Integrand size = 38, antiderivative size = 323 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (14 i A+11 B) \sqrt {a+i a \tan (c+d x)}}{99 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{693 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {4 a^2 (250 i A+253 B) \sqrt {a+i a \tan (c+d x)}}{1155 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 a^2 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3465 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {8 a^2 (2155 i A+2167 B) \sqrt {a+i a \tan (c+d x)}}{3465 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)} \] Output:
(-4+4*I)*a^(5/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan (d*x+c))^(1/2))/d-2/99*a^2*(14*I*A+11*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d* x+c)^(9/2)+2/693*a^2*(212*A-209*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c) ^(7/2)+4/1155*a^2*(250*I*A+253*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5 /2)-8/3465*a^2*(655*A-649*I*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2) -8/3465*a^2*(2155*I*A+2167*B)*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)- 2/11*a*A*(a+I*a*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(11/2)
Time = 18.22 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.02 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\frac {4 \sqrt {2} a^2 (i A+B) e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \text {arctanh}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-\frac {a^2 \csc ^3(c+d x) \sec ^2(c+d x) (66 (95 A-47 i B) \cos (c+d x)+(-5225 A+6743 i B) \cos (3 (c+d x))+3995 A \cos (5 (c+d x))-3641 i B \cos (5 (c+d x))+84810 i A \sin (c+d x)+84414 B \sin (c+d x)-42185 i A \sin (3 (c+d x))-43703 B \sin (3 (c+d x))+10925 i A \sin (5 (c+d x))+10571 B \sin (5 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{27720 d \tan ^{\frac {5}{2}}(c+d x)} \] Input:
Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x] ^(13/2),x]
Output:
(4*Sqrt[2]*a^2*(I*A + B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[(a*E^((2*I)*( c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^ ((2*I)*(c + d*x))]])/(d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x ))))/(1 + E^((2*I)*(c + d*x)))]) - (a^2*Csc[c + d*x]^3*Sec[c + d*x]^2*(66* (95*A - (47*I)*B)*Cos[c + d*x] + (-5225*A + (6743*I)*B)*Cos[3*(c + d*x)] + 3995*A*Cos[5*(c + d*x)] - (3641*I)*B*Cos[5*(c + d*x)] + (84810*I)*A*Sin[c + d*x] + 84414*B*Sin[c + d*x] - (42185*I)*A*Sin[3*(c + d*x)] - 43703*B*Si n[3*(c + d*x)] + (10925*I)*A*Sin[5*(c + d*x)] + 10571*B*Sin[5*(c + d*x)])* Sqrt[a + I*a*Tan[c + d*x]])/(27720*d*Tan[c + d*x]^(5/2))
Time = 2.16 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.10, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.553, Rules used = {3042, 4076, 27, 3042, 4076, 27, 3042, 4081, 27, 3042, 4081, 25, 3042, 4081, 27, 3042, 4081, 27, 3042, 4027, 218}
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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{13/2}}dx\) |
| \(\Big \downarrow \) 4076 |
| \(\displaystyle \frac {2}{11} \int \frac {(i \tan (c+d x) a+a)^{3/2} (a (14 i A+11 B)-a (8 A-11 i B) \tan (c+d x))}{2 \tan ^{\frac {11}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \int \frac {(i \tan (c+d x) a+a)^{3/2} (a (14 i A+11 B)-a (8 A-11 i B) \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \int \frac {(i \tan (c+d x) a+a)^{3/2} (a (14 i A+11 B)-a (8 A-11 i B) \tan (c+d x))}{\tan (c+d x)^{11/2}}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4076 |
| \(\displaystyle \frac {1}{11} \left (\frac {2}{9} \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((212 A-209 i B) a^2+(184 i A+187 B) \tan (c+d x) a^2\right )}{2 \tan ^{\frac {9}{2}}(c+d x)}dx-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (-\frac {1}{9} \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((212 A-209 i B) a^2+(184 i A+187 B) \tan (c+d x) a^2\right )}{\tan ^{\frac {9}{2}}(c+d x)}dx-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (-\frac {1}{9} \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((212 A-209 i B) a^2+(184 i A+187 B) \tan (c+d x) a^2\right )}{\tan (c+d x)^{9/2}}dx-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 \int \frac {3 \sqrt {i \tan (c+d x) a+a} \left (a^3 (250 i A+253 B)-a^3 (212 A-209 i B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (250 i A+253 B)-a^3 (212 A-209 i B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (250 i A+253 B)-a^3 (212 A-209 i B) \tan (c+d x)\right )}{\tan (c+d x)^{7/2}}dx}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((655 A-649 i B) a^4+2 (250 i A+253 B) \tan (c+d x) a^4\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((655 A-649 i B) a^4+2 (250 i A+253 B) \tan (c+d x) a^4\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((655 A-649 i B) a^4+2 (250 i A+253 B) \tan (c+d x) a^4\right )}{\tan (c+d x)^{5/2}}dx}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^5 (2155 i A+2167 B)-2 a^5 (655 A-649 i B) \tan (c+d x)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^5 (2155 i A+2167 B)-2 a^5 (655 A-649 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^5 (2155 i A+2167 B)-2 a^5 (655 A-649 i B) \tan (c+d x)\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \left (\frac {\frac {2 \int -\frac {3465 a^6 (A-i B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a^5 (2167 B+2155 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \left (\frac {-3465 a^5 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^5 (2167 B+2155 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \left (\frac {-3465 a^5 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^5 (2167 B+2155 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 \left (\frac {\frac {6930 i a^7 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a^5 (2167 B+2155 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {1}{11} \left (\frac {1}{9} \left (\frac {2 a^2 (212 A-209 i B) \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 \left (-\frac {2 a^3 (253 B+250 i A) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (\frac {-\frac {(3465-3465 i) a^{11/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^5 (2167 B+2155 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^4 (655 A-649 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{5 a}\right )}{7 a}\right )-\frac {2 a^2 (11 B+14 i A) \sqrt {a+i a \tan (c+d x)}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{11 d \tan ^{\frac {11}{2}}(c+d x)}\) |
Input:
Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(13/2 ),x]
Output:
(-2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(11*d*Tan[c + d*x]^(11/2)) + ((-2*a^ 2*((14*I)*A + 11*B)*Sqrt[a + I*a*Tan[c + d*x]])/(9*d*Tan[c + d*x]^(9/2)) + ((2*a^2*(212*A - (209*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(7*d*Tan[c + d*x] ^(7/2)) - (6*((-2*a^3*((250*I)*A + 253*B)*Sqrt[a + I*a*Tan[c + d*x]])/(5*d *Tan[c + d*x]^(5/2)) - (2*((-2*a^4*(655*A - (649*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + (((-3465 + 3465*I)*a^(11/2)*(A - I*B)* ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/ d - (2*a^5*((2155*I)*A + 2167*B)*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a)))/(5*a)))/(7*a))/9)/11
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/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1)) Int[ (a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b *d*(n + 1)))*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] && GtQ[m, 1] && 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[(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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 975 vs. \(2 (266 ) = 532\).
Time = 0.23 (sec) , antiderivative size = 976, normalized size of antiderivative = 3.02
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(976\) |
| default | \(\text {Expression too large to display}\) | \(976\) |
| parts | \(\text {Expression too large to display}\) | \(1045\) |
Input:
int((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x,method=_ RETURNVERBOSE)
Output:
-1/3465/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(11/2)*(17336*B*tan(d* x+c)^5*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-5192 *I*B*tan(d*x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a) ^(1/2)+6930*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^( 1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^6+5240*A*tan(d* x+c)^4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+3465 *I*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan (d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^6+13860*A *ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1 /2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^6+17240*I*A*tan(d*x+c)^5*(a* tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-3465*(I*a)^(1/ 2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1 /2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^6-3036*B*(I*a)^(1/2)* (-I*a)^(1/2)*tan(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-13860*I*B* ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/ 2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^6+6930*ln(1/2*(2*I*a*tan(d*x+ c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I *a)^(1/2)*a*tan(d*x+c)^6-2120*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*t an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+2090*I*B*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I *tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+1610*I*A*tan(d*x+c)*(a*tan...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 771 vs. \(2 (249) = 498\).
Time = 0.10 (sec) , antiderivative size = 771, normalized size of antiderivative = 2.39 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, a lgorithm="fricas")
Output:
2/3465*(3465*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) - 20*d*e ^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) - 6*d*e^(2*I*d*x + 2*I*c) + d)*log((I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^2)*sqrt(a/(e ^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2 *I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 3465*sqrt(2)*sqrt(-(I*A^ 2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 1 0*I*c) + 15*d*e^(8*I*d*x + 8*I*c) - 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I *d*x + 4*I*c) - 6*d*e^(2*I*d*x + 2*I*c) + d)*log((-I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2 *I*d*x + 2*I*c) + (-I*A - B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(( -I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/( (-I*A - B)*a^2)) + 2*sqrt(2)*(2*(3730*A - 3553*I*B)*a^2*e^(13*I*d*x + 13*I *c) - 9*(1805*A - 2013*I*B)*a^2*e^(11*I*d*x + 11*I*c) + 55*(397*A - 337*I* B)*a^2*e^(9*I*d*x + 9*I*c) + 66*(95*A - 47*I*B)*a^2*e^(7*I*d*x + 7*I*c) - 1386*(15*A - 16*I*B)*a^2*e^(5*I*d*x + 5*I*c) + 15015*(A - I*B)*a^2*e^(3*I* d*x + 3*I*c) - 3465*(A - I*B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2* I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/( d*e^(12*I*d*x + 12*I*c) - 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x +...
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(13/2),x)
Output:
Timed out
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, a lgorithm="maxima")
Output:
Timed out
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x, a lgorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeWarning, need to choose a branch for the root of a polynomial with par
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{13/2}} \,d x \] Input:
int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(13/ 2),x)
Output:
int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(13/ 2), x)
\[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {13}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a^{2} \left (3392 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{5} a i +3344 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{5} b -1696 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{4} a +1672 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{4} b i -1272 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3} a i -1254 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3} b +1060 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2} a -1045 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2} b i -805 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) a i -385 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) b -315 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, a -6930 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{4}}d x \right ) \tan \left (d x +c \right )^{6} a d i -6930 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{4}}d x \right ) \tan \left (d x +c \right )^{6} b d \right )}{3465 \tan \left (d x +c \right )^{6} d} \] Input:
int((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(13/2),x)
Output:
(2*sqrt(a)*a**2*(3392*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**5*a*i + 3344*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x )**5*b - 1696*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**4* a + 1672*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**4*b*i - 1272*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3*a*i - 12 54*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3*b + 1060*sq rt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2*a - 1045*sqrt(ta n(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2*b*i - 805*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*a*i - 385*sqrt(tan(c + d*x)) *sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*b - 315*sqrt(tan(c + d*x))*sqrt(tan (c + d*x)*i + 1)*a - 6930*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) )/tan(c + d*x)**4,x)*tan(c + d*x)**6*a*d*i - 6930*int((sqrt(tan(c + d*x))* sqrt(tan(c + d*x)*i + 1))/tan(c + d*x)**4,x)*tan(c + d*x)**6*b*d))/(3465*t an(c + d*x)**6*d)