3.85.41 $\int \frac{15-45x+45x^2-15x^3+(600x+210x^2-162x^3-6x^4+6x^5+e^{2x}(96x-216x^2+48x^3+192x^4-144x^5+24x^6)){\log }^2(-4+x)}{(4-13x+15x^2-7x^3+x^4){\log }^2(-4+x)}dx$

Optimal. Leaf size=35 $$3(4e^{2x}{x}^2+\frac{x^2(5+x{)}^2}{(-1+x{)}^2}+\frac{5}{\log (-4+x)})$$

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Rubi [A]  time = 0.71, antiderivative size = 57, normalized size of antiderivative = 1.63, number of steps used = 21, number of rules used = 10, integrand size = 108, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.093, Rules used = {6688, 6742, 37, 43, 2196, 2176, 2194, 2390, 2302, 30} $$\begin{array}{c}12e^{2x}{x}^2+\frac{75x^2}{(1-x{)}^2}+3x^2+36x-\frac{102}{1-x}+\frac{33}{(1-x{)}^2}+\frac{15}{\log (x-4)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(15 - 45*x + 45*x^2 - 15*x^3 + (600*x + 210*x^2 - 162*x^3 - 6*x^4 + 6*x^5 + E^(2*x)*(96*x - 216*x^2 + 48*x
^3 + 192*x^4 - 144*x^5 + 24*x^6))*Log[-4 + x]^2)/((4 - 13*x + 15*x^2 - 7*x^3 + x^4)*Log[-4 + x]^2),x]

[Out]

33/(1 - x)^2 - 102/(1 - x) + 36*x + 3*x^2 + 12*E^(2*x)*x^2 + (75*x^2)/(1 - x)^2 + 15/Log[-4 + x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2302

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

Rule 2390

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

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{6x(-25-15x+3x^2+x^3+4e^{2x}(-1+x{)}^3(1+x))}{(-1+x{)}^3}-\frac{15}{(-4+x){\log }^2(-4+x)})dx\\ & =6\int \frac{x(-25-15x+3x^2+x^3+4e^{2x}(-1+x{)}^3(1+x))}{(-1+x{)}^3}dx-15\int \frac{1}{(-4+x){\log }^2(-4+x)}dx\\ & =6\int (-\frac{25x}{(-1+x{)}^3}-\frac{15x^2}{(-1+x{)}^3}+\frac{3x^3}{(-1+x{)}^3}+\frac{x^4}{(-1+x{)}^3}+4e^{2x}x(1+x))dx-15\mathrm{Subst}(\int \frac{1}{x{\log }^2(x)}dx,x,-4+x)\\ & =6\int \frac{x^4}{(-1+x{)}^3}dx-15\mathrm{Subst}(\int \frac{1}{x^2}dx,x,\log (-4+x))+18\int \frac{x^3}{(-1+x{)}^3}dx+24\int e^{2x}x(1+x)dx-90\int \frac{x^2}{(-1+x{)}^3}dx-150\int \frac{x}{(-1+x{)}^3}dx\\ & =\frac{75x^2}{(1-x{)}^2}+\frac{15}{\log (-4+x)}+6\int (3+\frac{1}{(-1+x{)}^3}+\frac{4}{(-1+x{)}^2}+\frac{6}{-1+x}+x)dx+18\int (1+\frac{1}{(-1+x{)}^3}+\frac{3}{(-1+x{)}^2}+\frac{3}{-1+x})dx+24\int (e^{2x}x+e^{2x}{x}^2)dx-90\int (\frac{1}{(-1+x{)}^3}+\frac{2}{(-1+x{)}^2}+\frac{1}{-1+x})dx\\ & =\frac{33}{(1-x{)}^2}-\frac{102}{1-x}+36x+3x^2+\frac{75x^2}{(1-x{)}^2}+\frac{15}{\log (-4+x)}+24\int e^{2x}xdx+24\int e^{2x}{x}^{2}dx\\ & =\frac{33}{(1-x{)}^2}-\frac{102}{1-x}+36x+12e^{2x}x+3x^2+12e^{2x}{x}^2+\frac{75x^2}{(1-x{)}^2}+\frac{15}{\log (-4+x)}-12\int e^{2x}dx-24\int e^{2x}xdx\\ & =-6e^{2x}+\frac{33}{(1-x{)}^2}-\frac{102}{1-x}+36x+3x^2+12e^{2x}{x}^2+\frac{75x^2}{(1-x{)}^2}+\frac{15}{\log (-4+x)}+12\int e^{2x}dx\\ & =\frac{33}{(1-x{)}^2}-\frac{102}{1-x}+36x+3x^2+12e^{2x}{x}^2+\frac{75x^2}{(1-x{)}^2}+\frac{15}{\log (-4+x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.18, size = 57, normalized size = 1.63 $$\begin{array}{c}\frac{33}{(1-x{)}^2}-\frac{102}{1-x}+36x+3x^2+12e^{2x}{x}^2+\frac{75x^2}{(1-x{)}^2}+\frac{15}{\log (-4+x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(15 - 45*x + 45*x^2 - 15*x^3 + (600*x + 210*x^2 - 162*x^3 - 6*x^4 + 6*x^5 + E^(2*x)*(96*x - 216*x^2
+ 48*x^3 + 192*x^4 - 144*x^5 + 24*x^6))*Log[-4 + x]^2)/((4 - 13*x + 15*x^2 - 7*x^3 + x^4)*Log[-4 + x]^2),x]

[Out]

33/(1 - x)^2 - 102/(1 - x) + 36*x + 3*x^2 + 12*E^(2*x)*x^2 + (75*x^2)/(1 - x)^2 + 15/Log[-4 + x]

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fricas [B]  time = 0.64, size = 69, normalized size = 1.97 $$\begin{array}{c}\frac{3(5x^2+(x^4+10x^3-23x^2+4(x^4-2x^3+x^2)e^{\left(2x\right)}+96x-48)\log (x-4)-10x+5)}{(x^2-2x+1)\log (x-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*log(x-4)^
2-15*x^3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/log(x-4)^2,x, algorithm="fricas")

[Out]

3*(5*x^2 + (x^4 + 10*x^3 - 23*x^2 + 4*(x^4 - 2*x^3 + x^2)*e^(2*x) + 96*x - 48)*log(x - 4) - 10*x + 5)/((x^2 -
2*x + 1)*log(x - 4))

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giac [B]  time = 0.26, size = 112, normalized size = 3.20 $$\begin{array}{c}\frac{3(4x^4{e}^{\left(2x\right)}\log (x-4)+x^4\log (x-4)-8x^3{e}^{\left(2x\right)}\log (x-4)+10x^3\log (x-4)+4x^2{e}^{\left(2x\right)}\log (x-4)-23x^2\log (x-4)+5x^2+96x\log (x-4)-10x-48\log (x-4)+5)}{x^2\log (x-4)-2x\log (x-4)+\log (x-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*log(x-4)^
2-15*x^3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/log(x-4)^2,x, algorithm="giac")

[Out]

3*(4*x^4*e^(2*x)*log(x - 4) + x^4*log(x - 4) - 8*x^3*e^(2*x)*log(x - 4) + 10*x^3*log(x - 4) + 4*x^2*e^(2*x)*lo
g(x - 4) - 23*x^2*log(x - 4) + 5*x^2 + 96*x*log(x - 4) - 10*x - 48*log(x - 4) + 5)/(x^2*log(x - 4) - 2*x*log(x
- 4) + log(x - 4))

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maple [A]  time = 0.04, size = 41, normalized size = 1.17

method	result	size
default	$\frac{15}{\ln (x-4)}+12{\mathrm{e}}^{2x}{x}^2+3x^2+36x+\frac{108}{{(x-1)}^2}+\frac{252}{x-1}$	$41$
risch	$\frac{12{\mathrm{e}}^{2x}{x}^4-24{\mathrm{e}}^{2x}{x}^3+3x^4+12{\mathrm{e}}^{2x}{x}^2+30x^3-69x^2+288x-144}{x^2-2x+1}+\frac{15}{\ln (x-4)}$	$67$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*ln(x-4)^2-15*x^
3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/ln(x-4)^2,x,method=_RETURNVERBOSE)

[Out]

15/ln(x-4)+12*exp(x)^2*x^2+3*x^2+36*x+108/(x-1)^2+252/(x-1)

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maxima [B]  time = 0.46, size = 69, normalized size = 1.97 $$\begin{array}{c}\frac{3(5x^2+(x^4+10x^3-23x^2+4(x^4-2x^3+x^2)e^{\left(2x\right)}+96x-48)\log (x-4)-10x+5)}{(x^2-2x+1)\log (x-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x^6-144*x^5+192*x^4+48*x^3-216*x^2+96*x)*exp(x)^2+6*x^5-6*x^4-162*x^3+210*x^2+600*x)*log(x-4)^
2-15*x^3+45*x^2-45*x+15)/(x^4-7*x^3+15*x^2-13*x+4)/log(x-4)^2,x, algorithm="maxima")

[Out]

3*(5*x^2 + (x^4 + 10*x^3 - 23*x^2 + 4*(x^4 - 2*x^3 + x^2)*e^(2*x) + 96*x - 48)*log(x - 4) - 10*x + 5)/((x^2 -
2*x + 1)*log(x - 4))

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mupad [B]  time = 0.22, size = 42, normalized size = 1.20 $$\begin{array}{c}36x+\frac{252x-144}{x^2-2x+1}+12x^2{\mathrm{e}}^{2x}+\frac{15}{\ln (x-4)}+3x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*x^2 - 45*x - 15*x^3 + log(x - 4)^2*(600*x + 210*x^2 - 162*x^3 - 6*x^4 + 6*x^5 + exp(2*x)*(96*x - 216*x
^2 + 48*x^3 + 192*x^4 - 144*x^5 + 24*x^6)) + 15)/(log(x - 4)^2*(15*x^2 - 13*x - 7*x^3 + x^4 + 4)),x)

[Out]

36*x + (252*x - 144)/(x^2 - 2*x + 1) + 12*x^2*exp(2*x) + 15/log(x - 4) + 3*x^2

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sympy [A]  time = 0.47, size = 37, normalized size = 1.06 $$\begin{array}{c}12x^2{e}^{2x}+3x^2+36x+\frac{252x-144}{x^2-2x+1}+\frac{15}{\log (x-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((24*x**6-144*x**5+192*x**4+48*x**3-216*x**2+96*x)*exp(x)**2+6*x**5-6*x**4-162*x**3+210*x**2+600*x)
*ln(x-4)**2-15*x**3+45*x**2-45*x+15)/(x**4-7*x**3+15*x**2-13*x+4)/ln(x-4)**2,x)

[Out]

12*x**2*exp(2*x) + 3*x**2 + 36*x + (252*x - 144)/(x**2 - 2*x + 1) + 15/log(x - 4)

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