Optimal. Leaf size=71 \[ -\frac {3 i \, _2F_1\left (-\frac {5}{6},\frac {17}{6};\frac {1}{6};\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}} \]
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Rubi [A]
time = 0.15, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3586, 3604, 72,
71} \begin {gather*} -\frac {3 i (1+i \tan (e+f x))^{5/6} \, _2F_1\left (-\frac {5}{6},\frac {17}{6};\frac {1}{6};\frac {1}{2} (1-i \tan (e+f x))\right )}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 3586
Rule 3604
Rubi steps
\begin {align*} \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))} \, dx &=\frac {\left ((a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \int \frac {1}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{11/6}} \, dx}{(d \sec (e+f x))^{5/3}}\\ &=\frac {\left (a^2 (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \text {Subst}\left (\int \frac {1}{(a-i a x)^{11/6} (a+i a x)^{17/6}} \, dx,x,\tan (e+f x)\right )}{f (d \sec (e+f x))^{5/3}}\\ &=\frac {\left ((a-i a \tan (e+f x))^{5/6} \left (\frac {a+i a \tan (e+f x)}{a}\right )^{5/6}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {1}{2}+\frac {i x}{2}\right )^{17/6} (a-i a x)^{11/6}} \, dx,x,\tan (e+f x)\right )}{4\ 2^{5/6} f (d \sec (e+f x))^{5/3}}\\ &=-\frac {3 i \, _2F_1\left (-\frac {5}{6},\frac {17}{6};\frac {1}{6};\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}}\\ \end {align*}
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Mathematica [A]
time = 0.91, size = 119, normalized size = 1.68 \begin {gather*} -\frac {3 \sec ^2(e+f x) \left (-26+6 \cos (2 (e+f x))+\frac {128 e^{2 i (e+f x)} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (e+f x)}\right )}{\left (1+e^{2 i (e+f x)}\right )^{2/3}}+16 i \sin (2 (e+f x))\right )}{220 a f (d \sec (e+f x))^{5/3} (-i+\tan (e+f x))} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.45, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +i a \tan \left (f x +e \right )\right )}\, dx\]
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 [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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}} \tan {\left (e + f x \right )} - i \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx}{a} \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 [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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