Integrand size = 28, antiderivative size = 659 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {i a (e+f x)^2}{\left (a^2-b^2\right ) d}-\frac {4 i b f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {i b^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {i b^2 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^2}+\frac {2 i b f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}-\frac {2 i b f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {2 b^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {2 b^2 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {i a f^2 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{\left (a^2-b^2\right ) d^3}+\frac {2 i b^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {2 i b^2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^3}-\frac {b (e+f x)^2 \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a (e+f x)^2 \tan (c+d x)}{\left (a^2-b^2\right ) d} \]
-I*a*(f*x+e)^2/(a^2-b^2)/d-4*I*b*f*(f*x+e)*arctan(exp(I*(d*x+c)))/(a^2-b^2 )/d^2+2*a*f*(f*x+e)*ln(1+exp(2*I*(d*x+c)))/(a^2-b^2)/d^2+I*b^2*(f*x+e)^2*l n(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d-I*b^2*(f*x+e )^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d+2*I*b*f ^2*polylog(2,-I*exp(I*(d*x+c)))/(a^2-b^2)/d^3-2*I*b*f^2*polylog(2,I*exp(I* (d*x+c)))/(a^2-b^2)/d^3-I*a*f^2*polylog(2,-exp(2*I*(d*x+c)))/(a^2-b^2)/d^3 +2*b^2*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^ 2)^(3/2)/d^2-2*b^2*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/ 2)))/(a^2-b^2)^(3/2)/d^2+2*I*b^2*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2- b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^3-2*I*b^2*f^2*polylog(3,I*b*exp(I*(d*x+c))/ (a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^3-b*(f*x+e)^2*sec(d*x+c)/(a^2-b^2)/ d+a*(f*x+e)^2*tan(d*x+c)/(a^2-b^2)/d
Time = 7.65 (sec) , antiderivative size = 1122, normalized size of antiderivative = 1.70 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {i b^2 \left (-2 \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 )+2 \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 )-i \left (d^2 \left (2 \sqrt {-a^2+b^2} e^2 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+\sqrt {a^2-b^2} f x (2 e+f x) \left (\log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-\log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )\right )+2 \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 )-2 \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 )\right )\right )}{\sqrt {-\left (a^2-b^2\right )^2} \left (-a^2+b^2\right ) d^3}+\frac {b (e+f x)^2 \sec (c)}{\left (-a^2+b^2\right ) d}+\frac {2 a e f \sec (c) (\cos (c) \log (\cos (c) \cos (d x)-\sin (c) \sin (d x))+d x \sin (c))}{\left (a^2-b^2\right ) d^2 \left (\cos ^2(c)+\sin ^2(c)\right )}+\frac {4 i b e f \arctan \left (\frac {-i \sin (c)-i \cos (c) \tan \left (\frac {d x}{2}\right )}{\sqrt {\cos ^2(c)+\sin ^2(c)}}\right )}{\left (a^2-b^2\right ) d^2 \sqrt {\cos ^2(c)+\sin ^2(c)}}+\frac {a f^2 \csc (c) \left (d^2 e^{-i \arctan (\cot (c))} x^2-\frac {\cot (c) \left (i d x (-\pi -2 \arctan (\cot (c)))-\pi \log \left (1+e^{-2 i d x}\right )-2 (d x-\arctan (\cot (c))) \log \left (1-e^{2 i (d x-\arctan (\cot (c)))}\right )+\pi \log (\cos (d x))-2 \arctan (\cot (c)) \log (\sin (d x-\arctan (\cot (c))))+i \operatorname {PolyLog}\left (2,e^{2 i (d x-\arctan (\cot (c)))}\right )\right )}{\sqrt {1+\cot ^2(c)}}\right ) \sec (c)}{\left (a^2-b^2\right ) d^3 \sqrt {\csc ^2(c) \left (\cos ^2(c)+\sin ^2(c)\right )}}+\frac {2 b f^2 \left (-\frac {\csc (c) \left ((d x-\arctan (\cot (c))) \left (\log \left (1-e^{i (d x-\arctan (\cot (c)))}\right )-\log \left (1+e^{i (d x-\arctan (\cot (c)))}\right )\right )+i \left (\operatorname {PolyLog}\left (2,-e^{i (d x-\arctan (\cot (c)))}\right )-\operatorname {PolyLog}\left (2,e^{i (d x-\arctan (\cot (c)))}\right )\right )\right )}{\sqrt {1+\cot ^2(c)}}+\frac {2 \arctan (\cot (c)) \text {arctanh}\left (\frac {\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )}{\sqrt {\cos ^2(c)+\sin ^2(c)}}\right )}{\sqrt {\cos ^2(c)+\sin ^2(c)}}\right )}{\left (a^2-b^2\right ) d^3}+\frac {e^2 \sin \left (\frac {d x}{2}\right )+2 e f x \sin \left (\frac {d x}{2}\right )+f^2 x^2 \sin \left (\frac {d x}{2}\right )}{(a+b) d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {e^2 \sin \left (\frac {d x}{2}\right )+2 e f x \sin \left (\frac {d x}{2}\right )+f^2 x^2 \sin \left (\frac {d x}{2}\right )}{(a-b) d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]
(I*b^2*(-2*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])] + 2*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]))] - I*(d^2*(2*Sqrt[-a^2 + b^2 ]*e^2*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + Sqrt[a^2 - b^2]* f*x*(2*e + f*x)*(Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])])) + 2*Sqrt[a^2 - b ^2]*f^2*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 2*Sq rt[a^2 - b^2]*f^2*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2] ))])))/(Sqrt[-(a^2 - b^2)^2]*(-a^2 + b^2)*d^3) + (b*(e + f*x)^2*Sec[c])/(( -a^2 + b^2)*d) + (2*a*e*f*Sec[c]*(Cos[c]*Log[Cos[c]*Cos[d*x] - Sin[c]*Sin[ d*x]] + d*x*Sin[c]))/((a^2 - b^2)*d^2*(Cos[c]^2 + Sin[c]^2)) + ((4*I)*b*e* f*ArcTan[((-I)*Sin[c] - I*Cos[c]*Tan[(d*x)/2])/Sqrt[Cos[c]^2 + Sin[c]^2]]) /((a^2 - b^2)*d^2*Sqrt[Cos[c]^2 + Sin[c]^2]) + (a*f^2*Csc[c]*((d^2*x^2)/E^ (I*ArcTan[Cot[c]]) - (Cot[c]*(I*d*x*(-Pi - 2*ArcTan[Cot[c]]) - Pi*Log[1 + E^((-2*I)*d*x)] - 2*(d*x - ArcTan[Cot[c]])*Log[1 - E^((2*I)*(d*x - ArcTan[ Cot[c]]))] + Pi*Log[Cos[d*x]] - 2*ArcTan[Cot[c]]*Log[Sin[d*x - ArcTan[Cot[ c]]]] + I*PolyLog[2, E^((2*I)*(d*x - ArcTan[Cot[c]]))]))/Sqrt[1 + Cot[c]^2 ])*Sec[c])/((a^2 - b^2)*d^3*Sqrt[Csc[c]^2*(Cos[c]^2 + Sin[c]^2)]) + (2*b*f ^2*(-((Csc[c]*((d*x - ArcTan[Cot[c]])*(Log[1 - E^(I*(d*x - ArcTan[Cot[c]]) )] - Log[1 + E^(I*(d*x - ArcTan[Cot[c]]))]) + I*(PolyLog[2, -E^(I*(d*x ...
Time = 2.56 (sec) , antiderivative size = 561, normalized size of antiderivative = 0.85, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {5044, 3042, 3804, 2694, 27, 2620, 3011, 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.
\(\displaystyle \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 5044 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {b^2 \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 3804 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \int \frac {e^{i (c+d x)} (e+f x)^2}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{a^2-b^2}\) |
\(\Big \downarrow \) 2694 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {\int (e+f x)^2 \sec ^2(c+d x) (a-b \sin (c+d x))dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (a (e+f x)^2 \sec ^2(c+d x)-b (e+f x)^2 \sec (c+d x) \tan (c+d x)\right )dx}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {i a f^2 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{d^3}+\frac {2 a f (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{d^2}+\frac {a (e+f x)^2 \tan (c+d x)}{d}-\frac {i a (e+f x)^2}{d}-\frac {4 i b f (e+f x) \arctan \left (e^{i (c+d x)}\right )}{d^2}+\frac {2 i b f^2 \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{d^3}-\frac {2 i b f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{d^3}-\frac {b (e+f x)^2 \sec (c+d x)}{d}}{a^2-b^2}-\frac {2 b^2 \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\) |
(-2*b^2*(((-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]))/(a^2 - b^2) + (((-I)*a*(e + f*x)^2)/d - ((4*I)*b*f*(e + f*x)*ArcTa n[E^(I*(c + d*x))])/d^2 + (2*a*f*(e + f*x)*Log[1 + E^((2*I)*(c + d*x))])/d ^2 + ((2*I)*b*f^2*PolyLog[2, (-I)*E^(I*(c + d*x))])/d^3 - ((2*I)*b*f^2*Pol yLog[2, I*E^(I*(c + d*x))])/d^3 - (I*a*f^2*PolyLog[2, -E^((2*I)*(c + d*x)) ])/d^3 - (b*(e + f*x)^2*Sec[c + d*x])/d + (a*(e + f*x)^2*Tan[c + d*x])/d)/ (a^2 - b^2)
3.4.11.3.1 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]
\[\int \frac {\left (f x +e \right )^{2} \left (\sec ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2659 vs. \(2 (574) = 1148\).
Time = 0.58 (sec) , antiderivative size = 2659, normalized size of antiderivative = 4.03 \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
-1/2*(2*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) - 2*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*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) + 2*b^3*f^2*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c )*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) - 2*b^3*f^2*sqrt(-(a^2 - b^2)/b^2 )*cos(d*x + c)*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) + 2*(a^2*b - b^3)*d^ 2*f^2*x^2 + 4*(a^2*b - b^3)*d^2*e*f*x + 2*(a^2*b - b^3)*d^2*e^2 - 2*I*(a^3 - a^2*b - a*b^2 + b^3)*f^2*cos(d*x + c)*dilog(I*cos(d*x + c) + sin(d*x + c)) + 2*I*(a^3 + a^2*b - a*b^2 - b^3)*f^2*cos(d*x + c)*dilog(I*cos(d*x + c ) - sin(d*x + c)) + 2*I*(a^3 - a^2*b - a*b^2 + b^3)*f^2*cos(d*x + c)*dilog (-I*cos(d*x + c) + sin(d*x + c)) - 2*I*(a^3 + a^2*b - a*b^2 - b^3)*f^2*cos (d*x + c)*dilog(-I*cos(d*x + c) - sin(d*x + c)) + 2*(I*b^3*d*f^2*x + I*b^3 *d*e*f)*sqrt(-(a^2 - b^2)/b^2)*cos(d*x + c)*dilog((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)/b + 1) + 2*(-I*b^3*d*f^2*x - I*b^3*d*e*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*si n(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 2*(-I*b^3*d*f^2*x - I*...
\[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\left (e + f x\right )^{2} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sec \left (d x + c\right )^{2}}{b \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^2 \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]