Optimal. Leaf size=140 \[ \frac {2 a \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
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Rubi [A] time = 0.23, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3016, 2754, 12, 2659, 205} \[ \frac {2 a \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {a b \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 2659
Rule 2754
Rule 3016
Rubi steps
\begin {align*} \int \frac {a^2-b^2 \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx &=-\int \frac {-a+b \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx\\ &=-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\int \frac {2 \left (a^2+b^2\right )-2 a b \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\int -\frac {2 a \left (a^2+2 b^2\right )}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2+2 b^2\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (2 a \left (a^2+2 b^2\right )\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 )}{\left (a^2-b^2\right )^2 d}\\ &=\frac {2 a \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {a b \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (2 a^2+b^2\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.62, size = 116, normalized size = 0.83 \[ -\frac {\frac {2 a \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac {b \sin (c+d x) \left (3 a^3+b \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}}{d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.61, size = 598, normalized size = 4.27 \[ \left [-\frac {{\left (a^{5} + 2 \, a^{3} b^{2} + {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (3 \, a^{5} b - 3 \, a^{3} b^{3} + {\left (2 \, a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}, \frac {{\left (a^{5} + 2 \, a^{3} b^{2} + {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (3 \, a^{5} b - 3 \, a^{3} b^{3} + {\left (2 \, a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.64, size = 255, normalized size = 1.82 \[ -\frac {2 \, {\left (\frac {{\left (a^{3} + 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}\right )}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.16, size = 547, normalized size = 3.91 \[ -\frac {6 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 b^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 b^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {6 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )^{2} \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 a^{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 )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 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^{2}}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}} \]
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 = 4.14, size = 242, normalized size = 1.73 \[ \frac {2\,a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}\,\left (a^3+2\,a\,b^2\right )}\right )\,\left (a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2\,b+a\,b^2+b^3\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b-a\,b^2+b^3\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]
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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