) + 1/24*(60*A*tan(d*x + c)^(5/2 ) + 105*I*B*tan(d*x + c)^(5/2) - 98*I*A*tan(d*x + c)^(3/2) + 182*B*tan(d*x + c)^(3/2) - 42*A*sqrt(tan(d*x + c)) - 81*I*B*sqrt(tan(d*x + c)))/(a^3*d* (tan(d*x + c) - I)^3) - 2/3*(-I*B*a^6*d^2*tan(d*x + c)^(3/2) - 3*I*A*a^6*d ^2*sqrt(tan(d*x + c)) + 9*B*a^6*d^2*sqrt(tan(d*x + c)))/(a^9*d^3)
Time = 11.60 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{29\,A}\right )\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}}{19\,B}\right )\,\sqrt {\frac {B^2\,361{}\mathrm {i}}{16\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {49\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,7{}\mathrm {i}}{4\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,5{}\mathrm {i}}{2\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {\frac {27\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {35\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,91{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a^3\,d}-\frac {6\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d} \]
atan((a^3*d*tan(c + d*x)^(1/2)*((A^2*1i)/(256*a^6*d^2))^(1/2)*16i)/A)*((A^ 2*1i)/(256*a^6*d^2))^(1/2)*2i - atan((a^3*d*tan(c + d*x)^(1/2)*(-(A^2*841i )/(256*a^6*d^2))^(1/2)*16i)/(29*A))*(-(A^2*841i)/(256*a^6*d^2))^(1/2)*2i - atan((16*a^3*d*tan(c + d*x)^(1/2)*(-(B^2*1i)/(256*a^6*d^2))^(1/2))/B)*(-( B^2*1i)/(256*a^6*d^2))^(1/2)*2i - atan((4*a^3*d*tan(c + d*x)^(1/2)*((B^2*3 61i)/(16*a^6*d^2))^(1/2))/(19*B))*((B^2*361i)/(16*a^6*d^2))^(1/2)*2i - ((4 9*A*tan(c + d*x)^(3/2))/(12*a^3*d) - (A*tan(c + d*x)^(1/2)*7i)/(4*a^3*d) + (A*tan(c + d*x)^(5/2)*5i)/(2*a^3*d))/(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1) - ((27*B*tan(c + d*x)^(1/2))/(8*a^3*d) + (B*tan(c + d*x)^(3/2)*91i)/(12*a^3*d) - (35*B*tan(c + d*x)^(5/2))/(8*a^3*d))/(tan( c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1) + (A*tan(c + d*x)^ (1/2)*2i)/(a^3*d) - (6*B*tan(c + d*x)^(1/2))/(a^3*d) + (B*tan(c + d*x)^(3/ 2)*2i)/(3*a^3*d)