Optimal. Leaf size=62 \[ \frac {2 a (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a B (c-i c \tan (e+f x))^{5/2}}{5 c f} \]
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Rubi [A] time = 0.11, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3588, 43} \[ \frac {2 a (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a B (c-i c \tan (e+f x))^{5/2}}{5 c f} \]
Antiderivative was successfully verified.
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Rule 43
Rule 3588
Rubi steps
\begin {align*} \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int (A+B x) \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(a c) \operatorname {Subst}\left (\int \left ((A-i B) \sqrt {c-i c x}+\frac {i B (c-i c x)^{3/2}}{c}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {2 a (i A+B) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a B (c-i c \tan (e+f x))^{5/2}}{5 c f}\\ \end {align*}
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Mathematica [A] time = 3.35, size = 97, normalized size = 1.56 \[ \frac {2 a c (\cos (e)-i \sin (e)) (\cos (f x)-i \sin (f x)) \sqrt {c-i c \tan (e+f x)} (5 i A+3 i B \tan (e+f x)+2 B) (A+B \tan (e+f x))}{15 f (A \cos (e+f x)+B \sin (e+f x))} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.32, size = 79, normalized size = 1.27 \[ \frac {\sqrt {2} {\left ({\left (20 i \, A + 20 \, B\right )} a c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (20 i \, A - 4 \, B\right )} a c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [A] time = 0.37, size = 55, normalized size = 0.89 \[ \frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 48, normalized size = 0.77 \[ \frac {2 i \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B a + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a c\right )}}{15 \, c f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.95, size = 99, normalized size = 1.60 \[ \frac {4\,a\,c\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,5{}\mathrm {i}-B+A\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,5{}\mathrm {i}+5\,B\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}{15\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2} \]
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 \[ i a \left (\int \left (- i A c \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- i A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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