Optimal. Leaf size=265 \[ \frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}-\frac {3 \sqrt {2} a F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} \left (3 a^2+2 b^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{5 b^2 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
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Rubi [A]
time = 0.25, antiderivative size = 265, normalized size of antiderivative = 1.00, 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 = {3925, 4092,
3919, 144, 143} \begin {gather*} \frac {\sqrt {2} \left (3 a^2+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 )}{5 b^2 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac {3 \sqrt {2} a \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 )}{5 b^2 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {3 \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{5 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 3919
Rule 3925
Rule 4092
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {3 \int \frac {\sec (c+d x) \left (\frac {2 b}{3}-a \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{5 b}\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {1}{5} \left (2+\frac {3 a^2}{b^2}\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx-\frac {(3 a) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{5 b^2}\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {\left (\left (-2-\frac {3 a^2}{b^2}\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \sqrt [3]{a+b x}} \, dx,x,\sec (c+d x)\right )}{5 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {(3 a \tan (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{5 b^2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}+\frac {\left (3 a (a+b \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {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 )}{5 b^2 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 (\left (-2-\frac {3 a^2}{b^2}\right ) \sqrt [3]{-\frac {a+b \sec (c+d x)}{-a-b}} \tan (c+d x)\right ) \text {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 )}{5 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac {3 (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d}-\frac {3 \sqrt {2} a 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)}{5 b^2 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} \left (2+\frac {3 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)}{5 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(7195\) vs. \(2(265)=530\).
time = 36.43, size = 7195, normalized size = 27.15 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\sec ^{3}\left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]
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 + b \sec {\left (c + d x \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 {b}{\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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