Optimal. Leaf size=336 \[ -\frac {9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}-\frac {7\ 3^{3/4} F\left (\text {ArcCos}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}} \]
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Rubi [A]
time = 0.30, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3885, 4086,
3913, 3912, 65, 231} \begin {gather*} -\frac {7\ 3^{3/4} \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} F\left (\text {ArcCos}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{10 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} \sqrt [3]{a \sec (c+d x)+a}}+\frac {3 \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{5 a d}-\frac {9 \tan (c+d x)}{10 d \sqrt [3]{a \sec (c+d x)+a}} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 231
Rule 3885
Rule 3912
Rule 3913
Rule 4086
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx &=\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}+\frac {3 \int \frac {\sec (c+d x) \left (\frac {2 a}{3}-a \sec (c+d x)\right )}{\sqrt [3]{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=-\frac {9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}+\frac {7}{10} \int \frac {\sec (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\\ &=-\frac {9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}+\frac {\left (7 \sqrt [3]{1+\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{10 \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}-\frac {(7 \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{10 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}-\frac {(21 \tan (c+d x)) \text {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac {9 \tan (c+d x)}{10 d \sqrt [3]{a+a \sec (c+d x)}}+\frac {3 (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{5 a d}-\frac {7\ 3^{3/4} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right ) \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right ) \sqrt {\frac {2^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\sec (c+d x)}+(1+\sec (c+d x))^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}} \tan (c+d x)}{10 \sqrt [3]{2} d (1-\sec (c+d x)) \sqrt [3]{a+a \sec (c+d x)} \sqrt {-\frac {\sqrt [3]{1+\sec (c+d x)} \left (\sqrt [3]{2}-\sqrt [3]{1+\sec (c+d x)}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{1+\sec (c+d x)}\right )^2}}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in
optimal.
time = 0.21, size = 95, normalized size = 0.28 \begin {gather*} \frac {\left (7 \sqrt [6]{2} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right )+3 \sqrt [6]{1+\sec (c+d x)} (-1+2 \sec (c+d x))\right ) \tan (c+d x)}{10 d \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a (1+\sec (c+d x))}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\sec ^{3}\left (d x +c \right )}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{3}{\left (c + d x \right )}}{\sqrt [3]{a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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