Optimal. Leaf size=240 \[ \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (B+i A) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {(21 A+11 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(-25 B+39 i A) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.77, 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} \[ -\frac {(21 A+11 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(-25 B+39 i A) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (B+i A) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 3544
Rule 3596
Rule 3598
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {3}{2} a (3 A+i B)-3 a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3}{4} a^2 (21 A+11 i B)-3 a^2 (5 i A-3 B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{3 a^4}\\ &=\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(21 A+11 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{8} a^3 (39 i A-25 B)-\frac {3}{4} a^3 (21 A+11 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{9 a^5}\\ &=\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(21 A+11 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(39 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {4 \int -\frac {9 a^4 (A-i B) \sqrt {a+i a \tan (c+d x)}}{16 \sqrt {\tan (c+d x)}} \, dx}{9 a^6}\\ &=\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(21 A+11 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(39 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}-\frac {(A-i B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(21 A+11 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(39 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {(i A+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 )}{2 d}\\ &=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {A+i B}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {5 A+3 i B}{2 a d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(21 A+11 i B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(39 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 7.24, size = 244, normalized size = 1.02 \[ \frac {i e^{-2 i (c+d x)} \sec ^2(c+d x) \left (-3 i (A-i B) e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-i A \left (18 e^{2 i (c+d x)}-87 e^{4 i (c+d x)}+52 e^{6 i (c+d x)}+1\right )+B \left (12 e^{2 i (c+d x)}-51 e^{4 i (c+d x)}+38 e^{6 i (c+d x)}+1\right )\right )}{12 a d \left (-1+e^{2 i (c+d x)}\right ) \sqrt {\tan (c+d x)} (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.22, size = 569, normalized size = 2.37 \[ -\frac {3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} \log \left (\frac {2 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} 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 ) - 3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} \log \left (\frac {-2 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} 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 (2 \, {\left (26 \, A + 19 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - {\left (35 \, A + 13 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, {\left (23 \, A + 13 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (19 \, A + 13 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, 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}}}{12 \, {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.34, size = 1012, normalized size = 4.22 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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