Integrand size = 36, antiderivative size = 312 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(43 A+20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{5/2} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(23 A+13 i B) \cot ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(337 A+167 i B) \cot ^2(c+d x)}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {21 (2 i A-B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 a^3 d} \]
1/4*(43*A+20*I*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d-1/8* (A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^ (1/2)+1/60*(337*A+167*I*B)*cot(d*x+c)^2/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+21/ 4*(2*I*A-B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d-1/12*(85*A+41*I*B)*c ot(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+1/5*(A+I*B)*cot(d*x+c)^2/d/(a+I *a*tan(d*x+c))^(5/2)+1/30*(23*A+13*I*B)*cot(d*x+c)^2/a/d/(a+I*a*tan(d*x+c) )^(3/2)
Time = 8.70 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.96 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {12 a^{17/2} (A+i B) \cot ^2(c+d x)-\frac {1}{2} a^8 (i+\cot (c+d x)) \tan ^2(c+d x) \left (-\frac {1}{2} \sqrt {a} \csc ^3(c+d x) (-((961 A+461 i B) \cos (c+d x))+(793 A+413 i B) \cos (3 (c+d x))+18 (-57 i A+27 B+(83 i A-43 B) \cos (2 (c+d x))) \sin (c+d x))-30 (43 A+20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right ) (i+\cot (c+d x)) \sqrt {a+i a \tan (c+d x)}+15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (i+\cot (c+d x)) \sqrt {a+i a \tan (c+d x)}\right )}{60 a^{17/2} d (a+i a \tan (c+d x))^{5/2}} \]
(12*a^(17/2)*(A + I*B)*Cot[c + d*x]^2 - (a^8*(I + Cot[c + d*x])*Tan[c + d* x]^2*(-1/2*(Sqrt[a]*Csc[c + d*x]^3*(-((961*A + (461*I)*B)*Cos[c + d*x]) + (793*A + (413*I)*B)*Cos[3*(c + d*x)] + 18*((-57*I)*A + 27*B + ((83*I)*A - 43*B)*Cos[2*(c + d*x)])*Sin[c + d*x])) - 30*(43*A + (20*I)*B)*ArcTanh[Sqrt [a + I*a*Tan[c + d*x]]/Sqrt[a]]*(I + Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d* x]] + 15*Sqrt[2]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqr t[a])]*(I + Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]]))/2)/(60*a^(17/2)*d*( a + I*a*Tan[c + d*x])^(5/2))
Time = 2.31 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.08, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 3042, 4083, 3042, 3961, 219, 4082, 73, 221}
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.
\(\displaystyle \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^3 (a+i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\int \frac {\cot ^3(c+d x) (2 a (7 A+2 i B)-9 a (i A-B) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\cot ^3(c+d x) (2 a (7 A+2 i B)-9 a (i A-B) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 a (7 A+2 i B)-9 a (i A-B) \tan (c+d x)}{\tan (c+d x)^3 (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {\int \frac {\cot ^3(c+d x) \left (4 a^2 (44 A+19 i B)-7 a^2 (23 i A-13 B) \tan (c+d x)\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\cot ^3(c+d x) \left (4 a^2 (44 A+19 i B)-7 a^2 (23 i A-13 B) \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {4 a^2 (44 A+19 i B)-7 a^2 (23 i A-13 B) \tan (c+d x)}{\tan (c+d x)^3 \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {\frac {\int \frac {5}{2} \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^3 (85 A+41 i B)-a^3 (337 i A-167 B) \tan (c+d x)\right )dx}{a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {5 \int \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^3 (85 A+41 i B)-a^3 (337 i A-167 B) \tan (c+d x)\right )dx}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (4 a^3 (85 A+41 i B)-a^3 (337 i A-167 B) \tan (c+d x)\right )}{\tan (c+d x)^3}dx}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {\frac {\frac {5 \left (\frac {\int -6 \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (42 (2 i A-B) a^4+(85 A+41 i B) \tan (c+d x) a^4\right )dx}{2 a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (42 (2 i A-B) a^4+(85 A+41 i B) \tan (c+d x) a^4\right )dx}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (42 (2 i A-B) a^4+(85 A+41 i B) \tan (c+d x) a^4\right )}{\tan (c+d x)^2}dx}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (a^5 (43 A+20 i B)-21 a^5 (2 i A-B) \tan (c+d x)\right )dx}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^5 (43 A+20 i B)-21 a^5 (2 i A-B) \tan (c+d x)\right )}{\tan (c+d x)}dx}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4083 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {a^5 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+a^4 (43 A+20 i B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {a^5 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+a^4 (43 A+20 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {a^4 (43 A+20 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {2 i a^6 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {a^4 (43 A+20 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {i \sqrt {2} a^{11/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {\frac {a^6 (43 A+20 i B) \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i \sqrt {2} a^{11/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {\frac {5 \left (-\frac {3 \left (\frac {-\frac {2 i a^5 (43 A+20 i B) \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i \sqrt {2} a^{11/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\frac {a^2 (337 A+167 i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {5 \left (-\frac {2 a^3 (85 A+41 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {3 \left (\frac {-\frac {2 a^{11/2} (43 A+20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {i \sqrt {2} a^{11/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^4 (-B+2 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}\right )}{2 a^2}}{6 a^2}+\frac {a (23 A+13 i B) \cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
((A + I*B)*Cot[c + d*x]^2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((a*(23*A + (13*I)*B)*Cot[c + d*x]^2)/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((a^2*(33 7*A + (167*I)*B)*Cot[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (5*((-2* a^3*(85*A + (41*I)*B)*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d - (3*(( (-2*a^(11/2)*(43*A + (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] )/d - (I*Sqrt[2]*a^(11/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sq rt[2]*Sqrt[a])])/d)/a - (42*a^4*((2*I)*A - B)*Cot[c + d*x]*Sqrt[a + I*a*Ta n[c + d*x]])/d))/a))/(2*a^2))/(6*a^2))/(10*a^2)
3.2.11.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
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[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[( A*b + a*B)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A *d)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T an[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Time = 0.30 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.72
method | result | size |
derivativedivides | \(\frac {2 a^{3} \left (-\frac {17 i B +31 A}{8 a^{5} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5 i B +7 A}{12 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B +A}{10 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {11}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}-\frac {11 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {13}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (20 i B +43 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{5}}\right )}{d}\) | \(225\) |
default | \(\frac {2 a^{3} \left (-\frac {17 i B +31 A}{8 a^{5} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5 i B +7 A}{12 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B +A}{10 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {11}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}-\frac {11 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {13}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (20 i B +43 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{5}}\right )}{d}\) | \(225\) |
2/d*a^3*(-1/8/a^5*(17*I*B+31*A)/(a+I*a*tan(d*x+c))^(1/2)-1/12/a^4*(7*A+5*I *B)/(a+I*a*tan(d*x+c))^(3/2)-1/10/a^3*(A+I*B)/(a+I*a*tan(d*x+c))^(5/2)-1/1 6*(A-I*B)/a^(11/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^ (1/2))+1/a^5*(-((-1/2*I*B-11/8*A)*(a+I*a*tan(d*x+c))^(3/2)+(1/2*I*a*B+13/8 *a*A)*(a+I*a*tan(d*x+c))^(1/2))/a^2/tan(d*x+c)^2+1/8*(43*A+20*I*B)/a^(1/2) *arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 923 vs. \(2 (243) = 486\).
Time = 0.29 (sec) , antiderivative size = 923, normalized size of antiderivative = 2.96 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
1/240*(30*sqrt(1/2)*(a^3*d*e^(9*I*d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I* c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*log( -4*(sqrt(2)*sqrt(1/2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a/(e^(2 *I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a *e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 30*sqrt(1/2)*(a^3*d*e^(9*I *d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*s qrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(-I*a^3*d*e ^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I* c)/(I*A + B)) + 15*(a^3*d*e^(9*I*d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I*c ) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((1849*A^2 + 1720*I*A*B - 400*B^2)/(a^5 *d^2))*log(-16*(3*(43*I*A - 20*B)*a^2*e^(2*I*d*x + 2*I*c) + (43*I*A - 20*B )*a^2 + 2*sqrt(2)*(I*a^4*d*e^(3*I*d*x + 3*I*c) + I*a^4*d*e^(I*d*x + I*c))* sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((1849*A^2 + 1720*I*A*B - 400*B^2)/( a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/(-43*I*A + 20*B)) - 15*(a^3*d*e^(9*I*d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((1 849*A^2 + 1720*I*A*B - 400*B^2)/(a^5*d^2))*log(-16*(3*(43*I*A - 20*B)*a^2* e^(2*I*d*x + 2*I*c) + (43*I*A - 20*B)*a^2 + 2*sqrt(2)*(-I*a^4*d*e^(3*I*d*x + 3*I*c) - I*a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr t((1849*A^2 + 1720*I*A*B - 400*B^2)/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/(-...
\[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.91 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {a^{2} {\left (\frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} {\left (2 \, A + i \, B\right )} - 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} {\left (211 \, A + 104 i \, B\right )} a + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (337 \, A + 167 i \, B\right )} a^{2} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (23 \, A + 13 i \, B\right )} a^{3} + 12 \, {\left (A + i \, B\right )} a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{4} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{5} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{6}} - \frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {9}{2}}} + \frac {30 \, {\left (43 \, A + 20 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {9}{2}}}\right )}}{240 \, d} \]
-1/240*a^2*(4*(315*(I*a*tan(d*x + c) + a)^4*(2*A + I*B) - 5*(I*a*tan(d*x + c) + a)^3*(211*A + 104*I*B)*a + (I*a*tan(d*x + c) + a)^2*(337*A + 167*I*B )*a^2 + 2*(I*a*tan(d*x + c) + a)*(23*A + 13*I*B)*a^3 + 12*(A + I*B)*a^4)/( (I*a*tan(d*x + c) + a)^(9/2)*a^4 - 2*(I*a*tan(d*x + c) + a)^(7/2)*a^5 + (I *a*tan(d*x + c) + a)^(5/2)*a^6) - 15*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt( a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(9/2) + 30*(43*A + 20*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt (a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(9/2))/d
\[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 10.01 (sec) , antiderivative size = 3048, normalized size of antiderivative = 9.77 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
2*atanh((192*a*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((3699*A^2)/(256*a^5*d^2) - ((13667809*A^4*a^2)/(16*d^4) + (638401*B^4*a^2)/(16*d^4) - (8877585*A^2 *B^2*a^2)/(8*d^4) - (A*B^3*a^2*1375079i)/(4*d^4) + (A^3*B*a^2*6362537i)/(4 *d^4))^(1/2)/(64*a^6) - (801*B^2)/(256*a^5*d^2) + (A*B*1719i)/(128*a^5*d^2 ))^(1/2)*((13667809*A^4*a^2)/(16*d^4) + (638401*B^4*a^2)/(16*d^4) - (88775 85*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*1375079i)/(4*d^4) + (A^3*B*a^2*636253 7i)/(4*d^4))^(1/2))/(B^3*d*62322i - 643278*A^3*d + 407502*A*B^2*d - A^2*B* d*887274i + (680*A*d^3*((13667809*A^4*a^2)/(16*d^4) + (638401*B^4*a^2)/(16 *d^4) - (8877585*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*1375079i)/(4*d^4) + (A^ 3*B*a^2*6362537i)/(4*d^4))^(1/2))/a + (B*d^3*((13667809*A^4*a^2)/(16*d^4) + (638401*B^4*a^2)/(16*d^4) - (8877585*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*1 375079i)/(4*d^4) + (A^3*B*a^2*6362537i)/(4*d^4))^(1/2)*328i)/a) - (59152*A ^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((3699*A^2)/(256*a^5*d^2) - ((136 67809*A^4*a^2)/(16*d^4) + (638401*B^4*a^2)/(16*d^4) - (8877585*A^2*B^2*a^2 )/(8*d^4) - (A*B^3*a^2*1375079i)/(4*d^4) + (A^3*B*a^2*6362537i)/(4*d^4))^( 1/2)/(64*a^6) - (801*B^2)/(256*a^5*d^2) + (A*B*1719i)/(128*a^5*d^2))^(1/2) )/(B^3*d*62322i - 643278*A^3*d + 407502*A*B^2*d - A^2*B*d*887274i + (680*A *d^3*((13667809*A^4*a^2)/(16*d^4) + (638401*B^4*a^2)/(16*d^4) - (8877585*A ^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*1375079i)/(4*d^4) + (A^3*B*a^2*6362537i)/ (4*d^4))^(1/2))/a + (B*d^3*((13667809*A^4*a^2)/(16*d^4) + (638401*B^4*a...