Optimal. Leaf size=154 \[ -\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{3/2}}+\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]
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Rubi [A] time = 0.48, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2889, 3056, 3001, 3770, 2660, 618, 204} \[ -\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{3/2}}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 2660
Rule 2889
Rule 3001
Rule 3056
Rule 3770
Rubi steps
\begin {align*} \int \frac {\cos (c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \frac {\csc (c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^3} \, dx\\ &=\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\int \frac {\csc (c+d x) \left (2 \left (a^2-b^2\right )-\left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \frac {\csc (c+d x) \left (2 \left (a^2-b^2\right )^2-a b \left (a^2-b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\int \csc (c+d x) \, dx}{a^3}-\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \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 )}{a^3 \left (a^2-b^2\right ) d}\\ &=-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\left (2 b \left (3 a^2-2 b^2\right )\right ) \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 )}{a^3 \left (a^2-b^2\right ) d}\\ &=-\frac {b \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2} d}-\frac {\tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac {\cos (c+d x)}{2 a d (a+b \sin (c+d x))^2}+\frac {\left (a^2-2 b^2\right ) \cos (c+d x)}{2 a^2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.18, size = 154, normalized size = 1.00 \[ \frac {\frac {2 b \left (2 b^2-3 a^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {a \cos (c+d x) \left (2 a^3+b \left (a^2-2 b^2\right ) \sin (c+d x)-3 a b^2\right )}{(a-b) (a+b) (a+b \sin (c+d x))^2}+2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^3 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.93, size = 996, normalized size = 6.47 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 277, normalized size = 1.80 \[ -\frac {\frac {{\left (3 \, a^{2} b - 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{4} - 3 \, a^{2} b^{2}}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.72, size = 632, normalized size = 4.10 \[ \frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}+\frac {3 b \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 {4 b^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \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 a \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 {b^{2} \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} a \left (a^{2}-b^{2}\right )}-\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{4}}{d \,a^{3} \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 {5 b \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 {8 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \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 a}{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 {3 b^{2}}{d a \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 {3 b \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 a \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}+\frac {2 b^{3} \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 \,a^{3} \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 = 12.44, size = 1610, normalized size = 10.45 \[ \text {result too large to display} \]
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 \[ \int \frac {\cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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