\(\int f^{a+b x+c x^2} (d+e x) \, dx\) [446]

Optimal result

Integrand size = 18, antiderivative size = 90 \[ \int f^{a+b x+c x^2} (d+e x) \, dx=\frac {e f^{a+b x+c x^2}}{2 c \log (f)}+\frac {(2 c d-b e) f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}} \]

[Out]

1/2*e*f^(c*x^2+b*x+a)/c/ln(f)+1/4*(-b*e+2*c*d)*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^(3/2)/ln(f)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2272, 2266, 2235} \[ \int f^{a+b x+c x^2} (d+e x) \, dx=\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} (2 c d-b e) \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}+\frac {e f^{a+b x+c x^2}}{2 c \log (f)} \]

[In]

Int[f^(a + b*x + c*x^2)*(d + e*x),x]

[Out]

(e*f^(a + b*x + c*x^2))/(2*c*Log[f]) + ((2*c*d - b*e)*f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]
])/(2*Sqrt[c])])/(4*c^(3/2)*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 2272

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {e f^{a+b x+c x^2}}{2 c \log (f)}-\frac {(-2 c d+b e) \int f^{a+b x+c x^2} \, dx}{2 c} \\ & = \frac {e f^{a+b x+c x^2}}{2 c \log (f)}+\frac {\left ((2 c d-b e) f^{a-\frac {b^2}{4 c}}\right ) \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx}{2 c} \\ & = \frac {e f^{a+b x+c x^2}}{2 c \log (f)}+\frac {(2 c d-b e) 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 c^{3/2} \sqrt {\log (f)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int f^{a+b x+c x^2} (d+e x) \, dx=\frac {f^{a-\frac {b^2}{4 c}} \left (2 \sqrt {c} e f^{\frac {(b+2 c x)^2}{4 c}}+(2 c d-b e) \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)}\right )}{4 c^{3/2} \log (f)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.46

method	result	size
risch	\(-\frac {f^{a} d \sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{2 \sqrt {-c \ln \left (f \right )}}+\frac {f^{a} e \,f^{b x} f^{c \,x^{2}}}{2 \ln \left (f \right ) c}+\frac {f^{a} e b \sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 c \sqrt {-c \ln \left (f \right )}}\)	\(131\)

[In]

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

[Out]

-1/2*f^a*d*Pi^(1/2)*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))+1/2*
f^a*e/ln(f)/c*f^(b*x)*f^(c*x^2)+1/4*f^a*e*b/c*Pi^(1/2)*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 = 0.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int f^{a+b x+c x^2} (d+e x) \, dx=\frac {2 \, c e f^{c x^{2} + b x + a} - \frac {\sqrt {\pi } {\left (2 \, c d - b e\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{4 \, c^{2} \log \left (f\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (72) = 144\).

Time = 0.30 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.78 \[ \int f^{a+b x+c x^2} (d+e x) \, dx=-\frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{2}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\right )^{\frac {3}{2}}} - \frac {2 \, c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )}{\left (c \log \left (f\right )\right )^{\frac {3}{2}}}\right )} e f^{a - \frac {b^{2}}{4 \, c}}}{4 \, \sqrt {c \log \left (f\right )}} + \frac {\sqrt {\pi } d f^{a} \operatorname {erf}\left (\sqrt {-c \log \left (f\right )} x - \frac {b \log \left (f\right )}{2 \, \sqrt {-c \log \left (f\right )}}\right )}{2 \, \sqrt {-c \log \left (f\right )} f^{\frac {b^{2}}{4 \, c}}} \]

[In]

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

[Out]

-1/4*(sqrt(pi)*(2*c*x + b)*b*(erf(1/2*sqrt(-(2*c*x + b)^2*log(f)/c)) - 1)*log(f)^2/(sqrt(-(2*c*x + b)^2*log(f)
/c)*(c*log(f))^(3/2)) - 2*c*f^(1/4*(2*c*x + b)^2/c)*log(f)/(c*log(f))^(3/2))*e*f^(a - 1/4*b^2/c)/sqrt(c*log(f)
) + 1/2*sqrt(pi)*d*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 = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.99 \[ \int f^{a+b x+c x^2} (d+e x) \, dx=-\frac {\frac {\sqrt {\pi } {\left (2 \, c d - b e\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{\sqrt {-c \log \left (f\right )}} - \frac {2 \, e e^{\left (c x^{2} \log \left (f\right ) + b x \log \left (f\right ) + a \log \left (f\right )\right )}}{\log \left (f\right )}}{4 \, c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.89 \[ \int f^{a+b x+c x^2} (d+e x) \, dx=\frac {e\,f^a\,f^{c\,x^2}\,f^{b\,x}}{2\,c\,\ln \left (f\right )}-\frac {f^{a-\frac {b^2}{4\,c}}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \left (f\right )}{2}+c\,x\,\ln \left (f\right )}{\sqrt {c\,\ln \left (f\right )}}\right )\,\left (\frac {b\,e}{4}-\frac {c\,d}{2}\right )}{c\,\sqrt {c\,\ln \left (f\right )}} \]

[In]

int(f^(a + b*x + c*x^2)*(d + e*x),x)

[Out]

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