Integrand size = 28, antiderivative size = 618 \[ \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a (e+f x)^4}{4 b^2 f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}+\frac {(e+f x)^3 \cos (c+d x)}{b d}+\frac {i \sqrt {a^2-b^2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d}-\frac {i \sqrt {a^2-b^2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {3 \sqrt {a^2-b^2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2}-\frac {3 \sqrt {a^2-b^2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2}+\frac {6 i \sqrt {a^2-b^2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3}-\frac {6 i \sqrt {a^2-b^2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^3}-\frac {6 \sqrt {a^2-b^2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^4}+\frac {6 \sqrt {a^2-b^2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^4}+\frac {6 f^3 \sin (c+d x)}{b d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2} \]
1/4*a*(f*x+e)^4/b^2/f-6*f^2*(f*x+e)*cos(d*x+c)/b/d^3+(f*x+e)^3*cos(d*x+c)/ b/d+6*f^3*sin(d*x+c)/b/d^4-3*f*(f*x+e)^2*sin(d*x+c)/b/d^2+I*(f*x+e)^3*ln(1 -I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))*(a^2-b^2)^(1/2)/b^2/d-I*(f*x+e)^3 *ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))*(a^2-b^2)^(1/2)/b^2/d+3*f*(f *x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))*(a^2-b^2)^(1/2)/ b^2/d^2-3*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))*(a ^2-b^2)^(1/2)/b^2/d^2+6*I*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2 -b^2)^(1/2)))*(a^2-b^2)^(1/2)/b^2/d^3-6*I*f^2*(f*x+e)*polylog(3,I*b*exp(I* (d*x+c))/(a+(a^2-b^2)^(1/2)))*(a^2-b^2)^(1/2)/b^2/d^3-6*f^3*polylog(4,I*b* exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))*(a^2-b^2)^(1/2)/b^2/d^4+6*f^3*polylog( 4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))*(a^2-b^2)^(1/2)/b^2/d^4
Time = 2.31 (sec) , antiderivative size = 1025, normalized size of antiderivative = 1.66 \[ \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a d^4 x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )+4 b d (e+f x) \left (-6 f^2+d^2 (e+f x)^2\right ) \cos (c+d x)+\frac {4 \left (-a^2+b^2\right ) \left (2 \sqrt {-a^2+b^2} d^3 e^3 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+3 \sqrt {a^2-b^2} d^3 e^2 f x \log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+3 \sqrt {a^2-b^2} d^3 e f^2 x^2 \log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+\sqrt {a^2-b^2} d^3 f^3 x^3 \log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-3 \sqrt {a^2-b^2} d^3 e^2 f x \log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-3 \sqrt {a^2-b^2} d^3 e f^2 x^2 \log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-\sqrt {a^2-b^2} d^3 f^3 x^3 \log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-3 i \sqrt {a^2-b^2} d^2 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+3 i \sqrt {a^2-b^2} d^2 f (e+f x)^2 \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )+6 \sqrt {a^2-b^2} d e f^2 \operatorname {PolyLog}\left (3,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+6 \sqrt {a^2-b^2} d f^3 x \operatorname {PolyLog}\left (3,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-6 \sqrt {a^2-b^2} d e f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-6 \sqrt {a^2-b^2} d f^3 x \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )+6 i \sqrt {a^2-b^2} f^3 \operatorname {PolyLog}\left (4,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-6 i \sqrt {a^2-b^2} f^3 \operatorname {PolyLog}\left (4,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )}{\sqrt {-\left (a^2-b^2\right )^2}}-12 b f \left (-2 f^2+d^2 (e+f x)^2\right ) \sin (c+d x)}{4 b^2 d^4} \]
(a*d^4*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3) + 4*b*d*(e + f*x)*(-6 *f^2 + d^2*(e + f*x)^2)*Cos[c + d*x] + (4*(-a^2 + b^2)*(2*Sqrt[-a^2 + b^2] *d^3*e^3*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + 3*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sq rt[-a^2 + b^2])] + Sqrt[a^2 - b^2]*d^3*f^3*x^3*Log[1 - (b*E^(I*(c + d*x))) /((-I)*a + Sqrt[-a^2 + b^2])] - 3*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1 + (b*E ^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^ 2*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - Sqrt[a^2 - b^2]* d^3*f^3*x^3*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - (3*I)* Sqrt[a^2 - b^2]*d^2*f*(e + f*x)^2*PolyLog[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + (3*I)*Sqrt[a^2 - b^2]*d^2*f*(e + f*x)^2*PolyLog[2, - ((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] + 6*Sqrt[a^2 - b^2]*d*e*f^ 2*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 6*Sqrt[a^2 - b^2]*d*f^3*x*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2]) ] - 6*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt [-a^2 + b^2]))] - 6*Sqrt[a^2 - b^2]*d*f^3*x*PolyLog[3, -((b*E^(I*(c + d*x) ))/(I*a + Sqrt[-a^2 + b^2]))] + (6*I)*Sqrt[a^2 - b^2]*f^3*PolyLog[4, (b*E^ (I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - (6*I)*Sqrt[a^2 - b^2]*f^3*Po lyLog[4, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))]))/Sqrt[-(a^2 ...
Time = 2.61 (sec) , antiderivative size = 576, normalized size of antiderivative = 0.93, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {5036, 17, 3042, 3777, 3042, 3777, 25, 3042, 3777, 3042, 3117, 3804, 2694, 27, 2620, 3011, 7163, 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.
| \(\displaystyle \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 5036 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \int (e+f x)^3dx}{b^2}-\frac {\int (e+f x)^3 \sin (c+d x)dx}{b}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\int (e+f x)^3 \sin (c+d x)dx}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\int (e+f x)^3 \sin (c+d x)dx}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3777 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {3 f \int (e+f x)^2 \cos (c+d x)dx}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {3 f \int (e+f x)^2 \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3777 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {3 f \left (\frac {2 f \int -((e+f x) \sin (c+d x))dx}{d}+\frac {(e+f x)^2 \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \int (e+f x) \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \int (e+f x) \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3777 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \int \cos (c+d x)dx}{d}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \int \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}+\frac {a (e+f x)^4}{4 b^2 f}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle -\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 3804 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 2694 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle -\frac {2 \left (a^2-b^2\right ) \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b^2}+\frac {a (e+f x)^4}{4 b^2 f}-\frac {\frac {3 f \left (\frac {(e+f x)^2 \sin (c+d x)}{d}-\frac {2 f \left (\frac {f \sin (c+d x)}{d^2}-\frac {(e+f x) \cos (c+d x)}{d}\right )}{d}\right )}{d}-\frac {(e+f x)^3 \cos (c+d x)}{d}}{b}\) |
(a*(e + f*x)^4)/(4*b^2*f) - (2*(a^2 - b^2)*(((-1/2*I)*b*(((e + f*x)^3*Log[ 1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) - (3*f*((I*(e + f* x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d - ((2*I)*f *(((-I)*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])]) /d + (f*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d^2))/d)) /(b*d)))/Sqrt[a^2 - b^2] + ((I/2)*b*(((e + f*x)^3*Log[1 - (I*b*E^(I*(c + d *x)))/(a + Sqrt[a^2 - b^2])])/(b*d) - (3*f*((I*(e + f*x)^2*PolyLog[2, (I*b *E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/d - ((2*I)*f*(((-I)*(e + f*x)*Po lyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/d + (f*PolyLog[4, ( I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/d^2))/d))/(b*d)))/Sqrt[a^2 - b^2]))/b^2 - (-(((e + f*x)^3*Cos[c + d*x])/d) + (3*f*(((e + f*x)^2*Sin[c + d*x])/d - (2*f*(-(((e + f*x)*Cos[c + d*x])/d) + (f*Sin[c + d*x])/d^2))/d) )/d)/b
3.3.98.3.1 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[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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_.)/((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[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.) *Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[a/b^2 Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] + (-Simp[1/b Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)* Sin[c + d*x], x], x] - Simp[(a^2 - b^2)/b^2 Int[(e + f*x)^m*(Cos[c + d*x] ^(n - 2)/(a + b*Sin[c + d*x])), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 1] && NeQ[a^2 - b^2, 0] && IGtQ[m, 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[((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] - Simp[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]
\[\int \frac {\left (f x +e \right )^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2331 vs. \(2 (542) = 1084\).
Time = 0.48 (sec) , antiderivative size = 2331, normalized size of antiderivative = 3.77 \[ \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
1/4*(a*d^4*f^3*x^4 + 4*a*d^4*e*f^2*x^3 + 6*a*d^4*e^2*f*x^2 + 4*a*d^4*e^3*x + 12*I*b*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, -(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*I*b*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, -(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*I*b*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, -(-I*a*cos(d*x + c) + a*si n(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b ) + 12*I*b*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, -(-I*a*cos(d*x + c) + a*s in(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/ b) - 6*(I*b*d^2*f^3*x^2 + 2*I*b*d^2*e*f^2*x + I*b*d^2*e^2*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*(-I*b*d^2*f^3*x^2 - 2*I*b*d^2*e*f^2*x - I*b*d^2*e^2*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*(-I*b*d^2*f^3*x^2 - 2*I*b*d^2*e*f^2*x - I*b*d^ 2*e^2*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*(I*b*d^2*f^3*x^2 + 2*I*b*d^2*e*f^2*x + I*b*d^2*e^2*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*si n(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(b*d^3*e^3 - 3*b*c*d...
Timed out. \[ \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \cos \left (d x + c\right )^{2}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]