Optimal. Leaf size=307 \[ -\frac{a \left (3 a^2-4 b^2\right ) \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}+\frac{\left (3 a^2-2 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}-\frac{3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}} \]
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Rubi [A] time = 0.380614, antiderivative size = 307, 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 = {3839, 4007, 3834, 139, 138} \[ -\frac{a \left (3 a^2-4 b^2\right ) \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}+\frac{\left (3 a^2-2 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{\sqrt{2} b^2 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}-\frac{3 a^2 \tan (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}} \]
Antiderivative was successfully verified.
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Rule 3839
Rule 4007
Rule 3834
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx &=-\frac{3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac{3 \int \frac{\sec (c+d x) \left (-\frac{2 a b}{3}-\frac{1}{3} \left (3 a^2-2 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac{\left (a \left (3 a^2-4 b^2\right )\right ) \int \frac{\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac{\left (3 a^2-2 b^2\right ) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{\left (a \left (3 a^2-4 b^2\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} (a+b x)^{2/3}} \, dx,x,\sec (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}-\frac{\left (\left (3 a^2-2 b^2\right ) \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sec (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=-\frac{3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac{\left (\left (3 a^2-2 b^2\right ) \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 )}{2 b^2 \left (a^2-b^2\right ) 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{\left (a \left (3 a^2-4 b^2\right ) \left (-\frac{a+b \sec (c+d x)}{-a-b}\right )^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} \sqrt{1+x} \left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{2/3}} \, dx,x,\sec (c+d x)\right )}{2 b^2 \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}}\\ &=-\frac{3 a^2 \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{\left (3 a^2-2 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]{a+b \sec (c+d x)} \tan (c+d x)}{\sqrt{2} b^2 \left (a^2-b^2\right ) d \sqrt{1+\sec (c+d x)} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}-\frac{a \left (3 a^2-4 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 ) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{\sqrt{2} b^2 \left (a^2-b^2\right ) d \sqrt{1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}}\\ \end{align*}
Mathematica [B] time = 26.2113, size = 19126, normalized size = 62.3 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.111, 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{5}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}} \sec \left (d x + c\right )^{3}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, 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 )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{5}{3}}}\, 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{5}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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