Optimal. Leaf size=126 \[ \frac{2 b^3 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}-\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 b^2 d \sqrt{\cos (c+d x)}}+\frac{14 b \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b d \sqrt{b \cos (c+d x)}} \]
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Rubi [A] time = 0.0961565, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {16, 2636, 2640, 2639} \[ \frac{2 b^3 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}-\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 b^2 d \sqrt{\cos (c+d x)}}+\frac{14 b \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b d \sqrt{b \cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 16
Rule 2636
Rule 2640
Rule 2639
Rubi steps
\begin{align*} \int \frac{\sec ^4(c+d x)}{(b \cos (c+d x))^{3/2}} \, dx &=b^4 \int \frac{1}{(b \cos (c+d x))^{11/2}} \, dx\\ &=\frac{2 b^3 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{1}{9} \left (7 b^2\right ) \int \frac{1}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\frac{2 b^3 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{7}{15} \int \frac{1}{(b \cos (c+d x))^{3/2}} \, dx\\ &=\frac{2 b^3 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b d \sqrt{b \cos (c+d x)}}-\frac{7 \int \sqrt{b \cos (c+d x)} \, dx}{15 b^2}\\ &=\frac{2 b^3 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b d \sqrt{b \cos (c+d x)}}-\frac{\left (7 \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 b^2 \sqrt{\cos (c+d x)}}\\ &=-\frac{14 \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^2 d \sqrt{\cos (c+d x)}}+\frac{2 b^3 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 b \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b d \sqrt{b \cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.181281, size = 80, normalized size = 0.63 \[ \frac{42 \sin (c+d x)-42 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \tan (c+d x) \sec (c+d x) \left (5 \sec ^2(c+d x)+7\right )}{45 b d \sqrt{b \cos (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.647, size = 414, normalized size = 3.3 \begin{align*} -2\,{\frac{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{b\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d} \left ( -{\frac{\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}{144\,b \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{5}}}-{\frac{7\,\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}{180\,b \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{3}}}-{\frac{14\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) }{15\,\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}+{\frac{7\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{15\,\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}}-{\frac{7\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) }{15\,\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}} \right ) } \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{\sec \left (d x + c\right )^{4}}{\left (b \cos \left (d x + c\right )\right )^{\frac{3}{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{b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{4}}{b^{2} \cos \left (d x + c\right )^{2}}, 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{\sec \left (d x + c\right )^{4}}{\left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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