Integrand size = 26, antiderivative size = 268 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {e^{-2 i d-\frac {(2 e+i b \log (f))^2}{8 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {2 i e-b \log (f)+2 x (2 i f-c \log (f))}{2 \sqrt {2 i f-c \log (f)}}\right )}{8 \sqrt {2 i f-c \log (f)}}-\frac {e^{2 i d+\frac {(2 e-i b \log (f))^2}{8 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {2 i e+b \log (f)+2 x (2 i f+c \log (f))}{2 \sqrt {2 i f+c \log (f)}}\right )}{8 \sqrt {2 i f+c \log (f)}} \]
1/4*f^(a-1/4*b^2/c)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))*Pi^(1/2)/c^(1/ 2)/ln(f)^(1/2)-1/8*exp(-2*I*d-(2*e+I*b*ln(f))^2/(8*I*f-4*c*ln(f)))*f^a*erf (1/2*(2*I*e-b*ln(f)+2*x*(2*I*f-c*ln(f)))/(2*I*f-c*ln(f))^(1/2))*Pi^(1/2)/( 2*I*f-c*ln(f))^(1/2)-1/8*exp(2*I*d+(2*e-I*b*ln(f))^2/(8*I*f+4*c*ln(f)))*f^ a*erfi(1/2*(2*I*e+b*ln(f)+2*x*(2*I*f+c*ln(f)))/(2*I*f+c*ln(f))^(1/2))*Pi^( 1/2)/(2*I*f+c*ln(f))^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1120\) vs. \(2(268)=536\).
Time = 6.53 (sec) , antiderivative size = 1120, normalized size of antiderivative = 4.18 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx=\frac {f^a \sqrt {\pi } \left (8 \sqrt {c} f^{2-\frac {b^2}{4 c}} \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)}+2 c^{5/2} f^{-\frac {b^2}{4 c}} \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \log ^{\frac {5}{2}}(f)+2 \sqrt [4]{-1} c e^{\frac {i \left (-4 e^2+4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f-i c \log (f))}} f \cos (2 d) \text {erf}\left (\frac {(-1)^{3/4} (2 e+4 f x-i b \log (f)-2 i c x \log (f))}{2 \sqrt {2 f-i c \log (f)}}\right ) \log (f) \sqrt {2 f-i c \log (f)}+(-1)^{3/4} c^2 e^{\frac {i \left (-4 e^2+4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f-i c \log (f))}} \cos (2 d) \text {erf}\left (\frac {(-1)^{3/4} (2 e+4 f x-i b \log (f)-2 i c x \log (f))}{2 \sqrt {2 f-i c \log (f)}}\right ) \log ^2(f) \sqrt {2 f-i c \log (f)}+2 (-1)^{3/4} c e^{-\frac {i \left (-4 e^2-4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f+i c \log (f))}} f \cos (2 d) \text {erf}\left (\frac {\sqrt [4]{-1} (2 e+4 f x+i b \log (f)+2 i c x \log (f))}{2 \sqrt {2 f+i c \log (f)}}\right ) \log (f) \sqrt {2 f+i c \log (f)}+\sqrt [4]{-1} c^2 e^{-\frac {i \left (-4 e^2-4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f+i c \log (f))}} \cos (2 d) \text {erf}\left (\frac {\sqrt [4]{-1} (2 e+4 f x+i b \log (f)+2 i c x \log (f))}{2 \sqrt {2 f+i c \log (f)}}\right ) \log ^2(f) \sqrt {2 f+i c \log (f)}+2 (-1)^{3/4} c e^{\frac {i \left (-4 e^2+4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f-i c \log (f))}} f \text {erf}\left (\frac {(-1)^{3/4} (2 e+4 f x-i b \log (f)-2 i c x \log (f))}{2 \sqrt {2 f-i c \log (f)}}\right ) \log (f) \sqrt {2 f-i c \log (f)} \sin (2 d)-\sqrt [4]{-1} c^2 e^{\frac {i \left (-4 e^2+4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f-i c \log (f))}} \text {erf}\left (\frac {(-1)^{3/4} (2 e+4 f x-i b \log (f)-2 i c x \log (f))}{2 \sqrt {2 f-i c \log (f)}}\right ) \log ^2(f) \sqrt {2 f-i c \log (f)} \sin (2 d)+2 \sqrt [4]{-1} c e^{-\frac {i \left (-4 e^2-4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f+i c \log (f))}} f \text {erf}\left (\frac {\sqrt [4]{-1} (2 e+4 f x+i b \log (f)+2 i c x \log (f))}{2 \sqrt {2 f+i c \log (f)}}\right ) \log (f) \sqrt {2 f+i c \log (f)} \sin (2 d)-(-1)^{3/4} c^2 e^{-\frac {i \left (-4 e^2-4 i b e \log (f)+b^2 \log ^2(f)\right )}{4 (2 f+i c \log (f))}} \text {erf}\left (\frac {\sqrt [4]{-1} (2 e+4 f x+i b \log (f)+2 i c x \log (f))}{2 \sqrt {2 f+i c \log (f)}}\right ) \log ^2(f) \sqrt {2 f+i c \log (f)} \sin (2 d)\right )}{8 c \log (f) (2 f-i c \log (f)) (2 f+i c \log (f))} \]
(f^a*Sqrt[Pi]*(8*Sqrt[c]*f^(2 - b^2/(4*c))*Erfi[((b + 2*c*x)*Sqrt[Log[f]]) /(2*Sqrt[c])]*Sqrt[Log[f]] + (2*c^(5/2)*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2 *Sqrt[c])]*Log[f]^(5/2))/f^(b^2/(4*c)) + 2*(-1)^(1/4)*c*E^(((I/4)*(-4*e^2 + (4*I)*b*e*Log[f] + b^2*Log[f]^2))/(2*f - I*c*Log[f]))*f*Cos[2*d]*Erf[((- 1)^(3/4)*(2*e + 4*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*Sqrt[2*f - I*c* Log[f]])]*Log[f]*Sqrt[2*f - I*c*Log[f]] + (-1)^(3/4)*c^2*E^(((I/4)*(-4*e^2 + (4*I)*b*e*Log[f] + b^2*Log[f]^2))/(2*f - I*c*Log[f]))*Cos[2*d]*Erf[((-1 )^(3/4)*(2*e + 4*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*Sqrt[2*f - I*c*L og[f]])]*Log[f]^2*Sqrt[2*f - I*c*Log[f]] + (2*(-1)^(3/4)*c*f*Cos[2*d]*Erf[ ((-1)^(1/4)*(2*e + 4*f*x + I*b*Log[f] + (2*I)*c*x*Log[f]))/(2*Sqrt[2*f + I *c*Log[f]])]*Log[f]*Sqrt[2*f + I*c*Log[f]])/E^(((I/4)*(-4*e^2 - (4*I)*b*e* Log[f] + b^2*Log[f]^2))/(2*f + I*c*Log[f])) + ((-1)^(1/4)*c^2*Cos[2*d]*Erf [((-1)^(1/4)*(2*e + 4*f*x + I*b*Log[f] + (2*I)*c*x*Log[f]))/(2*Sqrt[2*f + I*c*Log[f]])]*Log[f]^2*Sqrt[2*f + I*c*Log[f]])/E^(((I/4)*(-4*e^2 - (4*I)*b *e*Log[f] + b^2*Log[f]^2))/(2*f + I*c*Log[f])) + 2*(-1)^(3/4)*c*E^(((I/4)* (-4*e^2 + (4*I)*b*e*Log[f] + b^2*Log[f]^2))/(2*f - I*c*Log[f]))*f*Erf[((-1 )^(3/4)*(2*e + 4*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*Sqrt[2*f - I*c*L og[f]])]*Log[f]*Sqrt[2*f - I*c*Log[f]]*Sin[2*d] - (-1)^(1/4)*c^2*E^(((I/4) *(-4*e^2 + (4*I)*b*e*Log[f] + b^2*Log[f]^2))/(2*f - I*c*Log[f]))*Erf[((-1) ^(3/4)*(2*e + 4*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*Sqrt[2*f - I*c...
Time = 0.79 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {4975, 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.
| \(\displaystyle \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx\) |
| \(\Big \downarrow \) 4975 |
| \(\displaystyle \int \left (-\frac {1}{4} e^{-2 i d-2 i e x-2 i f x^2} f^{a+b x+c x^2}-\frac {1}{4} e^{2 i d+2 i e x+2 i f x^2} f^{a+b x+c x^2}+\frac {1}{2} f^{a+b x+c x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a \exp \left (-\frac {(2 e+i b \log (f))^2}{-4 c \log (f)+8 i f}-2 i d\right ) \text {erf}\left (\frac {-b \log (f)+2 x (-c \log (f)+2 i f)+2 i e}{2 \sqrt {-c \log (f)+2 i f}}\right )}{8 \sqrt {-c \log (f)+2 i f}}-\frac {\sqrt {\pi } f^a \exp \left (\frac {(2 e-i b \log (f))^2}{4 c \log (f)+8 i f}+2 i d\right ) \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+2 i f)+2 i e}{2 \sqrt {c \log (f)+2 i f}}\right )}{8 \sqrt {c \log (f)+2 i f}}\) |
(f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/ (4*Sqrt[c]*Sqrt[Log[f]]) - (E^((-2*I)*d - (2*e + I*b*Log[f])^2/((8*I)*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[((2*I)*e - b*Log[f] + 2*x*((2*I)*f - c*Log[f ]))/(2*Sqrt[(2*I)*f - c*Log[f]])])/(8*Sqrt[(2*I)*f - c*Log[f]]) - (E^((2*I )*d + (2*e - I*b*Log[f])^2/((8*I)*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[((2*I )*e + b*Log[f] + 2*x*((2*I)*f + c*Log[f]))/(2*Sqrt[(2*I)*f + c*Log[f]])])/ (8*Sqrt[(2*I)*f + c*Log[f]])
3.2.1.3.1 Defintions of rubi rules used
Int[(F_)^(u_)*Sin[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[v]^n , x], x] /; FreeQ[F, x] && (LinearQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]
Time = 0.77 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-4 i \ln \left (f \right ) b e +8 i d \ln \left (f \right ) c +16 d f -4 e^{2}}{4 \left (c \ln \left (f \right )-2 i f \right )}} \operatorname {erf}\left (-x \sqrt {2 i f -c \ln \left (f \right )}+\frac {b \ln \left (f \right )-2 i e}{2 \sqrt {2 i f -c \ln \left (f \right )}}\right )}{8 \sqrt {2 i f -c \ln \left (f \right )}}+\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+4 i \ln \left (f \right ) b e -8 i d \ln \left (f \right ) c +16 d f -4 e^{2}}{4 \left (2 i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-2 i f}\, x +\frac {2 i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )-2 i f}}\right )}{8 \sqrt {-c \ln \left (f \right )-2 i f}}-\frac {\sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} f^{a} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\) | \(263\) |
1/8*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2-4*I*ln(f)*b*e+8*I*d*ln(f)*c+16*d*f- 4*e^2)/(c*ln(f)-2*I*f))/(2*I*f-c*ln(f))^(1/2)*erf(-x*(2*I*f-c*ln(f))^(1/2) +1/2*(b*ln(f)-2*I*e)/(2*I*f-c*ln(f))^(1/2))+1/8*Pi^(1/2)*f^a*exp(-1/4*(ln( f)^2*b^2+4*I*ln(f)*b*e-8*I*d*ln(f)*c+16*d*f-4*e^2)/(2*I*f+c*ln(f)))/(-c*ln (f)-2*I*f)^(1/2)*erf(-(-c*ln(f)-2*I*f)^(1/2)*x+1/2*(2*I*e+b*ln(f))/(-c*ln( f)-2*I*f)^(1/2))-1/4*Pi^(1/2)*f^(-1/4*b^2/c)*f^a/(-c*ln(f))^(1/2)*erf(-(-c *ln(f))^(1/2)*x+1/2*ln(f)*b/(-c*ln(f))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (199) = 398\).
Time = 0.26 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.75 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} - 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - 2 i \, f} \operatorname {erf}\left (\frac {{\left (8 \, f^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2} + 4 \, e f - 2 \, {\left (-i \, c e + i \, b f\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - 2 i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} + 8 i \, e^{2} f - 32 i \, d f^{2} + 2 \, {\left (-4 i \, c^{2} d + 2 i \, b c e - i \, b^{2} f\right )} \log \left (f\right )^{2} - 4 \, {\left (c e^{2} - 2 \, b e f + 4 \, a f^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right )} + \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + 2 i \, f} \operatorname {erf}\left (\frac {{\left (8 \, f^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2} + 4 \, e f - 2 \, {\left (i \, c e - i \, b f\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + 2 i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} - 8 i \, e^{2} f + 32 i \, d f^{2} + 2 \, {\left (4 i \, c^{2} d - 2 i \, b c e + i \, b^{2} f\right )} \log \left (f\right )^{2} - 4 \, {\left (c e^{2} - 2 \, b e f + 4 \, a f^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right )} - \frac {2 \, \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{8 \, {\left (c^{3} \log \left (f\right )^{3} + 4 \, c f^{2} \log \left (f\right )\right )}} \]
1/8*(sqrt(pi)*(c^2*log(f)^2 - 2*I*c*f*log(f))*sqrt(-c*log(f) - 2*I*f)*erf( 1/2*(8*f^2*x + (2*c^2*x + b*c)*log(f)^2 + 4*e*f - 2*(-I*c*e + I*b*f)*log(f ))*sqrt(-c*log(f) - 2*I*f)/(c^2*log(f)^2 + 4*f^2))*e^(-1/4*((b^2*c - 4*a*c ^2)*log(f)^3 + 8*I*e^2*f - 32*I*d*f^2 + 2*(-4*I*c^2*d + 2*I*b*c*e - I*b^2* f)*log(f)^2 - 4*(c*e^2 - 2*b*e*f + 4*a*f^2)*log(f))/(c^2*log(f)^2 + 4*f^2) ) + sqrt(pi)*(c^2*log(f)^2 + 2*I*c*f*log(f))*sqrt(-c*log(f) + 2*I*f)*erf(1 /2*(8*f^2*x + (2*c^2*x + b*c)*log(f)^2 + 4*e*f - 2*(I*c*e - I*b*f)*log(f)) *sqrt(-c*log(f) + 2*I*f)/(c^2*log(f)^2 + 4*f^2))*e^(-1/4*((b^2*c - 4*a*c^2 )*log(f)^3 - 8*I*e^2*f + 32*I*d*f^2 + 2*(4*I*c^2*d - 2*I*b*c*e + I*b^2*f)* log(f)^2 - 4*(c*e^2 - 2*b*e*f + 4*a*f^2)*log(f))/(c^2*log(f)^2 + 4*f^2)) - 2*sqrt(pi)*(c^2*log(f)^2 + 4*f^2)*sqrt(-c*log(f))*erf(1/2*(2*c*x + b)*sqr t(-c*log(f))/c)/f^(1/4*(b^2 - 4*a*c)/c))/(c^3*log(f)^3 + 4*c*f^2*log(f))
\[ \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \sin ^{2}{\left (d + e x + f x^{2} \right )}\, dx \]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.27 (sec) , antiderivative size = 1487, normalized size of antiderivative = 5.55 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \]
1/16*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 8*f^2)*((I*f^a*cos(-1/2*(4*e^2*f - 16 *d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2))*e^( c*e^2*log(f)/(c^2*log(f)^2 + 4*f^2) + 1/4*b^2*log(f)/c) + f^a*e^(c*e^2*log (f)/(c^2*log(f)^2 + 4*f^2) + 1/4*b^2*log(f)/c)*sin(-1/2*(4*e^2*f - 16*d*f^ 2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2)))*erf(1/2 *(2*(c*log(f) - 2*I*f)*x + b*log(f) - 2*I*e)*sqrt(-c*log(f) + 2*I*f)/(c*lo g(f) - 2*I*f)) + (-I*f^a*cos(-1/2*(4*e^2*f - 16*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2))*e^(c*e^2*log(f)/(c^2*log(f)^2 + 4*f^2) + 1/4*b^2*log(f)/c) + f^a*e^(c*e^2*log(f)/(c^2*log(f)^2 + 4*f^2) + 1/4*b^2*log(f)/c)*sin(-1/2*(4*e^2*f - 16*d*f^2 - (4*c^2*d - 2*b*c*e + b^ 2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2)))*erf(1/2*(2*(c*log(f) + 2*I*f)*x + b*log(f) + 2*I*e)*sqrt(-c*log(f) - 2*I*f)/(c*log(f) + 2*I*f)))*sqrt(c*log( f) + sqrt(c^2*log(f)^2 + 4*f^2))*sqrt(-c*log(f)) - sqrt(pi)*sqrt(2*c^2*log (f)^2 + 8*f^2)*((f^a*cos(-1/2*(4*e^2*f - 16*d*f^2 - (4*c^2*d - 2*b*c*e + b ^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2))*e^(c*e^2*log(f)/(c^2*log(f)^2 + 4* f^2) + 1/4*b^2*log(f)/c) - I*f^a*e^(c*e^2*log(f)/(c^2*log(f)^2 + 4*f^2) + 1/4*b^2*log(f)/c)*sin(-1/2*(4*e^2*f - 16*d*f^2 - (4*c^2*d - 2*b*c*e + b^2* f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2)))*erf(1/2*(2*(c*log(f) - 2*I*f)*x + b* log(f) - 2*I*e)*sqrt(-c*log(f) + 2*I*f)/(c*log(f) - 2*I*f)) + (f^a*cos(-1/ 2*(4*e^2*f - 16*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(...
\[ \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx=\int { f^{c x^{2} + b x + a} \sin \left (f x^{2} + e x + d\right )^{2} \,d x } \]
Timed out. \[ \int f^{a+b x+c x^2} \sin ^2\left (d+e x+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,{\sin \left (f\,x^2+e\,x+d\right )}^2 \,d x \]