\(\int \frac {f^{b x+c x^2}}{(b+2 c x)^2} \, dx\) [460]

Optimal result

Integrand size = 20, antiderivative size = 81 \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^2} \, dx=-\frac {f^{b x+c x^2}}{2 c (b+2 c x)}+\frac {f^{-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)}}{4 c^{3/2}} \]

[Out]

-1/2*f^(c*x^2+b*x)/c/(2*c*x+b)+1/4*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))*Pi^(1/2)*ln(f)^(1/2)/c^(3/2)/(f^(1/
4*b^2/c))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2271, 2266, 2235} \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^2} \, dx=\frac {\sqrt {\pi } \sqrt {\log (f)} f^{-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2}}-\frac {f^{b x+c x^2}}{2 c (b+2 c x)} \]

[In]

Int[f^(b*x + c*x^2)/(b + 2*c*x)^2,x]

[Out]

-1/2*f^(b*x + c*x^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 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 2271

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]

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

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^2} \, dx=\frac {f^{-\frac {b^2}{4 c}} \left (-2 \sqrt {c} f^{\frac {(b+2 c x)^2}{4 c}}+\sqrt {\pi } (b+2 c x) \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)}\right )}{4 c^{3/2} (b+2 c x)} \]

[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] (verified)

Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {f^{\frac {\left (2 x c +b \right )^{2}}{4 c}} f^{-\frac {b^{2}}{4 c}}}{2 c \left (2 x c +b \right )}+\frac {\ln \left (f \right ) \sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (\frac {\sqrt {-\frac {\ln \left (f \right )}{c}}\, \left (2 x c +b \right )}{2}\right )}{4 c^{2} \sqrt {-\frac {\ln \left (f \right )}{c}}}\)	\(87\)

[In]

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^2} \, dx=-\frac {2 \, c f^{c x^{2} + b x} + \frac {\sqrt {\pi } {\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2}}{4 \, c}}}}{4 \, {\left (2 \, c^{3} x + b c^{2}\right )}} \]

[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]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {f^{b x+c x^2}}{(b+2 c x)^2} \, dx=\frac {\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \left (f\right )}{2}+c\,x\,\ln \left (f\right )}{\sqrt {c\,\ln \left (f\right )}}\right )\,\ln \left (f\right )}{4\,c\,f^{\frac {b^2}{4\,c}}\,\sqrt {c\,\ln \left (f\right )}}-\frac {f^{c\,x^2}\,f^{b\,x}}{2\,c\,\left (b+2\,c\,x\right )} \]

[In]

int(f^(b*x + c*x^2)/(b + 2*c*x)^2,x)

[Out]

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