3.22.56 \(\int \frac {4+8 x+4 x^2}{5 x^2+4 x^3+x^4+e^6 (5+4 x+x^2)+e^3 (10 x+8 x^2+2 x^3)+(-10 x-8 x^2-2 x^3+e^3 (-10-8 x-2 x^2)) \log (50+40 x+10 x^2-(25+20 x+5 x^2) \log (\frac {25}{3}))+(5+4 x+x^2) \log ^2(50+40 x+10 x^2-(25+20 x+5 x^2) \log (\frac {25}{3}))} \, dx\)

Optimal. Leaf size=31 \[ \frac {4}{-e^3-x+\log \left (5 \left (1+(2+x)^2\right ) \left (2-\log \left (\frac {25}{3}\right )\right )\right )} \]

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Rubi [A]  time = 0.32, antiderivative size = 30, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, integrand size = 151, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {27, 12, 6688, 6686} \begin {gather*} -\frac {4}{-\log \left (5 \left (x^2+4 x+5\right ) \left (2-\log \left (\frac {25}{3}\right )\right )\right )+x+e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 8*x + 4*x^2)/(5*x^2 + 4*x^3 + x^4 + E^6*(5 + 4*x + x^2) + E^3*(10*x + 8*x^2 + 2*x^3) + (-10*x - 8*x^2
- 2*x^3 + E^3*(-10 - 8*x - 2*x^2))*Log[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]] + (5 + 4*x + x^2)*
Log[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]]^2),x]

[Out]

-4/(E^3 + x - Log[5*(5 + 4*x + x^2)*(2 - Log[25/3])])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 {4 (1+x)^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx\\ &=4 \int \frac {(1+x)^2}{5 x^2+4 x^3+x^4+e^6 \left (5+4 x+x^2\right )+e^3 \left (10 x+8 x^2+2 x^3\right )+\left (-10 x-8 x^2-2 x^3+e^3 \left (-10-8 x-2 x^2\right )\right ) \log \left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )+\left (5+4 x+x^2\right ) \log ^2\left (50+40 x+10 x^2-\left (25+20 x+5 x^2\right ) \log \left (\frac {25}{3}\right )\right )} \, dx\\ &=4 \int \frac {(1+x)^2}{\left (5+4 x+x^2\right ) \left (e^3+x-\log \left (-5 \left (5+4 x+x^2\right ) \left (-2+\log \left (\frac {25}{3}\right )\right )\right )\right )^2} \, dx\\ &=-\frac {4}{e^3+x-\log \left (5 \left (5+4 x+x^2\right ) \left (2-\log \left (\frac {25}{3}\right )\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.90 \begin {gather*} -\frac {4}{e^3+x-\log \left (-5 \left (5+4 x+x^2\right ) \left (-2+\log \left (\frac {25}{3}\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 8*x + 4*x^2)/(5*x^2 + 4*x^3 + x^4 + E^6*(5 + 4*x + x^2) + E^3*(10*x + 8*x^2 + 2*x^3) + (-10*x -
8*x^2 - 2*x^3 + E^3*(-10 - 8*x - 2*x^2))*Log[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]] + (5 + 4*x +
x^2)*Log[50 + 40*x + 10*x^2 - (25 + 20*x + 5*x^2)*Log[25/3]]^2),x]

[Out]

-4/(E^3 + x - Log[-5*(5 + 4*x + x^2)*(-2 + Log[25/3])])

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fricas [A]  time = 0.84, size = 33, normalized size = 1.06 \begin {gather*} -\frac {4}{x + e^{3} - \log \left (10 \, x^{2} + 5 \, {\left (x^{2} + 4 \, x + 5\right )} \log \left (\frac {3}{25}\right ) + 40 \, x + 50\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+8*x+4)/((x^2+4*x+5)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+40*x+50)^2+((-2*x^2-8*x-10)*exp(3)-2
*x^3-8*x^2-10*x)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+
x^4+4*x^3+5*x^2),x, algorithm="fricas")

[Out]

-4/(x + e^3 - log(10*x^2 + 5*(x^2 + 4*x + 5)*log(3/25) + 40*x + 50))

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giac [A]  time = 0.43, size = 40, normalized size = 1.29 \begin {gather*} -\frac {4}{x + e^{3} - \log \relax (5) - \log \left (x^{2} \log \left (\frac {3}{25}\right ) + 2 \, x^{2} + 4 \, x \log \left (\frac {3}{25}\right ) + 8 \, x + 5 \, \log \left (\frac {3}{25}\right ) + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+8*x+4)/((x^2+4*x+5)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+40*x+50)^2+((-2*x^2-8*x-10)*exp(3)-2
*x^3-8*x^2-10*x)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+
x^4+4*x^3+5*x^2),x, algorithm="giac")

[Out]

-4/(x + e^3 - log(5) - log(x^2*log(3/25) + 2*x^2 + 4*x*log(3/25) + 8*x + 5*log(3/25) + 10))

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maple [A]  time = 0.65, size = 35, normalized size = 1.13

method	result	size
norman	\(-\frac {4}{{\mathrm e}^{3}+x -\ln \left (\left (5 x^{2}+20 x +25\right ) \ln \left (\frac {3}{25}\right )+10 x^{2}+40 x +50\right )}\)	\(35\)
risch	\(-\frac {4}{{\mathrm e}^{3}+x -\ln \left (\left (5 x^{2}+20 x +25\right ) \left (\ln \relax (3)-2 \ln \relax (5)\right )+10 x^{2}+40 x +50\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+8*x+4)/((x^2+4*x+5)*ln((5*x^2+20*x+25)*ln(3/25)+10*x^2+40*x+50)^2+((-2*x^2-8*x-10)*exp(3)-2*x^3-8*x
^2-10*x)*ln((5*x^2+20*x+25)*ln(3/25)+10*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+x^4+4*x^3+
5*x^2),x,method=_RETURNVERBOSE)

[Out]

-4/(exp(3)+x-ln((5*x^2+20*x+25)*ln(3/25)+10*x^2+40*x+50))

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maxima [C]  time = 0.66, size = 37, normalized size = 1.19 \begin {gather*} \frac {4}{i \, \pi - x - e^{3} + \log \relax (5) + \log \left (x^{2} + 4 \, x + 5\right ) + \log \left (2 \, \log \relax (5) - \log \relax (3) - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+8*x+4)/((x^2+4*x+5)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+40*x+50)^2+((-2*x^2-8*x-10)*exp(3)-2
*x^3-8*x^2-10*x)*log((5*x^2+20*x+25)*log(3/25)+10*x^2+40*x+50)+(x^2+4*x+5)*exp(3)^2+(2*x^3+8*x^2+10*x)*exp(3)+
x^4+4*x^3+5*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.11, size = 34, normalized size = 1.10 \begin {gather*} -\frac {4}{x-\ln \left (40\,x+\ln \left (\frac {3}{25}\right )\,\left (5\,x^2+20\,x+25\right )+10\,x^2+50\right )+{\mathrm {e}}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 4*x^2 + 4)/(exp(3)*(10*x + 8*x^2 + 2*x^3) - log(40*x + log(3/25)*(20*x + 5*x^2 + 25) + 10*x^2 + 50)
*(10*x + exp(3)*(8*x + 2*x^2 + 10) + 8*x^2 + 2*x^3) + exp(6)*(4*x + x^2 + 5) + 5*x^2 + 4*x^3 + x^4 + log(40*x
+ log(3/25)*(20*x + 5*x^2 + 25) + 10*x^2 + 50)^2*(4*x + x^2 + 5)),x)

[Out]

-4/(x - log(40*x + log(3/25)*(20*x + 5*x^2 + 25) + 10*x^2 + 50) + exp(3))

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sympy [A]  time = 0.26, size = 32, normalized size = 1.03 \begin {gather*} \frac {4}{- x + \log {\left (10 x^{2} + 40 x + \left (5 x^{2} + 20 x + 25\right ) \log {\left (\frac {3}{25} \right )} + 50 \right )} - e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+8*x+4)/((x**2+4*x+5)*ln((5*x**2+20*x+25)*ln(3/25)+10*x**2+40*x+50)**2+((-2*x**2-8*x-10)*exp(
3)-2*x**3-8*x**2-10*x)*ln((5*x**2+20*x+25)*ln(3/25)+10*x**2+40*x+50)+(x**2+4*x+5)*exp(3)**2+(2*x**3+8*x**2+10*
x)*exp(3)+x**4+4*x**3+5*x**2),x)

[Out]

4/(-x + log(10*x**2 + 40*x + (5*x**2 + 20*x + 25)*log(3/25) + 50) - exp(3))

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