3.69.31 \(\int (1-6 e^9+9 e^{18}-\log (4)+(2-6 e^9) \log (x)+\log ^2(x)) \, dx\)

Optimal. Leaf size=22 \[ -3+x-x \log (4)+x \left (3 e^9-\log (x)\right )^2 \]

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Rubi [B]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 2.50, number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2295, 2296} \begin {gather*} -2 \left (1-3 e^9\right ) x+2 x+x \log ^2(x)+2 \left (1-3 e^9\right ) x \log (x)-2 x \log (x)+x \left (1-6 e^9+9 e^{18}-\log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - 6*E^9 + 9*E^18 - Log[4] + (2 - 6*E^9)*Log[x] + Log[x]^2,x]

[Out]

2*x - 2*(1 - 3*E^9)*x + x*(1 - 6*E^9 + 9*E^18 - Log[4]) - 2*x*Log[x] + 2*(1 - 3*E^9)*x*Log[x] + x*Log[x]^2

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x \left (1-6 e^9+9 e^{18}-\log (4)\right )+\left (2 \left (1-3 e^9\right )\right ) \int \log (x) \, dx+\int \log ^2(x) \, dx\\ &=-2 \left (1-3 e^9\right ) x+x \left (1-6 e^9+9 e^{18}-\log (4)\right )+2 \left (1-3 e^9\right ) x \log (x)+x \log ^2(x)-2 \int \log (x) \, dx\\ &=2 x-2 \left (1-3 e^9\right ) x+x \left (1-6 e^9+9 e^{18}-\log (4)\right )-2 x \log (x)+2 \left (1-3 e^9\right ) x \log (x)+x \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.23 \begin {gather*} x+9 e^{18} x-x \log (4)-6 e^9 x \log (x)+x \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - 6*E^9 + 9*E^18 - Log[4] + (2 - 6*E^9)*Log[x] + Log[x]^2,x]

[Out]

x + 9*E^18*x - x*Log[4] - 6*E^9*x*Log[x] + x*Log[x]^2

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fricas [A]  time = 1.03, size = 25, normalized size = 1.14 \begin {gather*} -6 \, x e^{9} \log \relax (x) + x \log \relax (x)^{2} + 9 \, x e^{18} - 2 \, x \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^2+(-6*exp(9)+2)*log(x)-2*log(2)+9*exp(9)^2-6*exp(9)+1,x, algorithm="fricas")

[Out]

-6*x*e^9*log(x) + x*log(x)^2 + 9*x*e^18 - 2*x*log(2) + x

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giac [B]  time = 0.12, size = 46, normalized size = 2.09 \begin {gather*} x \log \relax (x)^{2} - 2 \, {\left (x \log \relax (x) - x\right )} {\left (3 \, e^{9} - 1\right )} + 9 \, x e^{18} - 6 \, x e^{9} - 2 \, x \log \relax (2) - 2 \, x \log \relax (x) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^2+(-6*exp(9)+2)*log(x)-2*log(2)+9*exp(9)^2-6*exp(9)+1,x, algorithm="giac")

[Out]

x*log(x)^2 - 2*(x*log(x) - x)*(3*e^9 - 1) + 9*x*e^18 - 6*x*e^9 - 2*x*log(2) - 2*x*log(x) + 3*x

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maple [A]  time = 0.02, size = 26, normalized size = 1.18

method	result	size
risch	\(x +x \ln \relax (x )^{2}-6 x \,{\mathrm e}^{9} \ln \relax (x )+9 \,{\mathrm e}^{18} x -2 x \ln \relax (2)\)	\(26\)
default	\(x +x \ln \relax (x )^{2}-6 x \,{\mathrm e}^{9} \ln \relax (x )+9 \,{\mathrm e}^{18} x -2 x \ln \relax (2)\)	\(28\)
norman	\(x \ln \relax (x )^{2}+\left (1+9 \,{\mathrm e}^{18}-2 \ln \relax (2)\right ) x -6 x \,{\mathrm e}^{9} \ln \relax (x )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)^2+(-6*exp(9)+2)*ln(x)-2*ln(2)+9*exp(9)^2-6*exp(9)+1,x,method=_RETURNVERBOSE)

[Out]

x+x*ln(x)^2-6*x*exp(9)*ln(x)+9*exp(18)*x-2*x*ln(2)

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maxima [B]  time = 0.37, size = 45, normalized size = 2.05 \begin {gather*} {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x - 2 \, {\left (x \log \relax (x) - x\right )} {\left (3 \, e^{9} - 1\right )} + 9 \, x e^{18} - 6 \, x e^{9} - 2 \, x \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)^2+(-6*exp(9)+2)*log(x)-2*log(2)+9*exp(9)^2-6*exp(9)+1,x, algorithm="maxima")

[Out]

(log(x)^2 - 2*log(x) + 2)*x - 2*(x*log(x) - x)*(3*e^9 - 1) + 9*x*e^18 - 6*x*e^9 - 2*x*log(2) + x

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mupad [B]  time = 4.14, size = 22, normalized size = 1.00 \begin {gather*} x\,\left ({\ln \relax (x)}^2-6\,{\mathrm {e}}^9\,\ln \relax (x)+9\,{\mathrm {e}}^{18}-\ln \relax (4)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(9*exp(18) - 6*exp(9) - 2*log(2) + log(x)^2 - log(x)*(6*exp(9) - 2) + 1,x)

[Out]

x*(9*exp(18) - log(4) + log(x)^2 - 6*exp(9)*log(x) + 1)

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sympy [A]  time = 0.15, size = 29, normalized size = 1.32 \begin {gather*} x \log {\relax (x )}^{2} - 6 x e^{9} \log {\relax (x )} + x \left (- 2 \log {\relax (2 )} + 1 + 9 e^{18}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x)**2+(-6*exp(9)+2)*ln(x)-2*ln(2)+9*exp(9)**2-6*exp(9)+1,x)

[Out]

x*log(x)**2 - 6*x*exp(9)*log(x) + x*(-2*log(2) + 1 + 9*exp(18))

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