Optimal. Leaf size=214 \[ \frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {67 A-3 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {A+i B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.75, antiderivative size = 214, 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 = {4241, 3596, 12, 3544, 205} \[ \frac {67 A-3 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {A+i B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 3544
Rule 3596
Rule 4241
Rubi steps
\begin {align*} \int \frac {\sqrt {\cot (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)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac {A+i B}{5 d \sqrt {\cot (c+d x)} (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 (9 A-i B)-2 a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {A+i B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (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 (41 A-9 i B)-\frac {1}{2} a^2 (13 i A-3 B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac {A+i B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {67 A-3 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {15 a^3 (A-i B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac {A+i B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {67 A-3 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left ((A-i 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}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {67 A-3 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (i (A-i 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}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {67 A-3 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 6.96, size = 167, normalized size = 0.78 \[ \frac {\cot ^{\frac {3}{2}}(c+d x) \sec ^2(c+d x) \left (\frac {30 (A-i B) e^{3 i (c+d x)} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-1+e^{2 i (c+d x)}}}+2 ((86 A+6 i B) \cos (2 (c+d x))+80 i A \sin (2 (c+d x))+19 A+9 i B)\right )}{120 a^2 d (\cot (c+d x)+i)^2 \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.50, size = 479, normalized size = 2.24 \[ \frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} - {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (-83 i \, A + 3 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (64 i \, A + 6 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (16 i \, A - 6 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\cot \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 4.26, size = 1078, normalized size = 5.04 \[ \text {result too large to display} \]
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 \[ \int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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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