\(\int f^{b x+c x^2} (b+2 c x) \, dx\) [458]

Optimal result

Integrand size = 18, antiderivative size = 16 \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\frac {f^{b x+c x^2}}{\log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2268} \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\frac {f^{b x+c x^2}}{\log (f)} \]

[In]

Int[f^(b*x + c*x^2)*(b + 2*c*x),x]

[Out]

f^(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^{b x+c x^2}}{\log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\frac {f^{b x+c x^2}}{\log (f)} \]

[In]

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

[Out]

f^(b*x + c*x^2)/Log[f]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {f^{x \left (x c +b \right )}}{\ln \left (f \right )}\)	\(15\)
gosper	\(\frac {f^{c \,x^{2}+b x}}{\ln \left (f \right )}\)	\(17\)
derivativedivides	\(\frac {f^{c \,x^{2}+b x}}{\ln \left (f \right )}\)	\(17\)
default	\(\frac {f^{c \,x^{2}+b x}}{\ln \left (f \right )}\)	\(17\)
parallelrisch	\(\frac {f^{c \,x^{2}+b x}}{\ln \left (f \right )}\)	\(17\)
norman	\(\frac {{\mathrm e}^{\left (c \,x^{2}+b x \right ) \ln \left (f \right )}}{\ln \left (f \right )}\)	\(19\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\frac {f^{c x^{2} + b x}}{\log \left (f\right )} \]

[In]

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

[Out]

f^(c*x^2 + b*x)/log(f)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38 \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\begin {cases} \frac {f^{b x + c x^{2}}}{\log {\left (f \right )}} & \text {for}\: \log {\left (f \right )} \neq 0 \\b x + c x^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\frac {f^{c x^{2} + b x}}{\log \left (f\right )} \]

[In]

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

[Out]

f^(c*x^2 + b*x)/log(f)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\frac {f^{c x^{2} + b x}}{\log \left (f\right )} \]

[In]

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

[Out]

f^(c*x^2 + b*x)/log(f)

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int f^{b x+c x^2} (b+2 c x) \, dx=\frac {f^{c\,x^2+b\,x}}{\ln \left (f\right )} \]

[In]

int(f^(b*x + c*x^2)*(b + 2*c*x),x)

[Out]

f^(b*x + c*x^2)/log(f)