Optimal. Leaf size=80 \[ \frac{2 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{3 b}-\frac{d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{3 b \sqrt{d \tan (a+b x)}} \]
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Rubi [A] time = 0.0845474, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2611, 2614, 2573, 2641} \[ \frac{2 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{3 b}-\frac{d^2 \sqrt{\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{3 b \sqrt{d \tan (a+b x)}} \]
Antiderivative was successfully verified.
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Rule 2611
Rule 2614
Rule 2573
Rule 2641
Rubi steps
\begin{align*} \int \sec (a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac{2 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{3 b}-\frac{1}{3} d^2 \int \frac{\sec (a+b x)}{\sqrt{d \tan (a+b x)}} \, dx\\ &=\frac{2 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{3 b}-\frac{\left (d^2 \sqrt{\sin (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{3 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}}\\ &=\frac{2 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{3 b}-\frac{\left (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}{3 \sqrt{d \tan (a+b x)}}\\ &=-\frac{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)}}{3 b \sqrt{d \tan (a+b x)}}+\frac{2 d \sec (a+b x) \sqrt{d \tan (a+b x)}}{3 b}\\ \end{align*}
Mathematica [C] time = 0.299401, size = 69, normalized size = 0.86 \[ \frac{2 d \cos (a+b x) \sqrt{d \tan (a+b x)} \left (\sec ^2(a+b x)-\sqrt{\sec ^2(a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\tan ^2(a+b x)\right )\right )}{3 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.124, size = 186, normalized size = 2.3 \begin{align*}{\frac{\sqrt{2} \left ( \cos \left ( bx+a \right ) -1 \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{3\,b \left ( \sin \left ( bx+a \right ) \right ) ^{5}} \left ( \sin \left ( bx+a \right ) \cos \left ( bx+a \right ){\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{\cos \left ( bx+a \right ) -1}{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{1-\cos \left ( bx+a \right ) +\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 ) }}}+\cos \left ( bx+a \right ) \sqrt{2}-\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 )\,{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 ) \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan{\left (a + b x \right )}\right )^{\frac{3}{2}} \sec{\left (a + b x \right )}\, dx \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 )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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