3.57.61 \(\int \frac {e^x (-1+x)-4 x^2 \log (3) \log ^3(x)+(e^x (4 x-4 x^2)-x^2 \log (3)) \log ^4(x)+e^x (-4 x^2+4 x^3) \log ^8(x)}{x^2-4 x^3 \log ^4(x)+4 x^4 \log ^8(x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{2} \left (\frac {2 e^x}{x}-\frac {\log (3)}{1-2 x \log ^4(x)}\right ) \]

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Rubi [A]  time = 0.84, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6741, 6742, 2197, 6686} \begin {gather*} \frac {e^x}{x}-\frac {\log (3)}{2 \left (1-2 x \log ^4(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-1 + x) - 4*x^2*Log[3]*Log[x]^3 + (E^x*(4*x - 4*x^2) - x^2*Log[3])*Log[x]^4 + E^x*(-4*x^2 + 4*x^3)*L
og[x]^8)/(x^2 - 4*x^3*Log[x]^4 + 4*x^4*Log[x]^8),x]

[Out]

E^x/x - Log[3]/(2*(1 - 2*x*Log[x]^4))

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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^x (-1+x)-4 x^2 \log (3) \log ^3(x)+\left (e^x \left (4 x-4 x^2\right )-x^2 \log (3)\right ) \log ^4(x)+e^x \left (-4 x^2+4 x^3\right ) \log ^8(x)}{x^2 \left (1-2 x \log ^4(x)\right )^2} \, dx\\ &=\int \left (\frac {e^x (-1+x)}{x^2}-\frac {\log (3) \log ^3(x) (4+\log (x))}{\left (-1+2 x \log ^4(x)\right )^2}\right ) \, dx\\ &=-\left (\log (3) \int \frac {\log ^3(x) (4+\log (x))}{\left (-1+2 x \log ^4(x)\right )^2} \, dx\right )+\int \frac {e^x (-1+x)}{x^2} \, dx\\ &=\frac {e^x}{x}-\frac {\log (3)}{2 \left (1-2 x \log ^4(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 22, normalized size = 0.79 \begin {gather*} \frac {e^x}{x}+\frac {\log (3)}{-2+4 x \log ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-1 + x) - 4*x^2*Log[3]*Log[x]^3 + (E^x*(4*x - 4*x^2) - x^2*Log[3])*Log[x]^4 + E^x*(-4*x^2 + 4*
x^3)*Log[x]^8)/(x^2 - 4*x^3*Log[x]^4 + 4*x^4*Log[x]^8),x]

[Out]

E^x/x + Log[3]/(-2 + 4*x*Log[x]^4)

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fricas [A]  time = 0.84, size = 35, normalized size = 1.25 \begin {gather*} \frac {4 \, x e^{x} \log \relax (x)^{4} + x \log \relax (3) - 2 \, e^{x}}{2 \, {\left (2 \, x^{2} \log \relax (x)^{4} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3-4*x^2)*exp(x)*log(x)^8+((-4*x^2+4*x)*exp(x)-x^2*log(3))*log(x)^4-4*x^2*log(3)*log(x)^3+(x-1)
*exp(x))/(4*x^4*log(x)^8-4*x^3*log(x)^4+x^2),x, algorithm="fricas")

[Out]

1/2*(4*x*e^x*log(x)^4 + x*log(3) - 2*e^x)/(2*x^2*log(x)^4 - x)

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giac [A]  time = 0.61, size = 35, normalized size = 1.25 \begin {gather*} \frac {4 \, x e^{x} \log \relax (x)^{4} + x \log \relax (3) - 2 \, e^{x}}{2 \, {\left (2 \, x^{2} \log \relax (x)^{4} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3-4*x^2)*exp(x)*log(x)^8+((-4*x^2+4*x)*exp(x)-x^2*log(3))*log(x)^4-4*x^2*log(3)*log(x)^3+(x-1)
*exp(x))/(4*x^4*log(x)^8-4*x^3*log(x)^4+x^2),x, algorithm="giac")

[Out]

1/2*(4*x*e^x*log(x)^4 + x*log(3) - 2*e^x)/(2*x^2*log(x)^4 - x)

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maple [A]  time = 0.06, size = 23, normalized size = 0.82

method	result	size
risch	\(\frac {{\mathrm e}^{x}}{x}+\frac {\ln \relax (3)}{4 x \ln \relax (x )^{4}-2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3-4*x^2)*exp(x)*ln(x)^8+((-4*x^2+4*x)*exp(x)-x^2*ln(3))*ln(x)^4-4*x^2*ln(3)*ln(x)^3+(x-1)*exp(x))/(4
*x^4*ln(x)^8-4*x^3*ln(x)^4+x^2),x,method=_RETURNVERBOSE)

[Out]

exp(x)/x+1/2*ln(3)/(2*x*ln(x)^4-1)

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maxima [A]  time = 0.50, size = 35, normalized size = 1.25 \begin {gather*} \frac {2 \, {\left (2 \, x \log \relax (x)^{4} - 1\right )} e^{x} + x \log \relax (3)}{2 \, {\left (2 \, x^{2} \log \relax (x)^{4} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3-4*x^2)*exp(x)*log(x)^8+((-4*x^2+4*x)*exp(x)-x^2*log(3))*log(x)^4-4*x^2*log(3)*log(x)^3+(x-1)
*exp(x))/(4*x^4*log(x)^8-4*x^3*log(x)^4+x^2),x, algorithm="maxima")

[Out]

1/2*(2*(2*x*log(x)^4 - 1)*e^x + x*log(3))/(2*x^2*log(x)^4 - x)

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mupad [B]  time = 3.56, size = 21, normalized size = 0.75 \begin {gather*} \frac {{\mathrm {e}}^x}{x}+\frac {\ln \relax (3)}{2\,\left (2\,x\,{\ln \relax (x)}^4-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^4*(x^2*log(3) - exp(x)*(4*x - 4*x^2)) - exp(x)*(x - 1) + exp(x)*log(x)^8*(4*x^2 - 4*x^3) + 4*x^2*
log(3)*log(x)^3)/(4*x^4*log(x)^8 - 4*x^3*log(x)^4 + x^2),x)

[Out]

exp(x)/x + log(3)/(2*(2*x*log(x)^4 - 1))

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sympy [A]  time = 0.32, size = 17, normalized size = 0.61 \begin {gather*} \frac {\log {\relax (3 )}}{4 x \log {\relax (x )}^{4} - 2} + \frac {e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3-4*x**2)*exp(x)*ln(x)**8+((-4*x**2+4*x)*exp(x)-x**2*ln(3))*ln(x)**4-4*x**2*ln(3)*ln(x)**3+(x
-1)*exp(x))/(4*x**4*ln(x)**8-4*x**3*ln(x)**4+x**2),x)

[Out]

log(3)/(4*x*log(x)**4 - 2) + exp(x)/x

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