3.62.16 \(\int \frac {2 x-2 e^x x \log (2)+(4+2 x+e^x (-2 x-2 x^2) \log (2)) \log (x)}{x} \, dx\)

Optimal. Leaf size=18 \[ 2 \log (x) \left (x \left (1-e^x \log (2)\right )+\log (x)\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 5, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {14, 2288, 2346, 2301, 2295} \begin {gather*} 2 \log ^2(x)+2 x \log (x)-2 e^x x \log (2) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*x*Log[x] - 2*E^x*x*Log[2]*Log[x] + 2*Log[x]^2

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 2288

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]

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 16, normalized size = 0.89 \begin {gather*} 2 \log (x) \left (x-e^x x \log (2)+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Log[x]*(x - E^x*x*Log[2] + Log[x])

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fricas [A]  time = 0.67, size = 21, normalized size = 1.17 \begin {gather*} -2 \, {\left (x e^{x} \log \relax (2) - x\right )} \log \relax (x) + 2 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(x*e^x*log(2) - x)*log(x) + 2*log(x)^2

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giac [A]  time = 0.13, size = 21, normalized size = 1.17 \begin {gather*} -2 \, x e^{x} \log \relax (2) \log \relax (x) + 2 \, x \log \relax (x) + 2 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x*e^x*log(2)*log(x) + 2*x*log(x) + 2*log(x)^2

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maple [A]  time = 0.07, size = 22, normalized size = 1.22

method	result	size
default	\(-2 x \ln \relax (2) {\mathrm e}^{x} \ln \relax (x )+2 \ln \relax (x )^{2}+2 x \ln \relax (x )\)	\(22\)
norman	\(-2 x \ln \relax (2) {\mathrm e}^{x} \ln \relax (x )+2 \ln \relax (x )^{2}+2 x \ln \relax (x )\)	\(22\)
risch	\(2 \ln \relax (x )^{2}+\left (-2 x \ln \relax (2) {\mathrm e}^{x}+2 x \right ) \ln \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, e^{x} \log \relax (2) \log \relax (x) - 2 \, {\left ({\left (x - 1\right )} e^{x} \log \relax (x) - \int \frac {{\left (x - 1\right )} e^{x}}{x}\,{d x}\right )} \log \relax (2) + 2 \, {\rm Ei}\relax (x) \log \relax (2) - 2 \, e^{x} \log \relax (2) + 2 \, x \log \relax (x) + 2 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*e^x*log(2)*log(x) - 2*((x - 1)*e^x*log(x) - integrate((x - 1)*e^x/x, x))*log(2) + 2*Ei(x)*log(2) - 2*e^x*lo
g(2) + 2*x*log(x) + 2*log(x)^2

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mupad [B]  time = 4.95, size = 15, normalized size = 0.83 \begin {gather*} 2\,\ln \relax (x)\,\left (x+\ln \relax (x)-x\,{\mathrm {e}}^x\,\ln \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.33, size = 26, normalized size = 1.44 \begin {gather*} - 2 x e^{x} \log {\relax (2 )} \log {\relax (x )} + 2 x \log {\relax (x )} + 2 \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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