\(\int f^{a+b x+c x^2} (b+2 c x) \, dx\) [452]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 17 \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\frac {f^{a+b x+c x^2}}{\log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2268} \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\frac {f^{a+b x+c x^2}}{\log (f)} \]

[In]

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

[Out]

f^(a + b*x + c*x^2)/Log[f]

Rule 2268

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

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x+c x^2}}{\log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\frac {f^{a+b x+c x^2}}{\log (f)} \]

[In]

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

[Out]

f^(a + b*x + c*x^2)/Log[f]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method result size
gosper \(\frac {f^{c \,x^{2}+b x +a}}{\ln \left (f \right )}\) \(18\)
derivativedivides \(\frac {f^{c \,x^{2}+b x +a}}{\ln \left (f \right )}\) \(18\)
default \(\frac {f^{c \,x^{2}+b x +a}}{\ln \left (f \right )}\) \(18\)
risch \(\frac {f^{c \,x^{2}+b x +a}}{\ln \left (f \right )}\) \(18\)
parallelrisch \(\frac {f^{c \,x^{2}+b x +a}}{\ln \left (f \right )}\) \(18\)
norman \(\frac {{\mathrm e}^{\left (c \,x^{2}+b x +a \right ) \ln \left (f \right )}}{\ln \left (f \right )}\) \(20\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\frac {f^{c x^{2} + b x + a}}{\log \left (f\right )} \]

[In]

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

[Out]

f^(c*x^2 + b*x + a)/log(f)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\begin {cases} \frac {f^{a + b x + c x^{2}}}{\log {\left (f \right )}} & \text {for}\: \log {\left (f \right )} \neq 0 \\b x + c x^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((f**(a + b*x + c*x**2)/log(f), Ne(log(f), 0)), (b*x + c*x**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\frac {f^{c x^{2} + b x + a}}{\log \left (f\right )} \]

[In]

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

[Out]

f^(c*x^2 + b*x + a)/log(f)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\frac {f^{c x^{2} + b x + a}}{\log \left (f\right )} \]

[In]

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

[Out]

f^(c*x^2 + b*x + a)/log(f)

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int f^{a+b x+c x^2} (b+2 c x) \, dx=\frac {f^{c\,x^2+b\,x+a}}{\ln \left (f\right )} \]

[In]

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

[Out]

f^(a + b*x + c*x^2)/log(f)