\(\int \frac {1}{144} (225+1000 \log (x)+400 \log ^2(x)) \, dx\) [1243]

Optimal result

Integrand size = 16, antiderivative size = 15 \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25}{81} x \left (\frac {3}{4}+3 \log (x)\right )^2 \]

[Out]

25/81*x*(3*ln(x)+3/4)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 2332, 2333} \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25 x}{144}+\frac {25}{9} x \log ^2(x)+\frac {25}{18} x \log (x) \]

[In]

Int[(225 + 1000*Log[x] + 400*Log[x]^2)/144,x]

[Out]

(25*x)/144 + (25*x*Log[x])/18 + (25*x*Log[x]^2)/9

Rule 12

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

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{144} \int \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx \\ & = \frac {25 x}{16}+\frac {25}{9} \int \log ^2(x) \, dx+\frac {125}{18} \int \log (x) \, dx \\ & = -\frac {775 x}{144}+\frac {125}{18} x \log (x)+\frac {25}{9} x \log ^2(x)-\frac {50}{9} \int \log (x) \, dx \\ & = \frac {25 x}{144}+\frac {25}{18} x \log (x)+\frac {25}{9} x \log ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25 x}{144}+\frac {25}{18} x \log (x)+\frac {25}{9} x \log ^2(x) \]

[In]

Integrate[(225 + 1000*Log[x] + 400*Log[x]^2)/144,x]

[Out]

(25*x)/144 + (25*x*Log[x])/18 + (25*x*Log[x]^2)/9

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {25 x}{144}+\frac {25 x \ln \left (x \right )^{2}}{9}+\frac {25 x \ln \left (x \right )}{18}\)	\(17\)
norman	\(\frac {25 x}{144}+\frac {25 x \ln \left (x \right )^{2}}{9}+\frac {25 x \ln \left (x \right )}{18}\)	\(17\)
risch	\(\frac {25 x}{144}+\frac {25 x \ln \left (x \right )^{2}}{9}+\frac {25 x \ln \left (x \right )}{18}\)	\(17\)
parallelrisch	\(\frac {25 x}{144}+\frac {25 x \ln \left (x \right )^{2}}{9}+\frac {25 x \ln \left (x \right )}{18}\)	\(17\)
parts	\(\frac {25 x}{144}+\frac {25 x \ln \left (x \right )^{2}}{9}+\frac {25 x \ln \left (x \right )}{18}\)	\(17\)

[In]

int(25/9*ln(x)^2+125/18*ln(x)+25/16,x,method=_RETURNVERBOSE)

[Out]

25/144*x+25/9*x*ln(x)^2+25/18*x*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25}{9} \, x \log \left (x\right )^{2} + \frac {25}{18} \, x \log \left (x\right ) + \frac {25}{144} \, x \]

[In]

integrate(25/9*log(x)^2+125/18*log(x)+25/16,x, algorithm="fricas")

[Out]

25/9*x*log(x)^2 + 25/18*x*log(x) + 25/144*x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25 x \log {\left (x \right )}^{2}}{9} + \frac {25 x \log {\left (x \right )}}{18} + \frac {25 x}{144} \]

[In]

integrate(25/9*ln(x)**2+125/18*ln(x)+25/16,x)

[Out]

25*x*log(x)**2/9 + 25*x*log(x)/18 + 25*x/144

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25}{9} \, {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x + \frac {125}{18} \, x \log \left (x\right ) - \frac {775}{144} \, x \]

[In]

integrate(25/9*log(x)^2+125/18*log(x)+25/16,x, algorithm="maxima")

[Out]

25/9*(log(x)^2 - 2*log(x) + 2)*x + 125/18*x*log(x) - 775/144*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25}{9} \, x \log \left (x\right )^{2} + \frac {25}{18} \, x \log \left (x\right ) + \frac {25}{144} \, x \]

[In]

integrate(25/9*log(x)^2+125/18*log(x)+25/16,x, algorithm="giac")

[Out]

25/9*x*log(x)^2 + 25/18*x*log(x) + 25/144*x

Mupad [B] (verification not implemented)

Time = 7.98 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {1}{144} \left (225+1000 \log (x)+400 \log ^2(x)\right ) \, dx=\frac {25\,x\,{\left (4\,\ln \left (x\right )+1\right )}^2}{144} \]

[In]

int((125*log(x))/18 + (25*log(x)^2)/9 + 25/16,x)

[Out]

(25*x*(4*log(x) + 1)^2)/144