\(\int f^{a+b x+c x^2} \cos (d+e x) \, dx\) [125]

Optimal result

Integrand size = 19, antiderivative size = 172 \[ \int f^{a+b x+c x^2} \cos (d+e x) \, dx=-\frac {e^{-i d+\frac {(e+i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-b \log (f)-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{i d+\frac {(e-i b \log (f))^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {4561, 2325, 2266, 2235} \[ \int f^{a+b x+c x^2} \cos (d+e x) \, dx=\frac {\sqrt {\pi } f^a e^{\frac {(e-i b \log (f))^2}{4 c \log (f)}+i d} \text {erfi}\left (\frac {b \log (f)+2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {\sqrt {\pi } f^a e^{\frac {(e+i b \log (f))^2}{4 c \log (f)}-i d} \text {erfi}\left (\frac {-b \log (f)-2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

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

[Out]

-1/4*(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[Log[f]]) + (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]])])/(4*Sqrt[c]*Sqrt[Log[f]])

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2325

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rule 4561

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} e^{-i d-i e x} f^{a+b x+c x^2}+\frac {1}{2} e^{i d+i e x} f^{a+b x+c x^2}\right ) \, dx \\ & = \frac {1}{2} \int e^{-i d-i e x} f^{a+b x+c x^2} \, dx+\frac {1}{2} \int e^{i d+i e x} f^{a+b x+c x^2} \, dx \\ & = \frac {1}{2} \int \exp \left (-i d+a \log (f)+c x^2 \log (f)-x (i e-b \log (f))\right ) \, dx+\frac {1}{2} \int \exp \left (i d+a \log (f)+c x^2 \log (f)+x (i e+b \log (f))\right ) \, dx \\ & = \frac {1}{2} \left (e^{i d+\frac {(e-i b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx+\frac {1}{2} \left (e^{-i d+\frac {(e+i b \log (f))^2}{4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(-i e+b \log (f)+2 c x \log (f))^2}{4 c \log (f)}\right ) \, dx \\ & = -\frac {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 {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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.88 \[ \int f^{a+b x+c x^2} \cos (d+e x) \, dx=\frac {e^{\frac {e (e-2 i b \log (f))}{4 c \log (f)}} f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \left (e^{\frac {i b e}{c}} \text {erfi}\left (\frac {-i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)-i \sin (d))+\text {erfi}\left (\frac {i e+(b+2 c x) \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)+i \sin (d))\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} {\mathrm e}^{-\frac {2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(170\)

[In]

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

[Out]

-1/4*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))-1/4*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))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.02 \[ \int f^{a+b x+c x^2} \cos (d+e x) \, dx=-\frac {\sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (2 i \, c d - i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} + \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac {{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} + 2 \, {\left (-2 i \, c d + i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )}}{4 \, c \log \left (f\right )} \]

[In]

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

[Out]

-1/4*(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))) + 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]

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

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

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

Time = 0.25 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.06 \[ \int f^{a+b x+c x^2} \cos (d+e x) \, dx=-\frac {\sqrt {\pi } {\left (f^{a} {\left (\cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) - i \, \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - \frac {1}{2} \, {\left (b \log \left (f\right ) + i \, e\right )} \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (\cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) + i \, \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - \frac {1}{2} \, {\left (b \log \left (f\right ) - i \, e\right )} \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (\cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) - i \, \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + b \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (\cos \left (-\frac {2 \, c d - b e}{2 \, c}\right ) + i \, \sin \left (-\frac {2 \, c d - b e}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + b \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )}\right )} \sqrt {-c \log \left (f\right )}}{8 \, c f^{\frac {b^{2}}{4 \, c}} \log \left (f\right )} \]

[In]

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

[Out]

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

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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