3.50.37 \(\int \frac {-8-12 x-4 x^2-100 x^4-50 x^5+(-4-4 x-300 x^3-150 x^4) \log (x)+(-300 x^2-150 x^3) \log ^2(x)+(-100 x-50 x^2) \log ^3(x)}{25 x^4+75 x^3 \log (x)+75 x^2 \log ^2(x)+25 x \log ^3(x)} \, dx\)

Optimal. Leaf size=30 \[ -4-4 x-x^2-\log (3)+\frac {1}{25} \left (1+\frac {2}{x+\log (x)}\right )^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.52, antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 4, integrand size = 100, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {6688, 12, 6742, 6686} \begin {gather*} -(x+2)^2+\frac {4}{25 (x+\log (x))}+\frac {4}{25 (x+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 - 12*x - 4*x^2 - 100*x^4 - 50*x^5 + (-4 - 4*x - 300*x^3 - 150*x^4)*Log[x] + (-300*x^2 - 150*x^3)*Log[x
]^2 + (-100*x - 50*x^2)*Log[x]^3)/(25*x^4 + 75*x^3*Log[x] + 75*x^2*Log[x]^2 + 25*x*Log[x]^3),x]

[Out]

-(2 + x)^2 + 4/(25*(x + Log[x])^2) + 4/(25*(x + 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 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 {2 \left (-4-6 x-2 x^2-50 x^4-25 x^5-\left (2+2 x+150 x^3+75 x^4\right ) \log (x)-75 x^2 (2+x) \log ^2(x)-25 x (2+x) \log ^3(x)\right )}{25 x (x+\log (x))^3} \, dx\\ &=\frac {2}{25} \int \frac {-4-6 x-2 x^2-50 x^4-25 x^5-\left (2+2 x+150 x^3+75 x^4\right ) \log (x)-75 x^2 (2+x) \log ^2(x)-25 x (2+x) \log ^3(x)}{x (x+\log (x))^3} \, dx\\ &=\frac {2}{25} \int \left (-25 (2+x)-\frac {4 (1+x)}{x (x+\log (x))^3}-\frac {2 (1+x)}{x (x+\log (x))^2}\right ) \, dx\\ &=-(2+x)^2-\frac {4}{25} \int \frac {1+x}{x (x+\log (x))^2} \, dx-\frac {8}{25} \int \frac {1+x}{x (x+\log (x))^3} \, dx\\ &=-(2+x)^2+\frac {4}{25 (x+\log (x))^2}+\frac {4}{25 (x+\log (x))}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.23, size = 28, normalized size = 0.93 \begin {gather*} -\frac {2}{25} \left (50 x+\frac {25 x^2}{2}-\frac {2 (1+x+\log (x))}{(x+\log (x))^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 12*x - 4*x^2 - 100*x^4 - 50*x^5 + (-4 - 4*x - 300*x^3 - 150*x^4)*Log[x] + (-300*x^2 - 150*x^3)
*Log[x]^2 + (-100*x - 50*x^2)*Log[x]^3)/(25*x^4 + 75*x^3*Log[x] + 75*x^2*Log[x]^2 + 25*x*Log[x]^3),x]

[Out]

(-2*(50*x + (25*x^2)/2 - (2*(1 + x + Log[x]))/(x + Log[x])^2))/25

________________________________________________________________________________________

fricas [B]  time = 0.68, size = 61, normalized size = 2.03 \begin {gather*} -\frac {25 \, x^{4} + 100 \, x^{3} + 25 \, {\left (x^{2} + 4 \, x\right )} \log \relax (x)^{2} + 2 \, {\left (25 \, x^{3} + 100 \, x^{2} - 2\right )} \log \relax (x) - 4 \, x - 4}{25 \, {\left (x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x^2-100*x)*log(x)^3+(-150*x^3-300*x^2)*log(x)^2+(-150*x^4-300*x^3-4*x-4)*log(x)-50*x^5-100*x^4
-4*x^2-12*x-8)/(25*x*log(x)^3+75*x^2*log(x)^2+75*x^3*log(x)+25*x^4),x, algorithm="fricas")

[Out]

-1/25*(25*x^4 + 100*x^3 + 25*(x^2 + 4*x)*log(x)^2 + 2*(25*x^3 + 100*x^2 - 2)*log(x) - 4*x - 4)/(x^2 + 2*x*log(
x) + log(x)^2)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 31, normalized size = 1.03 \begin {gather*} -x^{2} - 4 \, x + \frac {4 \, {\left (x + \log \relax (x) + 1\right )}}{25 \, {\left (x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x^2-100*x)*log(x)^3+(-150*x^3-300*x^2)*log(x)^2+(-150*x^4-300*x^3-4*x-4)*log(x)-50*x^5-100*x^4
-4*x^2-12*x-8)/(25*x*log(x)^3+75*x^2*log(x)^2+75*x^3*log(x)+25*x^4),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 23, normalized size = 0.77

method	result	size
risch	\(-x^{2}-4 x +\frac {\frac {4}{25}+\frac {4 x}{25}+\frac {4 \ln \relax (x )}{25}}{\left (x +\ln \relax (x )\right )^{2}}\)	\(23\)
norman	\(\frac {\frac {4}{25}+2 \ln \relax (x )^{3}-6 x^{2} \ln \relax (x )+\frac {4 x}{25}-4 x^{3}-x^{4}-x^{2} \ln \relax (x )^{2}-2 x^{3} \ln \relax (x )+\frac {4 \ln \relax (x )}{25}}{\left (x +\ln \relax (x )\right )^{2}}-2 \ln \relax (x )\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-50*x^2-100*x)*ln(x)^3+(-150*x^3-300*x^2)*ln(x)^2+(-150*x^4-300*x^3-4*x-4)*ln(x)-50*x^5-100*x^4-4*x^2-12
*x-8)/(25*x*ln(x)^3+75*x^2*ln(x)^2+75*x^3*ln(x)+25*x^4),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.39, size = 61, normalized size = 2.03 \begin {gather*} -\frac {25 \, x^{4} + 100 \, x^{3} + 25 \, {\left (x^{2} + 4 \, x\right )} \log \relax (x)^{2} + 2 \, {\left (25 \, x^{3} + 100 \, x^{2} - 2\right )} \log \relax (x) - 4 \, x - 4}{25 \, {\left (x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x^2-100*x)*log(x)^3+(-150*x^3-300*x^2)*log(x)^2+(-150*x^4-300*x^3-4*x-4)*log(x)-50*x^5-100*x^4
-4*x^2-12*x-8)/(25*x*log(x)^3+75*x^2*log(x)^2+75*x^3*log(x)+25*x^4),x, algorithm="maxima")

[Out]

-1/25*(25*x^4 + 100*x^3 + 25*(x^2 + 4*x)*log(x)^2 + 2*(25*x^3 + 100*x^2 - 2)*log(x) - 4*x - 4)/(x^2 + 2*x*log(
x) + log(x)^2)

________________________________________________________________________________________

mupad [B]  time = 4.43, size = 25, normalized size = 0.83 \begin {gather*} \frac {\frac {4\,x}{25}+\frac {4\,\ln \relax (x)}{25}+\frac {4}{25}}{{\left (x+\ln \relax (x)\right )}^2}-x^2-4\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x + log(x)^3*(100*x + 50*x^2) + log(x)^2*(300*x^2 + 150*x^3) + 4*x^2 + 100*x^4 + 50*x^5 + log(x)*(4*x
+ 300*x^3 + 150*x^4 + 4) + 8)/(25*x*log(x)^3 + 75*x^3*log(x) + 75*x^2*log(x)^2 + 25*x^4),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.16, size = 34, normalized size = 1.13 \begin {gather*} - x^{2} - 4 x + \frac {4 x + 4 \log {\relax (x )} + 4}{25 x^{2} + 50 x \log {\relax (x )} + 25 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*x**2-100*x)*ln(x)**3+(-150*x**3-300*x**2)*ln(x)**2+(-150*x**4-300*x**3-4*x-4)*ln(x)-50*x**5-10
0*x**4-4*x**2-12*x-8)/(25*x*ln(x)**3+75*x**2*ln(x)**2+75*x**3*ln(x)+25*x**4),x)

[Out]

-x**2 - 4*x + (4*x + 4*log(x) + 4)/(25*x**2 + 50*x*log(x) + 25*log(x)**2)

________________________________________________________________________________________