3.27.39 \(\int \frac {10 e^4+e^8 (-15 x+18 x^4)+12 e^8 x^3 \log (4)+2 e^8 x^2 \log ^2(4)+(-24 e^4 x^3-8 e^4 x^2 \log (4)) \log (x)+8 x^2 \log ^2(x)}{9 e^8 x^3+6 e^8 x^2 \log (4)+e^8 x \log ^2(4)+(-12 e^4 x^2-4 e^4 x \log (4)) \log (x)+4 x \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ 30+x^2+\frac {5}{x+\log (4)+2 \left (x-\frac {\log (x)}{e^4}\right )} \]

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Rubi [A]  time = 0.73, antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 3, integrand size = 130, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6688, 6742, 6686} \begin {gather*} x^2+\frac {5 e^4}{3 e^4 x-2 \log (x)+e^4 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*E^4 + E^8*(-15*x + 18*x^4) + 12*E^8*x^3*Log[4] + 2*E^8*x^2*Log[4]^2 + (-24*E^4*x^3 - 8*E^4*x^2*Log[4])
*Log[x] + 8*x^2*Log[x]^2)/(9*E^8*x^3 + 6*E^8*x^2*Log[4] + E^8*x*Log[4]^2 + (-12*E^4*x^2 - 4*E^4*x*Log[4])*Log[
x] + 4*x*Log[x]^2),x]

[Out]

x^2 + (5*E^4)/(3*E^4*x + E^4*Log[4] - 2*Log[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 6688

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

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 {10 e^4+e^8 x \left (-15+18 x^3+12 x^2 \log (4)+2 x \log ^2(4)\right )-8 e^4 x^2 (3 x+\log (4)) \log (x)+8 x^2 \log ^2(x)}{x \left (e^4 (3 x+\log (4))-2 \log (x)\right )^2} \, dx\\ &=\int \left (2 x-\frac {5 e^4 \left (-2+3 e^4 x\right )}{x \left (3 e^4 x+e^4 \log (4)-2 \log (x)\right )^2}\right ) \, dx\\ &=x^2-\left (5 e^4\right ) \int \frac {-2+3 e^4 x}{x \left (3 e^4 x+e^4 \log (4)-2 \log (x)\right )^2} \, dx\\ &=x^2+\frac {5 e^4}{3 e^4 x+e^4 \log (4)-2 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 26, normalized size = 1.08 \begin {gather*} x^2+\frac {5 e^4}{e^4 (3 x+\log (4))-2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E^4 + E^8*(-15*x + 18*x^4) + 12*E^8*x^3*Log[4] + 2*E^8*x^2*Log[4]^2 + (-24*E^4*x^3 - 8*E^4*x^2*L
og[4])*Log[x] + 8*x^2*Log[x]^2)/(9*E^8*x^3 + 6*E^8*x^2*Log[4] + E^8*x*Log[4]^2 + (-12*E^4*x^2 - 4*E^4*x*Log[4]
)*Log[x] + 4*x*Log[x]^2),x]

[Out]

x^2 + (5*E^4)/(E^4*(3*x + Log[4]) - 2*Log[x])

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fricas [A]  time = 0.58, size = 46, normalized size = 1.92 \begin {gather*} \frac {2 \, x^{2} e^{4} \log \relax (2) - 2 \, x^{2} \log \relax (x) + {\left (3 \, x^{3} + 5\right )} e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \relax (2) - 2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*log(x)^2+(-16*x^2*exp(4)*log(2)-24*x^3*exp(4))*log(x)+8*x^2*exp(4)^2*log(2)^2+24*x^3*exp(4)^2
*log(2)+(18*x^4-15*x)*exp(4)^2+10*exp(4))/(4*x*log(x)^2+(-8*x*exp(4)*log(2)-12*x^2*exp(4))*log(x)+4*x*exp(4)^2
*log(2)^2+12*x^2*exp(4)^2*log(2)+9*x^3*exp(4)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.29, size = 47, normalized size = 1.96 \begin {gather*} \frac {3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} \log \relax (2) - 2 \, x^{2} \log \relax (x) + 5 \, e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \relax (2) - 2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*log(x)^2+(-16*x^2*exp(4)*log(2)-24*x^3*exp(4))*log(x)+8*x^2*exp(4)^2*log(2)^2+24*x^3*exp(4)^2
*log(2)+(18*x^4-15*x)*exp(4)^2+10*exp(4))/(4*x*log(x)^2+(-8*x*exp(4)*log(2)-12*x^2*exp(4))*log(x)+4*x*exp(4)^2
*log(2)^2+12*x^2*exp(4)^2*log(2)+9*x^3*exp(4)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.12, size = 27, normalized size = 1.12

method	result	size
risch	\(x^{2}+\frac {5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \relax (2)+3 x \,{\mathrm e}^{4}-2 \ln \relax (x )}\)	\(27\)
norman	\(\frac {-2 x^{2} \ln \relax (x )+3 x^{3} {\mathrm e}^{4}+2 x^{2} {\mathrm e}^{4} \ln \relax (2)+5 \,{\mathrm e}^{4}}{2 \,{\mathrm e}^{4} \ln \relax (2)+3 x \,{\mathrm e}^{4}-2 \ln \relax (x )}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2*ln(x)^2+(-16*x^2*exp(4)*ln(2)-24*x^3*exp(4))*ln(x)+8*x^2*exp(4)^2*ln(2)^2+24*x^3*exp(4)^2*ln(2)+(18
*x^4-15*x)*exp(4)^2+10*exp(4))/(4*x*ln(x)^2+(-8*x*exp(4)*ln(2)-12*x^2*exp(4))*ln(x)+4*x*exp(4)^2*ln(2)^2+12*x^
2*exp(4)^2*ln(2)+9*x^3*exp(4)^2),x,method=_RETURNVERBOSE)

[Out]

x^2+5*exp(4)/(2*exp(4)*ln(2)+3*x*exp(4)-2*ln(x))

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maxima [B]  time = 0.85, size = 47, normalized size = 1.96 \begin {gather*} \frac {3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} \log \relax (2) - 2 \, x^{2} \log \relax (x) + 5 \, e^{4}}{3 \, x e^{4} + 2 \, e^{4} \log \relax (2) - 2 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*log(x)^2+(-16*x^2*exp(4)*log(2)-24*x^3*exp(4))*log(x)+8*x^2*exp(4)^2*log(2)^2+24*x^3*exp(4)^2
*log(2)+(18*x^4-15*x)*exp(4)^2+10*exp(4))/(4*x*log(x)^2+(-8*x*exp(4)*log(2)-12*x^2*exp(4))*log(x)+4*x*exp(4)^2
*log(2)^2+12*x^2*exp(4)^2*log(2)+9*x^3*exp(4)^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {10\,{\mathrm {e}}^4-{\mathrm {e}}^8\,\left (15\,x-18\,x^4\right )-\ln \relax (x)\,\left (24\,{\mathrm {e}}^4\,x^3+16\,{\mathrm {e}}^4\,\ln \relax (2)\,x^2\right )+8\,x^2\,{\ln \relax (x)}^2+24\,x^3\,{\mathrm {e}}^8\,\ln \relax (2)+8\,x^2\,{\mathrm {e}}^8\,{\ln \relax (2)}^2}{4\,x\,{\ln \relax (x)}^2+9\,x^3\,{\mathrm {e}}^8-\ln \relax (x)\,\left (12\,{\mathrm {e}}^4\,x^2+8\,{\mathrm {e}}^4\,\ln \relax (2)\,x\right )+4\,x\,{\mathrm {e}}^8\,{\ln \relax (2)}^2+12\,x^2\,{\mathrm {e}}^8\,\ln \relax (2)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(4) - exp(8)*(15*x - 18*x^4) - log(x)*(24*x^3*exp(4) + 16*x^2*exp(4)*log(2)) + 8*x^2*log(x)^2 + 24*
x^3*exp(8)*log(2) + 8*x^2*exp(8)*log(2)^2)/(4*x*log(x)^2 + 9*x^3*exp(8) - log(x)*(12*x^2*exp(4) + 8*x*exp(4)*l
og(2)) + 4*x*exp(8)*log(2)^2 + 12*x^2*exp(8)*log(2)),x)

[Out]

int((10*exp(4) - exp(8)*(15*x - 18*x^4) - log(x)*(24*x^3*exp(4) + 16*x^2*exp(4)*log(2)) + 8*x^2*log(x)^2 + 24*
x^3*exp(8)*log(2) + 8*x^2*exp(8)*log(2)^2)/(4*x*log(x)^2 + 9*x^3*exp(8) - log(x)*(12*x^2*exp(4) + 8*x*exp(4)*l
og(2)) + 4*x*exp(8)*log(2)^2 + 12*x^2*exp(8)*log(2)), x)

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sympy [A]  time = 0.14, size = 27, normalized size = 1.12 \begin {gather*} x^{2} - \frac {5 e^{4}}{- 3 x e^{4} + 2 \log {\relax (x )} - 2 e^{4} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**2*ln(x)**2+(-16*x**2*exp(4)*ln(2)-24*x**3*exp(4))*ln(x)+8*x**2*exp(4)**2*ln(2)**2+24*x**3*exp(
4)**2*ln(2)+(18*x**4-15*x)*exp(4)**2+10*exp(4))/(4*x*ln(x)**2+(-8*x*exp(4)*ln(2)-12*x**2*exp(4))*ln(x)+4*x*exp
(4)**2*ln(2)**2+12*x**2*exp(4)**2*ln(2)+9*x**3*exp(4)**2),x)

[Out]

x**2 - 5*exp(4)/(-3*x*exp(4) + 2*log(x) - 2*exp(4)*log(2))

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