3.93.26 \(\int \frac {40+15 x+e^{4 e^e} (20 x+5 x^2)}{e^{8 e^e} (4 x^3+x^4)+e^{4 e^e} (-8 x^2-2 x^3) \log (-\frac {2}{4 x^2+x^3})+(4 x+x^2) \log ^2(-\frac {2}{4 x^2+x^3})} \, dx\)

Optimal. Leaf size=28 \[ \frac {5}{-e^{4 e^e} x+\log \left (\frac {2}{(-4-x) x^2}\right )} \]

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Rubi [A]  time = 0.54, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, integrand size = 100, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {5}{e^{4 e^e} x-\log \left (-\frac {2}{x^2 (x+4)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 + 15*x + E^(4*E^E)*(20*x + 5*x^2))/(E^(8*E^E)*(4*x^3 + x^4) + E^(4*E^E)*(-8*x^2 - 2*x^3)*Log[-2/(4*x^2
+ x^3)] + (4*x + x^2)*Log[-2/(4*x^2 + x^3)]^2),x]

[Out]

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

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 26, normalized size = 0.93 \begin {gather*} \frac {5}{-e^{4 e^e} x+\log \left (-\frac {2}{x^2 (4+x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 + 15*x + E^(4*E^E)*(20*x + 5*x^2))/(E^(8*E^E)*(4*x^3 + x^4) + E^(4*E^E)*(-8*x^2 - 2*x^3)*Log[-2/
(4*x^2 + x^3)] + (4*x + x^2)*Log[-2/(4*x^2 + x^3)]^2),x]

[Out]

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

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fricas [A]  time = 0.82, size = 29, normalized size = 1.04 \begin {gather*} -\frac {5}{x e^{\left (4 \, e^{e}\right )} - \log \left (-\frac {2}{x^{3} + 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+20*x)*exp(4*exp(exp(1)))+15*x+40)/((x^4+4*x^3)*exp(4*exp(exp(1)))^2+(-2*x^3-8*x^2)*log(-2/(x
^3+4*x^2))*exp(4*exp(exp(1)))+(x^2+4*x)*log(-2/(x^3+4*x^2))^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.37, size = 29, normalized size = 1.04 \begin {gather*} -\frac {5}{x e^{\left (4 \, e^{e}\right )} - \log \left (-\frac {2}{x^{3} + 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+20*x)*exp(4*exp(exp(1)))+15*x+40)/((x^4+4*x^3)*exp(4*exp(exp(1)))^2+(-2*x^3-8*x^2)*log(-2/(x
^3+4*x^2))*exp(4*exp(exp(1)))+(x^2+4*x)*log(-2/(x^3+4*x^2))^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.21, size = 30, normalized size = 1.07

method	result	size
norman	\(-\frac {5}{x \,{\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}}}-\ln \left (-\frac {2}{x^{3}+4 x^{2}}\right )}\)	\(30\)
risch	\(-\frac {5}{x \,{\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}}}-\ln \left (-\frac {2}{x^{3}+4 x^{2}}\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2+20*x)*exp(4*exp(exp(1)))+15*x+40)/((x^4+4*x^3)*exp(4*exp(exp(1)))^2+(-2*x^3-8*x^2)*ln(-2/(x^3+4*x^
2))*exp(4*exp(exp(1)))+(x^2+4*x)*ln(-2/(x^3+4*x^2))^2),x,method=_RETURNVERBOSE)

[Out]

-5/(x*exp(4*exp(exp(1)))-ln(-2/(x^3+4*x^2)))

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maxima [A]  time = 0.53, size = 27, normalized size = 0.96 \begin {gather*} -\frac {5}{x e^{\left (4 \, e^{e}\right )} - \log \relax (2) + 2 \, \log \relax (x) + \log \left (-x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2+20*x)*exp(4*exp(exp(1)))+15*x+40)/((x^4+4*x^3)*exp(4*exp(exp(1)))^2+(-2*x^3-8*x^2)*log(-2/(x
^3+4*x^2))*exp(4*exp(exp(1)))+(x^2+4*x)*log(-2/(x^3+4*x^2))^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 10.16, size = 28, normalized size = 1.00 \begin {gather*} \frac {5}{\ln \left (-\frac {2}{x^3+4\,x^2}\right )-x\,{\mathrm {e}}^{4\,{\mathrm {e}}^{\mathrm {e}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x + exp(4*exp(exp(1)))*(20*x + 5*x^2) + 40)/(exp(8*exp(exp(1)))*(4*x^3 + x^4) + log(-2/(4*x^2 + x^3))^
2*(4*x + x^2) - exp(4*exp(exp(1)))*log(-2/(4*x^2 + x^3))*(8*x^2 + 2*x^3)),x)

[Out]

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

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sympy [A]  time = 0.21, size = 24, normalized size = 0.86 \begin {gather*} \frac {5}{- x e^{4 e^{e}} + \log {\left (- \frac {2}{x^{3} + 4 x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2+20*x)*exp(4*exp(exp(1)))+15*x+40)/((x**4+4*x**3)*exp(4*exp(exp(1)))**2+(-2*x**3-8*x**2)*ln(
-2/(x**3+4*x**2))*exp(4*exp(exp(1)))+(x**2+4*x)*ln(-2/(x**3+4*x**2))**2),x)

[Out]

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

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