Integrand size = 36, antiderivative size = 383 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {2 \left (2 i A n (5+2 n)+B \left (15+10 n+4 n^2\right )\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}+\frac {2 (i A+B) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d}-\frac {2 \left (4 B n \left (9+8 n+2 n^2\right )+i A \left (15+36 n+32 n^2+8 n^3\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (1+2 n) (3+2 n) (5+2 n)}-\frac {2 (2 i B n-A (5+2 n)) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (3+2 n) (5+2 n)}+\frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (5+2 n)} \]
-2*(2*I*A*n*(5+2*n)+B*(4*n^2+10*n+15))*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c)) ^n/d/(5+2*n)/(4*n^2+8*n+3)+2*(I*A+B)*AppellF1(1/2,1-n,1,3/2,-I*tan(d*x+c), I*tan(d*x+c))*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n/d/((1+I*tan(d*x+c))^n) -2*(4*B*n*(2*n^2+8*n+9)+I*A*(8*n^3+32*n^2+36*n+15))*hypergeom([1/2, 1-n],[ 3/2],-I*tan(d*x+c))*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^n/d/(5+2*n)/(4*n^2 +8*n+3)/((1+I*tan(d*x+c))^n)-2*(2*I*B*n-A*(5+2*n))*tan(d*x+c)^(3/2)*(a+I*a *tan(d*x+c))^n/d/(3+2*n)/(5+2*n)+2*B*tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^n /d/(5+2*n)
\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]
Time = 2.06 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.10, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {3042, 4080, 27, 3042, 4080, 27, 3042, 4080, 27, 3042, 4084, 3042, 4047, 25, 27, 148, 27, 334, 333, 4082, 76, 74}
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.
| \(\displaystyle \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^{5/2} (a+i a \tan (c+d x))^n (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {2 \int -\frac {1}{2} \tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^n (5 a B+a (2 i B n-A (2 n+5)) \tan (c+d x))dx}{a (2 n+5)}+\frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\int \tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^n (5 a B+a (2 i B n-A (2 n+5)) \tan (c+d x))dx}{a (2 n+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\int \tan (c+d x)^{3/2} (i \tan (c+d x) a+a)^n (5 a B+a (2 i B n-A (2 n+5)) \tan (c+d x))dx}{a (2 n+5)}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 \int -\frac {1}{2} \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^n \left (3 a^2 (2 i B n-A (2 n+5))-a^2 \left (2 i A n (2 n+5)+B \left (4 n^2+10 n+15\right )\right ) \tan (c+d x)\right )dx}{a (2 n+3)}+\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^n \left (3 a^2 (2 i B n-A (2 n+5))-a^2 \left (2 i A n (2 n+5)+B \left (4 n^2+10 n+15\right )\right ) \tan (c+d x)\right )dx}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^n \left (3 a^2 (2 i B n-A (2 n+5))-a^2 \left (2 i A n (2 n+5)+B \left (4 n^2+10 n+15\right )\right ) \tan (c+d x)\right )dx}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {2 \int \frac {(i \tan (c+d x) a+a)^n \left (\left (2 i A n (2 n+5)+B \left (4 n^2+10 n+15\right )\right ) a^3+\left (4 i B n \left (2 n^2+8 n+9\right )-A \left (8 n^3+32 n^2+36 n+15\right )\right ) \tan (c+d x) a^3\right )}{2 \sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\int \frac {(i \tan (c+d x) a+a)^n \left (\left (2 i A n (2 n+5)+B \left (4 n^2+10 n+15\right )\right ) a^3+\left (4 i B n \left (2 n^2+8 n+9\right )-A \left (8 n^3+32 n^2+36 n+15\right )\right ) \tan (c+d x) a^3\right )}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\int \frac {(i \tan (c+d x) a+a)^n \left (\left (2 i A n (2 n+5)+B \left (4 n^2+10 n+15\right )\right ) a^3+\left (4 i B n \left (2 n^2+8 n+9\right )-A \left (8 n^3+32 n^2+36 n+15\right )\right ) \tan (c+d x) a^3\right )}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {a^3 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \int \frac {(i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {a^3 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \int \frac {(i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 4047 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {i a^5 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \int -\frac {(i \tan (c+d x) a+a)^{n-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {-\frac {i a^5 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \int \frac {(i \tan (c+d x) a+a)^{n-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {-\frac {i a^4 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \int \frac {(i \tan (c+d x) a+a)^{n-1}}{\sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 148 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {2 a^5 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \int \frac {\left (a-i a^3 \tan ^2(c+d x)\right )^{n-1}}{a \left (i a^2 \tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {2 a^4 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \int \frac {\left (a-i a^3 \tan ^2(c+d x)\right )^{n-1}}{i a^2 \tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 334 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {2 a^3 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \left (a-i a^3 \tan ^2(c+d x)\right )^n \int \frac {\left (1-i a^2 \tan ^2(c+d x)\right )^{n-1}}{i a^2 \tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 333 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {2 i a^4 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \tan (c+d x) \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \left (a-i a^3 \tan ^2(c+d x)\right )^n \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}-a^2 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^n}{\sqrt {\tan (c+d x)}}dx}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {2 i a^4 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \tan (c+d x) \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \left (a-i a^3 \tan ^2(c+d x)\right )^n \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}-\frac {a^4 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \int \frac {(i \tan (c+d x) a+a)^{n-1}}{\sqrt {\tan (c+d x)}}d\tan (c+d x)}{d}}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 76 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {2 i a^4 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \tan (c+d x) \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \left (a-i a^3 \tan ^2(c+d x)\right )^n \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}-\frac {a^3 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \int \frac {(i \tan (c+d x)+1)^{n-1}}{\sqrt {\tan (c+d x)}}d\tan (c+d x)}{d}}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
| \(\Big \downarrow \) 74 |
| \(\displaystyle \frac {2 B \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+5)}-\frac {\frac {2 a (-A (2 n+5)+2 i B n) \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^n}{d (2 n+3)}-\frac {\frac {\frac {2 i a^4 \left (8 n^3+36 n^2+46 n+15\right ) (B+i A) \tan (c+d x) \left (1-i a^2 \tan ^2(c+d x)\right )^{-n} \left (a-i a^3 \tan ^2(c+d x)\right )^n \operatorname {AppellF1}\left (\frac {1}{2},1,1-n,\frac {3}{2},-i a^2 \tan ^2(c+d x),i a^2 \tan ^2(c+d x)\right )}{d}-\frac {2 a^3 \left (4 B n \left (2 n^2+8 n+9\right )+i A \left (8 n^3+32 n^2+36 n+15\right )\right ) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right )}{d}}{a (2 n+1)}-\frac {2 a^2 \left (B \left (4 n^2+10 n+15\right )+2 i A n (2 n+5)\right ) \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d (2 n+1)}}{a (2 n+3)}}{a (2 n+5)}\) |
(2*B*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^n)/(d*(5 + 2*n)) - ((2*a*(( 2*I)*B*n - A*(5 + 2*n))*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^n)/(d*(3 + 2*n)) - ((-2*a^2*((2*I)*A*n*(5 + 2*n) + B*(15 + 10*n + 4*n^2))*Sqrt[Tan [c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + 2*n)) + ((-2*a^3*(4*B*n*(9 + 8*n + 2*n^2) + I*A*(15 + 36*n + 32*n^2 + 8*n^3))*Hypergeometric2F1[1/2, 1 - n, 3/2, (-I)*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/ (d*(1 + I*Tan[c + d*x])^n) + ((2*I)*a^4*(I*A + B)*(15 + 46*n + 36*n^2 + 8* n^3)*AppellF1[1/2, 1, 1 - n, 3/2, (-I)*a^2*Tan[c + d*x]^2, I*a^2*Tan[c + d *x]^2]*Tan[c + d*x]*(a - I*a^3*Tan[c + d*x]^2)^n)/(d*(1 - I*a^2*Tan[c + d* x]^2)^n))/(a*(1 + 2*n)))/(a*(3 + 2*n)))/(a*(5 + 2*n))
3.3.26.3.1 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[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^IntPart [n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d* (x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !Integer Q[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0] && ((RationalQ[m] && !(EqQ[n, -2 ^(-1)] && EqQ[c^2 - d^2, 0])) || !RationalQ[n])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], 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] && 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]
\[\int \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]
\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
integral(-((A - I*B)*e^(6*I*d*x + 6*I*c) - (A - 3*I*B)*e^(4*I*d*x + 4*I*c) - (A + 3*I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^ (2*I*d*x + 2*I*c) + 1))^n*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))/(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1), x)
Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]