3.1.9 \(\int \frac {2400+1176 x+102 x^2-6 x^3+(800+392 x+34 x^2-2 x^3) \log (\frac {x+x^2+(-100-25 x) \log (x)}{100+25 x})}{-4 x^2-5 x^3-x^4+(400 x+200 x^2+25 x^3) \log (x)} \, dx\)

Optimal. Leaf size=24 \[ \left (3+\log \left (\frac {x+x^2}{25 (4+x)}-\log (x)\right )\right )^2 \]

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Rubi [A]  time = 0.28, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 12, 6686} \begin {gather*} \left (\log \left (\frac {x (x+1)}{25 (x+4)}-\log (x)\right )+3\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2400 + 1176*x + 102*x^2 - 6*x^3 + (800 + 392*x + 34*x^2 - 2*x^3)*Log[(x + x^2 + (-100 - 25*x)*Log[x])/(10
0 + 25*x)])/(-4*x^2 - 5*x^3 - x^4 + (400*x + 200*x^2 + 25*x^3)*Log[x]),x]

[Out]

(3 + Log[(x*(1 + x))/(25*(4 + x)) - Log[x]])^2

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 {2 \left (400+196 x+17 x^2-x^3\right ) \left (-3-\log \left (\frac {x (1+x)}{25 (4+x)}-\log (x)\right )\right )}{x (4+x) (x (1+x)-25 (4+x) \log (x))} \, dx\\ &=2 \int \frac {\left (400+196 x+17 x^2-x^3\right ) \left (-3-\log \left (\frac {x (1+x)}{25 (4+x)}-\log (x)\right )\right )}{x (4+x) (x (1+x)-25 (4+x) \log (x))} \, dx\\ &=\left (3+\log \left (\frac {x (1+x)}{25 (4+x)}-\log (x)\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 23, normalized size = 0.96 \begin {gather*} \left (3+\log \left (\frac {x (1+x)}{25 (4+x)}-\log (x)\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2400 + 1176*x + 102*x^2 - 6*x^3 + (800 + 392*x + 34*x^2 - 2*x^3)*Log[(x + x^2 + (-100 - 25*x)*Log[x
])/(100 + 25*x)])/(-4*x^2 - 5*x^3 - x^4 + (400*x + 200*x^2 + 25*x^3)*Log[x]),x]

[Out]

(3 + Log[(x*(1 + x))/(25*(4 + x)) - Log[x]])^2

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fricas [B]  time = 0.57, size = 45, normalized size = 1.88 \begin {gather*} \log \left (\frac {x^{2} - 25 \, {\left (x + 4\right )} \log \relax (x) + x}{25 \, {\left (x + 4\right )}}\right )^{2} + 6 \, \log \left (\frac {x^{2} - 25 \, {\left (x + 4\right )} \log \relax (x) + x}{25 \, {\left (x + 4\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+34*x^2+392*x+800)*log(((-25*x-100)*log(x)+x^2+x)/(25*x+100))-6*x^3+102*x^2+1176*x+2400)/((2
5*x^3+200*x^2+400*x)*log(x)-x^4-5*x^3-4*x^2),x, algorithm="fricas")

[Out]

log(1/25*(x^2 - 25*(x + 4)*log(x) + x)/(x + 4))^2 + 6*log(1/25*(x^2 - 25*(x + 4)*log(x) + x)/(x + 4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (3 \, x^{3} - 51 \, x^{2} + {\left (x^{3} - 17 \, x^{2} - 196 \, x - 400\right )} \log \left (\frac {x^{2} - 25 \, {\left (x + 4\right )} \log \relax (x) + x}{25 \, {\left (x + 4\right )}}\right ) - 588 \, x - 1200\right )}}{x^{4} + 5 \, x^{3} + 4 \, x^{2} - 25 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+34*x^2+392*x+800)*log(((-25*x-100)*log(x)+x^2+x)/(25*x+100))-6*x^3+102*x^2+1176*x+2400)/((2
5*x^3+200*x^2+400*x)*log(x)-x^4-5*x^3-4*x^2),x, algorithm="giac")

[Out]

integrate(2*(3*x^3 - 51*x^2 + (x^3 - 17*x^2 - 196*x - 400)*log(1/25*(x^2 - 25*(x + 4)*log(x) + x)/(x + 4)) - 5
88*x - 1200)/(x^4 + 5*x^3 + 4*x^2 - 25*(x^3 + 8*x^2 + 16*x)*log(x)), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-2 x^{3}+34 x^{2}+392 x +800\right ) \ln \left (\frac {\left (-25 x -100\right ) \ln \relax (x )+x^{2}+x}{25 x +100}\right )-6 x^{3}+102 x^{2}+1176 x +2400}{\left (25 x^{3}+200 x^{2}+400 x \right ) \ln \relax (x )-x^{4}-5 x^{3}-4 x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3+34*x^2+392*x+800)*ln(((-25*x-100)*ln(x)+x^2+x)/(25*x+100))-6*x^3+102*x^2+1176*x+2400)/((25*x^3+20
0*x^2+400*x)*ln(x)-x^4-5*x^3-4*x^2),x)

[Out]

int(((-2*x^3+34*x^2+392*x+800)*ln(((-25*x-100)*ln(x)+x^2+x)/(25*x+100))-6*x^3+102*x^2+1176*x+2400)/((25*x^3+20
0*x^2+400*x)*ln(x)-x^4-5*x^3-4*x^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, \int \frac {3 \, x^{3} - 51 \, x^{2} + {\left (x^{3} - 17 \, x^{2} - 196 \, x - 400\right )} \log \left (\frac {x^{2} - 25 \, {\left (x + 4\right )} \log \relax (x) + x}{25 \, {\left (x + 4\right )}}\right ) - 588 \, x - 1200}{x^{4} + 5 \, x^{3} + 4 \, x^{2} - 25 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3+34*x^2+392*x+800)*log(((-25*x-100)*log(x)+x^2+x)/(25*x+100))-6*x^3+102*x^2+1176*x+2400)/((2
5*x^3+200*x^2+400*x)*log(x)-x^4-5*x^3-4*x^2),x, algorithm="maxima")

[Out]

2*integrate((3*x^3 - 51*x^2 + (x^3 - 17*x^2 - 196*x - 400)*log(1/25*(x^2 - 25*(x + 4)*log(x) + x)/(x + 4)) - 5
88*x - 1200)/(x^4 + 5*x^3 + 4*x^2 - 25*(x^3 + 8*x^2 + 16*x)*log(x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {1176\,x+\ln \left (\frac {x-\ln \relax (x)\,\left (25\,x+100\right )+x^2}{25\,x+100}\right )\,\left (-2\,x^3+34\,x^2+392\,x+800\right )+102\,x^2-6\,x^3+2400}{4\,x^2+5\,x^3+x^4-\ln \relax (x)\,\left (25\,x^3+200\,x^2+400\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(1176*x + log((x - log(x)*(25*x + 100) + x^2)/(25*x + 100))*(392*x + 34*x^2 - 2*x^3 + 800) + 102*x^2 - 6*
x^3 + 2400)/(4*x^2 + 5*x^3 + x^4 - log(x)*(400*x + 200*x^2 + 25*x^3)),x)

[Out]

-int((1176*x + log((x - log(x)*(25*x + 100) + x^2)/(25*x + 100))*(392*x + 34*x^2 - 2*x^3 + 800) + 102*x^2 - 6*
x^3 + 2400)/(4*x^2 + 5*x^3 + x^4 - log(x)*(400*x + 200*x^2 + 25*x^3)), x)

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sympy [B]  time = 0.70, size = 41, normalized size = 1.71 \begin {gather*} \log {\left (\frac {x^{2} + x + \left (- 25 x - 100\right ) \log {\relax (x )}}{25 x + 100} \right )}^{2} + 6 \log {\left (\log {\relax (x )} + \frac {- x^{2} - x}{25 x + 100} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3+34*x**2+392*x+800)*ln(((-25*x-100)*ln(x)+x**2+x)/(25*x+100))-6*x**3+102*x**2+1176*x+2400)/
((25*x**3+200*x**2+400*x)*ln(x)-x**4-5*x**3-4*x**2),x)

[Out]

log((x**2 + x + (-25*x - 100)*log(x))/(25*x + 100))**2 + 6*log(log(x) + (-x**2 - x)/(25*x + 100))

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