Optimal. Leaf size=89 \[ -\frac {2 a (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}+\frac {x (b B-a C)}{b^2}+\frac {C \sin (c+d x)}{b d} \]
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Rubi [A] time = 0.13, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3023, 12, 2735, 2659, 205} \[ -\frac {2 a (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d \sqrt {a-b} \sqrt {a+b}}+\frac {x (b B-a C)}{b^2}+\frac {C \sin (c+d x)}{b d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 2659
Rule 2735
Rule 3023
Rubi steps
\begin {align*} \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx &=\frac {C \sin (c+d x)}{b d}+\frac {\int \frac {(b B-a C) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b}\\ &=\frac {C \sin (c+d x)}{b d}+\frac {(b B-a C) \int \frac {\cos (c+d x)}{a+b \cos (c+d x)} \, dx}{b}\\ &=\frac {(b B-a C) x}{b^2}+\frac {C \sin (c+d x)}{b d}-\frac {(a (b B-a C)) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^2}\\ &=\frac {(b B-a C) x}{b^2}+\frac {C \sin (c+d x)}{b d}-\frac {(2 a (b B-a C)) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 d}\\ &=\frac {(b B-a C) x}{b^2}-\frac {2 a (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {C \sin (c+d x)}{b d}\\ \end {align*}
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Mathematica [A] time = 0.21, size = 85, normalized size = 0.96 \[ \frac {-\frac {2 a (a C-b B) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+(c+d x) (b B-a C)+b C \sin (c+d x)}{b^2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 322, normalized size = 3.62 \[ \left [-\frac {2 \, {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x - {\left (C a^{2} - B a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac {{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x - {\left (C a^{2} - B a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 142, normalized size = 1.60 \[ -\frac {\frac {{\left (C a - B b\right )} {\left (d x + c\right )}}{b^{2}} - \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b} + \frac {2 \, {\left (C a^{2} - B a b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{2}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 172, normalized size = 1.93 \[ -\frac {2 a \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 a^{2} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \,b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d b}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C a}{d \,b^{2}} \]
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 = 2.42, size = 541, normalized size = 6.08 \[ \frac {2\,C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,\left (a^2-b^2\right )}-\frac {2\,B\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d\,\left (a^2-b^2\right )}-\frac {C\,b\,\sin \left (c+d\,x\right )}{d\,\left (a^2-b^2\right )}+\frac {2\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,\left (a^2-b^2\right )}-\frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d\,\left (a^2-b^2\right )}+\frac {B\,a\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,\sqrt {b^2-a^2}}-\frac {B\,a\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d\,\sqrt {b^2-a^2}}-\frac {C\,a^2\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d\,\sqrt {b^2-a^2}}+\frac {C\,a^2\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^2\,d\,\sqrt {b^2-a^2}}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{b\,d\,\left (a^2-b^2\right )} \]
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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