Optimal. Leaf size=413 \[ \frac {3 \tan (c+d x) (a \sec (c+d x)+a)^{8/3}}{11 a d}-\frac {9 \tan (c+d x) (a \sec (c+d x)+a)^{5/3}}{88 d}+\frac {147 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{440 d}+\frac {1029 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{880 d (\sec (c+d x)+1)}-\frac {343\ 3^{3/4} a \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{880 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
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Rubi [A] time = 0.51, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3800, 4001, 3828, 3827, 50, 63, 225} \[ \frac {3 \tan (c+d x) (a \sec (c+d x)+a)^{8/3}}{11 a d}-\frac {9 \tan (c+d x) (a \sec (c+d x)+a)^{5/3}}{88 d}+\frac {147 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{440 d}+\frac {1029 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3}}{880 d (\sec (c+d x)+1)}-\frac {343\ 3^{3/4} a \tan (c+d x) \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right ) \sqrt {\frac {(\sec (c+d x)+1)^{2/3}+\sqrt [3]{2} \sqrt [3]{\sec (c+d x)+1}+2^{2/3}}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}} (a \sec (c+d x)+a)^{2/3} F\left (\cos ^{-1}\left (\frac {\sqrt [3]{2}-\left (1-\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}{\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{880 \sqrt [3]{2} d (1-\sec (c+d x)) (\sec (c+d x)+1) \sqrt {-\frac {\sqrt [3]{\sec (c+d x)+1} \left (\sqrt [3]{2}-\sqrt [3]{\sec (c+d x)+1}\right )}{\left (\sqrt [3]{2}-\left (1+\sqrt {3}\right ) \sqrt [3]{\sec (c+d x)+1}\right )^2}}} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 225
Rule 3800
Rule 3827
Rule 3828
Rule 4001
Rubi steps
\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^{5/3} \, dx &=\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}+\frac {3 \int \sec (c+d x) \left (\frac {8 a}{3}-a \sec (c+d x)\right ) (a+a \sec (c+d x))^{5/3} \, dx}{11 a}\\ &=-\frac {9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}+\frac {49}{88} \int \sec (c+d x) (a+a \sec (c+d x))^{5/3} \, dx\\ &=-\frac {9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}+\frac {\left (49 a (a+a \sec (c+d x))^{2/3}\right ) \int \sec (c+d x) (1+\sec (c+d x))^{5/3} \, dx}{88 (1+\sec (c+d x))^{2/3}}\\ &=-\frac {9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac {\left (49 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {(1+x)^{7/6}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{88 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}-\frac {9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac {\left (343 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt [6]{1+x}}{\sqrt {1-x}} \, dx,x,\sec (c+d x)\right )}{440 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}+\frac {1029 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{880 d (1+\sec (c+d x))}-\frac {9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac {\left (343 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{880 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}+\frac {1029 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{880 d (1+\sec (c+d x))}-\frac {9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac {\left (1029 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-x^6}} \, dx,x,\sqrt [6]{1+\sec (c+d x)}\right )}{440 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac {147 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{440 d}+\frac {1029 a (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{880 d (1+\sec (c+d x))}-\frac {9 (a+a \sec (c+d x))^{5/3} \tan (c+d x)}{88 d}+\frac {3 (a+a \sec (c+d x))^{8/3} \tan (c+d x)}{11 a d}-\frac {343\ 3^{3/4} a 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 ) (a+a \sec (c+d x))^{2/3} \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)}{880 \sqrt [3]{2} d (1-\sec (c+d x)) (1+\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] time = 0.34, size = 96, normalized size = 0.23 \[ \frac {a \tan (c+d x) (a (\sec (c+d x)+1))^{2/3} \left (196 \sqrt [6]{2} \, _2F_1\left (-\frac {7}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sec (c+d x))\right )+3 (8 \sec (c+d x)+5) (\sec (c+d x)+1)^{13/6}\right )}{88 d (\sec (c+d x)+1)^{7/6}} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right )^{4} + a \sec \left (d x + c\right )^{3}\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}, x\right ) \]
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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.72, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}\, dx \]
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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \sec \left (d x + c\right )^{3}\,{d x} \]
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 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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