Optimal. Leaf size=157 \[ \frac {(3 B+2 i A) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(3 B+2 i A) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}} \]
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Rubi [A] time = 0.24, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ \frac {(3 B+2 i A) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(3 B+2 i A) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 37
Rule 45
Rule 78
Rule 3588
Rubi steps
\begin {align*} \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^{7/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {((2 A-3 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {((2 A-3 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{15 a f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 5.19, size = 106, normalized size = 0.68 \[ \frac {\sec ^2(e+f x) \sqrt {c-i c \tan (e+f x)} ((6 A-9 i B) \sin (2 (e+f x))+(-6 B-9 i A) \cos (2 (e+f x))-5 i A)}{30 a^2 f (\tan (e+f x)-i)^2 \sqrt {a+i a \tan (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [A] time = 2.29, size = 109, normalized size = 0.69 \[ \frac {{\left ({\left (15 i \, A + 15 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (25 i \, A + 15 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (13 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{60 \, a^{3} f} \]
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 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.43, size = 127, normalized size = 0.81 \[ -\frac {i \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \left (\tan ^{3}\left (f x +e \right )\right )-12 i B \left (\tan ^{2}\left (f x +e \right )\right )+3 B \left (\tan ^{3}\left (f x +e \right )\right )-13 i A \tan \left (f x +e \right )+8 A \left (\tan ^{2}\left (f x +e \right )\right )+3 i B -12 B \tan \left (f x +e \right )-7 A \right )}{15 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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 = 10.96, size = 246, normalized size = 1.57 \[ \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,15{}\mathrm {i}+15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,25{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,13{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+15\,B\,\cos \left (2\,e+2\,f\,x\right )-3\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+25\,A\,\sin \left (2\,e+2\,f\,x\right )+13\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{120\,a^3\,f} \]
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 \[ \int \frac {\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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