Optimal. Leaf size=152 \[ -\frac {i A+B}{f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(2 i A+B) \sqrt {c-i c \tan (e+f x)}}{3 c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 i A+B) \sqrt {c-i c \tan (e+f x)}}{3 a c f \sqrt {a+i a \tan (e+f x)}} \]
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Rubi [A]
time = 0.16, antiderivative size = 152, 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 = {3669, 79, 47,
37} \begin {gather*} -\frac {B+i A}{f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(B+2 i A) \sqrt {c-i c \tan (e+f x)}}{3 a c f \sqrt {a+i a \tan (e+f x)}}+\frac {(B+2 i A) \sqrt {c-i c \tan (e+f x)}}{3 c f (a+i a \tan (e+f x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 37
Rule 47
Rule 79
Rule 3669
Rubi steps
\begin {align*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^{5/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i A+B}{f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(a (2 A-i B)) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i A+B}{f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(2 i A+B) \sqrt {c-i c \tan (e+f x)}}{3 c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 A-i B) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{3 f}\\ &=-\frac {i A+B}{f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {(2 i A+B) \sqrt {c-i c \tan (e+f x)}}{3 c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 i A+B) \sqrt {c-i c \tan (e+f x)}}{3 a c f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 1.39, size = 85, normalized size = 0.56 \begin {gather*} -\frac {i (-3 A+(A-2 i B) \cos (2 (e+f x))+(2 i A+B) \sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}}{6 a c f \sqrt {a+i a \tan (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.43, size = 152, normalized size = 1.00
method | result | size |
derivativedivides | \(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \left (\tan ^{4}\left (f x +e \right )\right )-i B \left (\tan ^{3}\left (f x +e \right )\right )+B \left (\tan ^{4}\left (f x +e \right )\right )+3 i A \left (\tan ^{2}\left (f x +e \right )\right )+2 A \left (\tan ^{3}\left (f x +e \right )\right )-i B \tan \left (f x +e \right )+i A +2 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(152\) |
default | \(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \left (\tan ^{4}\left (f x +e \right )\right )-i B \left (\tan ^{3}\left (f x +e \right )\right )+B \left (\tan ^{4}\left (f x +e \right )\right )+3 i A \left (\tan ^{2}\left (f x +e \right )\right )+2 A \left (\tan ^{3}\left (f x +e \right )\right )-i B \tan \left (f x +e \right )+i A +2 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(152\) |
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 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.92, size = 153, normalized size = 1.01 \begin {gather*} -\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, {\left (i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 3 \, {\left (-i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, {\left (i \, A - B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} - {\left (7 i \, A - B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, A + 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 (-3 i \, f x - 3 i \, e\right )}}{12 \, a^{2} c f} \end {gather*}
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 \begin {gather*} \int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 9.77, size = 170, normalized size = 1.12 \begin {gather*} \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 (6\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,6{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-B\,\cos \left (4\,e+4\,f\,x\right )-A\,3{}\mathrm {i}+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{12\,a^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}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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