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

Optimal. Leaf size=81 \[ \frac{\sqrt{\pi } \sqrt{\log (f)} f^{-\frac{b^2}{4 c}} \text{Erfi}\left (\frac{\sqrt{\log (f)} (b+2 c x)}{2 \sqrt{c}}\right )}{4 c^{3/2}}-\frac{f^{b x+c x^2}}{2 c (b+2 c x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0485183, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2239, 2234, 2204} \[ \frac{\sqrt{\pi } \sqrt{\log (f)} f^{-\frac{b^2}{4 c}} \text{Erfi}\left (\frac{\sqrt{\log (f)} (b+2 c x)}{2 \sqrt{c}}\right )}{4 c^{3/2}}-\frac{f^{b x+c x^2}}{2 c (b+2 c x)} \]

Antiderivative was successfully verified.

[In]

Int[f^(b*x + c*x^2)/(b + 2*c*x)^2,x]

[Out]

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

Rule 2239

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

Rule 2234

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 2204

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]

Rubi steps

\begin{align*} \int \frac{f^{b x+c x^2}}{(b+2 c x)^2} \, dx &=-\frac{f^{b x+c x^2}}{2 c (b+2 c x)}+\frac{\log (f) \int f^{b x+c x^2} \, dx}{2 c}\\ &=-\frac{f^{b x+c x^2}}{2 c (b+2 c x)}+\frac{\left (f^{-\frac{b^2}{4 c}} \log (f)\right ) \int f^{\frac{(b+2 c x)^2}{4 c}} \, dx}{2 c}\\ &=-\frac{f^{b x+c x^2}}{2 c (b+2 c x)}+\frac{f^{-\frac{b^2}{4 c}} \sqrt{\pi } \text{erfi}\left (\frac{(b+2 c x) \sqrt{\log (f)}}{2 \sqrt{c}}\right ) \sqrt{\log (f)}}{4 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0589249, size = 94, normalized size = 1.16 \[ \frac{f^{-\frac{b^2}{4 c}} \left (\sqrt{\pi } \sqrt{\log (f)} (b+2 c x) \text{Erfi}\left (\frac{\sqrt{\log (f)} (b+2 c x)}{2 \sqrt{c}}\right )-2 \sqrt{c} f^{\frac{(b+2 c x)^2}{4 c}}\right )}{4 c^{3/2} (b+2 c x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.034, size = 87, normalized size = 1.1 \begin{align*} -{\frac{1}{2\,c \left ( 2\,cx+b \right ) }{f}^{{\frac{ \left ( 2\,cx+b \right ) ^{2}}{4\,c}}}{f}^{-{\frac{{b}^{2}}{4\,c}}}}+{\frac{\ln \left ( f \right ) \sqrt{\pi }}{4\,{c}^{2}}{f}^{-{\frac{{b}^{2}}{4\,c}}}{\it Erf} \left ({\frac{2\,cx+b}{2}\sqrt{-{\frac{\ln \left ( f \right ) }{c}}}} \right ){\frac{1}{\sqrt{-{\frac{\ln \left ( f \right ) }{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.62962, size = 186, normalized size = 2.3 \begin{align*} -\frac{2 \, c f^{c x^{2} + b x} + \frac{\sqrt{\pi }{\left (2 \, c x + b\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 \, c}}}}{4 \,{\left (2 \, c^{3} x + b c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*c*f^(c*x^2 + b*x) + sqrt(pi)*(2*c*x + b)*sqrt(-c*log(f))*erf(1/2*(2*c*x + b)*sqrt(-c*log(f))/c)/f^(1/4
*b^2/c))/(2*c^3*x + b*c^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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