Integrand size = 26, antiderivative size = 268 \[ \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{-2 i d-\frac {(2 e+i b \log (f))^2}{8 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {2 i e-b \log (f)+2 x (2 i f-c \log (f))}{2 \sqrt {2 i f-c \log (f)}}\right )}{8 \sqrt {2 i f-c \log (f)}}+\frac {e^{2 i d+\frac {(2 e-i b \log (f))^2}{8 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {2 i e+b \log (f)+2 x (2 i f+c \log (f))}{2 \sqrt {2 i f+c \log (f)}}\right )}{8 \sqrt {2 i f+c \log (f)}} \] Output:
1/4*f^(a-1/4*b^2/c)*Pi^(1/2)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(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*Pi^ (1/2)*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))/( 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*Pi^(1/2)*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))/(2*I*f+c*ln(f))^(1/2)
Time = 6.32 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.56 \[ \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx=\frac {1}{8} f^a \sqrt {\pi } \left (\frac {2 f^{-\frac {b^2}{4 c}} \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{\sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt [4]{-1} e^{-\frac {1}{4} i \left (\frac {4 e^2}{2 f-i c \log (f)}+\frac {b^2 \log ^2(f)}{2 f+i c \log (f)}\right )} f^{-\frac {4 b e f}{4 f^2+c^2 \log ^2(f)}} \left (-e^{\frac {4 i e^2 f}{4 f^2+c^2 \log ^2(f)}} f^{\frac {b e}{2 f-i c \log (f)}} \text {erfi}\left (\frac {(-1)^{3/4} (2 (e+2 f x)+i (b+2 c x) \log (f))}{2 \sqrt {2 f+i c \log (f)}}\right ) (2 f-i c \log (f)) \sqrt {2 f+i c \log (f)} (\cos (2 d)-i \sin (2 d))+e^{\frac {i b^2 f \log ^2(f)}{4 f^2+c^2 \log ^2(f)}} f^{\frac {b e}{2 f+i c \log (f)}} \text {erfi}\left (\frac {\sqrt [4]{-1} (2 (e+2 f x)-i (b+2 c x) \log (f))}{2 \sqrt {2 f-i c \log (f)}}\right ) \sqrt {2 f-i c \log (f)} (2 f+i c \log (f)) (-i \cos (2 d)+\sin (2 d))\right )}{4 f^2+c^2 \log ^2(f)}\right ) \] Input:
Integrate[f^(a + b*x + c*x^2)*Cos[d + e*x + f*x^2]^2,x]
Output:
(f^a*Sqrt[Pi]*((2*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(Sqrt[c]*f ^(b^2/(4*c))*Sqrt[Log[f]]) + ((-1)^(1/4)*(-(E^(((4*I)*e^2*f)/(4*f^2 + c^2* Log[f]^2))*f^((b*e)/(2*f - I*c*Log[f]))*Erfi[((-1)^(3/4)*(2*(e + 2*f*x) + I*(b + 2*c*x)*Log[f]))/(2*Sqrt[2*f + I*c*Log[f]])]*(2*f - I*c*Log[f])*Sqrt [2*f + I*c*Log[f]]*(Cos[2*d] - I*Sin[2*d])) + E^((I*b^2*f*Log[f]^2)/(4*f^2 + c^2*Log[f]^2))*f^((b*e)/(2*f + I*c*Log[f]))*Erfi[((-1)^(1/4)*(2*(e + 2* f*x) - I*(b + 2*c*x)*Log[f]))/(2*Sqrt[2*f - I*c*Log[f]])]*Sqrt[2*f - I*c*L og[f]]*(2*f + I*c*Log[f])*((-I)*Cos[2*d] + Sin[2*d])))/(E^((I/4)*((4*e^2)/ (2*f - I*c*Log[f]) + (b^2*Log[f]^2)/(2*f + I*c*Log[f])))*f^((4*b*e*f)/(4*f ^2 + c^2*Log[f]^2))*(4*f^2 + c^2*Log[f]^2))))/8
Time = 0.81 (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 = {4976, 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.
| \(\displaystyle \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx\) |
| \(\Big \downarrow \) 4976 |
| \(\displaystyle \int \left (\frac {1}{4} e^{-2 i d-2 i e x-2 i f x^2} f^{a+b x+c x^2}+\frac {1}{4} e^{2 i d+2 i e x+2 i f x^2} f^{a+b x+c x^2}+\frac {1}{2} f^{a+b x+c x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt {\pi } f^a \exp \left (-\frac {(2 e+i b \log (f))^2}{-4 c \log (f)+8 i f}-2 i d\right ) \text {erf}\left (\frac {-b \log (f)+2 x (-c \log (f)+2 i f)+2 i e}{2 \sqrt {-c \log (f)+2 i f}}\right )}{8 \sqrt {-c \log (f)+2 i f}}+\frac {\sqrt {\pi } f^a \exp \left (\frac {(2 e-i b \log (f))^2}{4 c \log (f)+8 i f}+2 i d\right ) \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+2 i f)+2 i e}{2 \sqrt {c \log (f)+2 i f}}\right )}{8 \sqrt {c \log (f)+2 i f}}\) |
Input:
Int[f^(a + b*x + c*x^2)*Cos[d + e*x + f*x^2]^2,x]
Output:
(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]])
Int[Cos[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cos[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 = 1.62 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {16 d f -4 e^{2}-4 i \ln \left (f \right ) b e +8 i d \ln \left (f \right ) c +\ln \left (f \right )^{2} b^{2}}{4 \left (c \ln \left (f \right )-2 i f \right )}} \operatorname {erf}\left (-x \sqrt {2 i f -c \ln \left (f \right )}+\frac {b \ln \left (f \right )-2 i e}{2 \sqrt {2 i f -c \ln \left (f \right )}}\right )}{8 \sqrt {2 i f -c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {16 d f -4 e^{2}+4 i \ln \left (f \right ) b e -8 i d \ln \left (f \right ) c +\ln \left (f \right )^{2} b^{2}}{4 \left (2 i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-2 i f}\, x +\frac {2 i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )-2 i f}}\right )}{8 \sqrt {-c \ln \left (f \right )-2 i f}}-\frac {\sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} f^{a} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\) | \(263\) |
Input:
int(f^(c*x^2+b*x+a)*cos(f*x^2+e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
-1/8*Pi^(1/2)*f^a*exp(-1/4*(16*d*f-4*e^2-4*I*ln(f)*b*e+8*I*d*ln(f)*c+ln(f) ^2*b^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*(16 *d*f-4*e^2+4*I*ln(f)*b*e-8*I*d*ln(f)*c+ln(f)^2*b^2)/(2*I*f+c*ln(f)))/(-c*l n(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 = 0.09 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.75 \[ \int f^{a+b x+c x^2} \cos ^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 )}} \] Input:
integrate(f^(c*x^2+b*x+a)*cos(f*x^2+e*x+d)^2,x, algorithm="fricas")
Output:
-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)*sq rt(-c*log(f))/c)/f^(1/4*(b^2 - 4*a*c)/c))/(c^3*log(f)^3 + 4*c*f^2*log(f))
\[ \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \cos ^{2}{\left (d + e x + f x^{2} \right )}\, dx \] Input:
integrate(f**(c*x**2+b*x+a)*cos(f*x**2+e*x+d)**2,x)
Output:
Integral(f**(a + b*x + c*x**2)*cos(d + e*x + f*x**2)**2, x)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 1487, normalized size of antiderivative = 5.55 \[ \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(f^(c*x^2+b*x+a)*cos(f*x^2+e*x+d)^2,x, algorithm="maxima")
Output:
-1/16*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 8*f^2)*((I*f^a*cos(-1/2*(4*e^2*f - 1 6*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*lo g(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*l og(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*lo g(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...
\[ \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx=\int { f^{c x^{2} + b x + a} \cos \left (f x^{2} + e x + d\right )^{2} \,d x } \] Input:
integrate(f^(c*x^2+b*x+a)*cos(f*x^2+e*x+d)^2,x, algorithm="giac")
Output:
integrate(f^(c*x^2 + b*x + a)*cos(f*x^2 + e*x + d)^2, x)
Timed out. \[ \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,{\cos \left (f\,x^2+e\,x+d\right )}^2 \,d x \] Input:
int(f^(a + b*x + c*x^2)*cos(d + e*x + f*x^2)^2,x)
Output:
int(f^(a + b*x + c*x^2)*cos(d + e*x + f*x^2)^2, x)
\[ \int f^{a+b x+c x^2} \cos ^2\left (d+e x+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}+b x} \cos \left (f \,x^{2}+e x +d \right )^{2}d x \right ) \] Input:
int(f^(c*x^2+b*x+a)*cos(f*x^2+e*x+d)^2,x)
Output:
f**a*int(f**(b*x + c*x**2)*cos(d + e*x + f*x**2)**2,x)