Optimal. Leaf size=229 \[ \frac{\sqrt{2} (b B-a C) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{\sqrt{2} C (a+b) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}} \]
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Rubi [A] time = 0.246931, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {4062, 12, 3834, 139, 138} \[ \frac{\sqrt{2} (b B-a C) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{\sqrt{2} C (a+b) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{b d \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 4062
Rule 12
Rule 3834
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \sqrt [3]{a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{\int (b B-a C) \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{b}+\frac{C \int \sec (c+d x) (a+b \sec (c+d x))^{4/3} \, dx}{b}\\ &=\frac{(b B-a C) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{b}-\frac{(C \tan (c+d x)) \operatorname{Subst}\left (\int \frac{(a+b x)^{4/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=-\frac{((b B-a C) \tan (c+d x)) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}+\frac{\left ((-a-b) C \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{4/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{-\frac{a+b \sec (c+d x)}{-a-b}}}\\ &=\frac{\sqrt{2} (a+b) C F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt{1+\sec (c+d x)} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}-\frac{\left ((b B-a C) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{-\frac{a}{-a-b}-\frac{b x}{-a-b}}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{b d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{-\frac{a+b \sec (c+d x)}{-a-b}}}\\ &=\frac{\sqrt{2} (a+b) C F_1\left (\frac{1}{2};\frac{1}{2},-\frac{4}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt{1+\sec (c+d x)} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{\sqrt{2} (b B-a C) 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]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt{1+\sec (c+d x)} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}\\ \end{align*}
Mathematica [B] time = 26.6249, size = 21684, normalized size = 94.69 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.151, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{a+b\sec \left ( dx+c \right ) } \left ( B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \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{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )}{\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 ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )}{\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 \left (B + C \sec{\left (c + d x \right )}\right ) \sqrt [3]{a + b \sec{\left (c + d x \right )}} \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{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )}{\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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