\(\int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [250]

Optimal result

Integrand size = 26, antiderivative size = 2348 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {3 i a^2 (e+f x)^3}{2 b \left (a^2-b^2\right )^2 d}+\frac {i (e+f x)^3}{b \left (a^2-b^2\right ) d}-\frac {3 i a f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {9 a^2 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {3 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {3 i a^3 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i a (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 i a f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {9 a^2 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {3 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {3 i a^3 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {3 i a (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 a f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {9 i a^2 f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^2 d^3}+\frac {6 i f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {9 a^3 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}-\frac {9 a f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 a f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {9 i a^2 f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^2 d^3}+\frac {6 i f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {9 a^3 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{5/2} d^2}+\frac {9 a f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}+\frac {9 a^2 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^2 d^4}-\frac {6 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}+\frac {9 i a^3 f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^3}-\frac {9 i a f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {9 a^2 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^2 d^4}-\frac {6 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}-\frac {9 i a^3 f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^3}+\frac {9 i a f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {9 a^3 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^4}+\frac {9 a f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}+\frac {9 a^3 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d^4}-\frac {9 a f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {a (e+f x)^3 \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {3 a f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {3 a^2 (e+f x)^3 \cos (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]

[Out]

-3/2*a*f*(f*x+e)^2/b/(a^2-b^2)/d^2/(a+b*sin(d*x+c))+9/2*a^3*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b
^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2-9/2*a*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b
^2)^(3/2)/d^2-9/2*a^3*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^2+9/2*
a*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2+9/2*a^2*f*(f*x+e)^2*ln(1
-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2/d^2+9/2*a^2*f*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2
-b^2)^(1/2)))/b/(a^2-b^2)^2/d^2-3/2*I*a*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/
2)/d-3/2*I*a^3*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d+3*I*a*f^2*(f*x+e)*ln
(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^3+9*I*a^3*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+
c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^3+9*I*a*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/
2)))/b/(a^2-b^2)^(3/2)/d^3+3/2*I*a^3*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/
d+3/2*I*a*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d+6*I*f^2*(f*x+e)*polylog(2
,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^3+6*I*f^2*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-
b^2)^(1/2)))/b/(a^2-b^2)/d^3-3*I*a*f^2*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/
d^3-9*I*a^2*f^2*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2/d^3-9*I*a^2*f^2*(f*x+e
)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2/d^3-9*I*a*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d
*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^3-9*I*a^3*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2
)^(1/2)))/b/(a^2-b^2)^(5/2)/d^3+I*(f*x+e)^3/b/(a^2-b^2)/d+(f*x+e)^3*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))+3*
a*f^3*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^4+9*a^2*f^3*polylog(3,I*b*exp(I*(d
*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^2/d^4+9*a^2*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a
^2-b^2)^2/d^4-9*a^3*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^4+9*a*f^3*polylo
g(4,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^4+9*a^3*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a
^2-b^2)^(1/2)))/b/(a^2-b^2)^(5/2)/d^4-9*a*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3
/2)/d^4-3*f*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^2-3*f*(f*x+e)^2*ln(1-I*b*exp(
I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^2-3*a*f^3*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a
^2-b^2)^(3/2)/d^4-1/2*a*(f*x+e)^3*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^2-3/2*a^2*(f*x+e)^3*cos(d*x+c)/(a^2-
b^2)^2/d/(a+b*sin(d*x+c))-3/2*I*a^2*(f*x+e)^3/b/(a^2-b^2)^2/d-6*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^
(1/2)))/b/(a^2-b^2)/d^4-6*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^4

Rubi [A] (verified)

Time = 5.52 (sec) , antiderivative size = 2348, normalized size of antiderivative = 1.00, number of steps used = 92, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {6874, 3406, 3405, 3404, 2296, 2221, 2611, 6744, 2320, 6724, 4615, 4507, 2317, 2438} \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^3}{2 b \left (a^2-b^2\right )^{5/2} d}-\frac {3 i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^3}{2 b \left (a^2-b^2\right )^{5/2} d}+\frac {9 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^3}{2 b \left (a^2-b^2\right )^{5/2} d^2}-\frac {9 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^3}{2 b \left (a^2-b^2\right )^{5/2} d^2}+\frac {9 i f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^3}-\frac {9 i f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^3}-\frac {9 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^4}+\frac {9 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^3}{b \left (a^2-b^2\right )^{5/2} d^4}-\frac {3 i (e+f x)^3 a^2}{2 b \left (a^2-b^2\right )^2 d}+\frac {9 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{2 b \left (a^2-b^2\right )^2 d^2}+\frac {9 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{2 b \left (a^2-b^2\right )^2 d^2}-\frac {9 i f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^3}-\frac {9 i f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^3}+\frac {9 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^4}+\frac {9 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a^2}{b \left (a^2-b^2\right )^2 d^4}-\frac {3 (e+f x)^3 \cos (c+d x) a^2}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {3 i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{2 b \left (a^2-b^2\right )^{3/2} d}-\frac {3 i f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {3 i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{2 b \left (a^2-b^2\right )^{3/2} d}+\frac {3 i f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {3 f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {9 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{2 b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^4}+\frac {9 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {9 i f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {9 i f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {9 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {9 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) a}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {3 f (e+f x)^2 a}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {(e+f x)^3 \cos (c+d x) a}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {i (e+f x)^3}{b \left (a^2-b^2\right ) d}-\frac {3 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {3 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {6 i f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {6 i f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {6 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}-\frac {6 f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}+\frac {(e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]

[In]

Int[((e + f*x)^3*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

(((-3*I)/2)*a^2*(e + f*x)^3)/(b*(a^2 - b^2)^2*d) + (I*(e + f*x)^3)/(b*(a^2 - b^2)*d) - ((3*I)*a*f^2*(e + f*x)*
Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) + (9*a^2*f*(e + f*x)^2*Log[1 -
(I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^2*d^2) - (3*f*(e + f*x)^2*Log[1 - (I*b*E^(I*(c
+ d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^2) + (((3*I)/2)*a^3*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)
))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d) - (((3*I)/2)*a*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a
- Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) + ((3*I)*a*f^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt
[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) + (9*a^2*f*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 -
b^2])])/(2*b*(a^2 - b^2)^2*d^2) - (3*f*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a
^2 - b^2)*d^2) - (((3*I)/2)*a^3*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^
2)^(5/2)*d) + (((3*I)/2)*a*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3
/2)*d) - (3*a*f^3*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^4) - ((9*I)*
a^2*f^2*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^3) + ((6*I)*f^2*
(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^3) + (9*a^3*f*(e + f*x)^2*
PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(5/2)*d^2) - (9*a*f*(e + f*x)^2*Poly
Log[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(3/2)*d^2) + (3*a*f^3*PolyLog[2, (I*b*E^
(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^4) - ((9*I)*a^2*f^2*(e + f*x)*PolyLog[2, (I*b*E^
(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^3) + ((6*I)*f^2*(e + f*x)*PolyLog[2, (I*b*E^(I*(c +
d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^3) - (9*a^3*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a
+ Sqrt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(5/2)*d^2) + (9*a*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sq
rt[a^2 - b^2])])/(2*b*(a^2 - b^2)^(3/2)*d^2) + (9*a^2*f^3*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2
])])/(b*(a^2 - b^2)^2*d^4) - (6*f^3*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^
4) + ((9*I)*a^3*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d^
3) - ((9*I)*a*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3)
+ (9*a^2*f^3*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^2*d^4) - (6*f^3*PolyLog[
3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^4) - ((9*I)*a^3*f^2*(e + f*x)*PolyLog[3, (I*
b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d^3) + ((9*I)*a*f^2*(e + f*x)*PolyLog[3, (I*b*
E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) - (9*a^3*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)
))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(5/2)*d^4) + (9*a*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2
- b^2])])/(b*(a^2 - b^2)^(3/2)*d^4) + (9*a^3*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*
(a^2 - b^2)^(5/2)*d^4) - (9*a*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2
)*d^4) - (a*(e + f*x)^3*Cos[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^2) - (3*a*f*(e + f*x)^2)/(2*b*(a^2
- b^2)*d^2*(a + b*Sin[c + d*x])) - (3*a^2*(e + f*x)^3*Cos[c + d*x])/(2*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x]))
+ ((e + f*x)^3*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))

Rule 2221

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]
- Dist[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]

Rule 2296

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]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[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]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2611

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] + Dist[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]

Rule 3404

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[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]

Rule 3405

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(c + d*x)^m*(Cos[
e + f*x]/(f*(a^2 - b^2)*(a + b*Sin[e + f*x]))), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[b*d*(m/(f*(a^2 - b^2))), Int[(c + d*x)^(m - 1)*(Cos[e + f*x]/(a + b*Sin[e + f*x])), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3406

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*(c + d*x)^m
*Cos[e + f*x]*((a + b*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(a^2 - b^2))), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^
m*(a + b*Sin[e + f*x])^(n + 1), x], x] - Dist[b*((n + 2)/((n + 1)*(a^2 - b^2))), Int[(c + d*x)^m*Sin[e + f*x]*
(a + b*Sin[e + f*x])^(n + 1), x], x] + Dist[b*d*(m/(f*(n + 1)*(a^2 - b^2))), Int[(c + d*x)^(m - 1)*Cos[e + f*x
]*(a + b*Sin[e + f*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && ILtQ[n, -2] &&
IGtQ[m, 0]

Rule 4507

Int[Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol]
:> Simp[(e + f*x)^m*((a + b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] - Dist[f*(m/(b*d*(n + 1))), Int[(e + f*x
)^(m - 1)*(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 4615

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(-I)*((e + f*x)^(m + 1)/(b*f*(m + 1))), x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] - I*
b*E^(I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x)))), x])
/; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a (e+f x)^3}{b (a+b \sin (c+d x))^3}+\frac {(e+f x)^3}{b (a+b \sin (c+d x))^2}\right ) \, dx \\ & = \frac {\int \frac {(e+f x)^3}{(a+b \sin (c+d x))^2} \, dx}{b}-\frac {a \int \frac {(e+f x)^3}{(a+b \sin (c+d x))^3} \, dx}{b} \\ & = -\frac {a (e+f x)^3 \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}+\frac {(e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {a \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}+\frac {a \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}-\frac {a^2 \int \frac {(e+f x)^3}{(a+b \sin (c+d x))^2} \, dx}{b \left (a^2-b^2\right )}-\frac {(3 f) \int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right ) d}+\frac {(3 a f) \int \frac {(e+f x)^2 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 \left (a^2-b^2\right ) d} \\ & = \frac {i (e+f x)^3}{b \left (a^2-b^2\right ) d}-\frac {a (e+f x)^3 \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {3 a f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {a^2 (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {a^3 \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )^2}+\frac {a \int \left (-\frac {a (e+f x)^3}{b (a+b \sin (c+d x))^2}+\frac {(e+f x)^3}{b (a+b \sin (c+d x))}\right ) \, dx}{2 \left (a^2-b^2\right )}+\frac {(2 a) \int \frac {e^{i (c+d x)} (e+f x)^3}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b \left (a^2-b^2\right )}+\frac {\left (3 a^2 f\right ) \int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2 d}-\frac {(3 f) \int \frac {e^{i (c+d x)} (e+f x)^2}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d}-\frac {(3 f) \int \frac {e^{i (c+d x)} (e+f x)^2}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d}+\frac {\left (3 a f^2\right ) \int \frac {e+f x}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right ) d^2} \\ & = -\frac {i a^2 (e+f x)^3}{b \left (a^2-b^2\right )^2 d}+\frac {i (e+f x)^3}{b \left (a^2-b^2\right ) d}-\frac {3 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {3 f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {a (e+f x)^3 \cos (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^2}-\frac {3 a f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {a^2 (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {(e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (2 a^3\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b \left (a^2-b^2\right )^2}-\frac {(2 i a) \int \frac {e^{i (c+d x)} (e+f x)^3}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {(2 i a) \int \frac {e^{i (c+d x)} (e+f x)^3}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}+\frac {a \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}-\frac {a^2 \int \frac {(e+f x)^3}{(a+b \sin (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}+\frac {\left (3 a^2 f\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2 f\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^2 d}+\frac {\left (6 f^2\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right ) d^2}+\frac {\left (6 f^2\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right ) d^2}+\frac {\left (6 a f^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{b \left (a^2-b^2\right ) d^2} \\ & = \text {Too large to display} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(29732\) vs. \(2(2348)=4696\).

Time = 22.19 (sec) , antiderivative size = 29732, normalized size of antiderivative = 12.66 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \]

[In]

Integrate[((e + f*x)^3*Sin[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

Result too large to show

Maple [F]

\[\int \frac {\left (f x +e \right )^{3} \sin \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{3}}d x\]

[In]

int((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)

[Out]

int((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 10614 vs. \(2 (2046) = 4092\).

Time = 0.94 (sec) , antiderivative size = 10614, normalized size of antiderivative = 4.52 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((f*x+e)**3*sin(d*x+c)/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [F]

\[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((f*x + e)^3*sin(d*x + c)/(b*sin(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \]

[In]

int((sin(c + d*x)*(e + f*x)^3)/(a + b*sin(c + d*x))^3,x)

[Out]

\text{Hanged}