Optimal. Leaf size=92 \[ -\frac {2 i a^3}{c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.16, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3522, 3487, 43} \[ -\frac {2 i a^3}{c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 43
Rule 3487
Rule 3522
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^{11/2}} \, dx\\ &=\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {(c-x)^2}{(c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \left (\frac {4 c^2}{(c+x)^{7/2}}-\frac {4 c}{(c+x)^{5/2}}+\frac {1}{(c+x)^{3/2}}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=-\frac {8 i a^3}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {8 i a^3}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3}{c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 8.96, size = 98, normalized size = 1.07 \[ \frac {2 a^3 \cos (e+f x) \sqrt {c-i c \tan (e+f x)} (-5 i \sin (2 (e+f x))+11 \cos (2 (e+f x))-4) (\sin (3 (e+2 f x))-i \cos (3 (e+2 f x)))}{15 c^3 f (\cos (f x)+i \sin (f x))^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 76, normalized size = 0.83 \[ \frac {\sqrt {2} {\left (-3 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, c^{3} f} \]
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 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.22, size = 66, normalized size = 0.72 \[ \frac {2 i a^{3} \left (\frac {4 c}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {4 c^{2}}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.74, size = 65, normalized size = 0.71 \[ -\frac {2 i \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} - 20 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c + 12 \, a^{3} c^{2}\right )}}{15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{2} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.46, size = 121, normalized size = 1.32 \[ -\frac {a^3\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}-\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-4\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )-3\,\sin \left (6\,e+6\,f\,x\right )+8{}\mathrm {i}\right )}{15\,c^3\,f} \]
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 \[ - i a^{3} \left (\int \frac {i}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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