\(\int \frac {f^{b x+c x^2}}{b+2 c x} \, dx\) [459]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {f^{b x+c x^2}}{b+2 c x} \, dx=\frac {f^{-\frac {b^2}{4 c}} \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2270} \[ \int \frac {f^{b x+c x^2}}{b+2 c x} \, dx=\frac {f^{-\frac {b^2}{4 c}} \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c} \]

[In]

Int[f^(b*x + c*x^2)/(b + 2*c*x),x]

[Out]

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

Rule 2270

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

Rubi steps \begin{align*} \text {integral}& = \frac {f^{-\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] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {f^{b x+c x^2}}{b+2 c x} \, dx=\frac {f^{-\frac {b^2}{4 c}} \operatorname {ExpIntegralEi}\left (\frac {(b+2 c x)^2 \log (f)}{4 c}\right )}{4 c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {f^{-\frac {b^{2}}{4 c}} \operatorname {Ei}_{1}\left (-\frac {\left (2 x c +b \right )^{2} \ln \left (f \right )}{4 c}\right )}{4 c}\)	\(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \frac {f^{b x+c x^2}}{b+2 c x} \, dx=\frac {{\rm Ei}\left (\frac {{\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (f\right )}{4 \, c}\right )}{4 \, c f^{\frac {b^{2}}{4 \, c}}} \]

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {f^{b x+c x^2}}{b+2 c x} \, dx=\int \frac {f^{c\,x^2+b\,x}}{b+2\,c\,x} \,d x \]

[In]

int(f^(b*x + c*x^2)/(b + 2*c*x),x)

[Out]

int(f^(b*x + c*x^2)/(b + 2*c*x), x)