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

Optimal. Leaf size=17 \[ \frac {1}{5} x \left (4-e^x x \log (10 x)\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {12, 2176, 2194, 1593, 2196, 2554} \begin {gather*} \frac {4 x}{5}-\frac {1}{5} e^x x^2 \log (10 x) \end {gather*}

Antiderivative was successfully verified.

[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 1593

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 2176

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] &&  !$UseGamma === True

Rule 2194

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 2196

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] &&  !$UseGamma === True

Rule 2554

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 16, normalized size = 0.94 \begin {gather*} -\frac {1}{5} x \left (-4+e^x x \log (10 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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]))

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fricas [A]  time = 0.90, size = 15, normalized size = 0.88 \begin {gather*} -\frac {1}{5} \, x^{2} e^{x} \log \left (10 \, x\right ) + \frac {4}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.19, size = 15, normalized size = 0.88 \begin {gather*} -\frac {1}{5} \, x^{2} e^{x} \log \left (10 \, x\right ) + \frac {4}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.05, size = 16, normalized size = 0.94

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

Verification of antiderivative is not currently implemented for this CAS.

[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*exp(x)*ln(10*x)*x^2

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maxima [A]  time = 0.37, size = 15, normalized size = 0.88 \begin {gather*} -\frac {1}{5} \, x^{2} e^{x} \log \left (10 \, x\right ) + \frac {4}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.22, size = 15, normalized size = 0.88 \begin {gather*} \frac {4\,x}{5}-\frac {x^2\,\ln \left (10\,x\right )\,{\mathrm {e}}^x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.28, size = 17, normalized size = 1.00 \begin {gather*} - \frac {x^{2} e^{x} \log {\left (10 x \right )}}{5} + \frac {4 x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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