Optimal. Leaf size=260 \[ \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {(317 A+67 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {(151 A+41 i B) \sqrt {\cot (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.97, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4241, 3596, 3598, 12, 3544, 205} \[ -\frac {(317 A+67 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {(151 A+41 i B) \sqrt {\cot (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 3544
Rule 3596
Rule 3598
Rule 4241
Rubi steps
\begin {align*} \int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(151 A+41 i B) \sqrt {\cot (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(151 A+41 i B) \sqrt {\cot (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(151 A+41 i B) \sqrt {\cot (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {\left ((i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {(A+i B) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(151 A+41 i B) \sqrt {\cot (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}-\frac {\left (i (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \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 ) (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 ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(17 A+7 i B) \sqrt {\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(151 A+41 i B) \sqrt {\cot (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(317 A+67 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}\\ \end {align*}
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Mathematica [A] time = 9.20, size = 200, normalized size = 0.77 \[ \frac {\cot ^{\frac {3}{2}}(c+d x) \sec (c+d x) \left (-20 \csc (c+d x) ((23 A+4 i B) \cos (2 (c+d x))-17 A-4 i B)+\sec (c+d x) ((86 B-466 i A) \cos (2 (c+d x))-149 i A+19 B)+15 (A-i B) e^{2 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \csc (2 (c+d x)) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{60 a^2 d (\cot (c+d x)+i)^2 \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.52, size = 485, normalized size = 1.87 \[ \frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} - 8 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} - 8 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - \sqrt {2} {\left ({\left (463 \, A + 83 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (97 \, A + 32 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (13 \, A + 8 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac {3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 4.10, size = 764, normalized size = 2.94 \[ \text {result too large to display} \]
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 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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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