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]

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

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Rubi [A]  time = 0.04, antiderivative size = 39, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.05, size = 39, normalized size = 1.00 \[ \frac {f^{a-\frac {b^2}{4 c}} \text {Ei}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c} \]

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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fricas [A]  time = 0.42, size = 47, normalized size = 1.21 \[ \frac {{\rm Ei}\left (\frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \relax (f)}{4 \, c}\right )}{4 \, c f^{\frac {b^{2} - 4 \, a c}{4 \, c}}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{c x^{2} + b x + a}}{2 \, c x + b}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 40, normalized size = 1.03 \[ -\frac {f^{\frac {4 a c -b^{2}}{4 c}} \Ei \left (1, -\frac {\left (2 c x +b \right )^{2} \ln \relax (f )}{4 c}\right )}{4 c} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{c x^{2} + b x + a}}{2 \, c x + b}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {f^{c\,x^2+b\,x+a}}{b+2\,c\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x + c x^{2}}}{b + 2 c x}\, dx \]

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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