3.16.27 \(\int \frac {1}{25} (-180+81 x+162 x \log (x)) \, dx\)

Optimal. Leaf size=16 \[ 9 \left (3+x+\frac {9}{25} x (-5+x \log (x))\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2304} \begin {gather*} \frac {81}{25} x^2 \log (x)-\frac {36 x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-180 + 81*x + 162*x*Log[x])/25,x]

[Out]

(-36*x)/5 + (81*x^2*Log[x])/25

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}{25} \int (-180+81 x+162 x \log (x)) \, dx\\ &=-\frac {36 x}{5}+\frac {81 x^2}{50}+\frac {162}{25} \int x \log (x) \, dx\\ &=-\frac {36 x}{5}+\frac {81}{25} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.94 \begin {gather*} -\frac {36 x}{5}+\frac {81}{25} x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-180 + 81*x + 162*x*Log[x])/25,x]

[Out]

(-36*x)/5 + (81*x^2*Log[x])/25

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fricas [A]  time = 0.83, size = 11, normalized size = 0.69 \begin {gather*} \frac {81}{25} \, x^{2} \log \relax (x) - \frac {36}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(162/25*x*log(x)+81/25*x-36/5,x, algorithm="fricas")

[Out]

81/25*x^2*log(x) - 36/5*x

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giac [A]  time = 0.34, size = 11, normalized size = 0.69 \begin {gather*} \frac {81}{25} \, x^{2} \log \relax (x) - \frac {36}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(162/25*x*log(x)+81/25*x-36/5,x, algorithm="giac")

[Out]

81/25*x^2*log(x) - 36/5*x

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maple [A]  time = 0.02, size = 12, normalized size = 0.75

method	result	size
default	\(-\frac {36 x}{5}+\frac {81 x^{2} \ln \relax (x )}{25}\)	\(12\)
norman	\(-\frac {36 x}{5}+\frac {81 x^{2} \ln \relax (x )}{25}\)	\(12\)
risch	\(-\frac {36 x}{5}+\frac {81 x^{2} \ln \relax (x )}{25}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(162/25*x*ln(x)+81/25*x-36/5,x,method=_RETURNVERBOSE)

[Out]

-36/5*x+81/25*x^2*ln(x)

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maxima [A]  time = 0.46, size = 11, normalized size = 0.69 \begin {gather*} \frac {81}{25} \, x^{2} \log \relax (x) - \frac {36}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(162/25*x*log(x)+81/25*x-36/5,x, algorithm="maxima")

[Out]

81/25*x^2*log(x) - 36/5*x

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mupad [B]  time = 1.00, size = 10, normalized size = 0.62 \begin {gather*} \frac {9\,x\,\left (9\,x\,\ln \relax (x)-20\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((81*x)/25 + (162*x*log(x))/25 - 36/5,x)

[Out]

(9*x*(9*x*log(x) - 20))/25

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sympy [A]  time = 0.09, size = 14, normalized size = 0.88 \begin {gather*} \frac {81 x^{2} \log {\relax (x )}}{25} - \frac {36 x}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(162/25*x*ln(x)+81/25*x-36/5,x)

[Out]

81*x**2*log(x)/25 - 36*x/5

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