\(\int \frac {-25-19 x+(-50 x+12 x^2) \log (\frac {15 x}{-25+6 x})}{-25+6 x} \, dx\) [5552]

Optimal result

Integrand size = 34, antiderivative size = 20 \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=8+x+x^2 \log \left (\frac {3 x}{-5+\frac {6 x}{5}}\right ) \]

[Out]

ln(3/(6/5*x-5)*x)*x^2+8+x

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6874, 45, 2548} \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=x^2 \log \left (-\frac {15 x}{25-6 x}\right )+x \]

[In]

Int[(-25 - 19*x + (-50*x + 12*x^2)*Log[(15*x)/(-25 + 6*x)])/(-25 + 6*x),x]

[Out]

x + x^2*Log[(-15*x)/(25 - 6*x)]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2548

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))*((f_.) + (g_.)*(x_))^(m_.
), x_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])/(g*(m + 1))), x] - Dist[B*n*(
(b*c - a*d)/(g*(m + 1))), Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, A
, B, m, n}, x] && EqQ[n + mn, 0] && NeQ[b*c - a*d, 0] && NeQ[m, -1] &&  !(EqQ[m, -2] && IntegerQ[n])

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-25-19 x}{-25+6 x}+2 x \log \left (\frac {15 x}{-25+6 x}\right )\right ) \, dx \\ & = 2 \int x \log \left (\frac {15 x}{-25+6 x}\right ) \, dx+\int \frac {-25-19 x}{-25+6 x} \, dx \\ & = x^2 \log \left (-\frac {15 x}{25-6 x}\right )+25 \int \frac {x}{-25+6 x} \, dx+\int \left (-\frac {19}{6}-\frac {625}{6 (-25+6 x)}\right ) \, dx \\ & = -\frac {19 x}{6}-\frac {625}{36} \log (25-6 x)+x^2 \log \left (-\frac {15 x}{25-6 x}\right )+25 \int \left (\frac {1}{6}+\frac {25}{6 (-25+6 x)}\right ) \, dx \\ & = x+x^2 \log \left (-\frac {15 x}{25-6 x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=x+x^2 \log \left (-\frac {15 x}{25-6 x}\right ) \]

[In]

Integrate[(-25 - 19*x + (-50*x + 12*x^2)*Log[(15*x)/(-25 + 6*x)])/(-25 + 6*x),x]

[Out]

x + x^2*Log[(-15*x)/(25 - 6*x)]

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
norman	\(x +\ln \left (\frac {15 x}{6 x -25}\right ) x^{2}\)	\(18\)
risch	\(x +\ln \left (\frac {15 x}{6 x -25}\right ) x^{2}\)	\(18\)
parallelrisch	\(\ln \left (\frac {15 x}{6 x -25}\right ) x^{2}+\frac {25}{3}+x\)	\(19\)
derivativedivides	\(x -\frac {25}{6}+\frac {5 \ln \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (6 x -25\right )}{9}-\frac {\ln \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (-\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (6 x -25\right )^{2}}{225}\)	\(77\)
default	\(x -\frac {25}{6}+\frac {5 \ln \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (6 x -25\right )}{9}-\frac {\ln \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (-\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (6 x -25\right )^{2}}{225}\)	\(77\)
parts	\(x -\frac {625 \ln \left (6 x -25\right )}{36}-\frac {625 \ln \left (\frac {125}{6 x -25}\right )}{36}+\frac {5 \ln \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (6 x -25\right )}{9}-\frac {625}{36}-\frac {\ln \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (-\frac {5}{2}+\frac {125}{2 \left (6 x -25\right )}\right ) \left (6 x -25\right )^{2}}{225}\)	\(97\)

[In]

int(((12*x^2-50*x)*ln(15*x/(6*x-25))-19*x-25)/(6*x-25),x,method=_RETURNVERBOSE)

[Out]

x+ln(15*x/(6*x-25))*x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=x^{2} \log \left (\frac {15 \, x}{6 \, x - 25}\right ) + x \]

[In]

integrate(((12*x^2-50*x)*log(15*x/(6*x-25))-19*x-25)/(6*x-25),x, algorithm="fricas")

[Out]

x^2*log(15*x/(6*x - 25)) + x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=x^{2} \log {\left (\frac {15 x}{6 x - 25} \right )} + x \]

[In]

integrate(((12*x**2-50*x)*ln(15*x/(6*x-25))-19*x-25)/(6*x-25),x)

[Out]

x**2*log(15*x/(6*x - 25)) + x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (18) = 36\).

Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.00 \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=x^{2} {\left (\log \left (5\right ) + \log \left (3\right )\right )} + x^{2} \log \left (x\right ) - \frac {1}{36} \, {\left (36 \, x^{2} - 625\right )} \log \left (6 \, x - 25\right ) + x - \frac {625}{36} \, \log \left (6 \, x - 25\right ) \]

[In]

integrate(((12*x^2-50*x)*log(15*x/(6*x-25))-19*x-25)/(6*x-25),x, algorithm="maxima")

[Out]

x^2*(log(5) + log(3)) + x^2*log(x) - 1/36*(36*x^2 - 625)*log(6*x - 25) + x - 625/36*log(6*x - 25)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=x^{2} \log \left (\frac {15 \, x}{6 \, x - 25}\right ) + x \]

[In]

integrate(((12*x^2-50*x)*log(15*x/(6*x-25))-19*x-25)/(6*x-25),x, algorithm="giac")

[Out]

x^2*log(15*x/(6*x - 25)) + x

Mupad [B] (verification not implemented)

Time = 10.98 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-25-19 x+\left (-50 x+12 x^2\right ) \log \left (\frac {15 x}{-25+6 x}\right )}{-25+6 x} \, dx=x+x^2\,\ln \left (\frac {15\,x}{6\,x-25}\right ) \]

[In]

int(-(19*x + log((15*x)/(6*x - 25))*(50*x - 12*x^2) + 25)/(6*x - 25),x)

[Out]

x + x^2*log((15*x)/(6*x - 25))