3.22.95 \(\int \frac {1}{4} (45-86 x+6 x^2) \log (3) \, dx\)

Optimal. Leaf size=28 \[ \left (x+(1-x) \left (x+\frac {1}{4} x (1+3 (12-x)+x)\right )\right ) \log (3) \]

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Rubi [A]  time = 0.00, antiderivative size = 26, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12} \begin {gather*} \frac {1}{2} x^3 \log (3)-\frac {43}{4} x^2 \log (3)+\frac {45}{4} x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((45 - 86*x + 6*x^2)*Log[3])/4,x]

[Out]

(45*x*Log[3])/4 - (43*x^2*Log[3])/4 + (x^3*Log[3])/2

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \log (3) \int \left (45-86 x+6 x^2\right ) \, dx\\ &=\frac {45}{4} x \log (3)-\frac {43}{4} x^2 \log (3)+\frac {1}{2} x^3 \log (3)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 0.71 \begin {gather*} \frac {1}{4} \left (45 x-43 x^2+2 x^3\right ) \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((45 - 86*x + 6*x^2)*Log[3])/4,x]

[Out]

((45*x - 43*x^2 + 2*x^3)*Log[3])/4

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fricas [A]  time = 0.60, size = 18, normalized size = 0.64 \begin {gather*} \frac {1}{4} \, {\left (2 \, x^{3} - 43 \, x^{2} + 45 \, x\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*x^2-86*x+45)*log(3),x, algorithm="fricas")

[Out]

1/4*(2*x^3 - 43*x^2 + 45*x)*log(3)

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giac [A]  time = 0.93, size = 18, normalized size = 0.64 \begin {gather*} \frac {1}{4} \, {\left (2 \, x^{3} - 43 \, x^{2} + 45 \, x\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*x^2-86*x+45)*log(3),x, algorithm="giac")

[Out]

1/4*(2*x^3 - 43*x^2 + 45*x)*log(3)

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maple [A]  time = 0.02, size = 16, normalized size = 0.57

method	result	size
gosper	\(\frac {x \left (2 x^{2}-43 x +45\right ) \ln \relax (3)}{4}\)	\(16\)
default	\(\frac {\left (2 x^{3}-43 x^{2}+45 x \right ) \ln \relax (3)}{4}\)	\(19\)
norman	\(\frac {45 x \ln \relax (3)}{4}-\frac {43 x^{2} \ln \relax (3)}{4}+\frac {x^{3} \ln \relax (3)}{2}\)	\(21\)
risch	\(\frac {45 x \ln \relax (3)}{4}-\frac {43 x^{2} \ln \relax (3)}{4}+\frac {x^{3} \ln \relax (3)}{2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(6*x^2-86*x+45)*ln(3),x,method=_RETURNVERBOSE)

[Out]

1/4*x*(2*x^2-43*x+45)*ln(3)

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maxima [A]  time = 0.40, size = 18, normalized size = 0.64 \begin {gather*} \frac {1}{4} \, {\left (2 \, x^{3} - 43 \, x^{2} + 45 \, x\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*x^2-86*x+45)*log(3),x, algorithm="maxima")

[Out]

1/4*(2*x^3 - 43*x^2 + 45*x)*log(3)

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mupad [B]  time = 0.04, size = 15, normalized size = 0.54 \begin {gather*} \frac {x\,\ln \relax (3)\,\left (2\,x^2-43\,x+45\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3)*(6*x^2 - 86*x + 45))/4,x)

[Out]

(x*log(3)*(2*x^2 - 43*x + 45))/4

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sympy [A]  time = 0.05, size = 26, normalized size = 0.93 \begin {gather*} \frac {x^{3} \log {\relax (3 )}}{2} - \frac {43 x^{2} \log {\relax (3 )}}{4} + \frac {45 x \log {\relax (3 )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(6*x**2-86*x+45)*ln(3),x)

[Out]

x**3*log(3)/2 - 43*x**2*log(3)/4 + 45*x*log(3)/4

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