Integrand size = 28, antiderivative size = 592 \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {f^2 x}{4 b d^2}+\frac {a^2 (e+f x)^3}{3 b^3 f}+\frac {(e+f x)^3}{6 b f}-\frac {2 a f^2 \cos (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \cos (c+d x)}{b^2 d}+\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {2 a^3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}-\frac {2 a^3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}+\frac {2 i a^3 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^3}-\frac {2 i a^3 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^3}-\frac {2 a f (e+f x) \sin (c+d x)}{b^2 d^2}+\frac {f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f (e+f x) \sin ^2(c+d x)}{2 b d^2} \]
-1/4*f^2*x/b/d^2+1/3*a^2*(f*x+e)^3/b^3/f+1/6*(f*x+e)^3/b/f-2*a*f^2*cos(d*x +c)/b^2/d^3+a*(f*x+e)^2*cos(d*x+c)/b^2/d-2*a*f*(f*x+e)*sin(d*x+c)/b^2/d^2+ 1/4*f^2*cos(d*x+c)*sin(d*x+c)/b/d^3-1/2*(f*x+e)^2*cos(d*x+c)*sin(d*x+c)/b/ d+1/2*f*(f*x+e)*sin(d*x+c)^2/b/d^2+I*a^3*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c)) /(a-(a^2-b^2)^(1/2)))/b^3/d/(a^2-b^2)^(1/2)-I*a^3*(f*x+e)^2*ln(1-I*b*exp(I *(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d/(a^2-b^2)^(1/2)+2*a^3*f*(f*x+e)*polyl og(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^2/(a^2-b^2)^(1/2)-2*a^3 *f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^2/(a^2- b^2)^(1/2)+2*I*a^3*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b ^3/d^3/(a^2-b^2)^(1/2)-2*I*a^3*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^ 2)^(1/2)))/b^3/d^3/(a^2-b^2)^(1/2)
Time = 3.14 (sec) , antiderivative size = 1166, normalized size of antiderivative = 1.97 \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {24 a^2 \sqrt {-\left (a^2-b^2\right )^2} d^3 e^2 x+12 b^2 \sqrt {-\left (-a^2+b^2\right )^2} d^3 e^2 x+24 a^2 \sqrt {-\left (a^2-b^2\right )^2} d^3 e f x^2+12 b^2 \sqrt {-\left (-a^2+b^2\right )^2} d^3 e f x^2+8 a^2 \sqrt {-\left (a^2-b^2\right )^2} d^3 f^2 x^3+4 b^2 \sqrt {-\left (-a^2+b^2\right )^2} d^3 f^2 x^3-48 a^3 \sqrt {-a^2+b^2} d^2 e^2 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+24 a b \sqrt {-\left (a^2-b^2\right )^2} d^2 e^2 \cos (c+d x)-48 a b \sqrt {-\left (a^2-b^2\right )^2} f^2 \cos (c+d x)+48 a b \sqrt {-\left (a^2-b^2\right )^2} d^2 e f x \cos (c+d x)+24 a b \sqrt {-\left (a^2-b^2\right )^2} d^2 f^2 x^2 \cos (c+d x)-6 b^2 \sqrt {-\left (a^2-b^2\right )^2} d e f \cos (2 (c+d x))-6 b^2 \sqrt {-\left (a^2-b^2\right )^2} d f^2 x \cos (2 (c+d x))-48 a^3 \sqrt {a^2-b^2} d^2 e f x \log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-24 a^3 \sqrt {a^2-b^2} d^2 f^2 x^2 \log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+48 a^3 \sqrt {a^2-b^2} d^2 e f x \log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )+24 a^3 \sqrt {a^2-b^2} d^2 f^2 x^2 \log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )+48 i a^3 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-48 i a^3 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-48 a^3 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+48 a^3 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-48 a b \sqrt {-\left (a^2-b^2\right )^2} d e f \sin (c+d x)-48 a b \sqrt {-\left (a^2-b^2\right )^2} d f^2 x \sin (c+d x)-6 b^2 \sqrt {-\left (a^2-b^2\right )^2} d^2 e^2 \sin (2 (c+d x))+3 b^2 \sqrt {-\left (a^2-b^2\right )^2} f^2 \sin (2 (c+d x))-12 b^2 \sqrt {-\left (a^2-b^2\right )^2} d^2 e f x \sin (2 (c+d x))-6 b^2 \sqrt {-\left (a^2-b^2\right )^2} d^2 f^2 x^2 \sin (2 (c+d x))}{24 b^3 \sqrt {-\left (a^2-b^2\right )^2} d^3} \]
(24*a^2*Sqrt[-(a^2 - b^2)^2]*d^3*e^2*x + 12*b^2*Sqrt[-(-a^2 + b^2)^2]*d^3* e^2*x + 24*a^2*Sqrt[-(a^2 - b^2)^2]*d^3*e*f*x^2 + 12*b^2*Sqrt[-(-a^2 + b^2 )^2]*d^3*e*f*x^2 + 8*a^2*Sqrt[-(a^2 - b^2)^2]*d^3*f^2*x^3 + 4*b^2*Sqrt[-(- a^2 + b^2)^2]*d^3*f^2*x^3 - 48*a^3*Sqrt[-a^2 + b^2]*d^2*e^2*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + 24*a*b*Sqrt[-(a^2 - b^2)^2]*d^2*e^2* Cos[c + d*x] - 48*a*b*Sqrt[-(a^2 - b^2)^2]*f^2*Cos[c + d*x] + 48*a*b*Sqrt[ -(a^2 - b^2)^2]*d^2*e*f*x*Cos[c + d*x] + 24*a*b*Sqrt[-(a^2 - b^2)^2]*d^2*f ^2*x^2*Cos[c + d*x] - 6*b^2*Sqrt[-(a^2 - b^2)^2]*d*e*f*Cos[2*(c + d*x)] - 6*b^2*Sqrt[-(a^2 - b^2)^2]*d*f^2*x*Cos[2*(c + d*x)] - 48*a^3*Sqrt[a^2 - b^ 2]*d^2*e*f*x*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 24 *a^3*Sqrt[a^2 - b^2]*d^2*f^2*x^2*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqr t[-a^2 + b^2])] + 48*a^3*Sqrt[a^2 - b^2]*d^2*e*f*x*Log[1 + (b*E^(I*(c + d* x)))/(I*a + Sqrt[-a^2 + b^2])] + 24*a^3*Sqrt[a^2 - b^2]*d^2*f^2*x^2*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] + (48*I)*a^3*Sqrt[a^2 - b^ 2]*d*f*(e + f*x)*PolyLog[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2] )] - (48*I)*a^3*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -((b*E^(I*(c + d* x)))/(I*a + Sqrt[-a^2 + b^2]))] - 48*a^3*Sqrt[a^2 - b^2]*f^2*PolyLog[3, (b *E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 48*a^3*Sqrt[a^2 - b^2]*f^ 2*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] - 48*a*b*Sqr t[-(a^2 - b^2)^2]*d*e*f*Sin[c + d*x] - 48*a*b*Sqrt[-(a^2 - b^2)^2]*d*f^...
Time = 3.06 (sec) , antiderivative size = 540, normalized size of antiderivative = 0.91, number of steps used = 26, number of rules used = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.893, Rules used = {5026, 3042, 3792, 17, 3042, 3115, 24, 5026, 3042, 3777, 3042, 3777, 25, 3042, 3118, 5026, 17, 3042, 3804, 2694, 27, 2620, 3011, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 5026 |
\(\displaystyle \frac {\int (e+f x)^2 \sin ^2(c+d x)dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (e+f x)^2 \sin (c+d x)^2dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3792 |
\(\displaystyle \frac {-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {1}{2} \int (e+f x)^2dx+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {f^2 \int \sin (c+d x)^2dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {-\frac {f^2 \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 5026 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\int (e+f x)^2 \sin (c+d x)dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\int (e+f x)^2 \sin (c+d x)dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \int (e+f x) \cos (c+d x)dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \int (e+f x) \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \int -\sin (c+d x)dx}{d}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\) |
\(\Big \downarrow \) 5026 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {\int (e+f x)^2dx}{b}-\frac {a \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {a \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {a \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 3804 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \int \frac {e^{i (c+d x)} (e+f x)^2}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 2694 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\) |
((e + f*x)^3/(6*f) - ((e + f*x)^2*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (f*(e + f*x)*Sin[c + d*x]^2)/(2*d^2) - (f^2*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/ (2*d)))/(2*d^2))/b - (a*(-((a*((e + f*x)^3/(3*b*f) - (2*a*(((-1/2*I)*b*((( e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) - ( 2*f*((I*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])]) /d - (f*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d^2))/(b* d)))/Sqrt[a^2 - b^2] + ((I/2)*b*(((e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)) )/(a + Sqrt[a^2 - b^2])])/(b*d) - (2*f*((I*(e + f*x)*PolyLog[2, (I*b*E^(I* (c + d*x)))/(a + Sqrt[a^2 - b^2])])/d - (f*PolyLog[3, (I*b*E^(I*(c + d*x)) )/(a + Sqrt[a^2 - b^2])])/d^2))/(b*d)))/Sqrt[a^2 - b^2]))/b))/b) + (-(((e + f*x)^2*Cos[c + d*x])/d) + (2*f*((f*Cos[c + d*x])/d^2 + ((e + f*x)*Sin[c + d*x])/d))/d)/b))/b
3.3.29.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) *(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q) Int [(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q) Int[(f + g*x) ^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo l] :> Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Sim p[b*(c + d*x)^m*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^ 2*((n - 1)/n) Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[d^2 *m*((m - 1)/(f^2*n^2)) Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy mbol] :> Simp[2 Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x )) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ [a^2 - b^2, 0] && IGtQ[m, 0]
Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_. )*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/b Int[(e + f*x)^m*Sin[c + d*x]^(n - 1), x], x] - Simp[a/b Int[(e + f*x)^m*(Sin[c + d*x]^(n - 1)/(a + b*Sin[c + d*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] & & IGtQ[n, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
\[\int \frac {\left (f x +e \right )^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2050 vs. \(2 (522) = 1044\).
Time = 0.49 (sec) , antiderivative size = 2050, normalized size of antiderivative = 3.46 \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
1/12*(2*(2*a^4 - a^2*b^2 - b^4)*d^3*f^2*x^3 + 6*(2*a^4 - a^2*b^2 - b^4)*d^ 3*e*f*x^2 - 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^ 2)/b^2))/b) + 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*cos( d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*c os(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-( a^2 - b^2)/b^2))/b) - 6*((a^2*b^2 - b^4)*d*f^2*x + (a^2*b^2 - b^4)*d*e*f)* cos(d*x + c)^2 + 12*(-I*a^3*b*d*f^2*x - I*a^3*b*d*e*f)*sqrt(-(a^2 - b^2)/b ^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d *x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 12*(I*a^3*b*d*f^2*x + I*a^3* b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 12*(I*a^3*b*d*f^2*x + I*a^3*b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*co s(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a ^2 - b^2)/b^2) - b)/b + 1) + 12*(-I*a^3*b*d*f^2*x - I*a^3*b*d*e*f)*sqrt(-( a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c ) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*(a^3*b*d^2...
Timed out. \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
\[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]