Integrand size = 25, antiderivative size = 252 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 C\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \]
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Time = 0.53 (sec) , antiderivative size = 252, 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, 2833, 12, 2738, 211} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {a \left (a^2 (-C)+5 A b^2+6 b^2 C\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {\left (a^4 C-a^2 b^2 (11 A+10 C)-2 b^4 (2 A+3 C)\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]
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Rule 12
Rule 211
Rule 2738
Rule 2833
Rule 3101
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\int \frac {-3 a b (A+C)+\left (2 A b^2-a^2 C+3 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 C\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\int \frac {2 \left (b^3 (2 A+3 C)+\frac {1}{2} a^2 (6 A b+4 b C)\right )-a \left (5 A b^2-\left (a^2-6 b^2\right ) C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2} \\ & = -\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 C\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac {\int -\frac {3 a b \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )}{a+b \cos (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3} \\ & = -\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 C\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = -\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 C\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right )\right ) \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 )^3 d} \\ & = \frac {a \left (2 a^2 A+3 A b^2+a^2 C+4 b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {a \left (5 A b^2-a^2 C+6 b^2 C\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \\ \end{align*}
Time = 1.56 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.89 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {-\frac {24 a \left (a^2 (2 A+C)+b^2 (3 A+4 C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {2 \left (6 a \left (a^4 C-9 a^2 b^2 (A+C)-b^4 (A+2 C)\right ) \cos (c+d x)-b \left (2 b^4 (4 A+3 C)+a^2 b^2 (A+14 C)+a^4 (36 A+25 C)+\left (-a^4 C+2 b^4 (2 A+3 C)+a^2 b^2 (11 A+10 C)\right ) \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{(a+b \cos (c+d x))^3}}{24 \left (a^2-b^2\right )^3 d} \]
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Time = 2.36 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.46
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}+C \,a^{3}+6 C \,a^{2} b +2 C a \,b^{2}+2 C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {4 \left (9 A \,a^{2}+A \,b^{2}+7 a^{2} C +3 b^{2} C \right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-C \,a^{3}+6 C \,a^{2} b -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}+\frac {a \left (2 A \,a^{2}+3 A \,b^{2}+a^{2} C +4 b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(369\) |
default | \(\frac {\frac {-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}+C \,a^{3}+6 C \,a^{2} b +2 C a \,b^{2}+2 C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {4 \left (9 A \,a^{2}+A \,b^{2}+7 a^{2} C +3 b^{2} C \right ) b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-C \,a^{3}+6 C \,a^{2} b -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}+\frac {a \left (2 A \,a^{2}+3 A \,b^{2}+a^{2} C +4 b^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(369\) |
risch | \(\text {Expression too large to display}\) | \(1257\) |
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Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (236) = 472\).
Time = 0.35 (sec) , antiderivative size = 1105, normalized size of antiderivative = 4.38 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 689 vs. \(2 (236) = 472\).
Time = 0.35 (sec) , antiderivative size = 689, normalized size of antiderivative = 2.73 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {3 \, {\left (2 \, A a^{3} + C a^{3} + 3 \, A a b^{2} + 4 \, C 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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 28 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 32 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\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 )}^{3}}}{3 \, d} \]
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Time = 4.96 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.95 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\,\left (2\,A\,a^2+3\,A\,b^2+C\,a^2+4\,C\,b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (2\,A\,a^3+C\,a^3+3\,A\,a\,b^2+4\,C\,a\,b^2\right )}\right )\,\left (2\,A\,a^2+3\,A\,b^2+C\,a^2+4\,C\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^3+3\,C\,b^3+9\,A\,a^2\,b+7\,C\,a^2\,b\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b+2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \]
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