Integrand size = 28, antiderivative size = 370 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{4}+\frac {15 i}{8}\right ) d^{9/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^3 f}-\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {\left (\frac {7}{8}-\frac {15 i}{16}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^3 f}+\frac {15 i d^4 \sqrt {d \tan (e+f x)}}{4 a^3 f}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}+\frac {5 i d^2 (d \tan (e+f x))^{5/2}}{12 a f (a+i a \tan (e+f x))^2}+\frac {7 d^3 (d \tan (e+f x))^{3/2}}{6 f \left (a^3+i a^3 \tan (e+f x)\right )} \]
(7/8+15/16*I)*d^(9/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f *2^(1/2)-(7/8+15/16*I)*d^(9/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/ 2))/a^3/f*2^(1/2)+(-7/16+15/32*I)*d^(9/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e) )^(1/2)+d^(1/2)*tan(f*x+e))/a^3/f*2^(1/2)+(7/16-15/32*I)*d^(9/2)*ln(d^(1/2 )+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^3/f*2^(1/2)+15/4*I*d^ 4*(d*tan(f*x+e))^(1/2)/a^3/f-1/6*d*(d*tan(f*x+e))^(7/2)/f/(a+I*a*tan(f*x+e ))^3+5/12*I*d^2*(d*tan(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x+e))^2+7/6*d^3*(d*t an(f*x+e))^(3/2)/f/(a^3+I*a^3*tan(f*x+e))
Time = 1.84 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.55 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {d^4 \sec ^3(e+f x) \left (6 \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+174 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))+(33 \cos (e+f x)+147 \cos (3 (e+f x))+49 i \sin (e+f x)+145 i \sin (3 (e+f x))) \sqrt {d \tan (e+f x)}\right )}{48 a^3 f (-i+\tan (e+f x))^3} \]
-1/48*(d^4*Sec[e + f*x]^3*(6*(-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d* Tan[e + f*x]])/Sqrt[d]]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) + 174*(-1) ^(1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*(Cos[3*( e + f*x)] + I*Sin[3*(e + f*x)]) + (33*Cos[e + f*x] + 147*Cos[3*(e + f*x)] + (49*I)*Sin[e + f*x] + (145*I)*Sin[3*(e + f*x)])*Sqrt[d*Tan[e + f*x]]))/( a^3*f*(-I + Tan[e + f*x])^3)
Time = 1.39 (sec) , antiderivative size = 358, normalized size of antiderivative = 0.97, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4011, 3042, 4017, 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.
\(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3}dx\) |
\(\Big \downarrow \) 4041 |
\(\displaystyle -\frac {\int -\frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-13 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)^2}dx}{6 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-13 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-13 i a d^2 \tan (e+f x)\right )}{(i \tan (e+f x) a+a)^2}dx}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {2 (d \tan (e+f x))^{3/2} \left (25 i a^2 d^3+31 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{4 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (25 i a^2 d^3+31 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\int \frac {(d \tan (e+f x))^{3/2} \left (25 i a^2 d^3+31 a^2 \tan (e+f x) d^3\right )}{i \tan (e+f x) a+a}dx}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {-\frac {\int -6 \sqrt {d \tan (e+f x)} \left (14 a^3 d^4-15 i a^3 d^4 \tan (e+f x)\right )dx}{2 a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \int \sqrt {d \tan (e+f x)} \left (14 a^3 d^4-15 i a^3 d^4 \tan (e+f x)\right )dx}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \int \sqrt {d \tan (e+f x)} \left (14 a^3 d^4-15 i a^3 d^4 \tan (e+f x)\right )dx}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\int \frac {15 i a^3 d^5+14 a^3 \tan (e+f x) d^5}{\sqrt {d \tan (e+f x)}}dx-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\int \frac {15 i a^3 d^5+14 a^3 \tan (e+f x) d^5}{\sqrt {d \tan (e+f x)}}dx-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 4017 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 \int \frac {a^3 d^5 (14 \tan (e+f x) d+15 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \int \frac {14 \tan (e+f x) d+15 i d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 1482 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\frac {5 i a d^2 (d \tan (e+f x))^{5/2}}{f (a+i a \tan (e+f x))^2}-\frac {\frac {3 \left (\frac {2 a^3 d^5 \left (\left (7+\frac {15 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (7-\frac {15 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}-\frac {30 i a^3 d^4 \sqrt {d \tan (e+f x)}}{f}\right )}{a^2}-\frac {28 a^2 d^3 (d \tan (e+f x))^{3/2}}{f (a+i a \tan (e+f x))}}{2 a^2}}{12 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{6 f (a+i a \tan (e+f x))^3}\) |
-1/6*(d*(d*Tan[e + f*x])^(7/2))/(f*(a + I*a*Tan[e + f*x])^3) + (((5*I)*a*d ^2*(d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^2) - ((-28*a^2*d^3*(d *Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])) + (3*((2*a^3*d^5*((7 + (1 5*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqr t[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[ d])) - (7 - (15*I)/2)*(-1/2*Log[d + d*Tan[e + f*x] - Sqrt[2]*Sqrt[d]*Sqrt[ d*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[2]*Sqrt [d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]))))/f - ((30*I)*a^3*d^4*Sqrt[ d*Tan[e + f*x]])/f))/a^2)/(2*a^2))/(12*a^2)
3.2.79.3.1 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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(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 - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[m] || IntegersQ[2*m, 2*n])
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 = 0.86 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.38
method | result | size |
derivativedivides | \(\frac {2 d^{4} \left (i \sqrt {d \tan \left (f x +e \right )}+\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}-\frac {d \left (\frac {-20 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+14 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{16}\right )}{f \,a^{3}}\) | \(141\) |
default | \(\frac {2 d^{4} \left (i \sqrt {d \tan \left (f x +e \right )}+\frac {d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{16 \sqrt {i d}}-\frac {d \left (\frac {-20 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}+\frac {98 i d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+14 d^{2} \sqrt {d \tan \left (f x +e \right )}}{\left (d \tan \left (f x +e \right )-i d \right )^{3}}+\frac {29 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}\right )}{16}\right )}{f \,a^{3}}\) | \(141\) |
2/f/a^3*d^4*(I*(d*tan(f*x+e))^(1/2)+1/16*d/(I*d)^(1/2)*arctan((d*tan(f*x+e ))^(1/2)/(I*d)^(1/2))-1/16*d*((-20*(d*tan(f*x+e))^(5/2)+98/3*I*d*(d*tan(f* x+e))^(3/2)+14*d^2*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)-I*d)^3+29/(-I*d)^(1 /2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 592 vs. \(2 (276) = 552\).
Time = 0.27 (sec) , antiderivative size = 592, normalized size of antiderivative = 1.60 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {{\left (12 \, a^{3} \sqrt {\frac {i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} f\right )} \sqrt {\frac {i \, d^{9}}{64 \, a^{6} f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{4}}\right ) - 12 \, a^{3} \sqrt {\frac {i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} f\right )} \sqrt {\frac {i \, d^{9}}{64 \, a^{6} f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{4}}\right ) - 12 \, a^{3} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (29 \, d^{5} + 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) + 12 \, a^{3} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} f e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (29 \, d^{5} - 8 \, {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {841 i \, d^{9}}{64 \, a^{6} f^{2}}} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{3} f}\right ) - {\left (146 i \, d^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 41 i \, d^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 8 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d^{4}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \]
-1/48*(12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(-2*(I*d ^5*e^(2*I*f*x + 2*I*e) + 8*(I*a^3*f*e^(2*I*f*x + 2*I*e) + I*a^3*f)*sqrt(1/ 64*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2* I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^4) - 12*a^3*sqrt(1/64*I*d^9/(a^6*f^2))* f*e^(6*I*f*x + 6*I*e)*log(-2*(I*d^5*e^(2*I*f*x + 2*I*e) + 8*(-I*a^3*f*e^(2 *I*f*x + 2*I*e) - I*a^3*f)*sqrt(1/64*I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f* x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^4) - 12*a^3*sqrt(-841/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6*I*e)*log(1/8*(29*d^5 + 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-841/64*I*d^9/(a^6*f^2))*sqr t((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^3*f)) + 12*a^3*sqrt(-841/64*I*d^9/(a^6*f^2))*f*e^(6*I*f*x + 6 *I*e)*log(1/8*(29*d^5 - 8*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-841/64 *I*d^9/(a^6*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I* e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a^3*f)) - (146*I*d^4*e^(6*I*f*x + 6*I*e) + 41*I*d^4*e^(4*I*f*x + 4*I*e) - 8*I*d^4*e^(2*I*f*x + 2*I*e) + I*d^4)*sqrt( (-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
Timed out. \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.94 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.65 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {1}{24} \, d^{4} {\left (\frac {3 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {87 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{3} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {48 i \, \sqrt {d \tan \left (f x + e\right )}}{a^{3} f} + \frac {2 \, {\left (30 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )^{2} - 49 i \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 21 \, \sqrt {d \tan \left (f x + e\right )} d^{3}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} f}\right )} \]
1/24*d^4*(3*sqrt(2)*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*s qrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^3*f*(I*d/sqrt(d^2) + 1)) - 87*sqrt(2)*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2 )*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^3*f*(-I*d/sqrt(d^2) + 1)) + 4 8*I*sqrt(d*tan(f*x + e))/(a^3*f) + 2*(30*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)^2 - 49*I*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e) - 21*sqrt(d*tan(f*x + e))*d^3)/((d*tan(f*x + e) - I*d)^3*a^3*f))
Time = 6.84 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.65 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx=\mathrm {atan}\left (\frac {a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,16{}\mathrm {i}}{d^5}\right )\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^9\,841{}\mathrm {i}}{256\,a^6\,f^2}}\,16{}\mathrm {i}}{29\,d^5}\right )\,\sqrt {-\frac {d^9\,841{}\mathrm {i}}{256\,a^6\,f^2}}\,2{}\mathrm {i}+\frac {\frac {7\,d^7\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{4\,a^3\,f}-\frac {5\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{2\,a^3\,f}+\frac {d^6\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,49{}\mathrm {i}}{12\,a^3\,f}}{-d^3\,{\mathrm {tan}\left (e+f\,x\right )}^3+d^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,d^3\,\mathrm {tan}\left (e+f\,x\right )-d^3\,1{}\mathrm {i}}+\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{a^3\,f} \]
atan((a^3*f*(d*tan(e + f*x))^(1/2)*((d^9*1i)/(256*a^6*f^2))^(1/2)*16i)/d^5 )*((d^9*1i)/(256*a^6*f^2))^(1/2)*2i - atan((a^3*f*(d*tan(e + f*x))^(1/2)*( -(d^9*841i)/(256*a^6*f^2))^(1/2)*16i)/(29*d^5))*(-(d^9*841i)/(256*a^6*f^2) )^(1/2)*2i + ((7*d^7*(d*tan(e + f*x))^(1/2))/(4*a^3*f) + (d^6*(d*tan(e + f *x))^(3/2)*49i)/(12*a^3*f) - (5*d^5*(d*tan(e + f*x))^(5/2))/(2*a^3*f))/(3* d^3*tan(e + f*x) - d^3*1i + d^3*tan(e + f*x)^2*3i - d^3*tan(e + f*x)^3) + (d^4*(d*tan(e + f*x))^(1/2)*2i)/(a^3*f)