3.25.36 \(\int \frac {16 x^5+e^4 (-5+10 x^2+x^5)+(-32 x^3-2 e^4 x^3) \log (x)+(16 x+e^4 x) \log ^2(x)}{16 x^5-32 x^3 \log (x)+16 x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ x-\frac {1}{16} e^4 \left (-x+\frac {5}{x^2-\log (x)}\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 4, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6741, 12, 6742, 6686} \begin {gather*} \frac {1}{16} \left (16+e^4\right ) x-\frac {5 e^4}{16 \left (x^2-\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*x^5 + E^4*(-5 + 10*x^2 + x^5) + (-32*x^3 - 2*E^4*x^3)*Log[x] + (16*x + E^4*x)*Log[x]^2)/(16*x^5 - 32*x
^3*Log[x] + 16*x*Log[x]^2),x]

[Out]

((16 + E^4)*x)/16 - (5*E^4)/(16*(x^2 - Log[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 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 {16 x^5+e^4 \left (-5+10 x^2+x^5\right )+\left (-32 x^3-2 e^4 x^3\right ) \log (x)+\left (16 x+e^4 x\right ) \log ^2(x)}{16 x \left (x^2-\log (x)\right )^2} \, dx\\ &=\frac {1}{16} \int \frac {16 x^5+e^4 \left (-5+10 x^2+x^5\right )+\left (-32 x^3-2 e^4 x^3\right ) \log (x)+\left (16 x+e^4 x\right ) \log ^2(x)}{x \left (x^2-\log (x)\right )^2} \, dx\\ &=\frac {1}{16} \int \left (16 \left (1+\frac {e^4}{16}\right )+\frac {5 e^4 \left (-1+2 x^2\right )}{x \left (x^2-\log (x)\right )^2}\right ) \, dx\\ &=\frac {1}{16} \left (16+e^4\right ) x+\frac {1}{16} \left (5 e^4\right ) \int \frac {-1+2 x^2}{x \left (x^2-\log (x)\right )^2} \, dx\\ &=\frac {1}{16} \left (16+e^4\right ) x-\frac {5 e^4}{16 \left (x^2-\log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 27, normalized size = 1.08 \begin {gather*} \frac {1}{16} \left (\left (16+e^4\right ) x+\frac {5 e^4}{-x^2+\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*x^5 + E^4*(-5 + 10*x^2 + x^5) + (-32*x^3 - 2*E^4*x^3)*Log[x] + (16*x + E^4*x)*Log[x]^2)/(16*x^5
- 32*x^3*Log[x] + 16*x*Log[x]^2),x]

[Out]

((16 + E^4)*x + (5*E^4)/(-x^2 + Log[x]))/16

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fricas [A]  time = 1.17, size = 38, normalized size = 1.52 \begin {gather*} \frac {16 \, x^{3} + {\left (x^{3} - 5\right )} e^{4} - {\left (x e^{4} + 16 \, x\right )} \log \relax (x)}{16 \, {\left (x^{2} - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(4)+16*x)*log(x)^2+(-2*x^3*exp(4)-32*x^3)*log(x)+(x^5+10*x^2-5)*exp(4)+16*x^5)/(16*x*log(x)^2
-32*x^3*log(x)+16*x^5),x, algorithm="fricas")

[Out]

1/16*(16*x^3 + (x^3 - 5)*e^4 - (x*e^4 + 16*x)*log(x))/(x^2 - log(x))

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giac [A]  time = 0.26, size = 40, normalized size = 1.60 \begin {gather*} \frac {x^{3} e^{4} + 16 \, x^{3} - x e^{4} \log \relax (x) - 16 \, x \log \relax (x) - 5 \, e^{4}}{16 \, {\left (x^{2} - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(4)+16*x)*log(x)^2+(-2*x^3*exp(4)-32*x^3)*log(x)+(x^5+10*x^2-5)*exp(4)+16*x^5)/(16*x*log(x)^2
-32*x^3*log(x)+16*x^5),x, algorithm="giac")

[Out]

1/16*(x^3*e^4 + 16*x^3 - x*e^4*log(x) - 16*x*log(x) - 5*e^4)/(x^2 - log(x))

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maple [A]  time = 0.06, size = 22, normalized size = 0.88

method	result	size
risch	\(\frac {x \,{\mathrm e}^{4}}{16}+x -\frac {5 \,{\mathrm e}^{4}}{16 \left (x^{2}-\ln \relax (x )\right )}\)	\(22\)
norman	\(\frac {\left (\frac {{\mathrm e}^{4}}{16}+1\right ) x^{3}+\left (-\frac {{\mathrm e}^{4}}{16}-1\right ) x \ln \relax (x )-\frac {5 \,{\mathrm e}^{4}}{16}}{x^{2}-\ln \relax (x )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(4)+16*x)*ln(x)^2+(-2*x^3*exp(4)-32*x^3)*ln(x)+(x^5+10*x^2-5)*exp(4)+16*x^5)/(16*x*ln(x)^2-32*x^3*l
n(x)+16*x^5),x,method=_RETURNVERBOSE)

[Out]

1/16*x*exp(4)+x-5/16*exp(4)/(x^2-ln(x))

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maxima [A]  time = 0.48, size = 34, normalized size = 1.36 \begin {gather*} \frac {x^{3} {\left (e^{4} + 16\right )} - x {\left (e^{4} + 16\right )} \log \relax (x) - 5 \, e^{4}}{16 \, {\left (x^{2} - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(4)+16*x)*log(x)^2+(-2*x^3*exp(4)-32*x^3)*log(x)+(x^5+10*x^2-5)*exp(4)+16*x^5)/(16*x*log(x)^2
-32*x^3*log(x)+16*x^5),x, algorithm="maxima")

[Out]

1/16*(x^3*(e^4 + 16) - x*(e^4 + 16)*log(x) - 5*e^4)/(x^2 - log(x))

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mupad [B]  time = 1.53, size = 31, normalized size = 1.24 \begin {gather*} x\,\left (\frac {{\mathrm {e}}^4}{16}+1\right )+\frac {5\,x^2\,{\mathrm {e}}^4}{16\,x^2\,\ln \relax (x)-16\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(16*x + x*exp(4)) + exp(4)*(10*x^2 + x^5 - 5) - log(x)*(2*x^3*exp(4) + 32*x^3) + 16*x^5)/(16*x*l
og(x)^2 - 32*x^3*log(x) + 16*x^5),x)

[Out]

x*(exp(4)/16 + 1) + (5*x^2*exp(4))/(16*x^2*log(x) - 16*x^4)

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sympy [A]  time = 0.13, size = 22, normalized size = 0.88 \begin {gather*} x \left (1 + \frac {e^{4}}{16}\right ) + \frac {5 e^{4}}{- 16 x^{2} + 16 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(4)+16*x)*ln(x)**2+(-2*x**3*exp(4)-32*x**3)*ln(x)+(x**5+10*x**2-5)*exp(4)+16*x**5)/(16*x*ln(x
)**2-32*x**3*ln(x)+16*x**5),x)

[Out]

x*(1 + exp(4)/16) + 5*exp(4)/(-16*x**2 + 16*log(x))

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