3.85.6 $\int \frac{9x^2+3x^3+(-3x^2+3x^3)\log (-1+x)+e^{e^{x}x}(-25x+(-25+25x+e^x(25x-25x^3))\log (-1+x))}{-3x^2+3x^3}dx$

Optimal. Leaf size=22 $$(3-\frac{25e^{e^{x}x}}{3x}+x)\log (-1+x)$$

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Rubi [B]  time = 1.34, antiderivative size = 74, normalized size of antiderivative = 3.36, number of steps used = 10, number of rules used = 6, integrand size = 76, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.079, Rules used = {1593, 6742, 43, 2389, 2295, 2288} $$\begin{array}{c}-\frac{25e^{e^{x}x}(e^{x}x\log (x-1)-e^x{x}^3\log (x-1))}{3(1-x)x^2(e^{x}x+e^x)}+4\log (1-x)-(1-x)\log (x-1)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(9*x^2 + 3*x^3 + (-3*x^2 + 3*x^3)*Log[-1 + x] + E^(E^x*x)*(-25*x + (-25 + 25*x + E^x*(25*x - 25*x^3))*Log[
-1 + x]))/(-3*x^2 + 3*x^3),x]

[Out]

4*Log[1 - x] - (1 - x)*Log[-1 + x] - (25*E^(E^x*x)*(E^x*x*Log[-1 + x] - E^x*x^3*Log[-1 + x]))/(3*(1 - x)*x^2*(
E^x + E^x*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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{9x^2+3x^3+(-3x^2+3x^3)\log (-1+x)+e^{e^{x}x}(-25x+(-25+25x+e^x(25x-25x^3))\log (-1+x))}{x^2(-3+3x)}dx\\ & =\int (\frac{3+x-\log (-1+x)+x\log (-1+x)}{-1+x}-\frac{25e^{e^{x}x}(x+\log (-1+x)-x\log (-1+x)-e^{x}x\log (-1+x)+e^x{x}^3\log (-1+x))}{3(-1+x)x^2})dx\\ & =-(\frac{25}{3}\int \frac{e^{e^{x}x}(x+\log (-1+x)-x\log (-1+x)-e^{x}x\log (-1+x)+e^x{x}^3\log (-1+x))}{(-1+x)x^2}dx)+\int \frac{3+x-\log (-1+x)+x\log (-1+x)}{-1+x}dx\\ & =-\frac{25e^{e^{x}x}(e^{x}x\log (-1+x)-e^x{x}^3\log (-1+x))}{3(1-x)x^2(e^x+e^{x}x)}+\int (\frac{3+x}{-1+x}+\log (-1+x))dx\\ & =-\frac{25e^{e^{x}x}(e^{x}x\log (-1+x)-e^x{x}^3\log (-1+x))}{3(1-x)x^2(e^x+e^{x}x)}+\int \frac{3+x}{-1+x}dx+\int \log (-1+x)dx\\ & =-\frac{25e^{e^{x}x}(e^{x}x\log (-1+x)-e^x{x}^3\log (-1+x))}{3(1-x)x^2(e^x+e^{x}x)}+\int (1+\frac{4}{-1+x})dx+\mathrm{Subst}(\int \log (x)dx,x,-1+x)\\ & =4\log (1-x)+(-1+x)\log (-1+x)-\frac{25e^{e^{x}x}(e^{x}x\log (-1+x)-e^x{x}^3\log (-1+x))}{3(1-x)x^2(e^x+e^{x}x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.58, size = 37, normalized size = 1.68 $$\begin{array}{c}\frac{1}{3}(12\log (1-x)+\frac{(-25e^{e^{x}x}+3(-1+x)x)\log (-1+x)}{x})\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(9*x^2 + 3*x^3 + (-3*x^2 + 3*x^3)*Log[-1 + x] + E^(E^x*x)*(-25*x + (-25 + 25*x + E^x*(25*x - 25*x^3)
)*Log[-1 + x]))/(-3*x^2 + 3*x^3),x]

[Out]

(12*Log[1 - x] + ((-25*E^(E^x*x) + 3*(-1 + x)*x)*Log[-1 + x])/x)/3

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fricas [A]  time = 0.80, size = 30, normalized size = 1.36 $$\begin{array}{c}\frac{3(x^2+3x)\log (x-1)-25e^{\left(x{e}^x\right)}\log (x-1)}{3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*log(x-1)+3*x^3+9*x^2)/(
3*x^3-3*x^2),x, algorithm="fricas")

[Out]

1/3*(3*(x^2 + 3*x)*log(x - 1) - 25*e^(x*e^x)*log(x - 1))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{3x^3+9x^2-25(((x^3-x)e^x-x+1)\log (x-1)+x)e^{\left(x{e}^x\right)}+3(x^3-x^2)\log (x-1)}{3(x^3-x^2)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*log(x-1)+3*x^3+9*x^2)/(
3*x^3-3*x^2),x, algorithm="giac")

[Out]

integrate(1/3*(3*x^3 + 9*x^2 - 25*(((x^3 - x)*e^x - x + 1)*log(x - 1) + x)*e^(x*e^x) + 3*(x^3 - x^2)*log(x - 1
))/(x^3 - x^2), x)

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maple [A]  time = 0.38, size = 28, normalized size = 1.27

method	result	size
risch	$\ln (x-1)x+3\ln (x-1)-\frac{25\ln (x-1){\mathrm{e}}^{{\mathrm{e}}^{x}x}}{3x}$	$28$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-25*x^3+25*x)*exp(x)+25*x-25)*ln(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*ln(x-1)+3*x^3+9*x^2)/(3*x^3-3*
x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x-1)*x+3*ln(x-1)-25/3*ln(x-1)/x*exp(exp(x)*x)

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maxima [A]  time = 0.43, size = 40, normalized size = 1.82 $$\begin{array}{c}(x+\log (x-1))\log (x-1)-\log {(x-1)}^2-\frac{25e^{\left(x{e}^x\right)}\log (x-1)}{3x}+3\log (x-1)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x^3+25*x)*exp(x)+25*x-25)*log(x-1)-25*x)*exp(exp(x)*x)+(3*x^3-3*x^2)*log(x-1)+3*x^3+9*x^2)/(
3*x^3-3*x^2),x, algorithm="maxima")

[Out]

(x + log(x - 1))*log(x - 1) - log(x - 1)^2 - 25/3*e^(x*e^x)*log(x - 1)/x + 3*log(x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 $$\begin{array}{c}\int \frac{{\mathrm{e}}^{x{\mathrm{e}}^x}(25x-\ln (x-1)(25x+{\mathrm{e}}^x(25x-25x^3)-25))+\ln (x-1)(3x^2-3x^3)-9x^2-3x^3}{3x^2-3x^3}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x*exp(x))*(25*x - log(x - 1)*(25*x + exp(x)*(25*x - 25*x^3) - 25)) + log(x - 1)*(3*x^2 - 3*x^3) - 9*x
^2 - 3*x^3)/(3*x^2 - 3*x^3),x)

[Out]

int((exp(x*exp(x))*(25*x - log(x - 1)*(25*x + exp(x)*(25*x - 25*x^3) - 25)) + log(x - 1)*(3*x^2 - 3*x^3) - 9*x
^2 - 3*x^3)/(3*x^2 - 3*x^3), x)

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sympy [A]  time = 0.40, size = 29, normalized size = 1.32 $$\begin{array}{c}x\log (x-1)+3\log (x-1)-\frac{25e^{x{e}^x}\log (x-1)}{3x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-25*x**3+25*x)*exp(x)+25*x-25)*ln(x-1)-25*x)*exp(exp(x)*x)+(3*x**3-3*x**2)*ln(x-1)+3*x**3+9*x**2
)/(3*x**3-3*x**2),x)

[Out]

x*log(x - 1) + 3*log(x - 1) - 25*exp(x*exp(x))*log(x - 1)/(3*x)

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