\(\int f^{a+b x+c x^2} \sin (d+e x+f x^2) \, dx\) [112]

Optimal result

Integrand size = 24, antiderivative size = 212 \[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\frac {i e^{-i d-\frac {(e+i b \log (f))^2}{4 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {i e-b \log (f)+2 x (i f-c \log (f))}{2 \sqrt {i f-c \log (f)}}\right )}{4 \sqrt {i f-c \log (f)}}-\frac {i e^{i d+\frac {(e-i b \log (f))^2}{4 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 x (i f+c \log (f))}{2 \sqrt {i f+c \log (f)}}\right )}{4 \sqrt {i f+c \log (f)}} \] Output:

1/4*I*exp(-I*d-(e+I*b*ln(f))^2/(4*I*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(I* 
e-b*ln(f)+2*x*(I*f-c*ln(f)))/(I*f-c*ln(f))^(1/2))/(I*f-c*ln(f))^(1/2)-1/4* 
I*exp(I*d+(e-I*b*ln(f))^2/(4*I*f+4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/2*(I*e+b* 
ln(f)+2*x*(I*f+c*ln(f)))/(I*f+c*ln(f))^(1/2))/(I*f+c*ln(f))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.64 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.64 \[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt [4]{-1} e^{-\frac {1}{4} i \left (\frac {e^2}{f-i c \log (f)}+\frac {b^2 \log ^2(f)}{f+i c \log (f)}\right )} f^{\frac {f (-b e+a f)+a c^2 \log ^2(f)}{f^2+c^2 \log ^2(f)}} \sqrt {\pi } \left (e^{\frac {i b^2 f \log ^2(f)}{2 \left (f^2+c^2 \log ^2(f)\right )}} f^{\frac {b e}{2 f+2 i c \log (f)}} \text {erfi}\left (\frac {\sqrt [4]{-1} (e+2 f x-i (b+2 c x) \log (f))}{2 \sqrt {f-i c \log (f)}}\right ) \sqrt {f-i c \log (f)} (f+i c \log (f)) (\cos (d)+i \sin (d))+e^{\frac {i e^2 f}{2 \left (f^2+c^2 \log ^2(f)\right )}} f^{\frac {b e}{2 f-2 i c \log (f)}} \text {erfi}\left (\frac {(-1)^{3/4} (e+2 f x+i (b+2 c x) \log (f))}{2 \sqrt {f+i c \log (f)}}\right ) (f-i c \log (f)) \sqrt {f+i c \log (f)} (i \cos (d)+\sin (d))\right )}{4 \left (f^2+c^2 \log ^2(f)\right )} \] Input:

Integrate[f^(a + b*x + c*x^2)*Sin[d + e*x + f*x^2],x]
 

Output:

-1/4*((-1)^(1/4)*f^((f*(-(b*e) + a*f) + a*c^2*Log[f]^2)/(f^2 + c^2*Log[f]^ 
2))*Sqrt[Pi]*(E^(((I/2)*b^2*f*Log[f]^2)/(f^2 + c^2*Log[f]^2))*f^((b*e)/(2* 
f + (2*I)*c*Log[f]))*Erfi[((-1)^(1/4)*(e + 2*f*x - I*(b + 2*c*x)*Log[f]))/ 
(2*Sqrt[f - I*c*Log[f]])]*Sqrt[f - I*c*Log[f]]*(f + I*c*Log[f])*(Cos[d] + 
I*Sin[d]) + E^(((I/2)*e^2*f)/(f^2 + c^2*Log[f]^2))*f^((b*e)/(2*f - (2*I)*c 
*Log[f]))*Erfi[((-1)^(3/4)*(e + 2*f*x + I*(b + 2*c*x)*Log[f]))/(2*Sqrt[f + 
 I*c*Log[f]])]*(f - I*c*Log[f])*Sqrt[f + I*c*Log[f]]*(I*Cos[d] + Sin[d]))) 
/(E^((I/4)*(e^2/(f - I*c*Log[f]) + (b^2*Log[f]^2)/(f + I*c*Log[f])))*(f^2 
+ c^2*Log[f]^2))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

\(\displaystyle \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx\)

\(\Big \downarrow \) 4975

\(\displaystyle \int \left (\frac {1}{2} i e^{-i d-i e x-i f x^2} f^{a+b x+c x^2}-\frac {1}{2} i e^{i d+i e x+i f x^2} f^{a+b x+c x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i \sqrt {\pi } f^a \exp \left (-\frac {(e+i b \log (f))^2}{-4 c \log (f)+4 i f}-i d\right ) \text {erf}\left (\frac {-b \log (f)+2 x (-c \log (f)+i f)+i e}{2 \sqrt {-c \log (f)+i f}}\right )}{4 \sqrt {-c \log (f)+i f}}-\frac {i \sqrt {\pi } f^a \exp \left (\frac {(e-i b \log (f))^2}{4 c \log (f)+4 i f}+i d\right ) \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+i f)+i e}{2 \sqrt {c \log (f)+i f}}\right )}{4 \sqrt {c \log (f)+i f}}\)

Input:

Int[f^(a + b*x + c*x^2)*Sin[d + e*x + f*x^2],x]
 

Output:

((I/4)*E^((-I)*d - (e + I*b*Log[f])^2/((4*I)*f - 4*c*Log[f]))*f^a*Sqrt[Pi] 
*Erf[(I*e - b*Log[f] + 2*x*(I*f - c*Log[f]))/(2*Sqrt[I*f - c*Log[f]])])/Sq 
rt[I*f - c*Log[f]] - ((I/4)*E^(I*d + (e - I*b*Log[f])^2/((4*I)*f + 4*c*Log 
[f]))*f^a*Sqrt[Pi]*Erfi[(I*e + b*Log[f] + 2*x*(I*f + c*Log[f]))/(2*Sqrt[I* 
f + c*Log[f]])])/Sqrt[I*f + c*Log[f]]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4975

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]
 

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c +4 d f -e^{2}}{4 \left (i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-i f}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )-i f}}\right )}{4 \sqrt {-c \ln \left (f \right )-i f}}-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-2 i \ln \left (f \right ) b e +4 i d \ln \left (f \right ) c +4 d f -e^{2}}{4 \left (-i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-x \sqrt {i f -c \ln \left (f \right )}+\frac {b \ln \left (f \right )-i e}{2 \sqrt {i f -c \ln \left (f \right )}}\right )}{4 \sqrt {i f -c \ln \left (f \right )}}\)	\(216\)

Input:

int(f^(c*x^2+b*x+a)*sin(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
 

Output:

1/4*I*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2+2*I*ln(f)*b*e-4*I*d*ln(f)*c+4*d*f 
-e^2)/(I*f+c*ln(f)))/(-c*ln(f)-I*f)^(1/2)*erf(-(-c*ln(f)-I*f)^(1/2)*x+1/2* 
(I*e+b*ln(f))/(-c*ln(f)-I*f)^(1/2))-1/4*I*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b 
^2-2*I*ln(f)*b*e+4*I*d*ln(f)*c+4*d*f-e^2)/(-I*f+c*ln(f)))/(I*f-c*ln(f))^(1 
/2)*erf(-x*(I*f-c*ln(f))^(1/2)+1/2*(b*ln(f)-I*e)/(I*f-c*ln(f))^(1/2))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (155) = 310\).

Time = 0.09 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.77 \[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {\pi } {\left (i \, c \log \left (f\right ) + f\right )} \sqrt {-c \log \left (f\right ) - i \, f} \operatorname {erf}\left (\frac {{\left (2 \, f^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2} + e f + {\left (i \, c e - i \, b f\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} + i \, e^{2} f - 4 i \, d f^{2} - {\left (4 i \, c^{2} d - 2 i \, b c e + i \, b^{2} f\right )} \log \left (f\right )^{2} - {\left (c e^{2} - 2 \, b e f + 4 \, a f^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right )} + \sqrt {\pi } {\left (-i \, c \log \left (f\right ) + f\right )} \sqrt {-c \log \left (f\right ) + i \, f} \operatorname {erf}\left (\frac {{\left (2 \, f^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2} + e f + {\left (-i \, c e + i \, b f\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} - i \, e^{2} f + 4 i \, d f^{2} - {\left (-4 i \, c^{2} d + 2 i \, b c e - i \, b^{2} f\right )} \log \left (f\right )^{2} - {\left (c e^{2} - 2 \, b e f + 4 \, a f^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right )}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}} \] Input:

integrate(f^(c*x^2+b*x+a)*sin(f*x^2+e*x+d),x, algorithm="fricas")
 

Output:

1/4*(sqrt(pi)*(I*c*log(f) + f)*sqrt(-c*log(f) - I*f)*erf(1/2*(2*f^2*x + (2 
*c^2*x + b*c)*log(f)^2 + e*f + (I*c*e - I*b*f)*log(f))*sqrt(-c*log(f) - I* 
f)/(c^2*log(f)^2 + f^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 + I*e^2*f - 4 
*I*d*f^2 - (4*I*c^2*d - 2*I*b*c*e + I*b^2*f)*log(f)^2 - (c*e^2 - 2*b*e*f + 
 4*a*f^2)*log(f))/(c^2*log(f)^2 + f^2)) + sqrt(pi)*(-I*c*log(f) + f)*sqrt( 
-c*log(f) + I*f)*erf(1/2*(2*f^2*x + (2*c^2*x + b*c)*log(f)^2 + e*f + (-I*c 
*e + I*b*f)*log(f))*sqrt(-c*log(f) + I*f)/(c^2*log(f)^2 + f^2))*e^(-1/4*(( 
b^2*c - 4*a*c^2)*log(f)^3 - I*e^2*f + 4*I*d*f^2 - (-4*I*c^2*d + 2*I*b*c*e 
- I*b^2*f)*log(f)^2 - (c*e^2 - 2*b*e*f + 4*a*f^2)*log(f))/(c^2*log(f)^2 + 
f^2)))/(c^2*log(f)^2 + f^2)
 

Sympy [F]

\[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \sin {\left (d + e x + f x^{2} \right )}\, dx \] Input:

integrate(f**(c*x**2+b*x+a)*sin(f*x**2+e*x+d),x)
                                                                                    
                                                                                    
 

Output:

Integral(f**(a + b*x + c*x**2)*sin(d + e*x + f*x**2), x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1007 vs. \(2 (155) = 310\).

Time = 0.08 (sec) , antiderivative size = 1007, normalized size of antiderivative = 4.75 \[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:

integrate(f^(c*x^2+b*x+a)*sin(f*x^2+e*x+d),x, algorithm="maxima")
 

Output:

1/8*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 2*f^2)*((f^(1/4*c*e^2/(c^2*log(f)^2 + 
f^2))*f^a*cos(-1/4*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2 
)/(c^2*log(f)^2 + f^2)) - I*f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*sin(-1/ 
4*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 
 f^2)))*erf(1/2*(2*(c*log(f) - I*f)*x + b*log(f) - I*e)*sqrt(-c*log(f) + I 
*f)/(c*log(f) - I*f)) + (f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(-1/4*( 
e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^ 
2)) + I*f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*sin(-1/4*(e^2*f - 4*d*f^2 - 
 (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*( 
c*log(f) + I*f)*x + b*log(f) + I*e)*sqrt(-c*log(f) - I*f)/(c*log(f) + I*f) 
))*sqrt(c*log(f) + sqrt(c^2*log(f)^2 + f^2)) + sqrt(pi)*sqrt(2*c^2*log(f)^ 
2 + 2*f^2)*((I*f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(-1/4*(e^2*f - 4* 
d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)) + f^(1 
/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*sin(-1/4*(e^2*f - 4*d*f^2 - (4*c^2*d - 
2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) - I 
*f)*x + b*log(f) - I*e)*sqrt(-c*log(f) + I*f)/(c*log(f) - I*f)) + (-I*f^(1 
/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(-1/4*(e^2*f - 4*d*f^2 - (4*c^2*d - 
2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)) + f^(1/4*c*e^2/(c^2*log(f 
)^2 + f^2))*f^a*sin(-1/4*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*lo 
g(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) + I*f)*x + b*log(f)...
 

Giac [F]

\[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\int { f^{c x^{2} + b x + a} \sin \left (f x^{2} + e x + d\right ) \,d x } \] Input:

integrate(f^(c*x^2+b*x+a)*sin(f*x^2+e*x+d),x, algorithm="giac")
 

Output:

integrate(f^(c*x^2 + b*x + a)*sin(f*x^2 + e*x + d), x)
 

Mupad [F(-1)]

Timed out. \[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,\sin \left (f\,x^2+e\,x+d\right ) \,d x \] Input:

int(f^(a + b*x + c*x^2)*sin(d + e*x + f*x^2),x)
 

Output:

int(f^(a + b*x + c*x^2)*sin(d + e*x + f*x^2), x)
 

Reduce [F]

\[ \int f^{a+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}+b x} \sin \left (f \,x^{2}+e x +d \right )d x \right ) \] Input:

int(f^(c*x^2+b*x+a)*sin(f*x^2+e*x+d),x)
 

Output:

f**a*int(f**(b*x + c*x**2)*sin(d + e*x + f*x**2),x)