3.1.4 \(\int \frac {-30+30 x+(8-4 x+12 x^2) \log (3)}{\log (3)} \, dx\)

Optimal. Leaf size=27 \[ 2-\left (-5+2 x-x^2\right ) \left (x+3 \left (2+x+\frac {5}{\log (3)}\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12} \begin {gather*} 4 x^3-2 x^2+\frac {15 x^2}{\log (3)}+8 x-\frac {30 x}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30 + 30*x + (8 - 4*x + 12*x^2)*Log[3])/Log[3],x]

[Out]

8*x - 2*x^2 + 4*x^3 - (30*x)/Log[3] + (15*x^2)/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 \left (-30+30 x+\left (8-4 x+12 x^2\right ) \log (3)\right ) \, dx}{\log (3)}\\ &=-\frac {30 x}{\log (3)}+\frac {15 x^2}{\log (3)}+\int \left (8-4 x+12 x^2\right ) \, dx\\ &=8 x-2 x^2+4 x^3-\frac {30 x}{\log (3)}+\frac {15 x^2}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 33, normalized size = 1.22 \begin {gather*} \frac {-30 x+15 x^2+8 x \log (3)-2 x^2 \log (3)+4 x^3 \log (3)}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 + 30*x + (8 - 4*x + 12*x^2)*Log[3])/Log[3],x]

[Out]

(-30*x + 15*x^2 + 8*x*Log[3] - 2*x^2*Log[3] + 4*x^3*Log[3])/Log[3]

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fricas [A]  time = 0.54, size = 32, normalized size = 1.19 \begin {gather*} \frac {15 \, x^{2} + 2 \, {\left (2 \, x^{3} - x^{2} + 4 \, x\right )} \log \relax (3) - 30 \, x}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2-4*x+8)*log(3)+30*x-30)/log(3),x, algorithm="fricas")

[Out]

(15*x^2 + 2*(2*x^3 - x^2 + 4*x)*log(3) - 30*x)/log(3)

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giac [A]  time = 0.20, size = 32, normalized size = 1.19 \begin {gather*} \frac {15 \, x^{2} + 2 \, {\left (2 \, x^{3} - x^{2} + 4 \, x\right )} \log \relax (3) - 30 \, x}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2-4*x+8)*log(3)+30*x-30)/log(3),x, algorithm="giac")

[Out]

(15*x^2 + 2*(2*x^3 - x^2 + 4*x)*log(3) - 30*x)/log(3)

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maple [A]  time = 0.07, size = 28, normalized size = 1.04

method	result	size
gosper	\(\frac {x \left (4 x^{2} \ln \relax (3)-2 x \ln \relax (3)+8 \ln \relax (3)+15 x -30\right )}{\ln \relax (3)}\)	\(28\)
risch	\(4 x^{3}-2 x^{2}+8 x +\frac {15 x^{2}}{\ln \relax (3)}-\frac {30 x}{\ln \relax (3)}\)	\(31\)
default	\(\frac {\ln \relax (3) \left (4 x^{3}-2 x^{2}+8 x \right )+15 x^{2}-30 x}{\ln \relax (3)}\)	\(32\)
norman	\(4 x^{3}-\frac {\left (2 \ln \relax (3)-15\right ) x^{2}}{\ln \relax (3)}+\frac {2 \left (4 \ln \relax (3)-15\right ) x}{\ln \relax (3)}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*x^2-4*x+8)*ln(3)+30*x-30)/ln(3),x,method=_RETURNVERBOSE)

[Out]

x*(4*x^2*ln(3)-2*x*ln(3)+8*ln(3)+15*x-30)/ln(3)

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maxima [A]  time = 0.50, size = 32, normalized size = 1.19 \begin {gather*} \frac {15 \, x^{2} + 2 \, {\left (2 \, x^{3} - x^{2} + 4 \, x\right )} \log \relax (3) - 30 \, x}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x^2-4*x+8)*log(3)+30*x-30)/log(3),x, algorithm="maxima")

[Out]

(15*x^2 + 2*(2*x^3 - x^2 + 4*x)*log(3) - 30*x)/log(3)

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mupad [B]  time = 0.17, size = 31, normalized size = 1.15 \begin {gather*} 4\,x^3-\frac {\left (\ln \left (81\right )-30\right )\,x^2}{2\,\ln \relax (3)}+\frac {\left (8\,\ln \relax (3)-30\right )\,x}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*x + log(3)*(12*x^2 - 4*x + 8) - 30)/log(3),x)

[Out]

4*x^3 + (x*(8*log(3) - 30))/log(3) - (x^2*(log(81) - 30))/(2*log(3))

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sympy [A]  time = 0.05, size = 29, normalized size = 1.07 \begin {gather*} 4 x^{3} + \frac {x^{2} \left (15 - 2 \log {\relax (3 )}\right )}{\log {\relax (3 )}} + \frac {x \left (-30 + 8 \log {\relax (3 )}\right )}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*x**2-4*x+8)*ln(3)+30*x-30)/ln(3),x)

[Out]

4*x**3 + x**2*(15 - 2*log(3))/log(3) + x*(-30 + 8*log(3))/log(3)

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