\(\int \frac {1}{5} (4-e^x x+e^x (-2 x-x^2) \log (10 x)) \, dx\) [4724]

Optimal result

Integrand size = 29, antiderivative size = 17 \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=\frac {1}{5} x \left (4-e^x x \log (10 x)\right ) \]

[Out]

1/5*(4-exp(x)*x*ln(10*x))*x

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {12, 2207, 2225, 1607, 2227, 2634} \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=\frac {4 x}{5}-\frac {1}{5} e^x x^2 \log (10 x) \]

[In]

Int[(4 - E^x*x + E^x*(-2*x - x^2)*Log[10*x])/5,x]

[Out]

(4*x)/5 - (E^x*x^2*Log[10*x])/5

Rule 12

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

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx \\ & = \frac {4 x}{5}-\frac {1}{5} \int e^x x \, dx+\frac {1}{5} \int e^x \left (-2 x-x^2\right ) \log (10 x) \, dx \\ & = \frac {4 x}{5}-\frac {e^x x}{5}+\frac {\int e^x \, dx}{5}+\frac {1}{5} \int e^x (-2-x) x \log (10 x) \, dx \\ & = \frac {e^x}{5}+\frac {4 x}{5}-\frac {e^x x}{5}-\frac {1}{5} e^x x^2 \log (10 x)+\frac {1}{5} \int e^x x \, dx \\ & = \frac {e^x}{5}+\frac {4 x}{5}-\frac {1}{5} e^x x^2 \log (10 x)-\frac {\int e^x \, dx}{5} \\ & = \frac {4 x}{5}-\frac {1}{5} e^x x^2 \log (10 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=-\frac {1}{5} x \left (-4+e^x x \log (10 x)\right ) \]

[In]

Integrate[(4 - E^x*x + E^x*(-2*x - x^2)*Log[10*x])/5,x]

[Out]

-1/5*(x*(-4 + E^x*x*Log[10*x]))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {4 x}{5}-\frac {\ln \left (10 x \right ) {\mathrm e}^{x} x^{2}}{5}\)	\(16\)
norman	\(\frac {4 x}{5}-\frac {\ln \left (10 x \right ) {\mathrm e}^{x} x^{2}}{5}\)	\(16\)
risch	\(\frac {4 x}{5}-\frac {\ln \left (10 x \right ) {\mathrm e}^{x} x^{2}}{5}\)	\(16\)
parallelrisch	\(\frac {4 x}{5}-\frac {\ln \left (10 x \right ) {\mathrm e}^{x} x^{2}}{5}\)	\(16\)
parts	\(\frac {4 x}{5}-\frac {\ln \left (10 x \right ) {\mathrm e}^{x} x^{2}}{5}\)	\(16\)

[In]

int(1/5*(-x^2-2*x)*exp(x)*ln(10*x)-1/5*exp(x)*x+4/5,x,method=_RETURNVERBOSE)

[Out]

4/5*x-1/5*ln(10*x)*exp(x)*x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=-\frac {1}{5} \, x^{2} e^{x} \log \left (10 \, x\right ) + \frac {4}{5} \, x \]

[In]

integrate(1/5*(-x^2-2*x)*exp(x)*log(10*x)-1/5*exp(x)*x+4/5,x, algorithm="fricas")

[Out]

-1/5*x^2*e^x*log(10*x) + 4/5*x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=- \frac {x^{2} e^{x} \log {\left (10 x \right )}}{5} + \frac {4 x}{5} \]

[In]

integrate(1/5*(-x**2-2*x)*exp(x)*ln(10*x)-1/5*exp(x)*x+4/5,x)

[Out]

-x**2*exp(x)*log(10*x)/5 + 4*x/5

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=-\frac {1}{5} \, x^{2} e^{x} \log \left (10 \, x\right ) + \frac {4}{5} \, x \]

[In]

integrate(1/5*(-x^2-2*x)*exp(x)*log(10*x)-1/5*exp(x)*x+4/5,x, algorithm="maxima")

[Out]

-1/5*x^2*e^x*log(10*x) + 4/5*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=-\frac {1}{5} \, x^{2} e^{x} \log \left (10 \, x\right ) + \frac {4}{5} \, x \]

[In]

integrate(1/5*(-x^2-2*x)*exp(x)*log(10*x)-1/5*exp(x)*x+4/5,x, algorithm="giac")

[Out]

-1/5*x^2*e^x*log(10*x) + 4/5*x

Mupad [B] (verification not implemented)

Time = 10.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{5} \left (4-e^x x+e^x \left (-2 x-x^2\right ) \log (10 x)\right ) \, dx=\frac {4\,x}{5}-\frac {x^2\,\ln \left (10\,x\right )\,{\mathrm {e}}^x}{5} \]

[In]

int(4/5 - (log(10*x)*exp(x)*(2*x + x^2))/5 - (x*exp(x))/5,x)

[Out]

(4*x)/5 - (x^2*log(10*x)*exp(x))/5