Optimal. Leaf size=174 \[ \frac{a \sin (c+d x) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac{2}{3}}(c+d x) F_1\left (\frac{1}{2};-\frac{2}{3},1;\frac{3}{2};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac{b \sin (c+d x) F_1\left (\frac{1}{2};-\frac{1}{6},1;\frac{3}{2};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}} \]
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Rubi [A] time = 0.240486, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3869, 2823, 3189, 429} \[ \frac{a \sin (c+d x) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac{2}{3}}(c+d x) F_1\left (\frac{1}{2};-\frac{2}{3},1;\frac{3}{2};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right )}-\frac{b \sin (c+d x) F_1\left (\frac{1}{2};-\frac{1}{6},1;\frac{3}{2};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{d \left (a^2-b^2\right ) \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 3869
Rule 2823
Rule 3189
Rule 429
Rubi steps
\begin{align*} \int \frac{1}{\sqrt [3]{\sec (c+d x)} (a+b \sec (c+d x))} \, dx &=\left (\cos ^{\frac{2}{3}}(c+d x) \sec ^{\frac{2}{3}}(c+d x)\right ) \int \frac{\cos ^{\frac{4}{3}}(c+d x)}{b+a \cos (c+d x)} \, dx\\ &=-\left (\left (a \cos ^{\frac{2}{3}}(c+d x) \sec ^{\frac{2}{3}}(c+d x)\right ) \int \frac{\cos ^{\frac{7}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)} \, dx\right )+\left (b \cos ^{\frac{2}{3}}(c+d x) \sec ^{\frac{2}{3}}(c+d x)\right ) \int \frac{\cos ^{\frac{4}{3}}(c+d x)}{b^2-a^2 \cos ^2(c+d x)} \, dx\\ &=\frac{b \operatorname{Subst}\left (\int \frac{\sqrt [6]{1-x^2}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (c+d x)\right )}{d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}-\frac{\left (a \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac{2}{3}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{2/3}}{-a^2+b^2+a^2 x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{b F_1\left (\frac{1}{2};-\frac{1}{6},1;\frac{3}{2};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [6]{\cos ^2(c+d x)} \sqrt [3]{\sec (c+d x)}}+\frac{a F_1\left (\frac{1}{2};-\frac{2}{3},1;\frac{3}{2};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [3]{\cos ^2(c+d x)} \sec ^{\frac{2}{3}}(c+d x) \sin (c+d x)}{\left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [B] time = 28.4726, size = 7542, normalized size = 43.34 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.103, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{a+b\sec \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{\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{1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{1}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \sec{\left (c + d x \right )}\right ) \sqrt [3]{\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{1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac{1}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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