3.75.49 \(\int \frac {1+7 x+13 x^2+x^3+(x+2 x^2) \log (2 x)+(x+2 x^2) \log (e^{-1+x} x)}{x} \, dx\)

Optimal. Leaf size=23 \[ \log (x)+\left (x+x^2\right ) \left (5+\log (2 x)+\log \left (e^{-1+x} x\right )\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 46, normalized size of antiderivative = 2.00, number of steps used = 10, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14, 2313, 2551} \begin {gather*} 5 x^2+\left (x^2+x\right ) \log (2 x)+\frac {19 x}{4}+\frac {3 \log (x)}{4}+\frac {1}{4} (2 x+1)^2 \log \left (e^{x-1} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 7*x + 13*x^2 + x^3 + (x + 2*x^2)*Log[2*x] + (x + 2*x^2)*Log[E^(-1 + x)*x])/x,x]

[Out]

(19*x)/4 + 5*x^2 + (3*Log[x])/4 + (x + x^2)*Log[2*x] + ((1 + 2*x)^2*Log[E^(-1 + x)*x])/4

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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

Rule 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1+7 x+13 x^2+x^3+x \log (2 x)+2 x^2 \log (2 x)}{x}+(1+2 x) \log \left (e^{-1+x} x\right )\right ) \, dx\\ &=\int \frac {1+7 x+13 x^2+x^3+x \log (2 x)+2 x^2 \log (2 x)}{x} \, dx+\int (1+2 x) \log \left (e^{-1+x} x\right ) \, dx\\ &=\frac {1}{4} (1+2 x)^2 \log \left (e^{-1+x} x\right )-\frac {1}{4} \int \left (5+\frac {1}{x}+8 x+4 x^2\right ) \, dx+\int \left (\frac {1+7 x+13 x^2+x^3}{x}+(1+2 x) \log (2 x)\right ) \, dx\\ &=-\frac {5 x}{4}-x^2-\frac {x^3}{3}-\frac {\log (x)}{4}+\frac {1}{4} (1+2 x)^2 \log \left (e^{-1+x} x\right )+\int \frac {1+7 x+13 x^2+x^3}{x} \, dx+\int (1+2 x) \log (2 x) \, dx\\ &=-\frac {5 x}{4}-x^2-\frac {x^3}{3}-\frac {\log (x)}{4}+\left (x+x^2\right ) \log (2 x)+\frac {1}{4} (1+2 x)^2 \log \left (e^{-1+x} x\right )-\int (1+x) \, dx+\int \left (7+\frac {1}{x}+13 x+x^2\right ) \, dx\\ &=\frac {19 x}{4}+5 x^2+\frac {3 \log (x)}{4}+\left (x+x^2\right ) \log (2 x)+\frac {1}{4} (1+2 x)^2 \log \left (e^{-1+x} x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.96 \begin {gather*} \log (x)+x (1+x) \left (5+\log (2 x)+\log \left (e^{-1+x} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 7*x + 13*x^2 + x^3 + (x + 2*x^2)*Log[2*x] + (x + 2*x^2)*Log[E^(-1 + x)*x])/x,x]

[Out]

Log[x] + x*(1 + x)*(5 + Log[2*x] + Log[E^(-1 + x)*x])

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fricas [A]  time = 0.51, size = 36, normalized size = 1.57 \begin {gather*} x^{3} + 5 \, x^{2} - {\left (x^{2} + x\right )} \log \relax (2) + {\left (2 \, x^{2} + 2 \, x + 1\right )} \log \left (2 \, x\right ) + 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+x)*log(x*exp(x-1))+(2*x^2+x)*log(2*x)+x^3+13*x^2+7*x+1)/x,x, algorithm="fricas")

[Out]

x^3 + 5*x^2 - (x^2 + x)*log(2) + (2*x^2 + 2*x + 1)*log(2*x) + 4*x

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giac [A]  time = 0.17, size = 29, normalized size = 1.26 \begin {gather*} x^{3} + x^{2} {\left (\log \relax (2) + 5\right )} + x {\left (\log \relax (2) + 4\right )} + 2 \, {\left (x^{2} + x\right )} \log \relax (x) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+x)*log(x*exp(x-1))+(2*x^2+x)*log(2*x)+x^3+13*x^2+7*x+1)/x,x, algorithm="giac")

[Out]

x^3 + x^2*(log(2) + 5) + x*(log(2) + 4) + 2*(x^2 + x)*log(x) + log(x)

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maple [B]  time = 0.02, size = 47, normalized size = 2.04

method	result	size
default	\(\ln \left (x \,{\mathrm e}^{x -1}\right ) x^{2}+\ln \left (x \,{\mathrm e}^{x -1}\right ) x +5 x^{2}+5 x +\frac {7}{3}+x^{2} \ln \left (2 x \right )+x \ln \left (2 x \right )+\ln \relax (x )\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+x)*ln(x*exp(x-1))+(2*x^2+x)*ln(2*x)+x^3+13*x^2+7*x+1)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x*exp(x-1))*x^2+ln(x*exp(x-1))*x+5*x^2+5*x+7/3+x^2*ln(2*x)+x*ln(2*x)+ln(x)

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maxima [B]  time = 0.39, size = 45, normalized size = 1.96 \begin {gather*} x^{2} \log \left (x e^{\left (x - 1\right )}\right ) + x^{2} \log \left (2 \, x\right ) + 5 \, x^{2} + x \log \left (x e^{\left (x - 1\right )}\right ) + x \log \left (2 \, x\right ) + 5 \, x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+x)*log(x*exp(x-1))+(2*x^2+x)*log(2*x)+x^3+13*x^2+7*x+1)/x,x, algorithm="maxima")

[Out]

x^2*log(x*e^(x - 1)) + x^2*log(2*x) + 5*x^2 + x*log(x*e^(x - 1)) + x*log(2*x) + 5*x + log(x)

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mupad [B]  time = 4.75, size = 36, normalized size = 1.57 \begin {gather*} 4\,x+\ln \relax (x)+2\,x^2\,\ln \relax (x)+x\,\ln \relax (2)+x^2\,\ln \relax (2)+2\,x\,\ln \relax (x)+5\,x^2+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((7*x + log(2*x)*(x + 2*x^2) + 13*x^2 + x^3 + log(x*exp(x - 1))*(x + 2*x^2) + 1)/x,x)

[Out]

4*x + log(x) + 2*x^2*log(x) + x*log(2) + x^2*log(2) + 2*x*log(x) + 5*x^2 + x^3

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sympy [A]  time = 0.34, size = 34, normalized size = 1.48 \begin {gather*} x^{3} + x^{2} \left (5 - \log {\relax (2 )}\right ) + x \left (4 - \log {\relax (2 )}\right ) + \left (2 x^{2} + 2 x\right ) \log {\left (2 x \right )} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+x)*ln(x*exp(x-1))+(2*x**2+x)*ln(2*x)+x**3+13*x**2+7*x+1)/x,x)

[Out]

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

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