Integrand size = 25, antiderivative size = 117 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}+\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
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Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3101, 12, 2833, 2738, 211} \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {a \sin (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2} \]
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Rule 12
Rule 211
Rule 2738
Rule 2833
Rule 3101
Rubi steps \begin{align*} \text {integral}& = \frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {\int \frac {\left (a^2-b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )} \\ & = \frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {\int \frac {\cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b} \\ & = \frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {b}{a+b \cos (c+d x)} \, dx}{2 b \left (a^2-b^2\right )} \\ & = \frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )} \\ & = \frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right ) d} \\ & = \frac {\arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}+\frac {\sin (c+d x)}{2 b d (a+b \cos (c+d x))^2}-\frac {a \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \\ \end{align*}
Time = 0.54 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {2 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {(b+a \cos (c+d x)) \sin (c+d x)}{(a+b \cos (c+d x))^2}}{2 (a-b) (a+b) d} \]
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Time = 1.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {\frac {\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a +8 b}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a -b}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(127\) |
default | \(\frac {\frac {\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a +8 b}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a -b}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{2}}+\frac {\arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{2}-b^{2}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(127\) |
risch | \(-\frac {i \left (2 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+b^{3} {\mathrm e}^{i \left (d x +c \right )}+a \,b^{2}\right )}{b^{2} \left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(300\) |
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Time = 0.30 (sec) , antiderivative size = 449, normalized size of antiderivative = 3.84 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\left [\frac {{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (a^{2} b - b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d\right )}}, \frac {{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{2} b - b^{3} + {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d\right )}}\right ] \]
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Timed out. \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.33 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.51 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2} {\left (a^{2} - b^{2}\right )}}}{d} \]
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Time = 2.97 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.26 \[ \int \frac {1-\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a-b}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a+b}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]
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