Optimal. Leaf size=101 \[ \frac {2 a^2 (3 B+i A)}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {4 a^2 (B+i A)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^2 B \sqrt {c-i c \tan (e+f x)}}{c^2 f} \]
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Rubi [A] time = 0.18, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {3588, 77} \[ \frac {2 a^2 (3 B+i A)}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {4 a^2 (B+i A)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^2 B \sqrt {c-i c \tan (e+f x)}}{c^2 f} \]
Antiderivative was successfully verified.
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Rule 77
Rule 3588
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x) (A+B x)}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left (\frac {2 a (A-i B)}{(c-i c x)^{5/2}}-\frac {a (A-3 i B)}{c (c-i c x)^{3/2}}-\frac {i a B}{c^2 \sqrt {c-i c x}}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {4 a^2 (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^2 (i A+3 B)}{c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a^2 B \sqrt {c-i c \tan (e+f x)}}{c^2 f}\\ \end {align*}
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Mathematica [A] time = 8.77, size = 112, normalized size = 1.11 \[ \frac {a^2 \sqrt {c-i c \tan (e+f x)} (\cos (2 (e+2 f x))+i \sin (2 (e+2 f x))) (3 (A-5 i B) \sin (2 (e+f x))+(13 B+i A) \cos (2 (e+f x))+i A+7 B)}{3 c^2 f (\cos (f x)+i \sin (f x))^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.96, size = 80, normalized size = 0.79 \[ \frac {\sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, A + 7 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (2 i \, A + 14 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, c^{2} f} \]
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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 80, normalized size = 0.79 \[ -\frac {2 i a^{2} \left (i B \sqrt {c -i c \tan \left (f x +e \right )}+\frac {2 c^{2} \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {c \left (-3 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 79, normalized size = 0.78 \[ -\frac {2 i \, {\left (\frac {3 i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B a^{2}}{c} - \frac {3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 3 i \, B\right )} a^{2} - 2 \, {\left (A - i \, B\right )} a^{2} c}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\right )}}{3 \, c f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.15, size = 158, normalized size = 1.56 \[ \frac {a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,2{}\mathrm {i}+14\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+7\,B\,\cos \left (2\,e+2\,f\,x\right )-B\,\cos \left (4\,e+4\,f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,7{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{3\,c^2\,f} \]
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 \[ - a^{2} \left (\int \left (- \frac {A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {B \tan ^{3}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {2 i A \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \left (- \frac {2 i B \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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