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

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

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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 {gather*} -\frac {e^x (3 x-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 {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-2+2 x+e^x \left (-x-3 x^2\right )+\left (1+e^x x^2\right ) \log (x)}{x^2 \log (2) (3-\log (x))^2} \, dx\\ &=\frac {\int \frac {-2+2 x+e^x \left (-x-3 x^2\right )+\left (1+e^x x^2\right ) \log (x)}{x^2 (3-\log (x))^2} \, dx}{\log (2)}\\ &=\frac {\int \left (\frac {-2+2 x+\log (x)}{x^2 (-3+\log (x))^2}+\frac {e^x (-1-3 x+x \log (x))}{x (-3+\log (x))^2}\right ) \, dx}{\log (2)}\\ &=\frac {\int \frac {-2+2 x+\log (x)}{x^2 (-3+\log (x))^2} \, dx}{\log (2)}+\frac {\int \frac {e^x (-1-3 x+x \log (x))}{x (-3+\log (x))^2} \, dx}{\log (2)}\\ &=-\frac {e^x (3 x-x \log (x))}{x \log (2) (3-\log (x))^2}+\frac {\int \left (\frac {1+2 x}{x^2 (-3+\log (x))^2}+\frac {1}{x^2 (-3+\log (x))}\right ) \, dx}{\log (2)}\\ &=-\frac {e^x (3 x-x \log (x))}{x \log (2) (3-\log (x))^2}+\frac {\int \frac {1+2 x}{x^2 (-3+\log (x))^2} \, dx}{\log (2)}+\frac {\int \frac {1}{x^2 (-3+\log (x))} \, dx}{\log (2)}\\ &=-\frac {e^x (3 x-x \log (x))}{x \log (2) (3-\log (x))^2}+\frac {\int \left (\frac {1}{x^2 (-3+\log (x))^2}+\frac {2}{x (-3+\log (x))^2}\right ) \, dx}{\log (2)}+\frac {\operatorname {Subst}\left (\int \frac {e^{-x}}{-3+x} \, dx,x,\log (x)\right )}{\log (2)}\\ &=\frac {\text {Ei}(3-\log (x))}{e^3 \log (2)}-\frac {e^x (3 x-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 (3 x-x \log (x))}{x \log (2) (3-\log (x))^2}-\frac {\int \frac {1}{x^2 (-3+\log (x))} \, dx}{\log (2)}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-3+\log (x)\right )}{\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 (3 x-x \log (x))}{x \log (2) (3-\log (x))^2}-\frac {\operatorname {Subst}\left (\int \frac {e^{-x}}{-3+x} \, dx,x,\log (x)\right )}{\log (2)}\\ &=\frac {2}{\log (2) (3-\log (x))}+\frac {1}{x \log (2) (3-\log (x))}-\frac {e^x (3 x-x \log (x))}{x \log (2) (3-\log (x))^2}\\ \end {aligned} \end {gather*}

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

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 {gather*} \frac {x e^{x} - 2 \, x - 1}{x \log \relax (2) \log \relax (x) - 3 \, x \log \relax (2)} \end {gather*}

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 {gather*} \frac {x e^{x} - 2 \, x - 1}{x \log \relax (2) \log \relax (x) - 3 \, x \log \relax (2)} \end {gather*}

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	\(\frac {{\mathrm e}^{x} x -2 x -1}{x \ln \relax (2) \left (\ln \relax (x )-3\right )}\)	\(24\)
norman	\(\frac {-\frac {2 x}{\ln \relax (2)}+\frac {x \,{\mathrm e}^{x}}{\ln \relax (2)}-\frac {1}{\ln \relax (2)}}{x \left (\ln \relax (x )-3\right )}\)	\(33\)
default	\(\frac {-\frac {2 x}{\ln \relax (2)}-\frac {1}{\ln \relax (2)}}{x \left (\ln \relax (x )-3\right )}+\frac {{\mathrm e}^{x}}{\ln \relax (2) \left (\ln \relax (x )-3\right )}\)	\(39\)

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 {gather*} \frac {x e^{x} - 2 \, x - 1}{x \log \relax (2) \log \relax (x) - 3 \, x \log \relax (2)} \end {gather*}

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 {gather*} -\frac {2\,x-x\,{\mathrm {e}}^x+1}{x\,\ln \relax (2)\,\left (\ln \relax (x)-3\right )} \end {gather*}

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 {gather*} \frac {- 2 x - 1}{x \log {\relax (2 )} \log {\relax (x )} - 3 x \log {\relax (2 )}} + \frac {e^{x}}{\log {\relax (2 )} \log {\relax (x )} - 3 \log {\relax (2 )}} \end {gather*}

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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