Optimal. Leaf size=155 \[ -\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(2 i A-3 B) \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(2 i A-3 B) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}} \]
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Rubi [A]
time = 0.16, antiderivative size = 155, 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 {(-3 B+2 i A) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {(-3 B+2 i A) \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 37
Rule 47
Rule 79
Rule 3669
Rubi steps
\begin {align*} \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{\sqrt {a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {(a (2 A+3 i B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(2 i A-3 B) \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}+\frac {(a (2 A+3 i B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 c f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(2 i A-3 B) \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(2 i A-3 B) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 2.74, size = 114, normalized size = 0.74 \begin {gather*} \frac {\cos (e+f x) (-5 i A+(-9 i A+6 B) \cos (2 (e+f x))-3 (2 A+3 i B) \sin (2 (e+f x))) (\cos (3 (e+f x))+i \sin (3 (e+f x))) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{30 c^3 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.42, size = 125, normalized size = 0.81
method | result | size |
risch | \(-\frac {\sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+3 B \,{\mathrm e}^{4 i \left (f x +e \right )}+10 i A \,{\mathrm e}^{2 i \left (f x +e \right )}+15 i A -15 B \right )}{60 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(98\) |
derivativedivides | \(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i A \left (\tan ^{3}\left (f x +e \right )\right )-12 i B \left (\tan ^{2}\left (f x +e \right )\right )-3 B \left (\tan ^{3}\left (f x +e \right )\right )-13 i A \tan \left (f x +e \right )-8 A \left (\tan ^{2}\left (f x +e \right )\right )+3 i B +12 B \tan \left (f x +e \right )+7 A \right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(125\) |
default | \(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i A \left (\tan ^{3}\left (f x +e \right )\right )-12 i B \left (\tan ^{2}\left (f x +e \right )\right )-3 B \left (\tan ^{3}\left (f x +e \right )\right )-13 i A \tan \left (f x +e \right )-8 A \left (\tan ^{2}\left (f x +e \right )\right )+3 i B +12 B \tan \left (f x +e \right )+7 A \right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(125\) |
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 = 9.58, size = 119, normalized size = 0.77 \begin {gather*} -\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} - {\left (-13 i \, A - 3 \, B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 5 \, {\left (5 i \, A - 3 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 15 \, {\left (i \, A - B\right )} e^{\left (i \, f x + 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}}}{60 \, c^{3} 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 {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \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 \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 = 9.86, size = 171, normalized size = 1.10 \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 (A\,15{}\mathrm {i}-15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+3\,B\,\cos \left (4\,e+4\,f\,x\right )-10\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{60\,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}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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