Optimal. Leaf size=214 \[ -\frac {(7 A+15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(5 A+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{10 a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {(5 A+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a \sec (c+d x)+a}} \]
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Rubi [A] time = 0.63, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4085, 4021, 4010, 4001, 3795, 203} \[ -\frac {(7 A+15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(5 A+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{10 a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {(5 A+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 3795
Rule 4001
Rule 4010
Rule 4021
Rule 4085
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {\sec ^3(c+d x) \left (a (A+3 C)-\frac {1}{2} a (5 A+9 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\sec ^2(c+d x) \left (-a^2 (5 A+9 C)+\frac {3}{4} a^2 (5 A+13 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac {2 \int \frac {\sec (c+d x) \left (\frac {3}{8} a^3 (5 A+13 C)-\frac {3}{4} a^3 (15 A+31 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac {(7 A+15 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}+\frac {(7 A+15 C) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d}\\ &=-\frac {(7 A+15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}\\ \end {align*}
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Mathematica [A] time = 5.06, size = 189, normalized size = 0.88 \[ \frac {\sin (c+d x) \cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (\sec ^3(c+d x) ((75 A+131 C) \cos (c+d x)+8 (5 A+9 C) \cos (2 (c+d x))+25 A \cos (3 (c+d x))+40 A+49 C \cos (3 (c+d x))+88 C)-10 \sqrt {2} (7 A+15 C) \cot ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)-1} \tan ^{-1}\left (\frac {\sqrt {\sec (c+d x)-1}}{\sqrt {2}}\right )\right )}{20 d (a (\sec (c+d x)+1))^{3/2} (A \cos (2 (c+d x))+A+2 C)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.53, size = 492, normalized size = 2.30 \[ \left [-\frac {5 \, \sqrt {2} {\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {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 ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (25 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac {5 \, \sqrt {2} {\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (25 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.36, size = 311, normalized size = 1.45 \[ -\frac {\frac {5 \, \sqrt {2} {\left (7 \, A + 15 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {{\left ({\left ({\left (\frac {5 \, \sqrt {2} {\left (A a^{3} + C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {\sqrt {2} {\left (55 \, A a^{3} + 127 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, \sqrt {2} {\left (19 \, A a^{3} + 35 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {5 \, \sqrt {2} {\left (9 \, A a^{3} + 17 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{20 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.73, size = 784, normalized size = 3.66 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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