Optimal. Leaf size=115 \[ \frac {\tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {a \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2} \]
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Rubi [A] time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2693, 2754, 12, 2660, 618, 204} \[ \frac {\tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {a \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 2660
Rule 2693
Rule 2754
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {\int \frac {\sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 b}\\ &=-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \cos (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {b}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \cos (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \cos (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right ) d}\\ &=-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \cos (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right ) d}\\ &=\frac {\tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \cos (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 93, normalized size = 0.81 \[ \frac {\frac {2 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {\cos (c+d x) (a \sin (c+d x)+b)}{(a+b \sin (c+d x))^2}}{2 d (a-b) (a+b)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 501, normalized size = 4.36 \[ \left [-\frac {2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (d x + c\right ) - {\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} d\right )}}, -\frac {{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (d x + c\right ) - {\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.55, size = 207, normalized size = 1.80 \[ \frac {\frac {\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.30, size = 443, normalized size = 3.85 \[ -\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} a \left (a^{2}-b^{2}\right )}+\frac {b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {2 b^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right ) a^{2}}+\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} a \left (a^{2}-b^{2}\right )}+\frac {b}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )^{2} \left (a^{2}-b^{2}\right )}+\frac {\arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]
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 = 7.37, size = 282, normalized size = 2.45 \[ \frac {\frac {b}{a^2-b^2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2-2\,b^2\right )}{a\,\left (a^2-b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^2\right )}{a\,\left (a^2-b^2\right )}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+2\,b^2\right )}{a^2\,\left (a^2-b^2\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {\mathrm {atan}\left (\left (a^2-b^2\right )\,\left (\frac {a^2\,b-b^3}{{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]
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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