3.17.51 $\int \frac{-2+2x+e^x(-x-3x^2)+(1+e^x{x}^2)\log (x)}{9x^2\log (2)-6x^2\log (2)\log (x)+x^2\log (2){\log }^2(x)}dx$

Optimal. Leaf size=26 $$-\frac{1+(2-e^x)x}{x\log (2)(-3+\log (x))}$$

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Rubi [B]  time = 0.82, antiderivative size = 60, normalized size of antiderivative = 2.31, number of steps used = 16, number of rules used = 10, integrand size = 60, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.167, Rules used = {6741, 12, 6742, 2353, 2306, 2309, 2178, 2302, 30, 2288} $$\begin{array}{c}-\frac{e^x(3x-x\log (x))}{x\log (2)(3-\log (x){)}^2}+\frac{1}{x\log (2)(3-\log (x))}+\frac{2}{\log (2)(3-\log (x))}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-2 + 2*x + E^x*(-x - 3*x^2) + (1 + E^x*x^2)*Log[x])/(9*x^2*Log[2] - 6*x^2*Log[2]*Log[x] + x^2*Log[2]*Log[
x]^2),x]

[Out]

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

Rule 12

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

Rule 30

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 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 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6741

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

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 24, normalized size = 0.92 $$\begin{array}{c}\frac{-1-2x+e^{x}x}{x\log (2)(-3+\log (x))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 2*x + E^x*(-x - 3*x^2) + (1 + E^x*x^2)*Log[x])/(9*x^2*Log[2] - 6*x^2*Log[2]*Log[x] + x^2*Log[2
]*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.77, size = 24, normalized size = 0.92 $$\begin{array}{c}\frac{x{e}^x-2x-1}{x\log (2)\log (x)-3x\log (2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^2+1)*log(x)+(-3*x^2-x)*exp(x)+2*x-2)/(x^2*log(2)*log(x)^2-6*x^2*log(2)*log(x)+9*x^2*log(2
)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.34, size = 24, normalized size = 0.92 $$\begin{array}{c}\frac{x{e}^x-2x-1}{x\log (2)\log (x)-3x\log (2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^2+1)*log(x)+(-3*x^2-x)*exp(x)+2*x-2)/(x^2*log(2)*log(x)^2-6*x^2*log(2)*log(x)+9*x^2*log(2
)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.25, size = 24, normalized size = 0.92

method	result	size
risch	$\frac{{\mathrm{e}}^{x}x-2x-1}{x\ln (2)(\ln (x)-3)}$	$24$
norman	$\frac{-\frac{2x}{\ln (2)}+\frac{x{\mathrm{e}}^x}{\ln (2)}-\frac{1}{\ln (2)}}{x(\ln (x)-3)}$	$33$
default	$\frac{-\frac{2x}{\ln (2)}-\frac{1}{\ln (2)}}{x(\ln (x)-3)}+\frac{{\mathrm{e}}^x}{\ln (2)(\ln (x)-3)}$	$39$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x^2+1)*ln(x)+(-3*x^2-x)*exp(x)+2*x-2)/(x^2*ln(2)*ln(x)^2-6*x^2*ln(2)*ln(x)+9*x^2*ln(2)),x,method=
_RETURNVERBOSE)

[Out]

1/x*(exp(x)*x-2*x-1)/ln(2)/(ln(x)-3)

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maxima [A]  time = 0.66, size = 24, normalized size = 0.92 $$\begin{array}{c}\frac{x{e}^x-2x-1}{x\log (2)\log (x)-3x\log (2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x^2+1)*log(x)+(-3*x^2-x)*exp(x)+2*x-2)/(x^2*log(2)*log(x)^2-6*x^2*log(2)*log(x)+9*x^2*log(2
)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.26, size = 25, normalized size = 0.96 $$\begin{array}{c}-\frac{2x-x{\mathrm{e}}^x+1}{x\ln (2)(\ln (x)-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - exp(x)*(x + 3*x^2) + log(x)*(x^2*exp(x) + 1) - 2)/(9*x^2*log(2) - 6*x^2*log(2)*log(x) + x^2*log(2)*
log(x)^2),x)

[Out]

-(2*x - x*exp(x) + 1)/(x*log(2)*(log(x) - 3))

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sympy [A]  time = 0.31, size = 36, normalized size = 1.38 $$\begin{array}{c}\frac{-2x-1}{x\log (2)\log (x)-3x\log (2)}+\frac{e^x}{\log (2)\log (x)-3\log (2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x**2+1)*ln(x)+(-3*x**2-x)*exp(x)+2*x-2)/(x**2*ln(2)*ln(x)**2-6*x**2*ln(2)*ln(x)+9*x**2*ln(2
)),x)

[Out]

(-2*x - 1)/(x*log(2)*log(x) - 3*x*log(2)) + exp(x)/(log(2)*log(x) - 3*log(2))

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