Optimal. Leaf size=240 \[ \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.80, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3596, 3598, 12, 3544, 205} \[ -\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rule 205
Rule 3544
Rule 3596
Rule 3598
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\frac {1}{2} a (11 A+i B)-3 a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{4} a^2 (83 A+13 i B)-a^2 (17 i A-7 B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{8} a^3 (317 A+67 i B)-\frac {1}{4} a^3 (151 i A-41 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {2 \int \frac {15 a^4 (i A+B) \sqrt {a+i a \tan (c+d x)}}{16 \sqrt {\tan (c+d x)}} \, dx}{15 a^7}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {(i A+B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {(A-i B) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {A+i B}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17 A+7 i B}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151 A+41 i B}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 10.73, size = 288, normalized size = 1.20 \[ \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x)) \left (\frac {\sqrt {2} (B+i A) e^{3 i (c+d x)} \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}}}+\frac {2 ((19 B-149 i A) \tan (c+d x)+\cos (2 (c+d x)) ((86 B-466 i A) \tan (c+d x)-20 (23 A+4 i B))+340 A+80 i B)}{15 \sqrt {\tan (c+d x)} \sqrt {\sec (c+d x)}}\right )}{8 d (a+i a \tan (c+d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.60, size = 528, normalized size = 2.20 \[ \frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left ({\left (-463 i \, A + 83 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (-269 i \, A + 19 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (220 i \, A - 80 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (29 i \, A - 19 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.29, size = 1158, normalized size = 4.82 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________