Optimal. Leaf size=305 \[ -\frac{\left (6 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{20 \sqrt{2} b^2 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac{3 a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{10 \sqrt{2} b^2 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}-\frac{9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{40 b d} \]
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Rubi [A] time = 0.47028, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3840, 4002, 4007, 3834, 139, 138} \[ -\frac{\left (6 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{20 \sqrt{2} b^2 d \sqrt{\sec (c+d x)+1} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac{3 a \left (a^2-b^2\right ) \tan (c+d x) \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{10 \sqrt{2} b^2 d \sqrt{\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}+\frac{3 \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{8 b d}-\frac{9 a \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{40 b d} \]
Antiderivative was successfully verified.
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Rule 3840
Rule 4002
Rule 4007
Rule 3834
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^{2/3} \, dx &=\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac{3 \int \sec (c+d x) \left (\frac{5 b}{3}-a \sec (c+d x)\right ) (a+b \sec (c+d x))^{2/3} \, dx}{8 b}\\ &=-\frac{9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac{9 \int \frac{\sec (c+d x) \left (\frac{19 a b}{9}-\frac{1}{9} \left (6 a^2-25 b^2\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{40 b}\\ &=-\frac{9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac{1}{40} \left (25-\frac{6 a^2}{b^2}\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx+\frac{\left (3 a \left (a^2-b^2\right )\right ) \int \frac{\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{20 b^2}\\ &=-\frac{9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac{\left (\left (-25+\frac{6 a^2}{b^2}\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{2/3}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{40 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}-\frac{\left (3 a \left (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 )}{20 b^2 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=-\frac{9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac{\left (\left (-25+\frac{6 a^2}{b^2}\right ) (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 )}{40 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 (3 a \left (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 )}{20 b^2 d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=-\frac{9 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{40 b d}+\frac{3 (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{8 b d}+\frac{\left (25-\frac{6 a^2}{b^2}\right ) 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)}{20 \sqrt{2} d \sqrt{1+\sec (c+d x)} \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac{3 a \left (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)}{10 \sqrt{2} b^2 d \sqrt{1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end{align*}
Mathematica [B] time = 26.3891, size = 18991, normalized size = 62.27 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.115, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{3} \left ( a+b\sec \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}\, 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 (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{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 (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{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 (a + b \sec{\left (c + d x \right )}\right )^{\frac{2}{3}} \sec ^{3}{\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 (b \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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