3.453 \(\int \frac{f^{a+b x+c x^2}}{b+2 c x} \, dx\)

Optimal. Leaf size=39 \[ \frac{f^{a-\frac{b^2}{4 c}} \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.0371917, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {2238} \[ \frac{f^{a-\frac{b^2}{4 c}} \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x + c*x^2)/(b + 2*c*x),x]

[Out]

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

Rule 2238

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(1*F^(a - b^2/(4*c))*ExpI
ntegralEi[((b + 2*c*x)^2*Log[F])/(4*c)])/(2*e), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rubi steps

\begin{align*} \int \frac{f^{a+b x+c x^2}}{b+2 c x} \, dx &=\frac{f^{a-\frac{b^2}{4 c}} \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c}\\ \end{align*}

Mathematica [A]  time = 0.0545218, size = 39, normalized size = 1. \[ \frac{f^{a-\frac{b^2}{4 c}} \text{Ei}\left (\frac{(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 40, normalized size = 1. \begin{align*} -{\frac{1}{4\,c}{f}^{{\frac{4\,ac-{b}^{2}}{4\,c}}}{\it Ei} \left ( 1,-{\frac{ \left ( 2\,cx+b \right ) ^{2}\ln \left ( f \right ) }{4\,c}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c*x^2+b*x+a)/(2*c*x+b),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{c x^{2} + b x + a}}{2 \, c x + b}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.50717, size = 105, normalized size = 2.69 \begin{align*} \frac{{\rm Ei}\left (\frac{{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (f\right )}{4 \, c}\right )}{4 \, c f^{\frac{b^{2} - 4 \, a c}{4 \, c}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + b x + c x^{2}}}{b + 2 c x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{c x^{2} + b x + a}}{2 \, c x + b}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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