3.27.61 \(\int \frac {1}{5} (20+120 x-42 x^2-45 x^2 \log (x)) \, dx\)

Optimal. Leaf size=29 \[ 4+e^2-x-x \left (-5+3 x^2 \left (\frac {3}{5}-\frac {4}{x}+\log (x)\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2304} \begin {gather*} -\frac {9 x^3}{5}-3 x^3 \log (x)+12 x^2+4 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + 120*x - 42*x^2 - 45*x^2*Log[x])/5,x]

[Out]

4*x + 12*x^2 - (9*x^3)/5 - 3*x^3*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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (20+120 x-42 x^2-45 x^2 \log (x)\right ) \, dx\\ &=4 x+12 x^2-\frac {14 x^3}{5}-9 \int x^2 \log (x) \, dx\\ &=4 x+12 x^2-\frac {9 x^3}{5}-3 x^3 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 0.79 \begin {gather*} 4 x+12 x^2-\frac {9 x^3}{5}-3 x^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + 120*x - 42*x^2 - 45*x^2*Log[x])/5,x]

[Out]

4*x + 12*x^2 - (9*x^3)/5 - 3*x^3*Log[x]

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fricas [A]  time = 0.58, size = 21, normalized size = 0.72 \begin {gather*} -3 \, x^{3} \log \relax (x) - \frac {9}{5} \, x^{3} + 12 \, x^{2} + 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*x^2*log(x)-42/5*x^2+24*x+4,x, algorithm="fricas")

[Out]

-3*x^3*log(x) - 9/5*x^3 + 12*x^2 + 4*x

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giac [A]  time = 0.51, size = 21, normalized size = 0.72 \begin {gather*} -3 \, x^{3} \log \relax (x) - \frac {9}{5} \, x^{3} + 12 \, x^{2} + 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*x^2*log(x)-42/5*x^2+24*x+4,x, algorithm="giac")

[Out]

-3*x^3*log(x) - 9/5*x^3 + 12*x^2 + 4*x

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maple [A]  time = 0.01, size = 22, normalized size = 0.76

method	result	size
default	\(12 x^{2}+4 x -\frac {9 x^{3}}{5}-3 x^{3} \ln \relax (x )\)	\(22\)
norman	\(12 x^{2}+4 x -\frac {9 x^{3}}{5}-3 x^{3} \ln \relax (x )\)	\(22\)
risch	\(12 x^{2}+4 x -\frac {9 x^{3}}{5}-3 x^{3} \ln \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-9*x^2*ln(x)-42/5*x^2+24*x+4,x,method=_RETURNVERBOSE)

[Out]

12*x^2+4*x-9/5*x^3-3*x^3*ln(x)

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maxima [A]  time = 0.42, size = 21, normalized size = 0.72 \begin {gather*} -3 \, x^{3} \log \relax (x) - \frac {9}{5} \, x^{3} + 12 \, x^{2} + 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*x^2*log(x)-42/5*x^2+24*x+4,x, algorithm="maxima")

[Out]

-3*x^3*log(x) - 9/5*x^3 + 12*x^2 + 4*x

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mupad [B]  time = 1.48, size = 21, normalized size = 0.72 \begin {gather*} 4\,x-3\,x^3\,\ln \relax (x)+12\,x^2-\frac {9\,x^3}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(24*x - 9*x^2*log(x) - (42*x^2)/5 + 4,x)

[Out]

4*x - 3*x^3*log(x) + 12*x^2 - (9*x^3)/5

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sympy [A]  time = 0.09, size = 22, normalized size = 0.76 \begin {gather*} - 3 x^{3} \log {\relax (x )} - \frac {9 x^{3}}{5} + 12 x^{2} + 4 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*x**2*ln(x)-42/5*x**2+24*x+4,x)

[Out]

-3*x**3*log(x) - 9*x**3/5 + 12*x**2 + 4*x

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