3.57.65 \(\int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx\)

Optimal. Leaf size=34 \[ 5-4 (2-x)+2 \left (\log (3)-x \left (-x+\frac {x^2}{x-x \log (3)}\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12} \begin {gather*} \frac {4 x}{1-\log (3)}-\frac {2 (x+1)^2 \log (3)}{1-\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + (4 + 4*x)*Log[3])/(-1 + Log[3]),x]

[Out]

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

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

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Mathematica [A]  time = 0.00, size = 25, normalized size = 0.74 \begin {gather*} \frac {4 \left (-x+x \log (3)+\frac {1}{2} x^2 \log (3)\right )}{-1+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + (4 + 4*x)*Log[3])/(-1 + Log[3]),x]

[Out]

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

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fricas [A]  time = 0.52, size = 22, normalized size = 0.65 \begin {gather*} \frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \relax (3) - 2 \, x\right )}}{\log \relax (3) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(3)-4)/(log(3)-1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 22, normalized size = 0.65 \begin {gather*} \frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \relax (3) - 2 \, x\right )}}{\log \relax (3) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(3)-4)/(log(3)-1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 18, normalized size = 0.53

method	result	size
norman	\(4 x +\frac {2 \ln \relax (3) x^{2}}{\ln \relax (3)-1}\)	\(18\)
gosper	\(\frac {2 x \left (x \ln \relax (3)+2 \ln \relax (3)-2\right )}{\ln \relax (3)-1}\)	\(20\)
default	\(\frac {\ln \relax (3) \left (2 x^{2}+4 x \right )-4 x}{\ln \relax (3)-1}\)	\(24\)
risch	\(\frac {2 \ln \relax (3) x^{2}}{\ln \relax (3)-1}+\frac {4 x \ln \relax (3)}{\ln \relax (3)-1}-\frac {4 x}{\ln \relax (3)-1}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+4)*ln(3)-4)/(ln(3)-1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 22, normalized size = 0.65 \begin {gather*} \frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \relax (3) - 2 \, x\right )}}{\log \relax (3) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*log(3)-4)/(log(3)-1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.46, size = 24, normalized size = 0.71 \begin {gather*} \frac {{\left (\ln \relax (3)\,\left (4\,x+4\right )-4\right )}^2}{4\,\ln \relax (3)\,\left (\ln \relax (9)-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3)*(4*x + 4) - 4)/(log(3) - 1),x)

[Out]

(log(3)*(4*x + 4) - 4)^2/(4*log(3)*(log(9) - 2))

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sympy [A]  time = 0.05, size = 15, normalized size = 0.44 \begin {gather*} \frac {2 x^{2} \log {\relax (3 )}}{-1 + \log {\relax (3 )}} + 4 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+4)*ln(3)-4)/(ln(3)-1),x)

[Out]

2*x**2*log(3)/(-1 + log(3)) + 4*x

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