3.7.42 \(\int \frac {1}{2} (-45+4 x-\log ^2(3)-2 \log (x)) \, dx\)

Optimal. Leaf size=21 \[ x \left (-20+x+\frac {1}{2} \left (-3-\log ^2(3)\right )-\log (x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 2295} \begin {gather*} x^2+x-\frac {1}{2} x \left (45+\log ^2(3)\right )-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 + 4*x - Log[3]^2 - 2*Log[x])/2,x]

[Out]

x + x^2 - (x*(45 + Log[3]^2))/2 - x*Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (-45+4 x-\log ^2(3)-2 \log (x)\right ) \, dx\\ &=x^2-\frac {1}{2} x \left (45+\log ^2(3)\right )-\int \log (x) \, dx\\ &=x+x^2-\frac {1}{2} x \left (45+\log ^2(3)\right )-x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 1.10 \begin {gather*} -\frac {43 x}{2}+x^2-\frac {1}{2} x \log ^2(3)-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 + 4*x - Log[3]^2 - 2*Log[x])/2,x]

[Out]

(-43*x)/2 + x^2 - (x*Log[3]^2)/2 - x*Log[x]

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fricas [A]  time = 0.62, size = 19, normalized size = 0.90 \begin {gather*} -\frac {1}{2} \, x \log \relax (3)^{2} + x^{2} - x \log \relax (x) - \frac {43}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)-1/2*log(3)^2+2*x-45/2,x, algorithm="fricas")

[Out]

-1/2*x*log(3)^2 + x^2 - x*log(x) - 43/2*x

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giac [A]  time = 0.48, size = 19, normalized size = 0.90 \begin {gather*} -\frac {1}{2} \, x \log \relax (3)^{2} + x^{2} - x \log \relax (x) - \frac {43}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)-1/2*log(3)^2+2*x-45/2,x, algorithm="giac")

[Out]

-1/2*x*log(3)^2 + x^2 - x*log(x) - 43/2*x

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maple [A]  time = 0.02, size = 20, normalized size = 0.95

method	result	size
default	\(-\frac {43 x}{2}+x^{2}-\frac {x \ln \relax (3)^{2}}{2}-x \ln \relax (x )\)	\(20\)
norman	\(x^{2}+\left (-\frac {43}{2}-\frac {\ln \relax (3)^{2}}{2}\right ) x -x \ln \relax (x )\)	\(20\)
risch	\(-\frac {43 x}{2}+x^{2}-\frac {x \ln \relax (3)^{2}}{2}-x \ln \relax (x )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-ln(x)-1/2*ln(3)^2+2*x-45/2,x,method=_RETURNVERBOSE)

[Out]

-43/2*x+x^2-1/2*x*ln(3)^2-x*ln(x)

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maxima [A]  time = 0.51, size = 19, normalized size = 0.90 \begin {gather*} -\frac {1}{2} \, x \log \relax (3)^{2} + x^{2} - x \log \relax (x) - \frac {43}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-log(x)-1/2*log(3)^2+2*x-45/2,x, algorithm="maxima")

[Out]

-1/2*x*log(3)^2 + x^2 - x*log(x) - 43/2*x

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mupad [B]  time = 0.54, size = 16, normalized size = 0.76 \begin {gather*} -\frac {x\,\left (2\,\ln \relax (x)-2\,x+{\ln \relax (3)}^2+43\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - log(x) - log(3)^2/2 - 45/2,x)

[Out]

-(x*(2*log(x) - 2*x + log(3)^2 + 43))/2

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sympy [A]  time = 0.09, size = 20, normalized size = 0.95 \begin {gather*} x^{2} - x \log {\relax (x )} + x \left (- \frac {43}{2} - \frac {\log {\relax (3 )}^{2}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-ln(x)-1/2*ln(3)**2+2*x-45/2,x)

[Out]

x**2 - x*log(x) + x*(-43/2 - log(3)**2/2)

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