3.53.27 \(\int \frac {-19-e^6+e^3 (-8-2 x)-9 x-x^2+(20+19 x+5 x^2) \log (5)+(-9 x-2 e^3 x-2 x^2+(19 x+10 x^2) \log (5)) \log (x)}{x} \, dx\)

Optimal. Leaf size=32 \[ \left (-3-x-\left (4+e^3+x\right )^2+\left (-x+5 (2+x)^2\right ) \log (5)\right ) \log (x) \]

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Rubi [A]  time = 0.10, antiderivative size = 47, normalized size of antiderivative = 1.47, number of steps used = 6, number of rules used = 2, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {14, 2313} \begin {gather*} -\left (\left (x^2 (1-5 \log (5))+x \left (9+2 e^3-19 \log (5)\right )\right ) \log (x)\right )-\left (19+8 e^3+e^6-20 \log (5)\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-19 - E^6 + E^3*(-8 - 2*x) - 9*x - x^2 + (20 + 19*x + 5*x^2)*Log[5] + (-9*x - 2*E^3*x - 2*x^2 + (19*x + 1
0*x^2)*Log[5])*Log[x])/x,x]

[Out]

-((x*(9 + 2*E^3 - 19*Log[5]) + x^2*(1 - 5*Log[5]))*Log[x]) - (19 + 8*E^3 + E^6 - 20*Log[5])*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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-19-8 e^3-e^6-x \left (9+2 e^3-19 \log (5)\right )-x^2 (1-5 \log (5))+20 \log (5)}{x}+\left (-9-2 e^3-2 x (1-5 \log (5))+19 \log (5)\right ) \log (x)\right ) \, dx\\ &=\int \frac {-19-8 e^3-e^6-x \left (9+2 e^3-19 \log (5)\right )-x^2 (1-5 \log (5))+20 \log (5)}{x} \, dx+\int \left (-9-2 e^3-2 x (1-5 \log (5))+19 \log (5)\right ) \log (x) \, dx\\ &=-\left (\left (x \left (9+2 e^3-19 \log (5)\right )+x^2 (1-5 \log (5))\right ) \log (x)\right )-\int \left (-9-2 e^3+19 \log (5)+x (-1+5 \log (5))\right ) \, dx+\int \left (-9-2 e^3-x (1-5 \log (5))+19 \log (5)+\frac {-19-8 e^3-e^6+20 \log (5)}{x}\right ) \, dx\\ &=-\left (\left (x \left (9+2 e^3-19 \log (5)\right )+x^2 (1-5 \log (5))\right ) \log (x)\right )-\left (19+8 e^3+e^6-20 \log (5)\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 61, normalized size = 1.91 \begin {gather*} -19 \log (x)-8 e^3 \log (x)-e^6 \log (x)-9 x \log (x)-2 e^3 x \log (x)-x^2 \log (x)+20 \log (5) \log (x)+19 x \log (5) \log (x)+5 x^2 \log (5) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-19 - E^6 + E^3*(-8 - 2*x) - 9*x - x^2 + (20 + 19*x + 5*x^2)*Log[5] + (-9*x - 2*E^3*x - 2*x^2 + (19
*x + 10*x^2)*Log[5])*Log[x])/x,x]

[Out]

-19*Log[x] - 8*E^3*Log[x] - E^6*Log[x] - 9*x*Log[x] - 2*E^3*x*Log[x] - x^2*Log[x] + 20*Log[5]*Log[x] + 19*x*Lo
g[5]*Log[x] + 5*x^2*Log[5]*Log[x]

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fricas [A]  time = 0.47, size = 35, normalized size = 1.09 \begin {gather*} -{\left (x^{2} + 2 \, {\left (x + 4\right )} e^{3} - {\left (5 \, x^{2} + 19 \, x + 20\right )} \log \relax (5) + 9 \, x + e^{6} + 19\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^2+19*x)*log(5)-2*x*exp(3)-2*x^2-9*x)*log(x)+(5*x^2+19*x+20)*log(5)-exp(3)^2+(-2*x-8)*exp(3)-
x^2-9*x-19)/x,x, algorithm="fricas")

[Out]

-(x^2 + 2*(x + 4)*e^3 - (5*x^2 + 19*x + 20)*log(5) + 9*x + e^6 + 19)*log(x)

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giac [A]  time = 0.17, size = 58, normalized size = 1.81 \begin {gather*} 5 \, x^{2} \log \relax (5) \log \relax (x) - x^{2} \log \relax (x) - 2 \, x e^{3} \log \relax (x) + 19 \, x \log \relax (5) \log \relax (x) - 9 \, x \log \relax (x) - e^{6} \log \relax (x) - 8 \, e^{3} \log \relax (x) + 20 \, \log \relax (5) \log \relax (x) - 19 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^2+19*x)*log(5)-2*x*exp(3)-2*x^2-9*x)*log(x)+(5*x^2+19*x+20)*log(5)-exp(3)^2+(-2*x-8)*exp(3)-
x^2-9*x-19)/x,x, algorithm="giac")

[Out]

5*x^2*log(5)*log(x) - x^2*log(x) - 2*x*e^3*log(x) + 19*x*log(5)*log(x) - 9*x*log(x) - e^6*log(x) - 8*e^3*log(x
) + 20*log(5)*log(x) - 19*log(x)

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maple [A]  time = 0.05, size = 47, normalized size = 1.47

method	result	size
norman	\(\left (-{\mathrm e}^{6}-8 \,{\mathrm e}^{3}+20 \ln \relax (5)-19\right ) \ln \relax (x )+\left (5 \ln \relax (5)-1\right ) x^{2} \ln \relax (x )+\left (-2 \,{\mathrm e}^{3}+19 \ln \relax (5)-9\right ) x \ln \relax (x )\)	\(47\)
risch	\(\left (5 x^{2} \ln \relax (5)-2 x \,{\mathrm e}^{3}+19 x \ln \relax (5)-x^{2}-9 x \right ) \ln \relax (x )-8 \ln \relax (x ) {\mathrm e}^{3}-\ln \relax (x ) {\mathrm e}^{6}+20 \ln \relax (5) \ln \relax (x )-19 \ln \relax (x )\)	\(53\)
default	\(10 \ln \relax (5) \left (\frac {x^{2} \ln \relax (x )}{2}-\frac {x^{2}}{4}\right )-2 \,{\mathrm e}^{3} \left (x \ln \relax (x )-x \right )+19 \ln \relax (5) \left (x \ln \relax (x )-x \right )-x^{2} \ln \relax (x )+\frac {5 x^{2} \ln \relax (5)}{2}-9 x \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{6}-2 x \,{\mathrm e}^{3}+19 x \ln \relax (5)-8 \ln \relax (x ) {\mathrm e}^{3}+20 \ln \relax (5) \ln \relax (x )-19 \ln \relax (x )\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x^2+19*x)*ln(5)-2*x*exp(3)-2*x^2-9*x)*ln(x)+(5*x^2+19*x+20)*ln(5)-exp(3)^2+(-2*x-8)*exp(3)-x^2-9*x-1
9)/x,x,method=_RETURNVERBOSE)

[Out]

(-exp(3)^2-8*exp(3)+20*ln(5)-19)*ln(x)+(5*ln(5)-1)*x^2*ln(x)+(-2*exp(3)+19*ln(5)-9)*x*ln(x)

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maxima [B]  time = 0.35, size = 93, normalized size = 2.91 \begin {gather*} \frac {5}{2} \, x^{2} \log \relax (5) - x^{2} \log \relax (x) - 2 \, {\left (x \log \relax (x) - x\right )} e^{3} - 2 \, x e^{3} + \frac {5}{2} \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \relax (5) + 19 \, {\left (x \log \relax (x) - x\right )} \log \relax (5) + 19 \, x \log \relax (5) - 9 \, x \log \relax (x) - e^{6} \log \relax (x) - 8 \, e^{3} \log \relax (x) + 20 \, \log \relax (5) \log \relax (x) - 19 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^2+19*x)*log(5)-2*x*exp(3)-2*x^2-9*x)*log(x)+(5*x^2+19*x+20)*log(5)-exp(3)^2+(-2*x-8)*exp(3)-
x^2-9*x-19)/x,x, algorithm="maxima")

[Out]

5/2*x^2*log(5) - x^2*log(x) - 2*(x*log(x) - x)*e^3 - 2*x*e^3 + 5/2*(2*x^2*log(x) - x^2)*log(5) + 19*(x*log(x)
- x)*log(5) + 19*x*log(5) - 9*x*log(x) - e^6*log(x) - 8*e^3*log(x) + 20*log(5)*log(x) - 19*log(x)

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mupad [B]  time = 3.46, size = 44, normalized size = 1.38 \begin {gather*} x^2\,\ln \relax (x)\,\left (5\,\ln \relax (5)-1\right )-\ln \relax (x)\,\left (8\,{\mathrm {e}}^3+{\mathrm {e}}^6-20\,\ln \relax (5)+19\right )-x\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^3-19\,\ln \relax (5)+9\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x + exp(6) + log(x)*(9*x - log(5)*(19*x + 10*x^2) + 2*x*exp(3) + 2*x^2) - log(5)*(19*x + 5*x^2 + 20) +
x^2 + exp(3)*(2*x + 8) + 19)/x,x)

[Out]

x^2*log(x)*(5*log(5) - 1) - log(x)*(8*exp(3) + exp(6) - 20*log(5) + 19) - x*log(x)*(2*exp(3) - 19*log(5) + 9)

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sympy [A]  time = 0.16, size = 49, normalized size = 1.53 \begin {gather*} \left (- x^{2} + 5 x^{2} \log {\relax (5 )} - 2 x e^{3} - 9 x + 19 x \log {\relax (5 )}\right ) \log {\relax (x )} + \left (- e^{6} - 8 e^{3} - 19 + 20 \log {\relax (5 )}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x**2+19*x)*ln(5)-2*x*exp(3)-2*x**2-9*x)*ln(x)+(5*x**2+19*x+20)*ln(5)-exp(3)**2+(-2*x-8)*exp(3)
-x**2-9*x-19)/x,x)

[Out]

(-x**2 + 5*x**2*log(5) - 2*x*exp(3) - 9*x + 19*x*log(5))*log(x) + (-exp(6) - 8*exp(3) - 19 + 20*log(5))*log(x)

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