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

Optimal. Leaf size=20 \[ 8+x+x^2 \log \left (\frac {3 x}{-5+\frac {6 x}{5}}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6742, 43, 2492} \begin {gather*} x^2 \log \left (-\frac {15 x}{25-6 x}\right )+x \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2492

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*((g_.) + (h_.)*(x_))^
(m_.), x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(h*(m + 1)), x] - Dist[(p*
r*s*(b*c - a*d))/(h*(m + 1)), Int[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a + b*x)*
(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0]
&& IGtQ[s, 0] && NeQ[m, -1]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 0.85 \begin {gather*} x+x^2 \log \left (-\frac {15 x}{25-6 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[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)]

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fricas [A]  time = 0.68, size = 17, normalized size = 0.85 \begin {gather*} x^{2} \log \left (\frac {15 \, x}{6 \, x - 25}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.17, size = 17, normalized size = 0.85 \begin {gather*} x^{2} \log \left (\frac {15 \, x}{6 \, x - 25}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.24, size = 18, normalized size = 0.90

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [B]  time = 0.51, size = 40, normalized size = 2.00 \begin {gather*} x^{2} {\left (\log \relax (5) + \log \relax (3)\right )} + x^{2} \log \relax (x) - \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.54, size = 17, normalized size = 0.85 \begin {gather*} x+x^2\,\ln \left (\frac {15\,x}{6\,x-25}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.17, size = 14, normalized size = 0.70 \begin {gather*} x^{2} \log {\left (\frac {15 x}{6 x - 25} \right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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