Optimal. Leaf size=412 \[ \frac {3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac {3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac {\left (36 a^4+164 a^2 b^2+605 b^4\right ) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{616 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {a \left (18 a^4+79 a^2 b^2-97 b^4\right ) F_1\left (\frac {1}{2};\frac {1}{2},\frac {1}{3};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{308 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
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Rubi [A]
time = 0.59, antiderivative size = 412, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3950, 4167,
4087, 4092, 3919, 144, 143} \begin {gather*} \frac {3 \left (18 a^2+121 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{5/3}}{1232 b^2 d}+\frac {3 a \left (18 a^2+97 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{2/3}}{1232 b^2 d}+\frac {\left (36 a^4+164 a^2 b^2+605 b^4\right ) \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 )}{616 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {a \left (18 a^4+79 a^2 b^2-97 b^4\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 )}{308 \sqrt {2} b^3 d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}-\frac {9 a \tan (c+d x) (a+b \sec (c+d x))^{8/3}}{77 b^2 d}+\frac {3 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{8/3}}{14 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 3919
Rule 3950
Rule 4087
Rule 4092
Rule 4167
Rubi steps
\begin {align*} \int \sec ^4(c+d x) (a+b \sec (c+d x))^{5/3} \, dx &=\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac {3 \int \sec (c+d x) (a+b \sec (c+d x))^{5/3} \left (a+\frac {11}{3} b \sec (c+d x)-2 a \sec ^2(c+d x)\right ) \, dx}{14 b}\\ &=-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac {9 \int \sec (c+d x) (a+b \sec (c+d x))^{5/3} \left (-\frac {5 a b}{3}+\frac {1}{9} \left (18 a^2+121 b^2\right ) \sec (c+d x)\right ) \, dx}{154 b^2}\\ &=\frac {3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac {27 \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \left (-\frac {5}{27} b \left (6 a^2-121 b^2\right )+\frac {5}{27} a \left (18 a^2+97 b^2\right ) \sec (c+d x)\right ) \, dx}{1232 b^2}\\ &=\frac {3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac {3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac {81 \int \frac {\sec (c+d x) \left (\frac {5}{81} a b \left (6 a^2+799 b^2\right )+\frac {5}{81} \left (36 a^4+164 a^2 b^2+605 b^4\right ) \sec (c+d x)\right )}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{6160 b^2}\\ &=\frac {3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac {3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}-\frac {\left (a \left (18 a^4+79 a^2 b^2-97 b^4\right )\right ) \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx}{616 b^3}+\frac {\left (36 a^4+164 a^2 b^2+605 b^4\right ) \int \sec (c+d x) (a+b \sec (c+d x))^{2/3} \, dx}{1232 b^3}\\ &=\frac {3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac {3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac {\left (a \left (18 a^4+79 a^2 b^2-97 b^4\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 )}{616 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {\left (\left (36 a^4+164 a^2 b^2+605 b^4\right ) \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sec (c+d x)\right )}{1232 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\\ &=\frac {3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac {3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}-\frac {\left (\left (36 a^4+164 a^2 b^2+605 b^4\right ) (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 )}{1232 b^3 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 (a \left (18 a^4+79 a^2 b^2-97 b^4\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 )}{616 b^3 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}}\\ &=\frac {3 a \left (18 a^2+97 b^2\right ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{1232 b^2 d}+\frac {3 \left (18 a^2+121 b^2\right ) (a+b \sec (c+d x))^{5/3} \tan (c+d x)}{1232 b^2 d}-\frac {9 a (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{77 b^2 d}+\frac {3 \sec (c+d x) (a+b \sec (c+d x))^{8/3} \tan (c+d x)}{14 b d}+\frac {\left (36 a^4+164 a^2 b^2+605 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 ) (a+b \sec (c+d x))^{2/3} \tan (c+d x)}{616 \sqrt {2} b^3 d \sqrt {1+\sec (c+d x)} \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3}}-\frac {a \left (18 a^4+79 a^2 b^2-97 b^4\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)}{308 \sqrt {2} b^3 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. \(28057\) vs. \(2(412)=824\).
time = 38.67, size = 28057, normalized size = 68.10 \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 \[\int \left (\sec ^{4}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{\frac {5}{3}}\, dx\]
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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/3}}{{\cos \left (c+d\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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