3.79.90 \(\int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+(-16 x-8 e^x x) \log (x)+(8 x+4 e^x x) \log ^2(x)}{4 x-8 x \log (x)+4 x \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ -4+e^x+e^{\frac {1}{4 (-1+\log (x))}}+2 x \]

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Rubi [A]  time = 1.48, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 32, number of rules used = 9, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6741, 12, 6742, 2209, 2297, 2299, 2178, 2360, 2194} \begin {gather*} 2 x+e^x+e^{-\frac {1}{4 (1-\log (x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^(-4 + 4*Log[x])^(-1) + 8*x + 4*E^x*x + (-16*x - 8*E^x*x)*Log[x] + (8*x + 4*E^x*x)*Log[x]^2)/(4*x - 8*x
*Log[x] + 4*x*Log[x]^2),x]

[Out]

E^x + E^(-1/4*1/(1 - Log[x])) + 2*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 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 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 2209

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

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2299

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

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

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 {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{4 x (1-\log (x))^2} \, dx\\ &=\frac {1}{4} \int \frac {-e^{\frac {1}{-4+4 \log (x)}}+8 x+4 e^x x+\left (-16 x-8 e^x x\right ) \log (x)+\left (8 x+4 e^x x\right ) \log ^2(x)}{x (1-\log (x))^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {e^{\frac {1}{4 (-1+\log (x))}}}{x (1-\log (x))^2}+\frac {8}{(-1+\log (x))^2}+\frac {4 e^x}{(-1+\log (x))^2}-\frac {8 \left (2+e^x\right ) \log (x)}{(-1+\log (x))^2}+\frac {4 \left (2+e^x\right ) \log ^2(x)}{(-1+\log (x))^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {e^{\frac {1}{4 (-1+\log (x))}}}{x (1-\log (x))^2} \, dx\right )+2 \int \frac {1}{(-1+\log (x))^2} \, dx-2 \int \frac {\left (2+e^x\right ) \log (x)}{(-1+\log (x))^2} \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx+\int \frac {\left (2+e^x\right ) \log ^2(x)}{(-1+\log (x))^2} \, dx\\ &=\frac {2 x}{1-\log (x)}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {e^{\frac {1}{4 (-1+x)}}}{(-1+x)^2} \, dx,x,\log (x)\right )+2 \int \frac {1}{-1+\log (x)} \, dx-2 \int \left (\frac {2 \log (x)}{(-1+\log (x))^2}+\frac {e^x \log (x)}{(-1+\log (x))^2}\right ) \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx+\int \left (\frac {2 \log ^2(x)}{(-1+\log (x))^2}+\frac {e^x \log ^2(x)}{(-1+\log (x))^2}\right ) \, dx\\ &=e^{-\frac {1}{4 (1-\log (x))}}+\frac {2 x}{1-\log (x)}-2 \int \frac {e^x \log (x)}{(-1+\log (x))^2} \, dx+2 \int \frac {\log ^2(x)}{(-1+\log (x))^2} \, dx+2 \operatorname {Subst}\left (\int \frac {e^x}{-1+x} \, dx,x,\log (x)\right )-4 \int \frac {\log (x)}{(-1+\log (x))^2} \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx+\int \frac {e^x \log ^2(x)}{(-1+\log (x))^2} \, dx\\ &=e^{-\frac {1}{4 (1-\log (x))}}+2 e \text {Ei}(-1+\log (x))+\frac {2 x}{1-\log (x)}+2 \int \left (1+\frac {1}{(-1+\log (x))^2}+\frac {2}{-1+\log (x)}\right ) \, dx-2 \int \left (\frac {e^x}{(-1+\log (x))^2}+\frac {e^x}{-1+\log (x)}\right ) \, dx-4 \int \left (\frac {1}{(-1+\log (x))^2}+\frac {1}{-1+\log (x)}\right ) \, dx+\int \left (e^x+\frac {e^x}{(-1+\log (x))^2}+\frac {2 e^x}{-1+\log (x)}\right ) \, dx+\int \frac {e^x}{(-1+\log (x))^2} \, dx\\ &=e^{-\frac {1}{4 (1-\log (x))}}+2 x+2 e \text {Ei}(-1+\log (x))+\frac {2 x}{1-\log (x)}+2 \int \frac {1}{(-1+\log (x))^2} \, dx-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx-4 \int \frac {1}{(-1+\log (x))^2} \, dx+\int e^x \, dx+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx\\ &=e^x+e^{-\frac {1}{4 (1-\log (x))}}+2 x+2 e \text {Ei}(-1+\log (x))-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx+2 \int \frac {1}{-1+\log (x)} \, dx-4 \int \frac {1}{-1+\log (x)} \, dx+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx\\ &=e^x+e^{-\frac {1}{4 (1-\log (x))}}+2 x+2 e \text {Ei}(-1+\log (x))-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx+2 \operatorname {Subst}\left (\int \frac {e^x}{-1+x} \, dx,x,\log (x)\right )-4 \operatorname {Subst}\left (\int \frac {e^x}{-1+x} \, dx,x,\log (x)\right )+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx\\ &=e^x+e^{-\frac {1}{4 (1-\log (x))}}+2 x-2 \int \frac {e^x}{(-1+\log (x))^2} \, dx+2 \int \frac {e^x}{(-1+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 27, normalized size = 1.35 \begin {gather*} \frac {1}{4} \left (4 e^x+4 e^{\frac {1}{4 (-1+\log (x))}}+8 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^(-4 + 4*Log[x])^(-1) + 8*x + 4*E^x*x + (-16*x - 8*E^x*x)*Log[x] + (8*x + 4*E^x*x)*Log[x]^2)/(4*x
- 8*x*Log[x] + 4*x*Log[x]^2),x]

[Out]

(4*E^x + 4*E^(1/(4*(-1 + Log[x]))) + 8*x)/4

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fricas [A]  time = 0.66, size = 15, normalized size = 0.75 \begin {gather*} 2 \, x + e^{x} + e^{\left (\frac {1}{4 \, {\left (\log \relax (x) - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(1/(4*log(x)-4))+(4*exp(x)*x+8*x)*log(x)^2+(-8*exp(x)*x-16*x)*log(x)+4*exp(x)*x+8*x)/(4*x*log(x
)^2-8*x*log(x)+4*x),x, algorithm="fricas")

[Out]

2*x + e^x + e^(1/4/(log(x) - 1))

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giac [A]  time = 0.18, size = 15, normalized size = 0.75 \begin {gather*} 2 \, x + e^{x} + e^{\left (\frac {1}{4 \, {\left (\log \relax (x) - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(1/(4*log(x)-4))+(4*exp(x)*x+8*x)*log(x)^2+(-8*exp(x)*x-16*x)*log(x)+4*exp(x)*x+8*x)/(4*x*log(x
)^2-8*x*log(x)+4*x),x, algorithm="giac")

[Out]

2*x + e^x + e^(1/4/(log(x) - 1))

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maple [A]  time = 0.04, size = 16, normalized size = 0.80

method	result	size
risch	\(2 x +{\mathrm e}^{x}+{\mathrm e}^{\frac {1}{4 \ln \relax (x )-4}}\)	\(16\)
default	\(2 x +{\mathrm e}^{x}+\frac {4 \ln \relax (x ) {\mathrm e}^{\frac {1}{4 \ln \relax (x )-4}}-4 \,{\mathrm e}^{\frac {1}{4 \ln \relax (x )-4}}}{4 \ln \relax (x )-4}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(1/(4*ln(x)-4))+(4*exp(x)*x+8*x)*ln(x)^2+(-8*exp(x)*x-16*x)*ln(x)+4*exp(x)*x+8*x)/(4*x*ln(x)^2-8*x*ln
(x)+4*x),x,method=_RETURNVERBOSE)

[Out]

2*x+exp(x)+exp(1/4/(ln(x)-1))

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maxima [A]  time = 0.37, size = 15, normalized size = 0.75 \begin {gather*} 2 \, x + e^{x} + e^{\left (\frac {1}{4 \, {\left (\log \relax (x) - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(1/(4*log(x)-4))+(4*exp(x)*x+8*x)*log(x)^2+(-8*exp(x)*x-16*x)*log(x)+4*exp(x)*x+8*x)/(4*x*log(x
)^2-8*x*log(x)+4*x),x, algorithm="maxima")

[Out]

2*x + e^x + e^(1/4/(log(x) - 1))

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mupad [B]  time = 6.53, size = 15, normalized size = 0.75 \begin {gather*} 2\,x+{\mathrm {e}}^{\frac {1}{4\,\ln \relax (x)-4}}+{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - exp(1/(4*log(x) - 4)) - log(x)*(16*x + 8*x*exp(x)) + log(x)^2*(8*x + 4*x*exp(x)) + 4*x*exp(x))/(4*x
+ 4*x*log(x)^2 - 8*x*log(x)),x)

[Out]

2*x + exp(1/(4*log(x) - 4)) + exp(x)

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sympy [A]  time = 0.51, size = 15, normalized size = 0.75 \begin {gather*} 2 x + e^{x} + e^{\frac {1}{4 \log {\relax (x )} - 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(1/(4*ln(x)-4))+(4*exp(x)*x+8*x)*ln(x)**2+(-8*exp(x)*x-16*x)*ln(x)+4*exp(x)*x+8*x)/(4*x*ln(x)**
2-8*x*ln(x)+4*x),x)

[Out]

2*x + exp(x) + exp(1/(4*log(x) - 4))

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