\(\int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [147]

Optimal result

Integrand size = 36, antiderivative size = 310 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {((5-7 i) A+(28+30 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+6 i) A-(29+i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((6+i) A+(1+29 i) B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:

-1/32*((5-7*I)*A+(28+30*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/ 
a^3/d+(1/32+1/32*I)*((1+6*I)*A-(29+I)*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2) 
)*2^(1/2)/a^3/d-(1/32+1/32*I)*((6+I)*A+(1+29*I)*B)*arctanh(2^(1/2)*tan(d*x 
+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/a^3/d+5/8*(A+6*I*B)*tan(d*x+c)^(1/2)/a^3 
/d+1/6*(I*A-B)*tan(d*x+c)^(7/2)/d/(a+I*a*tan(d*x+c))^3+1/12*(2*A+5*I*B)*ta 
n(d*x+c)^(5/2)/a/d/(a+I*a*tan(d*x+c))^2-7/24*(I*A-4*B)*tan(d*x+c)^(3/2)/d/ 
(a^3+I*a^3*tan(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 5.95 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (-i \left (((7+5 i) A-(30-28 i) B) \arcsin (\cos (c+d x)-\sin (c+d x))+(1-i) ((6+i) A+(1+29 i) B) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right )\right ) \sec (c+d x) (\cos (3 c)+i \sin (3 c)) \sqrt {\sin (2 (c+d x))}+\frac {2}{3} (\cos (3 d x)-i \sin (3 d x)) ((9 A+33 i B) \cos (c+d x)+21 (A+7 i B) \cos (3 (c+d x))+2 i (19 A+97 i B+(19 A+145 i B) \cos (2 (c+d x))) \sin (c+d x)) \tan (c+d x)\right )}{32 d (A \cos (c+d x)+B \sin (c+d x)) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3} \] Input:

Integrate[(Tan[c + d*x]^(7/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x]) 
^3,x]
 

Output:

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^3*(A + B*Tan[c + d*x])*((-I)*(((7 
+ 5*I)*A - (30 - 28*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]] + (1 - I)*(( 
6 + I)*A + (1 + 29*I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + 
 d*x)]]])*Sec[c + d*x]*(Cos[3*c] + I*Sin[3*c])*Sqrt[Sin[2*(c + d*x)]] + (2 
*(Cos[3*d*x] - I*Sin[3*d*x])*((9*A + (33*I)*B)*Cos[c + d*x] + 21*(A + (7*I 
)*B)*Cos[3*(c + d*x)] + (2*I)*(19*A + (97*I)*B + (19*A + (145*I)*B)*Cos[2* 
(c + d*x)])*Sin[c + d*x])*Tan[c + d*x])/3))/(32*d*(A*Cos[c + d*x] + B*Sin[ 
c + d*x])*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.09, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.639, Rules used = {3042, 4078, 27, 3042, 4078, 27, 3042, 4078, 27, 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.

\(\displaystyle \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{7/2} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) (7 a (i A-B)+a (A+13 i B) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^2}dx}{6 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) (7 a (i A-B)+a (A+13 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{12 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x)^{5/2} (7 a (i A-B)+a (A+13 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{12 a^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {-\frac {\int -\frac {2 \tan ^{\frac {3}{2}}(c+d x) \left (5 a^2 (2 A+5 i B)-a^2 (4 i A-31 B) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (5 a^2 (2 A+5 i B)-a^2 (4 i A-31 B) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\int \frac {\tan (c+d x)^{3/2} \left (5 a^2 (2 A+5 i B)-a^2 (4 i A-31 B) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int 3 \sqrt {\tan (c+d x)} \left (7 (i A-4 B) a^3+5 (A+6 i B) \tan (c+d x) a^3\right )dx}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \int \sqrt {\tan (c+d x)} \left (7 (i A-4 B) a^3+5 (A+6 i B) \tan (c+d x) a^3\right )dx}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \int \sqrt {\tan (c+d x)} \left (7 (i A-4 B) a^3+5 (A+6 i B) \tan (c+d x) a^3\right )dx}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\int \frac {7 a^3 (i A-4 B) \tan (c+d x)-5 a^3 (A+6 i B)}{\sqrt {\tan (c+d x)}}dx+\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\int \frac {7 a^3 (i A-4 B) \tan (c+d x)-5 a^3 (A+6 i B)}{\sqrt {\tan (c+d x)}}dx+\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {2 \int -\frac {a^3 (5 (A+6 i B)-7 (i A-4 B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 \int \frac {a^3 (5 (A+6 i B)-7 (i A-4 B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \int \frac {5 (A+6 i B)-7 (i A-4 B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((5-7 i) A+(28+30 i) B) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((5-7 i) A+(28+30 i) B) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((5-7 i) A+(28+30 i) B) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((5-7 i) A+(28+30 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} ((5-7 i) A+(28+30 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} ((5-7 i) A+(28+30 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} ((5-7 i) A+(28+30 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {7 a^2 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {3 \left (\frac {10 a^3 (A+6 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^3 \left (\frac {1}{2} ((5-7 i) A+(28+30 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) ((6+i) A+(1+29 i) B) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}\right )}{2 a^2}}{2 a^2}-\frac {a (2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{d (a+i a \tan (c+d x))^2}}{12 a^2}\)

Input:

Int[(Tan[c + d*x]^(7/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]
 

Output:

((I*A - B)*Tan[c + d*x]^(7/2))/(6*d*(a + I*a*Tan[c + d*x])^3) - (-((a*(2*A 
 + (5*I)*B)*Tan[c + d*x]^(5/2))/(d*(a + I*a*Tan[c + d*x])^2)) + ((-3*((-2* 
a^3*((((5 - 7*I)*A + (28 + 30*I)*B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x 
]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + (1/2 + 
 I/2)*((6 + I)*A + (1 + 29*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 + (10*a^3*(A + (6*I)*B)*Sqrt[Tan[c + d*x]])/d))/(2*a^2 
) + (7*a^2*(I*A - 4*B)*Tan[c + d*x]^(3/2))/(d*(a + I*a*Tan[c + d*x])))/(2* 
a^2))/(12*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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])
 

rule 1082

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]
 

rule 1103

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]
 

rule 1476

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]
 

rule 1479

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]
 

rule 1482

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011

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]
 

rule 4017

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]
 

rule 4078

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]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(\frac {2 i B \sqrt {\tan \left (d x +c \right )}-\frac {i \left (\frac {-i \left (9 i A -20 B \right ) \tan \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {38 i A}{3}+\frac {98 B}{3}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}+\left (-14 i B -5 A \right ) \sqrt {\tan \left (d x +c \right )}}{\left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {2 \left (29 i B +6 A \right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {i A}{16}+\frac {B}{16}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(181\)
default	\(\frac {2 i B \sqrt {\tan \left (d x +c \right )}-\frac {i \left (\frac {-i \left (9 i A -20 B \right ) \tan \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {38 i A}{3}+\frac {98 B}{3}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}+\left (-14 i B -5 A \right ) \sqrt {\tan \left (d x +c \right )}}{\left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {2 \left (29 i B +6 A \right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {i A}{16}+\frac {B}{16}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(181\)

Input:

int(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x,method=_RETUR 
NVERBOSE)
 

Output:

1/d/a^3*(2*I*B*tan(d*x+c)^(1/2)-1/8*I*((-I*(9*I*A-20*B)*tan(d*x+c)^(5/2)+( 
-38/3*I*A+98/3*B)*tan(d*x+c)^(3/2)+(-5*A-14*I*B)*tan(d*x+c)^(1/2))/(-I+tan 
(d*x+c))^3-2*(29*I*B+6*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*I*A+1/16*B)/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d 
*x+c)^(1/2)/(2^(1/2)+I*2^(1/2))))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 685 vs. \(2 (233) = 466\).

Time = 0.12 (sec) , antiderivative size = 685, normalized size of antiderivative = 2.21 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algori 
thm="fricas")
 

Output:

-1/96*(3*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c 
)*log(-2*((I*a^3*d*e^(2*I*d*x + 2*I*c) + I*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*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*( 
(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*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*sqrt 
((36*I*A^2 - 348*A*B - 841*I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*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((36*I*A^2 - 348*A*B - 841*I*B^2)/(a^6*d^2)) + 
 6*I*A - 29*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 3*a^3*d*sqrt((36*I*A^2 - 34 
8*A*B - 841*I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*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((36*I*A^2 - 348*A*B - 841*I*B^2)/(a^6*d^2)) - 6*I*A + 29*B)*e 
^(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*(2*(10*A + 73*I*B)*e^(6*I*d*x + 6*I*c) + 
(14*A + 41*I*B)*e^(4*I*d*x + 4*I*c) - (5*A + 8*I*B)*e^(2*I*d*x + 2*I*c) + 
A + I*B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^( 
-6*I*d*x - 6*I*c)/(a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algori 
thm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.49 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {3 \, \sqrt {2} {\left (-\left (6 i - 6\right ) \, A + \left (29 i + 29\right ) \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + 3 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - 96 i \, B \sqrt {\tan \left (d x + c\right )} - \frac {2 \, {\left (-27 i \, A \tan \left (d x + c\right )^{\frac {5}{2}} + 60 \, B \tan \left (d x + c\right )^{\frac {5}{2}} - 38 \, A \tan \left (d x + c\right )^{\frac {3}{2}} - 98 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} + 15 i \, A \sqrt {\tan \left (d x + c\right )} - 42 \, B \sqrt {\tan \left (d x + c\right )}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{48 \, a^{3} d} \] Input:

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algori 
thm="giac")
 

Output:

-1/48*(3*sqrt(2)*(-(6*I - 6)*A + (29*I + 29)*B)*arctan((1/2*I + 1/2)*sqrt( 
2)*sqrt(tan(d*x + c))) + 3*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(-(1/2*I 
 - 1/2)*sqrt(2)*sqrt(tan(d*x + c))) - 96*I*B*sqrt(tan(d*x + c)) - 2*(-27*I 
*A*tan(d*x + c)^(5/2) + 60*B*tan(d*x + c)^(5/2) - 38*A*tan(d*x + c)^(3/2) 
- 98*I*B*tan(d*x + c)^(3/2) + 15*I*A*sqrt(tan(d*x + c)) - 42*B*sqrt(tan(d* 
x + c)))/(tan(d*x + c) - I)^3)/(a^3*d)
 

Mupad [B] (verification not implemented)

Time = 4.61 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.27 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,A}\right )\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{29\,B}\right )\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}+\frac {\frac {5\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {9\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {\frac {49\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,7{}\mathrm {i}}{4\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,5{}\mathrm {i}}{2\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a^3\,d} \] Input:

int((tan(c + d*x)^(7/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)
 

Output:

atan((8*a^3*d*tan(c + d*x)^(1/2)*((A^2*9i)/(64*a^6*d^2))^(1/2))/(3*A))*((A 
^2*9i)/(64*a^6*d^2))^(1/2)*2i + atan((16*a^3*d*tan(c + d*x)^(1/2)*(-(A^2*1 
i)/(256*a^6*d^2))^(1/2))/A)*(-(A^2*1i)/(256*a^6*d^2))^(1/2)*2i + atan((a^3 
*d*tan(c + d*x)^(1/2)*((B^2*1i)/(256*a^6*d^2))^(1/2)*16i)/B)*((B^2*1i)/(25 
6*a^6*d^2))^(1/2)*2i - atan((a^3*d*tan(c + d*x)^(1/2)*(-(B^2*841i)/(256*a^ 
6*d^2))^(1/2)*16i)/(29*B))*(-(B^2*841i)/(256*a^6*d^2))^(1/2)*2i + ((5*A*ta 
n(c + d*x)^(1/2))/(8*a^3*d) + (A*tan(c + d*x)^(3/2)*19i)/(12*a^3*d) - (9*A 
*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) - ((49*B*tan(c + d*x)^(3/2))/(12*a^3*d) - (B*tan(c + d* 
x)^(1/2)*7i)/(4*a^3*d) + (B*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) + (B*tan(c + d*x)^(1/2)*2 
i)/(a^3*d)
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {-\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) b -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) a}{a^{3}} \] Input:

int(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)
 

Output:

( - (int((sqrt(tan(c + d*x))*tan(c + d*x)**4)/(tan(c + d*x)**3*i + 3*tan(c 
 + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*b + int((sqrt(tan(c + d*x))*tan(c + 
d*x)**3)/(tan(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x) 
*a))/a**3