\(\int 2 e^{-10+e^3+e^{20 x}} (1+20 e^{20 x} x) \, dx\) [7190]

Optimal result

Integrand size = 24, antiderivative size = 15 \[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=2 e^{-10+e^3+e^{20 x}} x \]

[Out]

exp(exp(20*x)+ln(2)+exp(3)-10)*x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2326} \[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=2 e^{e^{20 x}-10+e^3} x \]

[In]

Int[2*E^(-10 + E^3 + E^(20*x))*(1 + 20*E^(20*x)*x),x]

[Out]

2*E^(-10 + E^3 + E^(20*x))*x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = 2 \int e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx \\ & = 2 e^{-10+e^3+e^{20 x}} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=2 e^{-10+e^3+e^{20 x}} x \]

[In]

Integrate[2*E^(-10 + E^3 + E^(20*x))*(1 + 20*E^(20*x)*x),x]

[Out]

2*E^(-10 + E^3 + E^(20*x))*x

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
risch	\(2 x \,{\mathrm e}^{{\mathrm e}^{20 x}+{\mathrm e}^{3}-10}\)	\(13\)
norman	\({\mathrm e}^{{\mathrm e}^{20 x}+\ln \left (2\right )+{\mathrm e}^{3}-10} x\)	\(14\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{20 x}+\ln \left (2\right )+{\mathrm e}^{3}-10} x\)	\(14\)

[In]

int((20*x*exp(20*x)+1)*exp(exp(20*x)+ln(2)+exp(3)-10),x,method=_RETURNVERBOSE)

[Out]

2*x*exp(exp(20*x)+exp(3)-10)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=x e^{\left (e^{3} + e^{\left (20 \, x\right )} + \log \left (2\right ) - 10\right )} \]

[In]

integrate((20*x*exp(20*x)+1)*exp(exp(20*x)+log(2)+exp(3)-10),x, algorithm="fricas")

[Out]

x*e^(e^3 + e^(20*x) + log(2) - 10)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=2 x e^{e^{20 x} - 10 + e^{3}} \]

[In]

integrate((20*x*exp(20*x)+1)*exp(exp(20*x)+ln(2)+exp(3)-10),x)

[Out]

2*x*exp(exp(20*x) - 10 + exp(3))

Maxima [F]

\[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=\int { {\left (20 \, x e^{\left (20 \, x\right )} + 1\right )} e^{\left (e^{3} + e^{\left (20 \, x\right )} + \log \left (2\right ) - 10\right )} \,d x } \]

[In]

integrate((20*x*exp(20*x)+1)*exp(exp(20*x)+log(2)+exp(3)-10),x, algorithm="maxima")

[Out]

2*x*e^(e^3 + e^(20*x) - 10) + 1/10*Ei(e^(20*x))*e^(e^3 - 10) - 2*integrate(e^(e^3 + e^(20*x) - 10), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=2 \, x e^{\left (e^{3} + e^{\left (20 \, x\right )} - 10\right )} \]

[In]

integrate((20*x*exp(20*x)+1)*exp(exp(20*x)+log(2)+exp(3)-10),x, algorithm="giac")

[Out]

2*x*e^(e^3 + e^(20*x) - 10)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int 2 e^{-10+e^3+e^{20 x}} \left (1+20 e^{20 x} x\right ) \, dx=2\,x\,{\mathrm {e}}^{-10}\,{\mathrm {e}}^{{\mathrm {e}}^{20\,x}}\,{\mathrm {e}}^{{\mathrm {e}}^3} \]

[In]

int(exp(exp(20*x) + exp(3) + log(2) - 10)*(20*x*exp(20*x) + 1),x)

[Out]

2*x*exp(-10)*exp(exp(20*x))*exp(exp(3))