Optimal. Leaf size=265 \[ \frac{\sqrt{2} \left (3 a^2+2 b^2\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac{3 \sqrt{2} a \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac{3 \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 b d} \]
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Rubi [A] time = 0.332234, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3840, 4007, 3834, 139, 138} \[ \frac{\sqrt{2} \left (3 a^2+2 b^2\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac{3 \sqrt{2} a \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{5 b^2 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac{3 \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 b d} \]
Antiderivative was successfully verified.
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Rule 3840
Rule 4007
Rule 3834
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx &=\frac{3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac{3 \int \frac{\sec (c+d x) \left (\frac{2 b}{3}-a \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{5 b}\\ &=\frac{3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac{1}{5} \left (2+\frac{3 a^2}{b^2}\right ) \int \frac{\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx-\frac{(3 a) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{5 b^2}\\ &=\frac{3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac{\left (\left (-2-\frac{3 a^2}{b^2}\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{5 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}+\frac{(3 a \tan (c+d x)) \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{5 b^2 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=\frac{3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac{\left (3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{5 b^2 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \left (-\frac{a+b \sec (c+d x)}{-a-b}\right )^{2/3}}+\frac{\left (\left (-2-\frac{3 a^2}{b^2}\right ) \sqrt [3]{-\frac{a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \sqrt [3]{-\frac{a}{-a-b}-\frac{b x}{-a-b}}} \, dx,x,\sec (c+d x)\right )}{5 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac{3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}-\frac{3 \sqrt{2} a F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d \sqrt{1+\sec (c+d x)} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac{\sqrt{2} \left (2+\frac{3 a^2}{b^2}\right ) F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{5 d \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end{align*}
Mathematica [B] time = 25.971, size = 7195, normalized size = 27.15 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.119, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{3}{\frac{1}{\sqrt [3]{a+b\sec \left ( dx+c \right ) }}}}\, dx \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 )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{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{\sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}, 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 \frac{\sec ^{3}{\left (c + d x \right )}}{\sqrt [3]{a + b \sec{\left (c + d 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 \frac{\sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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