Optimal. Leaf size=217 \[ \frac {\left (a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 d \left (a^2-b^2\right )} \]
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Rubi [A] time = 0.68, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2802, 3055, 3001, 3770, 2659, 205} \[ -\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac {\left (a^2-3 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac {b^2 \tan (c+d x) \sec (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 205
Rule 2659
Rule 2802
Rule 3001
Rule 3055
Rule 3770
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac {b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (a^2-3 b^2-a b \cos (c+d x)+2 b^2 \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (-2 b \left (2 a^2-3 b^2\right )+a \left (a^2+b^2\right ) \cos (c+d x)+b \left (a^2-3 b^2\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac {b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (a^4+5 a^2 b^2-6 b^4+a b \left (a^2-3 b^2\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac {b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (b^3 \left (4 a^2-3 b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )}+\frac {\left (a^2+6 b^2\right ) \int \sec (c+d x) \, dx}{2 a^4}\\ &=\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac {\left (2 b^3 \left (4 a^2-3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right ) d}\\ &=-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac {b \left (2 a^2-3 b^2\right ) \tan (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sec (c+d x) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 5.52, size = 285, normalized size = 1.31 \[ \frac {\frac {8 b^3 \left (3 b^2-4 a^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 a b^4 \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-8 a b \tan (c+d x)-12 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{4 a^4 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 3.27, size = 899, normalized size = 4.14 \[ \left [-\frac {2 \, {\left ({\left (4 \, a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, a^{3} b^{3} - 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left ({\left (a^{6} b + 4 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a^{6} b + 4 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} - 2 \, {\left (2 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d \cos \left (d x + c\right )^{3} + {\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d \cos \left (d x + c\right )^{2}\right )}}, -\frac {4 \, {\left ({\left (4 \, a^{2} b^{4} - 3 \, b^{6}\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, a^{3} b^{3} - 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left ({\left (a^{6} b + 4 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (a^{6} b + 4 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + 6 \, b^{7}\right )} \cos \left (d x + c\right )^{3} + {\left (a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 6 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} - 2 \, {\left (2 \, a^{5} b^{2} - 5 \, a^{3} b^{4} + 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d \cos \left (d x + c\right )^{3} + {\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} 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 = 0.96, size = 293, normalized size = 1.35 \[ \frac {\frac {4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} + \frac {4 \, {\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {{\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {{\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 401, normalized size = 1.85 \[ \frac {2 b^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {8 b^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {6 b^{5} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{4} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 b}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}}{d \,a^{4}}-\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}}{d \,a^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.93, size = 3699, normalized size = 17.05 \[ \text {result too large to display} \]
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 \[ \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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