Optimal. Leaf size=102 \[ -\frac {(-4 B+i A) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]
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Rubi [A] time = 0.23, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3588, 78, 37} \[ -\frac {(-4 B+i A) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 37
Rule 78
Rule 3588
Rubi steps
\begin {align*} \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x} (A+B x)}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {(a (A+4 i B)) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(i A-4 B) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}\\ \end {align*}
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Mathematica [A] time = 11.68, size = 117, normalized size = 1.15 \[ \frac {a \cos (e+f x) (\cos (f x)-i \sin (f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (\cos (4 e+5 f x)+i \sin (4 e+5 f x)) ((B-4 i A) \cos (e+f x)-(A+4 i B) \sin (e+f x))}{15 c^3 f} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.67, size = 97, normalized size = 0.95 \[ \frac {{\left ({\left (-3 i \, A - 3 \, B\right )} a e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-8 i \, A + 2 \, B\right )} a e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-5 i \, A + 5 \, B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]
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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.48, size = 90, normalized size = 0.88 \[ \frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (-4 A +i A \tan \left (f x +e \right )-i B -4 B \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.26, size = 155, normalized size = 1.52 \[ -\frac {{\left (90 \, {\left (A - i \, B\right )} a \cos \left (7 \, f x + 7 \, e\right ) + 60 \, {\left (4 \, A + i \, B\right )} a \cos \left (5 \, f x + 5 \, e\right ) + 150 \, {\left (A + i \, B\right )} a \cos \left (3 \, f x + 3 \, e\right ) + {\left (90 i \, A + 90 \, B\right )} a \sin \left (7 \, f x + 7 \, e\right ) + {\left (240 i \, A - 60 \, B\right )} a \sin \left (5 \, f x + 5 \, e\right ) + {\left (150 i \, A - 150 \, B\right )} a \sin \left (3 \, f x + 3 \, e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (-900 i \, c^{3} \cos \left (2 \, f x + 2 \, e\right ) + 900 \, c^{3} \sin \left (2 \, f x + 2 \, e\right ) - 900 i \, c^{3}\right )} f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.26, size = 190, normalized size = 1.86 \[ -\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-5\,B\,\cos \left (2\,e+2\,f\,x\right )+3\,B\,\cos \left (4\,e+4\,f\,x\right )-5\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
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 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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