\(\int f^{a+b x+c x^2} \sin (d+e x) \, dx\) [106]

Optimal result

Integrand size = 19, antiderivative size = 176 \[ \int f^{a+b x+c x^2} \sin (d+e x) \, dx=-\frac {i e^{-i d+\frac {(e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {i e^{i d+\frac {(e-i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.88 \[ \int f^{a+b x+c x^2} \sin (d+e x) \, dx=\frac {e^{\frac {e (e-2 i b \log (f))}{4 c \log (f)}} f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \left (i \text {erfi}\left (\frac {-i e-(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)+i \sin (d))+e^{\frac {i b e}{c}} \text {erfi}\left (\frac {-i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (i \cos (d)+\sin (d))\right )}{4 \sqrt {c} \sqrt {\log (f)}} \] Input:

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

Output:

(E^((e*(e - (2*I)*b*Log[f]))/(4*c*Log[f]))*f^(a - b^2/(4*c))*Sqrt[Pi]*(I*E 
rfi[((-I)*e - (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[d] + I*Si 
n[d]) + E^((I*b*e)/c)*Erfi[((-I)*e + (b + 2*c*x)*Log[f])/(2*Sqrt[c]*Sqrt[L 
og[f]])]*(I*Cos[d] + Sin[d])))/(4*Sqrt[c]*Sqrt[Log[f]])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, 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.

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

\(\Big \downarrow \) 4975

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((-1/4*I)*E^((-I)*d + (e + I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[( 
I*e - b*Log[f] - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(Sqrt[c]*Sqrt[Lo 
g[f]]) - ((I/4)*E^(I*d + (e - I*b*Log[f])^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erf 
i[(I*e + b*Log[f] + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(Sqrt[c]*Sqrt 
[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 = 0.65 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {i \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}-\frac {i \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(172\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.01 \[ \int f^{a+b x+c x^2} \sin (d+e x) \, dx=\frac {-i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (2 i \, c d - i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} + i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (-2 i \, c d + i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )}}{4 \, c \log \left (f\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.06 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.01 \[ \int f^{a+b x+c x^2} \sin (d+e x) \, dx=\frac {\sqrt {\pi } {\left (f^{a} {\left (i \, \cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) + \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - \frac {1}{2} \, {\left (b \log \left (f\right ) + i \, e\right )} \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (-i \, \cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) + \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - \frac {1}{2} \, {\left (b \log \left (f\right ) - i \, e\right )} \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (i \, \cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) + \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + b \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (-i \, \cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) + \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + b \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )}\right )} \sqrt {-c \log \left (f\right )}}{8 \, c f^{\frac {b^{2}}{4 \, c}} \log \left (f\right )} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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