Optimal. Leaf size=234 \[ -\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.19, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3523, 45, 37} \[ -\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 37
Rule 45
Rule 3523
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {(4 c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}+\frac {(4 c) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}+\frac {8 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}+\frac {8 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f}\\ &=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 6.02, size = 130, normalized size = 0.56 \[ \frac {\sec (e+f x) \sqrt {c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-40 i \sin (2 (e+f x))-4 i \sin (4 (e+f x))+20 \cos (2 (e+f x))+\cos (4 (e+f x))-45)}{120 a c^3 f (\tan (e+f x)-i) \sqrt {a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 133, normalized size = 0.57 \[ \frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 23 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 110 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 30 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 48 i \, e^{\left (3 i \, f x + 3 i \, e\right )} + 65 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{240 \, a^{2} c^{3} f} \]
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 \[ \int \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 130, normalized size = 0.56 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (8 i \left (\tan ^{5}\left (f x +e \right )\right )+8 \left (\tan ^{6}\left (f x +e \right )\right )+20 i \left (\tan ^{3}\left (f x +e \right )\right )+20 \left (\tan ^{4}\left (f x +e \right )\right )+12 i \tan \left (f x +e \right )+15 \left (\tan ^{2}\left (f x +e \right )\right )+3\right )}{15 f \,c^{3} a^{2} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{3}} \]
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 [B] time = 5.53, size = 140, normalized size = 0.60 \[ \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,20{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+40\,\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )-45{}\mathrm {i}\right )}{120\,a^2\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
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 \[ \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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