Integrand size = 28, antiderivative size = 923 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {i a (e+f x)^3}{\left (a^2-b^2\right ) d}-\frac {6 i b f (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {3 a f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {6 i b f^2 (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 i b f^2 (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {3 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 )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 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 )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {3 i a f^2 (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {6 b f^3 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 b f^3 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^4}+\frac {6 i 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 )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i 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 )}{\left (a^2-b^2\right )^{3/2} d^3}+\frac {3 a f^3 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 \left (a^2-b^2\right ) d^4}-\frac {6 b^2 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}+\frac {6 b^2 f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^4}-\frac {b (e+f x)^3 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^3 \tan (c+d x)}{\left (a^2-b^2\right ) d} \]
-6*I*b*f*(f*x+e)^2*arctan(exp(I*(d*x+c)))/(a^2-b^2)/d^2+6*I*b*f^2*(f*x+e)* polylog(2,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^3+3*a*f*(f*x+e)^2*ln(1+exp(2*I*(d *x+c)))/(a^2-b^2)/d^2+I*b^2*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2) ^(1/2)))/(a^2-b^2)^(3/2)/d-I*a*(f*x+e)^3/(a^2-b^2)/d-I*b^2*(f*x+e)^3*ln(1- I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d-3*I*a*f^2*(f*x+e )*polylog(2,-exp(2*I*(d*x+c)))/(a^2-b^2)/d^3+6*I*b^2*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)^(3/2)/d^3+3*b^2*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)^(3/2)/d^2 -3*b^2*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)^(3/2)/d^2-6*b*f^3*polylog(3,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^4+6*b*f^3* polylog(3,I*exp(I*(d*x+c)))/(a^2-b^2)/d^4+3/2*a*f^3*polylog(3,-exp(2*I*(d* x+c)))/(a^2-b^2)/d^4-6*I*b^2*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)^(3/2)/d^3-6*I*b*f^2*(f*x+e)*polylog(2,I*exp(I*( d*x+c)))/(a^2-b^2)/d^3-6*b^2*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a-(a^2-b^2) ^(1/2)))/(a^2-b^2)^(3/2)/d^4+6*b^2*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^ 2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^4-b*(f*x+e)^3*sec(d*x+c)/(a^2-b^2)/d+a*(f *x+e)^3*tan(d*x+c)/(a^2-b^2)/d
Time = 8.25 (sec) , antiderivative size = 1438, normalized size of antiderivative = 1.56 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \]
(f*(((2*I)*a*(e + f*x)^3)/f + (3*(a - b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log [1 - I/E^(I*(c + d*x))])/d + (3*(a + b)*(1 + E^((2*I)*c))*(e + f*x)^2*Log[ 1 + I/E^(I*(c + d*x))])/d + (6*(a + b)*(1 + E^((2*I)*c))*f*(I*d*(e + f*x)* PolyLog[2, (-I)/E^(I*(c + d*x))] + f*PolyLog[3, (-I)/E^(I*(c + d*x))]))/d^ 3 + (6*(a - b)*(1 + E^((2*I)*c))*f*(I*d*(e + f*x)*PolyLog[2, I/E^(I*(c + d *x))] + f*PolyLog[3, I/E^(I*(c + d*x))]))/d^3))/((a^2 - b^2)*d*(1 + E^((2* I)*c))) + (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 + 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*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 + Sqr t[-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*PolyLo g[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...
Time = 3.35 (sec) , antiderivative size = 774, normalized size of antiderivative = 0.84, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5044, 3042, 3804, 2694, 27, 2620, 3011, 7163, 2720, 7143, 7293, 2009}
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.
\(\displaystyle \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 5044 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \int \frac {(e+f x)^3}{a+b \sin (c+d x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3804 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \int \frac {e^{i (c+d x)} (e+f x)^3}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 2694 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^3}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \int (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {i f \int \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {\int (e+f x)^3 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (a (e+f x)^3 \sec ^2(c+d x)-b (e+f x)^3 \sec (c+d x) \tan (c+d x)\right )dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {3 a f^3 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right )}{2 d^4}-\frac {3 i a f^2 (e+f x) \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^3}+\frac {3 a f (e+f x)^2 \log \left (1+e^{2 i (c+d x)}\right )}{d^2}+\frac {a (e+f x)^3 \tan (c+d x)}{d}-\frac {i a (e+f x)^3}{d}-\frac {6 i b f (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{d^2}-\frac {6 b f^3 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{d^4}+\frac {6 b f^3 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{d^4}+\frac {6 i b f^2 (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^3}-\frac {6 i b f^2 (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^3}-\frac {b (e+f x)^3 \sec (c+d x)}{d}}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {3 f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}-\frac {i (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}\right )}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
(-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)*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]))/(a^2 - b^2) + (((-I)*a*(e + f*x)^3)/d - ((6*I)*b*f*(e + f*x)^2*ArcTan[E^(I*(c + d*x))])/d^2 + (3*a*f*( e + f*x)^2*Log[1 + E^((2*I)*(c + d*x))])/d^2 + ((6*I)*b*f^2*(e + f*x)*Poly Log[2, (-I)*E^(I*(c + d*x))])/d^3 - ((6*I)*b*f^2*(e + f*x)*PolyLog[2, I*E^ (I*(c + d*x))])/d^3 - ((3*I)*a*f^2*(e + f*x)*PolyLog[2, -E^((2*I)*(c + d*x ))])/d^3 - (6*b*f^3*PolyLog[3, (-I)*E^(I*(c + d*x))])/d^4 + (6*b*f^3*PolyL og[3, I*E^(I*(c + d*x))])/d^4 + (3*a*f^3*PolyLog[3, -E^((2*I)*(c + d*x))]) /(2*d^4) - (b*(e + f*x)^3*Sec[c + d*x])/d + (a*(e + f*x)^3*Tan[c + d*x])/d )/(a^2 - b^2)
3.4.10.3.1 Defintions of rubi rules used
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[((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_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_. )*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[-b^2/(a^2 - b^2) Int[(e + f *x)^m*(Sec[c + d*x]^(n - 2)/(a + b*Sin[c + d*x])), x], x] + Simp[1/(a^2 - b ^2) Int[(e + f*x)^m*Sec[c + d*x]^n*(a - b*Sin[c + d*x]), x], x] /; FreeQ[ {a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[a^2 - b^2, 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[((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]
\[\int \frac {\left (f x +e \right )^{3} \left (\sec ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4116 vs. \(2 (810) = 1620\).
Time = 0.72 (sec) , antiderivative size = 4116, normalized size of antiderivative = 4.46 \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
-1/2*(2*(a^2*b - b^3)*d^3*f^3*x^3 + 6*(a^2*b - b^3)*d^3*e*f^2*x^2 - 6*I*b^ 3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) + 6*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*polylog(4, -(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*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*poly log(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) - 6*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*co s(d*x + c)*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) + 6*(a^2*b - b^3)*d^3*e^ 2*f*x + 2*(a^2*b - b^3)*d^3*e^3 - 6*(a^3 - a^2*b - a*b^2 + b^3)*f^3*cos(d* x + c)*polylog(3, I*cos(d*x + c) + sin(d*x + c)) - 6*(a^3 + a^2*b - a*b^2 - b^3)*f^3*cos(d*x + c)*polylog(3, I*cos(d*x + c) - sin(d*x + c)) - 6*(a^3 - a^2*b - a*b^2 + b^3)*f^3*cos(d*x + c)*polylog(3, -I*cos(d*x + c) + sin( d*x + c)) - 6*(a^3 + a^2*b - a*b^2 - b^3)*f^3*cos(d*x + c)*polylog(3, -I*c os(d*x + c) - sin(d*x + c)) + 3*(I*b^3*d^2*f^3*x^2 + 2*I*b^3*d^2*e*f^2*x + I*b^3*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) + 3*(-I*b^3*d^2*f^3*x^2 - 2*I*b^3*d^2*e*f^2*x - I*b^3* d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog((I*a*cos(d*x + c) ...
\[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\left (e + f x\right )^{3} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \sec \left (d x + c\right )^{2}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^3 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]