\(\int \frac {1}{3} (-3+2 x+150 x \log (x)+150 x \log ^2(x)) \, dx\) [6405]

Optimal result

Integrand size = 21, antiderivative size = 21 \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=7-x+\frac {x^2}{3}+25 x^2 \log ^2(x) \]

[Out]

25*x^2*ln(x)^2-x+7+1/3*x^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2341, 2342} \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=\frac {x^2}{3}+25 x^2 \log ^2(x)-x \]

[In]

Int[(-3 + 2*x + 150*x*Log[x] + 150*x*Log[x]^2)/3,x]

[Out]

-x + x^2/3 + 25*x^2*Log[x]^2

Rule 12

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

Rule 2341

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
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx \\ & = -x+\frac {x^2}{3}+50 \int x \log (x) \, dx+50 \int x \log ^2(x) \, dx \\ & = -x-\frac {73 x^2}{6}+25 x^2 \log (x)+25 x^2 \log ^2(x)-50 \int x \log (x) \, dx \\ & = -x+\frac {x^2}{3}+25 x^2 \log ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=-x+\frac {x^2}{3}+25 x^2 \log ^2(x) \]

[In]

Integrate[(-3 + 2*x + 150*x*Log[x] + 150*x*Log[x]^2)/3,x]

[Out]

-x + x^2/3 + 25*x^2*Log[x]^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {x^{2}}{3}-x +25 x^{2} \ln \left (x \right )^{2}\)	\(19\)
norman	\(\frac {x^{2}}{3}-x +25 x^{2} \ln \left (x \right )^{2}\)	\(19\)
risch	\(\frac {x^{2}}{3}-x +25 x^{2} \ln \left (x \right )^{2}\)	\(19\)
parallelrisch	\(\frac {x^{2}}{3}-x +25 x^{2} \ln \left (x \right )^{2}\)	\(19\)
parts	\(\frac {x^{2}}{3}-x +25 x^{2} \ln \left (x \right )^{2}\)	\(19\)

[In]

int(50*x*ln(x)^2+50*x*ln(x)+2/3*x-1,x,method=_RETURNVERBOSE)

[Out]

1/3*x^2-x+25*x^2*ln(x)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=25 \, x^{2} \log \left (x\right )^{2} + \frac {1}{3} \, x^{2} - x \]

[In]

integrate(50*x*log(x)^2+50*x*log(x)+2/3*x-1,x, algorithm="fricas")

[Out]

25*x^2*log(x)^2 + 1/3*x^2 - x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=25 x^{2} \log {\left (x \right )}^{2} + \frac {x^{2}}{3} - x \]

[In]

integrate(50*x*ln(x)**2+50*x*ln(x)+2/3*x-1,x)

[Out]

25*x**2*log(x)**2 + x**2/3 - x

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=\frac {25}{2} \, {\left (2 \, \log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )} x^{2} + 25 \, x^{2} \log \left (x\right ) - \frac {73}{6} \, x^{2} - x \]

[In]

integrate(50*x*log(x)^2+50*x*log(x)+2/3*x-1,x, algorithm="maxima")

[Out]

25/2*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 25*x^2*log(x) - 73/6*x^2 - x

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=25 \, x^{2} \log \left (x\right )^{2} + \frac {1}{3} \, x^{2} - x \]

[In]

integrate(50*x*log(x)^2+50*x*log(x)+2/3*x-1,x, algorithm="giac")

[Out]

25*x^2*log(x)^2 + 1/3*x^2 - x

Mupad [B] (verification not implemented)

Time = 12.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1}{3} \left (-3+2 x+150 x \log (x)+150 x \log ^2(x)\right ) \, dx=\frac {x\,\left (75\,x\,{\ln \left (x\right )}^2+x-3\right )}{3} \]

[In]

int((2*x)/3 + 50*x*log(x)^2 + 50*x*log(x) - 1,x)

[Out]

(x*(x + 75*x*log(x)^2 - 3))/3