3.38.14 \(\int \frac {-3-12 x-10 x^2+e^x (-1-x-5 x^2)+(-x-e^x x) \log (x)}{x} \, dx\)

Optimal. Leaf size=22 \[ e^{25/4}+\left (3+e^x+x\right ) (4-5 x-\log (x)) \]

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Rubi [A]  time = 0.20, antiderivative size = 36, normalized size of antiderivative = 1.64, number of steps used = 17, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {14, 2295, 6742, 2199, 2194, 2178, 2176, 2554} \begin {gather*} -5 x^2-5 e^x x-11 x+4 e^x-x \log (x)-e^x \log (x)-3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 - 12*x - 10*x^2 + E^x*(-1 - x - 5*x^2) + (-x - E^x*x)*Log[x])/x,x]

[Out]

4*E^x - 11*x - 5*E^x*x - 5*x^2 - 3*Log[x] - E^x*Log[x] - x*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-3-12 x-10 x^2-x \log (x)}{x}-\frac {e^x \left (1+x+5 x^2+x \log (x)\right )}{x}\right ) \, dx\\ &=\int \frac {-3-12 x-10 x^2-x \log (x)}{x} \, dx-\int \frac {e^x \left (1+x+5 x^2+x \log (x)\right )}{x} \, dx\\ &=\int \left (\frac {-3-12 x-10 x^2}{x}-\log (x)\right ) \, dx-\int \left (\frac {e^x \left (1+x+5 x^2\right )}{x}+e^x \log (x)\right ) \, dx\\ &=\int \frac {-3-12 x-10 x^2}{x} \, dx-\int \frac {e^x \left (1+x+5 x^2\right )}{x} \, dx-\int \log (x) \, dx-\int e^x \log (x) \, dx\\ &=x-e^x \log (x)-x \log (x)+\int \left (-12-\frac {3}{x}-10 x\right ) \, dx+\int \frac {e^x}{x} \, dx-\int \left (e^x+\frac {e^x}{x}+5 e^x x\right ) \, dx\\ &=-11 x-5 x^2+\text {Ei}(x)-3 \log (x)-e^x \log (x)-x \log (x)-5 \int e^x x \, dx-\int e^x \, dx-\int \frac {e^x}{x} \, dx\\ &=-e^x-11 x-5 e^x x-5 x^2-3 \log (x)-e^x \log (x)-x \log (x)+5 \int e^x \, dx\\ &=4 e^x-11 x-5 e^x x-5 x^2-3 \log (x)-e^x \log (x)-x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 32, normalized size = 1.45 \begin {gather*} -11 x-5 x^2-e^x (-4+5 x)-3 \log (x)-\left (e^x+x\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 - 12*x - 10*x^2 + E^x*(-1 - x - 5*x^2) + (-x - E^x*x)*Log[x])/x,x]

[Out]

-11*x - 5*x^2 - E^x*(-4 + 5*x) - 3*Log[x] - (E^x + x)*Log[x]

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fricas [A]  time = 0.84, size = 27, normalized size = 1.23 \begin {gather*} -5 \, x^{2} - {\left (5 \, x - 4\right )} e^{x} - {\left (x + e^{x} + 3\right )} \log \relax (x) - 11 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*log(x)+(-5*x^2-x-1)*exp(x)-10*x^2-12*x-3)/x,x, algorithm="fricas")

[Out]

-5*x^2 - (5*x - 4)*e^x - (x + e^x + 3)*log(x) - 11*x

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giac [A]  time = 0.15, size = 33, normalized size = 1.50 \begin {gather*} -5 \, x^{2} - 5 \, x e^{x} - x \log \relax (x) - e^{x} \log \relax (x) - 11 \, x + 4 \, e^{x} - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*log(x)+(-5*x^2-x-1)*exp(x)-10*x^2-12*x-3)/x,x, algorithm="giac")

[Out]

-5*x^2 - 5*x*e^x - x*log(x) - e^x*log(x) - 11*x + 4*e^x - 3*log(x)

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maple [A]  time = 0.04, size = 34, normalized size = 1.55

method	result	size
default	\(-5 \,{\mathrm e}^{x} x -{\mathrm e}^{x} \ln \relax (x )+4 \,{\mathrm e}^{x}-5 x^{2}-11 x -3 \ln \relax (x )-x \ln \relax (x )\)	\(34\)
norman	\(-5 \,{\mathrm e}^{x} x -{\mathrm e}^{x} \ln \relax (x )+4 \,{\mathrm e}^{x}-5 x^{2}-11 x -3 \ln \relax (x )-x \ln \relax (x )\)	\(34\)
risch	\(\left (-{\mathrm e}^{x}-x \right ) \ln \relax (x )-5 x^{2}-5 \,{\mathrm e}^{x} x -3 \ln \relax (x )-11 x +4 \,{\mathrm e}^{x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*x-x)*ln(x)+(-5*x^2-x-1)*exp(x)-10*x^2-12*x-3)/x,x,method=_RETURNVERBOSE)

[Out]

-5*exp(x)*x-exp(x)*ln(x)+4*exp(x)-5*x^2-11*x-3*ln(x)-x*ln(x)

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maxima [B]  time = 0.41, size = 35, normalized size = 1.59 \begin {gather*} -5 \, x^{2} - 5 \, {\left (x - 1\right )} e^{x} - x \log \relax (x) - e^{x} \log \relax (x) - 11 \, x - e^{x} - 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*log(x)+(-5*x^2-x-1)*exp(x)-10*x^2-12*x-3)/x,x, algorithm="maxima")

[Out]

-5*x^2 - 5*(x - 1)*e^x - x*log(x) - e^x*log(x) - 11*x - e^x - 3*log(x)

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mupad [B]  time = 2.15, size = 33, normalized size = 1.50 \begin {gather*} 4\,{\mathrm {e}}^x-11\,x-3\,\ln \relax (x)-{\mathrm {e}}^x\,\ln \relax (x)-5\,x\,{\mathrm {e}}^x-x\,\ln \relax (x)-5\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x + exp(x)*(x + 5*x^2 + 1) + log(x)*(x + x*exp(x)) + 10*x^2 + 3)/x,x)

[Out]

4*exp(x) - 11*x - 3*log(x) - exp(x)*log(x) - 5*x*exp(x) - x*log(x) - 5*x^2

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sympy [A]  time = 0.36, size = 29, normalized size = 1.32 \begin {gather*} - 5 x^{2} - x \log {\relax (x )} - 11 x + \left (- 5 x - \log {\relax (x )} + 4\right ) e^{x} - 3 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*x-x)*ln(x)+(-5*x**2-x-1)*exp(x)-10*x**2-12*x-3)/x,x)

[Out]

-5*x**2 - x*log(x) - 11*x + (-5*x - log(x) + 4)*exp(x) - 3*log(x)

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