Optimal. Leaf size=284 \[ \frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]
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Rubi [A] time = 0.61, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3581, 3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 1168
Rule 3534
Rule 3581
Rule 3595
Rubi steps
\begin {align*} \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\int \frac {\cot ^{\frac {3}{2}}(c+d x) (B+A \cot (c+d x))}{(i a+a \cot (c+d x))^2} \, dx\\ &=\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {\sqrt {\cot (c+d x)} \left (-\frac {3}{2} a (i A-B)+\frac {1}{2} a (7 A-i B) \cot (c+d x)\right )}{i a+a \cot (c+d x)} \, dx}{4 a^2}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-\frac {1}{2} a^2 (5 i A-B)+\frac {3}{2} a^2 (3 A-i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a^2 (5 i A-B)-\frac {3}{2} a^2 (3 A-i B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^4 d}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {((9+5 i) A-(1+3 i) B) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^2 d}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac {((9+5 i) A-(1+3 i) B) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{32 \sqrt {2} a^2 d}-\frac {((9+5 i) A-(1+3 i) B) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{32 \sqrt {2} a^2 d}+\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}+\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac {((9+5 i) A-(1+3 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {((9-5 i) A+(1-3 i) B) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}\\ &=\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac {((9+5 i) A-(1+3 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}\\ \end {align*}
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Mathematica [A] time = 2.36, size = 243, normalized size = 0.86 \[ \frac {\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (4 \cos (c+d x) (\cos (2 d x)-i \sin (2 d x)) ((-B+5 i A) \sin (c+d x)+(7 A+3 i B) \cos (c+d x))+(-\sin (2 c)+i \cos (2 c)) \sqrt {\sin (2 (c+d x))} \csc (c+d x) \left (((5+9 i) A+(3+i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((2+7 i) A+(1-2 i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.08, size = 666, normalized size = 2.35 \[ \frac {{\left (2 \, a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \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^{4} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 2 \, a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \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^{4} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - a^{2} d \sqrt {\frac {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \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 {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} + 7 \, A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + a^{2} d \sqrt {\frac {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \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 {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} - 7 \, A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 2 \, {\left ({\left (-6 i \, A + 2 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (5 i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \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 (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} 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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\cot \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 4.47, size = 1517, normalized size = 5.34 \[ \text {result too large to display} \]
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 \[ \int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {A \sqrt {\cot {\left (c + d x \right )}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx + \int \frac {B \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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