3.95.63 \(\int \frac {22500 x-7800 x^2+100 x^3+(22500 x-7800 x^2+100 x^3) \log (2)+(-22500+37500 x-15100 x^2+100 x^3+(-22500+37500 x-15100 x^2+100 x^3) \log (2)) \log (-1+x)}{-5625+5775 x-151 x^2+x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac {4 (3-x) x (1+\log (2)) \log (-1+x)}{3-\frac {x}{25}} \]

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Rubi [A]  time = 0.33, antiderivative size = 45, normalized size of antiderivative = 1.80, number of steps used = 14, number of rules used = 10, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6688, 12, 6742, 142, 2418, 2389, 2295, 2395, 36, 31} \begin {gather*} 7300 (1+\log (2)) \log (1-x)-100 (1-x) (1+\log (2)) \log (x-1)-\frac {540000 (1+\log (2)) \log (x-1)}{75-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(22500*x - 7800*x^2 + 100*x^3 + (22500*x - 7800*x^2 + 100*x^3)*Log[2] + (-22500 + 37500*x - 15100*x^2 + 10
0*x^3 + (-22500 + 37500*x - 15100*x^2 + 100*x^3)*Log[2])*Log[-1 + x])/(-5625 + 5775*x - 151*x^2 + x^3),x]

[Out]

7300*(1 + Log[2])*Log[1 - x] - 100*(1 - x)*(1 + Log[2])*Log[-1 + x] - (540000*(1 + Log[2])*Log[-1 + x])/(75 -
x)

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 142

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h},
x] && (IGtQ[m, 0] || IntegersQ[m, n])

Rule 2295

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

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6688

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

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 \frac {100 (1+\log (2)) \left (-x \left (225-78 x+x^2\right )-\left (-225+375 x-151 x^2+x^3\right ) \log (-1+x)\right )}{(1-x) (75-x)^2} \, dx\\ &=(100 (1+\log (2))) \int \frac {-x \left (225-78 x+x^2\right )-\left (-225+375 x-151 x^2+x^3\right ) \log (-1+x)}{(1-x) (75-x)^2} \, dx\\ &=(100 (1+\log (2))) \int \left (\frac {(-3+x) x}{(-75+x) (-1+x)}+\frac {\left (225-150 x+x^2\right ) \log (-1+x)}{(-75+x)^2}\right ) \, dx\\ &=(100 (1+\log (2))) \int \frac {(-3+x) x}{(-75+x) (-1+x)} \, dx+(100 (1+\log (2))) \int \frac {\left (225-150 x+x^2\right ) \log (-1+x)}{(-75+x)^2} \, dx\\ &=(100 (1+\log (2))) \int \left (1+\frac {2700}{37 (-75+x)}+\frac {1}{37 (-1+x)}\right ) \, dx+(100 (1+\log (2))) \int \left (\log (-1+x)-\frac {5400 \log (-1+x)}{(-75+x)^2}\right ) \, dx\\ &=100 x (1+\log (2))+\frac {100}{37} (1+\log (2)) \log (1-x)+\frac {270000}{37} (1+\log (2)) \log (75-x)+(100 (1+\log (2))) \int \log (-1+x) \, dx-(540000 (1+\log (2))) \int \frac {\log (-1+x)}{(-75+x)^2} \, dx\\ &=100 x (1+\log (2))+\frac {100}{37} (1+\log (2)) \log (1-x)+\frac {270000}{37} (1+\log (2)) \log (75-x)-\frac {540000 (1+\log (2)) \log (-1+x)}{75-x}+(100 (1+\log (2))) \operatorname {Subst}(\int \log (x) \, dx,x,-1+x)-(540000 (1+\log (2))) \int \frac {1}{(-75+x) (-1+x)} \, dx\\ &=\frac {100}{37} (1+\log (2)) \log (1-x)+\frac {270000}{37} (1+\log (2)) \log (75-x)-100 (1-x) (1+\log (2)) \log (-1+x)-\frac {540000 (1+\log (2)) \log (-1+x)}{75-x}-\frac {1}{37} (270000 (1+\log (2))) \int \frac {1}{-75+x} \, dx+\frac {1}{37} (270000 (1+\log (2))) \int \frac {1}{-1+x} \, dx\\ &=7300 (1+\log (2)) \log (1-x)-100 (1-x) (1+\log (2)) \log (-1+x)-\frac {540000 (1+\log (2)) \log (-1+x)}{75-x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 53, normalized size = 2.12 \begin {gather*} \frac {100}{37} (1+\log (2)) \left (5400 \tanh ^{-1}\left (\frac {1}{37} (-38+x)\right )+\log (1-x)+2700 \log (75-x)+\frac {199800 \log (-1+x)}{-75+x}+37 (-1+x) \log (-1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(22500*x - 7800*x^2 + 100*x^3 + (22500*x - 7800*x^2 + 100*x^3)*Log[2] + (-22500 + 37500*x - 15100*x^
2 + 100*x^3 + (-22500 + 37500*x - 15100*x^2 + 100*x^3)*Log[2])*Log[-1 + x])/(-5625 + 5775*x - 151*x^2 + x^3),x
]

[Out]

(100*(1 + Log[2])*(5400*ArcTanh[(-38 + x)/37] + Log[1 - x] + 2700*Log[75 - x] + (199800*Log[-1 + x])/(-75 + x)
+ 37*(-1 + x)*Log[-1 + x]))/37

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fricas [A]  time = 0.96, size = 28, normalized size = 1.12 \begin {gather*} \frac {100 \, {\left (x^{2} + {\left (x^{2} - 3 \, x\right )} \log \relax (2) - 3 \, x\right )} \log \left (x - 1\right )}{x - 75} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x^3-15100*x^2+37500*x-22500)*log(2)+100*x^3-15100*x^2+37500*x-22500)*log(x-1)+(100*x^3-7800*x
^2+22500*x)*log(2)+100*x^3-7800*x^2+22500*x)/(x^3-151*x^2+5775*x-5625),x, algorithm="fricas")

[Out]

100*(x^2 + (x^2 - 3*x)*log(2) - 3*x)*log(x - 1)/(x - 75)

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giac [A]  time = 0.13, size = 35, normalized size = 1.40 \begin {gather*} 100 \, {\left (x {\left (\log \relax (2) + 1\right )} + \frac {5400 \, {\left (\log \relax (2) + 1\right )}}{x - 75}\right )} \log \left (x - 1\right ) + 7200 \, {\left (\log \relax (2) + 1\right )} \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x^3-15100*x^2+37500*x-22500)*log(2)+100*x^3-15100*x^2+37500*x-22500)*log(x-1)+(100*x^3-7800*x
^2+22500*x)*log(2)+100*x^3-7800*x^2+22500*x)/(x^3-151*x^2+5775*x-5625),x, algorithm="giac")

[Out]

100*(x*(log(2) + 1) + 5400*(log(2) + 1)/(x - 75))*log(x - 1) + 7200*(log(2) + 1)*log(x - 1)

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maple [A]  time = 0.10, size = 34, normalized size = 1.36

method	result	size
norman	\(\frac {\left (100+100 \ln \relax (2)\right ) x^{2} \ln \left (x -1\right )+\left (-300-300 \ln \relax (2)\right ) x \ln \left (x -1\right )}{x -75}\)	\(34\)
risch	\(\frac {100 \left (x^{2} \ln \relax (2)-75 x \ln \relax (2)+x^{2}+5400 \ln \relax (2)-75 x +5400\right ) \ln \left (x -1\right )}{x -75}+7200 \ln \relax (2) \ln \left (x -1\right )+7200 \ln \left (x -1\right )\)	\(50\)
derivativedivides	\(100 \ln \relax (2) \ln \left (x -1\right ) \left (x -1\right )+\frac {270000 \ln \relax (2) \ln \left (x -1\right ) \left (x -1\right )}{37 \left (x -75\right )}+\frac {100 \ln \relax (2) \ln \left (x -1\right )}{37}+100 \left (x -1\right ) \ln \left (x -1\right )+\frac {270000 \ln \left (x -1\right ) \left (x -1\right )}{37 \left (x -75\right )}+\frac {100 \ln \left (x -1\right )}{37}\)	\(66\)
default	\(100 \ln \relax (2) \ln \left (x -1\right ) \left (x -1\right )+\frac {270000 \ln \relax (2) \ln \left (x -1\right ) \left (x -1\right )}{37 \left (x -75\right )}+\frac {100 \ln \relax (2) \ln \left (x -1\right )}{37}+100 \left (x -1\right ) \ln \left (x -1\right )+\frac {270000 \ln \left (x -1\right ) \left (x -1\right )}{37 \left (x -75\right )}+\frac {100 \ln \left (x -1\right )}{37}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((100*x^3-15100*x^2+37500*x-22500)*ln(2)+100*x^3-15100*x^2+37500*x-22500)*ln(x-1)+(100*x^3-7800*x^2+22500
*x)*ln(2)+100*x^3-7800*x^2+22500*x)/(x^3-151*x^2+5775*x-5625),x,method=_RETURNVERBOSE)

[Out]

((100+100*ln(2))*x^2*ln(x-1)+(-300-300*ln(2))*x*ln(x-1))/(x-75)

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maxima [A]  time = 0.48, size = 27, normalized size = 1.08 \begin {gather*} \frac {100 \, {\left (x^{2} {\left (\log \relax (2) + 1\right )} - 3 \, x {\left (\log \relax (2) + 1\right )}\right )} \log \left (x - 1\right )}{x - 75} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x^3-15100*x^2+37500*x-22500)*log(2)+100*x^3-15100*x^2+37500*x-22500)*log(x-1)+(100*x^3-7800*x
^2+22500*x)*log(2)+100*x^3-7800*x^2+22500*x)/(x^3-151*x^2+5775*x-5625),x, algorithm="maxima")

[Out]

100*(x^2*(log(2) + 1) - 3*x*(log(2) + 1))*log(x - 1)/(x - 75)

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mupad [B]  time = 0.45, size = 19, normalized size = 0.76 \begin {gather*} \frac {100\,x\,\ln \left (x-1\right )\,\left (\ln \relax (2)+1\right )\,\left (x-3\right )}{x-75} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((22500*x + log(2)*(22500*x - 7800*x^2 + 100*x^3) + log(x - 1)*(37500*x + log(2)*(37500*x - 15100*x^2 + 100
*x^3 - 22500) - 15100*x^2 + 100*x^3 - 22500) - 7800*x^2 + 100*x^3)/(5775*x - 151*x^2 + x^3 - 5625),x)

[Out]

(100*x*log(x - 1)*(log(2) + 1)*(x - 3))/(x - 75)

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sympy [B]  time = 0.23, size = 49, normalized size = 1.96 \begin {gather*} \left (7200 \log {\relax (2 )} + 7200\right ) \log {\left (x - 1 \right )} + \frac {\left (100 x^{2} \log {\relax (2 )} + 100 x^{2} - 7500 x - 7500 x \log {\relax (2 )} + 540000 \log {\relax (2 )} + 540000\right ) \log {\left (x - 1 \right )}}{x - 75} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x**3-15100*x**2+37500*x-22500)*ln(2)+100*x**3-15100*x**2+37500*x-22500)*ln(x-1)+(100*x**3-780
0*x**2+22500*x)*ln(2)+100*x**3-7800*x**2+22500*x)/(x**3-151*x**2+5775*x-5625),x)

[Out]

(7200*log(2) + 7200)*log(x - 1) + (100*x**2*log(2) + 100*x**2 - 7500*x - 7500*x*log(2) + 540000*log(2) + 54000
0)*log(x - 1)/(x - 75)

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