Optimal. Leaf size=123 \[ -\frac {2 b^3 B \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {b B \tan (c+d x)}{a^2 d}+\frac {B \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 a d} \]
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Rubi [A] time = 0.35, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {21, 2802, 3055, 3001, 3770, 2659, 205} \[ -\frac {2 b^3 B \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {B \left (a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {b B \tan (c+d x)}{a^2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 a d} \]
Antiderivative was successfully verified.
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Rule 21
Rule 205
Rule 2659
Rule 2802
Rule 3001
Rule 3055
Rule 3770
Rubi steps
\begin {align*} \int \frac {(a B+b B \cos (c+d x)) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=B \int \frac {\sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx\\ &=\frac {B \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {B \int \frac {\left (-2 b+a \cos (c+d x)+b \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a}\\ &=-\frac {b B \tan (c+d x)}{a^2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {B \int \frac {\left (a^2+2 b^2+a b \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac {b B \tan (c+d x)}{a^2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\left (b^3 B\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^3}+\frac {\left (\left (a^2+2 b^2\right ) B\right ) \int \sec (c+d x) \, dx}{2 a^3}\\ &=\frac {\left (a^2+2 b^2\right ) B \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {b B \tan (c+d x)}{a^2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\left (2 b^3 B\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^3 d}\\ &=-\frac {2 b^3 B \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}+\frac {\left (a^2+2 b^2\right ) B \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac {b B \tan (c+d x)}{a^2 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 a d}\\ \end {align*}
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Mathematica [A] time = 1.05, size = 239, normalized size = 1.94 \[ \frac {B \left (\frac {8 b^3 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+\frac {a^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^2}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-2 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 a b \tan (c+d x)-4 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{4 a^3 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 487, normalized size = 3.96 \[ \left [-\frac {2 \, \sqrt {-a^{2} + b^{2}} B b^{3} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{4} - B a^{2} b^{2} - 2 \, {\left (B a^{3} b - B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, \sqrt {a^{2} - b^{2}} B b^{3} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{4} + B a^{2} b^{2} - 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (B a^{4} - B a^{2} b^{2} - 2 \, {\left (B a^{3} b - B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.75, size = 221, normalized size = 1.80 \[ -\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} B b^{3}}{\sqrt {a^{2} - b^{2}} a^{3}} - \frac {{\left (B a^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} + \frac {{\left (B a^{2} + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {2 \, {\left (B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 273, normalized size = 2.22 \[ -\frac {2 b^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {B b}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{2 a d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B \,b^{2}}{d \,a^{3}}-\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {B}{2 a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {B b}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{2 a d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B \,b^{2}}{d \,a^{3}} \]
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 = 1.83, size = 1099, normalized size = 8.93 \[ \text {result too large to display} \]
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 \[ B \int \frac {\sec ^{3}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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