3.74.23 \(\int \frac {2 e^x x^2+e (-1-2 x-2 x^2)+(-e x+e^x x) \log (x)}{2 x^2+x \log (x)} \, dx\)

Optimal. Leaf size=17 \[ -4+e^x-e (x+\log (2 x+\log (x))) \]

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Rubi [A]  time = 0.57, antiderivative size = 18, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 4, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {2561, 6742, 2194, 6684} \begin {gather*} -e x+e^x-e \log (2 x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x - E*x - E*Log[2*x + Log[x]]

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 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 \frac {2 e^x x^2+e \left (-1-2 x-2 x^2\right )+\left (-e x+e^x x\right ) \log (x)}{x (2 x+\log (x))} \, dx\\ &=\int \left (e^x-\frac {e \left (1+2 x+2 x^2+x \log (x)\right )}{x (2 x+\log (x))}\right ) \, dx\\ &=-\left (e \int \frac {1+2 x+2 x^2+x \log (x)}{x (2 x+\log (x))} \, dx\right )+\int e^x \, dx\\ &=e^x-e \int \left (1+\frac {1+2 x}{x (2 x+\log (x))}\right ) \, dx\\ &=e^x-e x-e \int \frac {1+2 x}{x (2 x+\log (x))} \, dx\\ &=e^x-e x-e \log (2 x+\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 18, normalized size = 1.06 \begin {gather*} e^x-e x-e \log (2 x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^x - E*x - E*Log[2*x + Log[x]]

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fricas [A]  time = 0.57, size = 19, normalized size = 1.12 \begin {gather*} -x e - e \log \left (2 \, x + \log \relax (x)\right ) + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x-x*exp(1))*log(x)+2*exp(x)*x^2+(-2*x^2-2*x-1)*exp(1))/(x*log(x)+2*x^2),x, algorithm="frica
s")

[Out]

-x*e - e*log(2*x + log(x)) + e^x

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giac [A]  time = 0.21, size = 19, normalized size = 1.12 \begin {gather*} -x e - e \log \left (2 \, x + \log \relax (x)\right ) + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*e - e*log(2*x + log(x)) + e^x

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maple [A]  time = 0.06, size = 20, normalized size = 1.18

method	result	size
default	\(-x \,{\mathrm e}-{\mathrm e} \ln \left (2 x +\ln \relax (x )\right )+{\mathrm e}^{x}\)	\(20\)
norman	\(-x \,{\mathrm e}-{\mathrm e} \ln \left (2 x +\ln \relax (x )\right )+{\mathrm e}^{x}\)	\(20\)
risch	\(-x \,{\mathrm e}-{\mathrm e} \ln \left (2 x +\ln \relax (x )\right )+{\mathrm e}^{x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*exp(1)-exp(1)*ln(2*x+ln(x))+exp(x)

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maxima [A]  time = 0.39, size = 19, normalized size = 1.12 \begin {gather*} -x e - e \log \left (2 \, x + \log \relax (x)\right ) + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x-x*exp(1))*log(x)+2*exp(x)*x^2+(-2*x^2-2*x-1)*exp(1))/(x*log(x)+2*x^2),x, algorithm="maxim
a")

[Out]

-x*e - e*log(2*x + log(x)) + e^x

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mupad [B]  time = 4.47, size = 19, normalized size = 1.12 \begin {gather*} {\mathrm {e}}^x-x\,\mathrm {e}-\mathrm {e}\,\ln \left (2\,x+\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1)*(2*x + 2*x^2 + 1) - 2*x^2*exp(x) + log(x)*(x*exp(1) - x*exp(x)))/(x*log(x) + 2*x^2),x)

[Out]

exp(x) - x*exp(1) - exp(1)*log(2*x + log(x))

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sympy [A]  time = 0.37, size = 19, normalized size = 1.12 \begin {gather*} - e x + e^{x} - e \log {\left (2 x + \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-E*x + exp(x) - E*log(2*x + log(x))

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