Integrand size = 21, antiderivative size = 112 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {3 a b x}{4}-\frac {\left (a^2+b^2\right ) \cos (e+f x)}{f}+\frac {\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac {b^2 \cos ^5(e+f x)}{5 f}-\frac {3 a b \cos (e+f x) \sin (e+f x)}{4 f}-\frac {a b \cos (e+f x) \sin ^3(e+f x)}{2 f} \]
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Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2868, 2715, 8, 3092, 380} \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac {\left (a^2+b^2\right ) \cos (e+f x)}{f}-\frac {a b \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac {3 a b \sin (e+f x) \cos (e+f x)}{4 f}+\frac {3 a b x}{4}-\frac {b^2 \cos ^5(e+f x)}{5 f} \]
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Rule 8
Rule 380
Rule 2715
Rule 2868
Rule 3092
Rubi steps \begin{align*} \text {integral}& = (2 a b) \int \sin ^4(e+f x) \, dx+\int \sin ^3(e+f x) \left (a^2+b^2 \sin ^2(e+f x)\right ) \, dx \\ & = -\frac {a b \cos (e+f x) \sin ^3(e+f x)}{2 f}+\frac {1}{2} (3 a b) \int \sin ^2(e+f x) \, dx-\frac {\text {Subst}\left (\int \left (1-x^2\right ) \left (a^2+b^2-b^2 x^2\right ) \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {3 a b \cos (e+f x) \sin (e+f x)}{4 f}-\frac {a b \cos (e+f x) \sin ^3(e+f x)}{2 f}+\frac {1}{4} (3 a b) \int 1 \, dx-\frac {\text {Subst}\left (\int \left (a^2 \left (1+\frac {b^2}{a^2}\right )-\left (a^2+2 b^2\right ) x^2+b^2 x^4\right ) \, dx,x,\cos (e+f x)\right )}{f} \\ & = \frac {3 a b x}{4}-\frac {\left (a^2+b^2\right ) \cos (e+f x)}{f}+\frac {\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac {b^2 \cos ^5(e+f x)}{5 f}-\frac {3 a b \cos (e+f x) \sin (e+f x)}{4 f}-\frac {a b \cos (e+f x) \sin ^3(e+f x)}{2 f} \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {-30 \left (6 a^2+5 b^2\right ) \cos (e+f x)+5 \left (4 a^2+5 b^2\right ) \cos (3 (e+f x))-3 b (b \cos (5 (e+f x))-5 a (12 (e+f x)-8 \sin (2 (e+f x))+\sin (4 (e+f x))))}{240 f} \]
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Time = 1.92 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+2 a b \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {b^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\) | \(95\) |
default | \(\frac {-\frac {a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+2 a b \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {b^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\) | \(95\) |
parts | \(-\frac {a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3 f}-\frac {b^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5 f}+\frac {2 a b \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\) | \(100\) |
parallelrisch | \(\frac {180 a b x f -3 b^{2} \cos \left (5 f x +5 e \right )+15 a b \sin \left (4 f x +4 e \right )+20 \cos \left (3 f x +3 e \right ) a^{2}+25 \cos \left (3 f x +3 e \right ) b^{2}-120 a b \sin \left (2 f x +2 e \right )-180 \cos \left (f x +e \right ) a^{2}-150 \cos \left (f x +e \right ) b^{2}-160 a^{2}-128 b^{2}}{240 f}\) | \(113\) |
risch | \(\frac {3 a b x}{4}-\frac {3 a^{2} \cos \left (f x +e \right )}{4 f}-\frac {5 b^{2} \cos \left (f x +e \right )}{8 f}-\frac {b^{2} \cos \left (5 f x +5 e \right )}{80 f}+\frac {a b \sin \left (4 f x +4 e \right )}{16 f}+\frac {\cos \left (3 f x +3 e \right ) a^{2}}{12 f}+\frac {5 \cos \left (3 f x +3 e \right ) b^{2}}{48 f}-\frac {a b \sin \left (2 f x +2 e \right )}{2 f}\) | \(118\) |
norman | \(\frac {-\frac {20 a^{2}+16 b^{2}}{15 f}+\frac {3 a b x}{4}-\frac {4 a^{2} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 \left (14 a^{2}+16 b^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {\left (20 a^{2}+16 b^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {3 a b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f}-\frac {7 a b \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {7 a b \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a b \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}+\frac {15 a b x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {15 a b x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {15 a b x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {15 a b x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {3 a b x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{5}}\) | \(262\) |
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Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.80 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=-\frac {12 \, b^{2} \cos \left (f x + e\right )^{5} - 45 \, a b f x - 20 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 60 \, {\left (a^{2} + b^{2}\right )} \cos \left (f x + e\right ) - 15 \, {\left (2 \, a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{60 \, f} \]
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Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (104) = 208\).
Time = 0.24 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.97 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=\begin {cases} - \frac {a^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {3 a b x \sin ^{4}{\left (e + f x \right )}}{4} + \frac {3 a b x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} + \frac {3 a b x \cos ^{4}{\left (e + f x \right )}}{4} - \frac {5 a b \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{4 f} - \frac {3 a b \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} - \frac {b^{2} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 b^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {8 b^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right )^{2} \sin ^{3}{\left (e \right )} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {80 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} + 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a b - 16 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} b^{2}}{240 \, f} \]
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Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.91 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {3}{4} \, a b x - \frac {b^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {a b \sin \left (4 \, f x + 4 \, e\right )}{16 \, f} - \frac {a b \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} + \frac {{\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac {{\left (6 \, a^{2} + 5 \, b^{2}\right )} \cos \left (f x + e\right )}{8 \, f} \]
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Time = 10.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.40 \[ \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {3\,a\,b\,x}{4}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {20\,a^2}{3}+\frac {16\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {28\,a^2}{3}+\frac {32\,b^2}{3}\right )+4\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\frac {4\,a^2}{3}+\frac {16\,b^2}{15}+7\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-7\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7-\frac {3\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{2}+\frac {3\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^5} \]
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