3.99.72 \(\int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+(-4-20 x+(1+5 x) \log ^2(x)) \log (x+5 x^2)}{(2 x^2+10 x^3) \log ^2(x)+(x+5 x^2) \log ^2(x) \log (x+5 x^2)} \, dx\)

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

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Rubi [A]  time = 0.25, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 3, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6688, 14, 6684} \begin {gather*} \log (x)-\log (2 x+\log (x (5 x+1)))+\frac {4}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*x - 40*x^2 + (-1 - 10*x)*Log[x]^2 + (-4 - 20*x + (1 + 5*x)*Log[x]^2)*Log[x + 5*x^2])/((2*x^2 + 10*x^3)
*Log[x]^2 + (x + 5*x^2)*Log[x]^2*Log[x + 5*x^2]),x]

[Out]

4/Log[x] + Log[x] - Log[2*x + Log[x*(1 + 5*x)]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {-\frac {4}{\log ^2(x)}+\frac {-1-10 x+(1+5 x) \log (x (1+5 x))}{(1+5 x) (2 x+\log (x (1+5 x)))}}{x} \, dx\\ &=\int \left (\frac {-4+\log ^2(x)}{x \log ^2(x)}+\frac {-1-12 x-10 x^2}{x (1+5 x) (2 x+\log (x (1+5 x)))}\right ) \, dx\\ &=\int \frac {-4+\log ^2(x)}{x \log ^2(x)} \, dx+\int \frac {-1-12 x-10 x^2}{x (1+5 x) (2 x+\log (x (1+5 x)))} \, dx\\ &=-\log (2 x+\log (x (1+5 x)))+\operatorname {Subst}\left (\int \frac {-4+x^2}{x^2} \, dx,x,\log (x)\right )\\ &=-\log (2 x+\log (x (1+5 x)))+\operatorname {Subst}\left (\int \left (1-\frac {4}{x^2}\right ) \, dx,x,\log (x)\right )\\ &=\frac {4}{\log (x)}+\log (x)-\log (2 x+\log (x (1+5 x)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 24, normalized size = 0.96 \begin {gather*} \frac {4}{\log (x)}+\log (x)-\log (2 x+\log (x (1+5 x))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x - 40*x^2 + (-1 - 10*x)*Log[x]^2 + (-4 - 20*x + (1 + 5*x)*Log[x]^2)*Log[x + 5*x^2])/((2*x^2 + 1
0*x^3)*Log[x]^2 + (x + 5*x^2)*Log[x]^2*Log[x + 5*x^2]),x]

[Out]

4/Log[x] + Log[x] - Log[2*x + Log[x*(1 + 5*x)]]

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fricas [A]  time = 0.61, size = 30, normalized size = 1.20 \begin {gather*} -\frac {\log \left (2 \, x + \log \left (5 \, x^{2} + x\right )\right ) \log \relax (x) - \log \relax (x)^{2} - 4}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+5*x)*log(x)^2-20*x-4)*log(5*x^2+x)+(-10*x-1)*log(x)^2-40*x^2-8*x)/((5*x^2+x)*log(x)^2*log(5*x^2
+x)+(10*x^3+2*x^2)*log(x)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 24, normalized size = 0.96 \begin {gather*} \frac {4}{\log \relax (x)} - \log \left (2 \, x + \log \left (5 \, x + 1\right ) + \log \relax (x)\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+5*x)*log(x)^2-20*x-4)*log(5*x^2+x)+(-10*x-1)*log(x)^2-40*x^2-8*x)/((5*x^2+x)*log(x)^2*log(5*x^2
+x)+(10*x^3+2*x^2)*log(x)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 25, normalized size = 1.00

method	result	size
default	\(\ln \relax (x )-\ln \left (2 x +\ln \left (5 x^{2}+x \right )\right )+\frac {4}{\ln \relax (x )}\)	\(25\)
risch	\(\frac {\ln \relax (x )^{2}+4}{\ln \relax (x )}-\ln \left (\ln \left (\frac {1}{5}+x \right )-\frac {i \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (\frac {1}{5}+x \right )\right ) \mathrm {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\frac {1}{5}+x \right )\right ) \mathrm {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{2}+\pi \mathrm {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{3}+4 i x +2 i \ln \relax (x )\right )}{2}\right )\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((1+5*x)*ln(x)^2-20*x-4)*ln(5*x^2+x)+(-10*x-1)*ln(x)^2-40*x^2-8*x)/((5*x^2+x)*ln(x)^2*ln(5*x^2+x)+(10*x^3
+2*x^2)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 24, normalized size = 0.96 \begin {gather*} \frac {4}{\log \relax (x)} - \log \left (2 \, x + \log \left (5 \, x + 1\right ) + \log \relax (x)\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+5*x)*log(x)^2-20*x-4)*log(5*x^2+x)+(-10*x-1)*log(x)^2-40*x^2-8*x)/((5*x^2+x)*log(x)^2*log(5*x^2
+x)+(10*x^3+2*x^2)*log(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.09, size = 24, normalized size = 0.96 \begin {gather*} \ln \relax (x)-\ln \left (2\,x+\ln \left (x\,\left (5\,x+1\right )\right )\right )+\frac {4}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x + 40*x^2 + log(x)^2*(10*x + 1) + log(x + 5*x^2)*(20*x - log(x)^2*(5*x + 1) + 4))/(log(x)^2*(2*x^2 +
10*x^3) + log(x + 5*x^2)*log(x)^2*(x + 5*x^2)),x)

[Out]

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

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sympy [A]  time = 0.45, size = 20, normalized size = 0.80 \begin {gather*} \log {\relax (x )} - \log {\left (2 x + \log {\left (5 x^{2} + x \right )} \right )} + \frac {4}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1+5*x)*ln(x)**2-20*x-4)*ln(5*x**2+x)+(-10*x-1)*ln(x)**2-40*x**2-8*x)/((5*x**2+x)*ln(x)**2*ln(5*x*
*2+x)+(10*x**3+2*x**2)*ln(x)**2),x)

[Out]

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

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