Integrand size = 32, antiderivative size = 89 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \]
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Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3045, 2729, 2727} \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac {7 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac {A x}{a^3} \]
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Rule 2727
Rule 2729
Rule 3045
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {A}{a^3}+\frac {2 A}{a^3 (1+\sin (c+d x))^3}-\frac {5 A}{a^3 (1+\sin (c+d x))^2}+\frac {4 A}{a^3 (1+\sin (c+d x))}\right ) \, dx \\ & = -\frac {A x}{a^3}+\frac {(2 A) \int \frac {1}{(1+\sin (c+d x))^3} \, dx}{a^3}+\frac {(4 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}-\frac {(5 A) \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{a^3} \\ & = -\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {5 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}-\frac {4 A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {(4 A) \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{5 a^3}-\frac {(5 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{3 a^3} \\ & = -\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {7 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}+\frac {(4 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{15 a^3} \\ & = -\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {A \sec (c+d x) \sqrt {1-\sin (c+d x)} \left (20 \sqrt {2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (1+\sin (c+d x))+\sqrt {1-\sin (c+d x)} \left (-4+3 \sin (c+d x)+\sin ^2(c+d x)\right )\right )}{15 a^3 d (1+\sin (c+d x))^2} \]
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Result contains complex when optimal does not.
Time = 0.66 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {A x}{a^{3}}-\frac {2 \left (-75 A \,{\mathrm e}^{2 i \left (d x +c \right )}+55 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+20 A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 i A \,{\mathrm e}^{i \left (d x +c \right )}+13 A \right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5}}\) | \(85\) |
derivativedivides | \(\frac {8 A \left (-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\right )}{d \,a^{3}}\) | \(96\) |
default | \(\frac {8 A \left (-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\right )}{d \,a^{3}}\) | \(96\) |
parallelrisch | \(-\frac {\left (\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d x +\left (5 d x +2\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (10 d x +10\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (10 d x +22\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (5 d x +14\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+d x +\frac {16}{5}\right ) A}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(112\) |
norman | \(\frac {-\frac {A x}{a}-\frac {16 A}{5 a d}-\frac {5 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {13 A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {25 A x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {38 A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {46 A x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {46 A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {38 A x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {25 A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {13 A x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {5 A x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {A x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {376 A \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {388 A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {158 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {28 A \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {52 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {72 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {44 A \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {10 A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 A \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {14 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(444\) |
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Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (83) = 166\).
Time = 0.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.29 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {{\left (5 \, A d x + 13 \, A\right )} \cos \left (d x + c\right )^{3} - 20 \, A d x + 3 \, {\left (5 \, A d x - 2 \, A\right )} \cos \left (d x + c\right )^{2} - {\left (10 \, A d x + 21 \, A\right )} \cos \left (d x + c\right ) - {\left (20 \, A d x - {\left (5 \, A d x - 13 \, A\right )} \cos \left (d x + c\right )^{2} + {\left (10 \, A d x + 19 \, A\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right ) - 2 \, A}{5 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1268 vs. \(2 (85) = 170\).
Time = 7.41 (sec) , antiderivative size = 1268, normalized size of antiderivative = 14.25 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]
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Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (83) = 166\).
Time = 0.34 (sec) , antiderivative size = 392, normalized size of antiderivative = 4.40 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (A {\left (\frac {\frac {95 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {145 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 22}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {2 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}\right )}}{15 \, d} \]
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Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {5 \, {\left (d x + c\right )} A}{a^{3}} + \frac {2 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 55 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{5 \, d} \]
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Time = 14.39 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.00 \[ \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {\left (5\,A\,\left (c+d\,x\right )-\frac {A\,\left (25\,c+25\,d\,x+10\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (10\,A\,\left (c+d\,x\right )-\frac {A\,\left (50\,c+50\,d\,x+50\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (10\,A\,\left (c+d\,x\right )-\frac {A\,\left (50\,c+50\,d\,x+110\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (5\,A\,\left (c+d\,x\right )-\frac {A\,\left (25\,c+25\,d\,x+70\right )}{5}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+A\,\left (c+d\,x\right )-\frac {A\,\left (5\,c+5\,d\,x+16\right )}{5}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5}-\frac {A\,x}{a^3} \]
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