Integrand size = 28, antiderivative size = 353 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {49}{16}+\frac {45 i}{16}\right ) d^{9/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {49}{32}-\frac {45 i}{32}\right ) d^{9/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {45 i d^4 \sqrt {d \tan (e+f x)}}{8 a^2 f}-\frac {49 d^3 (d \tan (e+f x))^{3/2}}{24 a^2 f}+\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2} \]
(-49/32-45/32*I)*d^(9/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^ 2/f*2^(1/2)+(49/32+45/32*I)*d^(9/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/ d^(1/2))/a^2/f*2^(1/2)+(49/64-45/64*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^2/f*2^(1/2)+(-49/64+45/64*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^2/f*2^(1/2)-45 /8*I*d^4*(d*tan(f*x+e))^(1/2)/a^2/f-49/24*d^3*(d*tan(f*x+e))^(3/2)/a^2/f+9 /8*I*d^2*(d*tan(f*x+e))^(5/2)/a^2/f/(1+I*tan(f*x+e))-1/4*d*(d*tan(f*x+e))^ (7/2)/f/(a+I*a*tan(f*x+e))^2
Time = 4.08 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.58 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {d^4 \left (6 (-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) \sec ^2(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+141 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) \sec ^2(e+f x) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))+\sqrt {d \tan (e+f x)} \left (-135 i+221 \tan (e+f x)+64 i \tan ^2(e+f x)+16 \tan ^3(e+f x)\right )\right )}{24 a^2 f (-i+\tan (e+f x))^2} \]
-1/24*(d^4*(6*(-1)^(3/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/ Sqrt[d]]*Sec[e + f*x]^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + 141*(-1) ^(1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*Sec[e + f*x]^2*((-I)*Cos[2*(e + f*x)] + Sin[2*(e + f*x)]) + Sqrt[d*Tan[e + f*x]]*( -135*I + 221*Tan[e + f*x] + (64*I)*Tan[e + f*x]^2 + 16*Tan[e + f*x]^3)))/( a^2*f*(-I + Tan[e + f*x])^2)
Time = 1.25 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.97, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 4041, 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.
| \(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-11 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)}dx}{4 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-11 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{5/2} \left (7 a d^2-11 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {\int (d \tan (e+f x))^{3/2} \left (45 i a^2 d^3+49 a^2 \tan (e+f x) d^3\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {\int (d \tan (e+f x))^{3/2} \left (45 i a^2 d^3+49 a^2 \tan (e+f x) d^3\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}+\int \sqrt {d \tan (e+f x)} \left (45 i a^2 d^4 \tan (e+f x)-49 a^2 d^4\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}+\int \sqrt {d \tan (e+f x)} \left (45 i a^2 d^4 \tan (e+f x)-49 a^2 d^4\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {\int \frac {-45 i a^2 d^5-49 a^2 \tan (e+f x) d^5}{\sqrt {d \tan (e+f x)}}dx+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {\int \frac {-45 i a^2 d^5-49 a^2 \tan (e+f x) d^5}{\sqrt {d \tan (e+f x)}}dx+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {\frac {2 \int -\frac {a^2 d^5 (49 \tan (e+f x) d+45 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 \int \frac {a^2 d^5 (49 \tan (e+f x) d+45 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \int \frac {49 \tan (e+f x) d+45 i d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {9 i d^2 (d \tan (e+f x))^{5/2}}{f (1+i \tan (e+f x))}-\frac {-\frac {2 a^2 d^5 \left (\left (\frac {49}{2}+\frac {45 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {49}{2}-\frac {45 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {90 i a^2 d^4 \sqrt {d \tan (e+f x)}}{f}+\frac {98 a^2 d^3 (d \tan (e+f x))^{3/2}}{3 f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{7/2}}{4 f (a+i a \tan (e+f x))^2}\) |
-1/4*(d*(d*Tan[e + f*x])^(7/2))/(f*(a + I*a*Tan[e + f*x])^2) + (((9*I)*d^2 *(d*Tan[e + f*x])^(5/2))/(f*(1 + I*Tan[e + f*x])) - ((-2*a^2*d^5*((49/2 + (45*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*S qrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqr t[d])) - (49/2 - (45*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 + ((90*I)*a^2*d^4* Sqrt[d*Tan[e + f*x]])/f + (98*a^2*d^3*(d*Tan[e + f*x])^(3/2))/(3*f))/(2*a^ 2))/(8*a^2)
3.2.71.3.1 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 = 0.76 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.40
| method | result | size |
| derivativedivides | \(\frac {2 d^{3} \left (-\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (\frac {\frac {15 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {13 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {47 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\) | \(142\) |
| default | \(\frac {2 d^{3} \left (-\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 i d \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (\frac {\frac {15 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {13 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {47 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {d^{2} \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\) | \(142\) |
2/f/a^2*d^3*(-1/3*(d*tan(f*x+e))^(3/2)-2*I*d*(d*tan(f*x+e))^(1/2)+1/8*d^2* ((15/2*(d*tan(f*x+e))^(3/2)-13/2*I*d*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*x+e) +d)^2+47/2/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))+1/8*d^2 /(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. 672 vs. \(2 (261) = 522\).
Time = 0.27 (sec) , antiderivative size = 672, normalized size of antiderivative = 1.90 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} f^{2}}} \log \left (-\frac {{\left (47 \, d^{5} + 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} 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^{2} f}\right ) - 12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} f^{2}}} \log \left (-\frac {{\left (47 \, d^{5} - 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {2209 i \, d^{9}}{64 \, a^{4} 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^{2} f}\right ) + 12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} f^{2}}} \log \left (-\frac {2 \, {\left (i \, d^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} f\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} 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 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} f^{2}}} \log \left (-\frac {2 \, {\left (i \, d^{5} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (-i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} f\right )} \sqrt {\frac {i \, d^{9}}{16 \, a^{4} 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 ) - {\left (-202 i \, d^{4} e^{\left (6 i \, f x + 6 i \, e\right )} - 305 i \, d^{4} e^{\left (4 i \, f x + 4 i \, e\right )} - 36 i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 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}}}{48 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]
-1/48*(12*(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*sqrt(-22 09/64*I*d^9/(a^4*f^2))*log(-1/8*(47*d^5 + 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a ^2*f)*sqrt(-2209/64*I*d^9/(a^4*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^2*f)) - 12*(a^2*f*e^( 6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*sqrt(-2209/64*I*d^9/(a^4*f^2 ))*log(-1/8*(47*d^5 - 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-2209/64* I*d^9/(a^4*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^2*f)) + 12*(a^2*f*e^(6*I*f*x + 6*I*e) + a ^2*f*e^(4*I*f*x + 4*I*e))*sqrt(1/16*I*d^9/(a^4*f^2))*log(-2*(I*d^5*e^(2*I* f*x + 2*I*e) + 4*(I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt(1/16*I*d^9/( a^4*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^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I *f*x + 4*I*e))*sqrt(1/16*I*d^9/(a^4*f^2))*log(-2*(I*d^5*e^(2*I*f*x + 2*I*e ) + 4*(-I*a^2*f*e^(2*I*f*x + 2*I*e) - I*a^2*f)*sqrt(1/16*I*d^9/(a^4*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) - (-202*I*d^4*e^(6*I*f*x + 6*I*e) - 305*I*d^4*e^(4*I*f*x + 4*I*e) - 36*I*d^4*e^(2*I*f*x + 2*I*e) + 3*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)))/(a^2*f*e^(6*I*f*x + 6*I*e) + a ^2*f*e^(4*I*f*x + 4*I*e))
\[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {9}{2}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
Exception generated. \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.65 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.72 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {1}{24} \, d^{3} {\left (\frac {6 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {141 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {3 \, {\left (15 \, \sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right ) - 13 i \, \sqrt {d \tan \left (f x + e\right )} d^{3}\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f} - \frac {16 \, {\left (\sqrt {d \tan \left (f x + e\right )} a^{4} d f^{2} \tan \left (f x + e\right ) + 6 i \, \sqrt {d \tan \left (f x + e\right )} a^{4} d f^{2}\right )}}{a^{6} f^{3}}\right )} \]
1/24*d^3*(6*sqrt(2)*d^(3/2)*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^2*f*(I*d/sqrt(d^2) + 1)) + 141*sqrt(2)*d^(3/2)*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^2*f*(-I*d/sqrt(d^2) + 1)) - 3*(15*sqrt(d*tan(f*x + e))*d^3*tan(f*x + e) - 13*I*sqrt(d*tan(f*x + e))*d^ 3)/((d*tan(f*x + e) - I*d)^2*a^2*f) - 16*(sqrt(d*tan(f*x + e))*a^4*d*f^2*t an(f*x + e) + 6*I*sqrt(d*tan(f*x + e))*a^4*d*f^2)/(a^6*f^3))
Time = 6.06 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.64 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {-\frac {15\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8\,a^2\,f}+\frac {d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,13{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}-\frac {2\,d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,a^2\,f}+\mathrm {atan}\left (\frac {a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,8{}\mathrm {i}}{d^5}\right )\,\sqrt {\frac {d^9\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^9\,2209{}\mathrm {i}}{256\,a^4\,f^2}}\,16{}\mathrm {i}}{47\,d^5}\right )\,\sqrt {-\frac {d^9\,2209{}\mathrm {i}}{256\,a^4\,f^2}}\,2{}\mathrm {i}-\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,4{}\mathrm {i}}{a^2\,f} \]
atan((a^2*f*(d*tan(e + f*x))^(1/2)*((d^9*1i)/(64*a^4*f^2))^(1/2)*8i)/d^5)* ((d^9*1i)/(64*a^4*f^2))^(1/2)*2i - ((d^6*(d*tan(e + f*x))^(1/2)*13i)/(8*a^ 2*f) - (15*d^5*(d*tan(e + f*x))^(3/2))/(8*a^2*f))/(d^2*tan(e + f*x)*2i + d ^2 - d^2*tan(e + f*x)^2) + atan((a^2*f*(d*tan(e + f*x))^(1/2)*(-(d^9*2209i )/(256*a^4*f^2))^(1/2)*16i)/(47*d^5))*(-(d^9*2209i)/(256*a^4*f^2))^(1/2)*2 i - (d^4*(d*tan(e + f*x))^(1/2)*4i)/(a^2*f) - (2*d^3*(d*tan(e + f*x))^(3/2 ))/(3*a^2*f)