Integrand size = 36, antiderivative size = 362 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(A+i B) \tan ^{1+m}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(i B (1-4 m)+A (11-4 m)) \tan ^{1+m}(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {\left (i B \left (13+12 m-16 m^2\right )-A \left (37-52 m+16 m^2\right )\right ) \tan ^{1+m}(c+d x)}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(i A+B) \operatorname {AppellF1}\left (\frac {1}{2},-m,1,\frac {3}{2},1+i \tan (c+d x),\frac {1}{2} (1+i \tan (c+d x))\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{8 a^3 d}+\frac {(1+2 m) \left (B \left (13+12 m-16 m^2\right )+i A \left (37-52 m+16 m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d} \] Output:
1/5*(A+I*B)*tan(d*x+c)^(1+m)/d/(a+I*a*tan(d*x+c))^(5/2)+1/30*(I*B*(1-4*m)+ A*(11-4*m))*tan(d*x+c)^(1+m)/a/d/(a+I*a*tan(d*x+c))^(3/2)-1/60*(I*B*(-16*m ^2+12*m+13)-A*(16*m^2-52*m+37))*tan(d*x+c)^(1+m)/a^2/d/(a+I*a*tan(d*x+c))^ (1/2)-1/8*(I*A+B)*AppellF1(1/2,-m,1,3/2,1+I*tan(d*x+c),1/2+1/2*I*tan(d*x+c ))*tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/((-I*tan(d*x+c))^m)+1/60*(1 +2*m)*(B*(-16*m^2+12*m+13)+I*A*(16*m^2-52*m+37))*hypergeom([1/2, -m],[3/2] ,1+I*tan(d*x+c))*tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/((-I*tan(d*x+ c))^m)
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx \] Input:
Integrate[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/ 2),x]
Output:
Integrate[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/ 2), x]
Time = 2.22 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.05, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4084, 3042, 4047, 25, 27, 152, 150, 4082, 77, 75}
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.
| \(\displaystyle \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^m (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {\tan ^m(c+d x) (2 a (A (4-m)-i B (m+1))-a (i A-B) (3-2 m) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\tan ^m(c+d x) (2 a (A (4-m)-i B (m+1))-a (i A-B) (3-2 m) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^m (2 a (A (4-m)-i B (m+1))-a (i A-B) (3-2 m) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int -\frac {\tan ^m(c+d x) \left (2 a^2 \left (i B \left (-4 m^2+3 m+7\right )-A \left (4 m^2-13 m+13\right )\right )-a^2 (B (1-4 m)-i A (11-4 m)) (1-2 m) \tan (c+d x)\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan ^m(c+d x) \left (2 a^2 \left (i B \left (-4 m^2+3 m+7\right )-A \left (4 m^2-13 m+13\right )\right )-a^2 (B (1-4 m)-i A (11-4 m)) (1-2 m) \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x)^m \left (2 a^2 \left (i B \left (-4 m^2+3 m+7\right )-A \left (4 m^2-13 m+13\right )\right )-a^2 (B (1-4 m)-i A (11-4 m)) (1-2 m) \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {1}{2} \tan ^m(c+d x) \sqrt {i \tan (c+d x) a+a} \left (2 a^3 \left (A \left (16 m^3-44 m^2+11 m+11\right )+i B \left (16 m^3-4 m^2-19 m+1\right )\right )-a^3 (2 m+1) \left (B \left (-16 m^2+12 m+13\right )+i A \left (16 m^2-52 m+37\right )\right ) \tan (c+d x)\right )dx}{a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \tan ^m(c+d x) \sqrt {i \tan (c+d x) a+a} \left (2 a^3 \left (A \left (16 m^3-44 m^2+11 m+11\right )+i B \left (16 m^3-4 m^2-19 m+1\right )\right )-a^3 (2 m+1) \left (B \left (-16 m^2+12 m+13\right )+i A \left (16 m^2-52 m+37\right )\right ) \tan (c+d x)\right )dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \tan (c+d x)^m \sqrt {i \tan (c+d x) a+a} \left (2 a^3 \left (A \left (16 m^3-44 m^2+11 m+11\right )+i B \left (16 m^3-4 m^2-19 m+1\right )\right )-a^3 (2 m+1) \left (B \left (-16 m^2+12 m+13\right )+i A \left (16 m^2-52 m+37\right )\right ) \tan (c+d x)\right )dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-15 a^3 (A-i B) \int \tan ^m(c+d x) \sqrt {i \tan (c+d x) a+a}dx-a^2 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-15 a^3 (A-i B) \int \tan (c+d x)^m \sqrt {i \tan (c+d x) a+a}dx-a^2 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4047 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-\frac {15 i a^5 (A-i B) \int -\frac {\tan ^m(c+d x)}{a (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}-a^2 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {15 i a^5 (A-i B) \int \frac {\tan ^m(c+d x)}{a (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}-a^2 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {15 i a^4 (A-i B) \int \frac {\tan ^m(c+d x)}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}-a^2 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 152 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {15 i a^4 (A-i B) \sqrt {1+i \tan (c+d x)} \int \frac {\tan ^m(c+d x)}{\sqrt {i \tan (c+d x)+1} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d \sqrt {a+i a \tan (c+d x)}}-a^2 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 150 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-a^2 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \tan (c+d x)^m (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx-\frac {15 a^4 (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-\frac {a^4 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \int \frac {\tan ^m(c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {15 a^4 (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 77 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-\frac {a^4 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \int \frac {(-i \tan (c+d x))^m}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {15 a^4 (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 75 |
| \(\displaystyle \frac {\frac {a (A (11-4 m)+i B (1-4 m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {a^2 \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \tan ^{m+1}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {\frac {2 i a^3 (2 m+1) \left (-A \left (16 m^2-52 m+37\right )+i B \left (-16 m^2+12 m+13\right )\right ) \sqrt {a+i a \tan (c+d x)} (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},i \tan (c+d x)+1\right )}{d}-\frac {15 a^4 (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{6 a^2}}{10 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
Input:
Int[(Tan[c + d*x]^m*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
((A + I*B)*Tan[c + d*x]^(1 + m))/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((a* (I*B*(1 - 4*m) + A*(11 - 4*m))*Tan[c + d*x]^(1 + m))/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) - ((a^2*(I*B*(13 + 12*m - 16*m^2) - A*(37 - 52*m + 16*m^2))* Tan[c + d*x]^(1 + m))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + ((-15*a^4*(A - I*B) *AppellF1[1 + m, 1/2, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(1 + m))/(d*(1 + m)*Sqrt[a + I*a*Tan[c + d* x]]) + ((2*I)*a^3*(1 + 2*m)*(I*B*(13 + 12*m - 16*m^2) - A*(37 - 52*m + 16* m^2))*Hypergeometric2F1[1/2, -m, 3/2, 1 + I*Tan[c + d*x]]*Tan[c + d*x]^m*S qrt[a + I*a*Tan[c + d*x]])/(d*((-I)*Tan[c + d*x])^m))/(2*a^2))/(6*a^2))/(1 0*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-b)*(c/ d))^IntPart[m]*((b*x)^FracPart[m]/((-d)*(x/c))^FracPart[m]) Int[((-d)*(x/ c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(b/f) Subst[Int[(a + x)^(m - 1)*(( c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d ^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(a*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[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_)])^(n_), x_Symbol] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), 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] && NeQ[A*b + a*B, 0]
\[\int \frac {\tan \left (d x +c \right )^{m} \left (A +B \tan \left (d x +c \right )\right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]
Input:
int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algori thm="fricas")
Output:
integral(1/8*sqrt(2)*((A - I*B)*e^(6*I*d*x + 6*I*c) + (3*A - I*B)*e^(4*I*d *x + 4*I*c) + (3*A + I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) )*e^(-5*I*d*x - 5*I*c)/a^3, x)
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(tan(d*x+c)**m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Integral((A + B*tan(c + d*x))*tan(c + d*x)**m/(I*a*(tan(c + d*x) - I))**(5 /2), x)
Exception generated. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algori thm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algori thm="giac")
Output:
Timed out
Timed out. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:
int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(5/2),x)
Output:
int((tan(c + d*x)^m*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(5/2), x )
\[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-\left (\int \frac {\tan \left (d x +c \right )^{m}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) a -\left (\int \frac {\tan \left (d x +c \right )^{m} \tan \left (d x +c \right )}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) b}{\sqrt {a}\, a^{2}} \] Input:
int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - (int(tan(c + d*x)**m/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqr t(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x)*a + in t((tan(c + d*x)**m*tan(c + d*x))/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x )*b))/(sqrt(a)*a**2)