Optimal. Leaf size=132 \[ -\frac{8 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{3 b^2 d^2 f \sqrt{b \tan (e+f x)}}-\frac{4 \sqrt{b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}} \]
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Rubi [A] time = 0.184395, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2609, 2612, 2616, 2642, 2641} \[ -\frac{8 \sqrt{\sin (e+f x)} F\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{d \sec (e+f x)}}{3 b^2 d^2 f \sqrt{b \tan (e+f x)}}-\frac{4 \sqrt{b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2609
Rule 2612
Rule 2616
Rule 2642
Rule 2641
Rubi steps
\begin{align*} \int \frac{1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx &=-\frac{2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac{2 \int \frac{1}{(d \sec (e+f x))^{3/2} \sqrt{b \tan (e+f x)}} \, dx}{b^2}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac{4 \sqrt{b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac{4 \int \frac{\sqrt{d \sec (e+f x)}}{\sqrt{b \tan (e+f x)}} \, dx}{3 b^2 d^2}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac{4 \sqrt{b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac{\left (4 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}\right ) \int \frac{1}{\sqrt{b \sin (e+f x)}} \, dx}{3 b^2 d^2 \sqrt{b \tan (e+f x)}}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac{4 \sqrt{b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}-\frac{\left (4 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}\right ) \int \frac{1}{\sqrt{\sin (e+f x)}} \, dx}{3 b^2 d^2 \sqrt{b \tan (e+f x)}}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac{8 F\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}{3 b^2 d^2 f \sqrt{b \tan (e+f x)}}-\frac{4 \sqrt{b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 2.37566, size = 112, normalized size = 0.85 \[ \frac{\left (-\tan ^2(e+f x)\right )^{3/4} \csc ^2(e+f x) \sqrt{b \tan (e+f x)} \left (\sqrt [4]{-\tan ^2(e+f x)} \left (\cos (2 (e+f x))+2 \csc ^2(e+f x)-1\right )-8 \, _2F_1\left (\frac{1}{4},\frac{3}{4};\frac{5}{4};\sec ^2(e+f x)\right )\right )}{3 b^3 f (d \sec (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.209, size = 335, normalized size = 2.5 \begin{align*}{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{4}} \left ( -4\,i\cos \left ( fx+e \right ) \sin \left ( fx+e \right ) \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{-{\frac{i\cos \left ( fx+e \right ) -i-\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }}}-4\,i\sin \left ( fx+e \right ) \sqrt{{\frac{-i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{-{\frac{i\cos \left ( fx+e \right ) -i-\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{i\cos \left ( fx+e \right ) -i+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{\sqrt{2}}{2}} \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{3}\sqrt{2}-2\,\cos \left ( fx+e \right ) \sqrt{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}{b^{3} d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right )^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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