\(\int f^{a+b x+c x^2} \, dx\) [429]

Optimal result

Integrand size = 12, antiderivative size = 56 \[ \int f^{a+b x+c x^2} \, dx=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{2 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

1/2*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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2266, 2235} \[ \int f^{a+b x+c x^2} \, dx=\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{2 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Int[f^(a + b*x + c*x^2),x]

[Out]

(f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(2*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]

Rubi steps \begin{align*} \text {integral}& = f^{a-\frac {b^2}{4 c}} \int f^{\frac {(b+2 c x)^2}{4 c}} \, 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 )}{2 \sqrt {c} \sqrt {\log (f)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int f^{a+b x+c x^2} \, dx=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{2 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Integrate[f^(a + b*x + c*x^2),x]

[Out]

(f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(2*Sqrt[c]*Sqrt[Log[f]])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{2 \sqrt {-c \ln \left (f \right )}}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.98 \[ \int f^{a+b x+c x^2} \, dx=-\frac {\sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{2 \, c f^{\frac {b^{2} - 4 \, a c}{4 \, c}} \log \left (f\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int f^{a+b x+c x^2} \, dx=\int f^{a + b x + c x^{2}}\, dx \]

[In]

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

[Out]

Integral(f**(a + b*x + c*x**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89 \[ \int f^{a+b x+c x^2} \, dx=\frac {\sqrt {\pi } f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )}{2 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89 \[ \int f^{a+b x+c x^2} \, dx=-\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{2 \, \sqrt {-c \log \left (f\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int f^{a+b x+c x^2} \, dx=-\frac {f^a\,\sqrt {\pi }\,{\mathrm {e}}^{-\frac {b^2\,\ln \left (f\right )}{4\,c}}\,\mathrm {erf}\left (\frac {b\,\ln \left (f\right )\,1{}\mathrm {i}+c\,x\,\ln \left (f\right )\,2{}\mathrm {i}}{2\,\sqrt {c\,\ln \left (f\right )}}\right )\,1{}\mathrm {i}}{2\,\sqrt {c\,\ln \left (f\right )}} \]

[In]

int(f^(a + b*x + c*x^2),x)

[Out]

-(f^a*pi^(1/2)*exp(-(b^2*log(f))/(4*c))*erf((b*log(f)*1i + c*x*log(f)*2i)/(2*(c*log(f))^(1/2)))*1i)/(2*(c*log(
f))^(1/2))