Optimal. Leaf size=136 \[ -\frac{4 d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{77 b \sqrt{d \tan (a+b x)}}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b} \]
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Rubi [A] time = 0.183455, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2611, 2613, 2614, 2573, 2641} \[ -\frac{4 d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{77 b \sqrt{d \tan (a+b x)}}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b} \]
Antiderivative was successfully verified.
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Rule 2611
Rule 2613
Rule 2614
Rule 2573
Rule 2641
Rubi steps
\begin{align*} \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{1}{11} d^2 \int \frac{\sec ^5(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{1}{77} \left (6 d^2\right ) \int \frac{\sec ^3(a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{1}{77} \left (4 d^2\right ) \int \frac{\sec (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{\left (4 d^2 \sqrt{\sin (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{77 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}-\frac{\left (4 d^2 \sec (a+b x) \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{77 \sqrt{d \tan (a+b x)}}\\ &=-\frac{4 d^2 F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt{\sin (2 a+2 b x)}}{77 b \sqrt{d \tan (a+b x)}}-\frac{4 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{77 b}-\frac{2 d \sec ^3(a+b x) \sqrt{d \tan (a+b x)}}{77 b}+\frac{2 d \sec ^5(a+b x) \sqrt{d \tan (a+b x)}}{11 b}\\ \end{align*}
Mathematica [C] time = 0.799281, size = 90, normalized size = 0.66 \[ -\frac{d \sec ^5(a+b x) \sqrt{d \tan (a+b x)} \left (16 \cos ^6(a+b x) \sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2(a+b x)\right )+6 \cos (2 (a+b x))+\cos (4 (a+b x))-23\right )}{154 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.135, size = 251, normalized size = 1.9 \begin{align*}{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{77\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4} \left ( \sin \left ( bx+a \right ) \right ) ^{5}} \left ( 4\,\sin \left ( bx+a \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{\cos \left ( bx+a \right ) -1+\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}} \left ( \cos \left ( bx+a \right ) \right ) ^{5}{\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( bx+a \right ) -1-\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) -2\,\sqrt{2} \left ( \cos \left ( bx+a \right ) \right ) ^{5}+2\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}\sqrt{2}- \left ( \cos \left ( bx+a \right ) \right ) ^{3}\sqrt{2}+ \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}+7\,\cos \left ( bx+a \right ) \sqrt{2}-7\,\sqrt{2} \right ) \left ({\frac{d\sin \left ( bx+a \right ) }{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{3}{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 \left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}} \sec \left (b x + a\right )^{5}\,{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 (\sqrt{d \tan \left (b x + a\right )} d \sec \left (b x + a\right )^{5} \tan \left (b x + a\right ), 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 \left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}} \sec \left (b x + a\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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