Optimal. Leaf size=289 \[ \frac {3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac {a F_1\left (\frac {1}{2};\frac {1}{2},-\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{\sqrt {2} b \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+\frac {\left (a^2-2 b^2\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{\sqrt {2} b \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}} \]
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Rubi [A]
time = 0.27, antiderivative size = 289, 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 = {3921, 4092,
3919, 144, 143} \begin {gather*} \frac {\left (a^2-2 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 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}-\frac {a \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 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 \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{2/3}} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 3919
Rule 3921
Rule 4092
Rubi steps
\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/3}} \, dx &=\frac {3 a \tan (c+d x)}{2 \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 b}{3}-\frac {1}{3} a \sec (c+d x)\right )}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac {3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac {a \int \sec (c+d x) \sqrt [3]{a+b \sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}+\frac {\left (a^2-2 b^2\right ) \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac {3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {(a \tan (c+d x)) \text {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{2 b \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {\left (\left (a^2-2 b^2\right ) \tan (c+d x)\right ) \text {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 \left (a^2-b^2\right ) d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=\frac {3 a \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}+\frac {\left (a \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)\right ) \text {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 \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 (a^2-2 b^2\right ) \left (-\frac {a+b \sec (c+d x)}{-a-b}\right )^{2/3} \tan (c+d x)\right ) \text {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 \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 \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{2/3}}-\frac {a 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 \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 (a^2-2 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 \left (a^2-b^2\right ) d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(7325\) vs. \(2(289)=578\).
time = 42.69, size = 7325, normalized size = 25.35 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\sec ^{2}\left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}}\, dx\]
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 ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{3}}}\, 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 )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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