Optimal. Leaf size=378 \[ -\frac{a \left (9 a^2-7 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt{2} b^3 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{\left (-10 a^2 b^2+9 a^4-b^4\right ) \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt{2} b^3 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}-\frac{3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}+\frac{3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{4 b^2 d \left (a^2-b^2\right )} \]
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Rubi [A] time = 0.560408, antiderivative size = 378, 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 = {3845, 4082, 4007, 3834, 139, 138} \[ -\frac{a \left (9 a^2-7 b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)} F_1\left (\frac{1}{2};\frac{1}{2},-\frac{1}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt{2} b^3 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} \sqrt [3]{\frac{a+b \sec (c+d x)}{a+b}}}+\frac{\left (-10 a^2 b^2+9 a^4-b^4\right ) \tan (c+d x) \left (\frac{a+b \sec (c+d x)}{a+b}\right )^{2/3} F_1\left (\frac{1}{2};\frac{1}{2},\frac{2}{3};\frac{3}{2};\frac{1}{2} (1-\sec (c+d x)),\frac{b (1-\sec (c+d x))}{a+b}\right )}{2 \sqrt{2} b^3 d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}-\frac{3 a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}}+\frac{3 \left (3 a^2-b^2\right ) \tan (c+d x) \sqrt [3]{a+b \sec (c+d x)}}{4 b^2 d \left (a^2-b^2\right )} \]
Antiderivative was successfully verified.
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Rule 3845
Rule 4082
Rule 4007
Rule 3834
Rule 139
Rule 138
Rubi steps
\begin{align*} \int \frac{\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx &=-\frac{3 a^2 \sec (c+d x) \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 (a^2-\frac{2}{3} a b \sec (c+d x)-\frac{2}{3} \left (3 a^2-b^2\right ) \sec ^2(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 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac{9 \int \frac{\sec (c+d x) \left (\frac{2}{9} b \left (3 a^2+b^2\right )+\frac{2}{9} a \left (9 a^2-7 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}} \, dx}{8 b^2 \left (a^2-b^2\right )}\\ &=-\frac{3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac{\left (a \left (9 a^2-7 b^2\right )\right ) \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{4 b^3 \left (a^2-b^2\right )}+\frac{\left (9 a^4-10 a^2 b^2-b^4\right ) \int \frac{\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx}{4 b^3 \left (a^2-b^2\right )}\\ &=-\frac{3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}+\frac{\left (a \left (9 a^2-7 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 )}{4 b^3 \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}-\frac{\left (\left (9 a^4-10 a^2 b^2-b^4\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 )}{4 b^3 \left (a^2-b^2\right ) d \sqrt{1-\sec (c+d x)} \sqrt{1+\sec (c+d x)}}\\ &=-\frac{3 a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}+\frac{\left (a \left (9 a^2-7 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 )}{4 b^3 \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 (\left (9 a^4-10 a^2 b^2-b^4\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 )}{4 b^3 \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 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac{3 \left (3 a^2-b^2\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{4 b^2 \left (a^2-b^2\right ) d}-\frac{a \left (9 a^2-7 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)}{2 \sqrt{2} b^3 \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{\left (9 a^4-10 a^2 b^2-b^4\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)}{2 \sqrt{2} b^3 \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 = 26.4083, size = 21987, normalized size = 58.17 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.124, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sec \left ( dx+c \right ) \right ) ^{4} \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 )^{4}}{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 ^{4}{\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 )^{4}}{{\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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