Optimal. Leaf size=121 \[ \frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}+\frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a^2 d}-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a d}-\frac {2 (a \sec (c+d x)+a)^{3/2}}{3 d}-\frac {2 a \sqrt {a \sec (c+d x)+a}}{d} \]
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Rubi [A] time = 0.10, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3880, 80, 50, 63, 207} \[ \frac {2 (a \sec (c+d x)+a)^{7/2}}{7 a^2 d}+\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {2 (a \sec (c+d x)+a)^{5/2}}{5 a d}-\frac {2 (a \sec (c+d x)+a)^{3/2}}{3 d}-\frac {2 a \sqrt {a \sec (c+d x)+a}}{d} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 80
Rule 207
Rule 3880
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^{3/2} \tan ^3(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(-a+a x) (a+a x)^{5/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{5/2}}{x} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {a \operatorname {Subst}\left (\int \frac {\sqrt {a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}\\ &=\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}-\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}\\ \end {align*}
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Mathematica [A] time = 0.27, size = 92, normalized size = 0.76 \[ \frac {2 (a (\sec (c+d x)+1))^{3/2} \left (\sqrt {\sec (c+d x)+1} \left (15 \sec ^3(c+d x)+24 \sec ^2(c+d x)-32 \sec (c+d x)-146\right )+105 \tanh ^{-1}\left (\sqrt {\sec (c+d x)+1}\right )\right )}{105 d (\sec (c+d x)+1)^{3/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.33, size = 290, normalized size = 2.40 \[ \left [\frac {105 \, a^{\frac {3}{2}} \cos \left (d x + c\right )^{3} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, {\left (146 \, a \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} - 24 \, a \cos \left (d x + c\right ) - 15 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{210 \, d \cos \left (d x + c\right )^{3}}, -\frac {105 \, \sqrt {-a} a \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{3} + 2 \, {\left (146 \, a \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} - 24 \, a \cos \left (d x + c\right ) - 15 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.13, size = 173, normalized size = 1.43 \[ -\frac {\sqrt {2} {\left (\frac {105 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} a^{2} - 70 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a^{3} + 84 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{4} + 120 \, a^{5}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{105 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.14, size = 291, normalized size = 2.40 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (105 \left (\cos ^{3}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}+315 \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}+315 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}+105 \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}-2336 \left (\cos ^{3}\left (d x +c \right )\right )-512 \left (\cos ^{2}\left (d x +c \right )\right )+384 \cos \left (d x +c \right )+240\right ) a}{840 d \cos \left (d x +c \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 124, normalized size = 1.02 \[ -\frac {105 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 70 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} - \frac {30 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a^{2}} + \frac {42 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a} + 210 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a}{105 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]
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 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \tan ^{3}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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