3.3.63 \(\int \frac {325 x^2+(8-75 x^2) \log (x)+3 \log ^2(x)-\log ^3(x)}{-100 x^3+25 x^3 \log (x)-4 x \log ^2(x)+x \log ^3(x)} \, dx\)

Optimal. Leaf size=23 \[ \log \left (\frac {-4+\log (x)}{x \left (-25 x^2-\log ^2(x)\right )}\right ) \]

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Rubi [A]  time = 0.57, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6741, 6742, 2302, 29, 2541} \begin {gather*} -\log \left (25 x^2+\log ^2(x)\right )-\log (x)+\log (4-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(325*x^2 + (8 - 75*x^2)*Log[x] + 3*Log[x]^2 - Log[x]^3)/(-100*x^3 + 25*x^3*Log[x] - 4*x*Log[x]^2 + x*Log[x
]^3),x]

[Out]

-Log[x] + Log[4 - Log[x]] - Log[25*x^2 + Log[x]^2]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2541

Int[(Log[(c_.)*(x_)^(n_.)]^(r_.)*(e_.) + (d_.)*(x_)^(m_.))/((x_)*(Log[(c_.)*(x_)^(n_.)]^(q_)*(b_.) + (a_.)*(x_
)^(m_.))), x_Symbol] :> Simp[(e*Log[a*x^m + b*Log[c*x^n]^q])/(b*n*q), x] /; FreeQ[{a, b, c, d, e, m, n, q, r},
x] && EqQ[r, q - 1] && EqQ[a*e*m - b*d*n*q, 0]

Rule 6741

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

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 {-325 x^2-\left (8-75 x^2\right ) \log (x)-3 \log ^2(x)+\log ^3(x)}{x (4-\log (x)) \left (25 x^2+\log ^2(x)\right )} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {1}{x (-4+\log (x))}-\frac {2 \left (25 x^2+\log (x)\right )}{x \left (25 x^2+\log ^2(x)\right )}\right ) \, dx\\ &=-\log (x)-2 \int \frac {25 x^2+\log (x)}{x \left (25 x^2+\log ^2(x)\right )} \, dx+\int \frac {1}{x (-4+\log (x))} \, dx\\ &=-\log (x)-\log \left (25 x^2+\log ^2(x)\right )+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-4+\log (x)\right )\\ &=-\log (x)+\log (4-\log (x))-\log \left (25 x^2+\log ^2(x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 25, normalized size = 1.09 \begin {gather*} -\log (x)+\log (4-\log (x))-\log \left (25 x^2+\log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(325*x^2 + (8 - 75*x^2)*Log[x] + 3*Log[x]^2 - Log[x]^3)/(-100*x^3 + 25*x^3*Log[x] - 4*x*Log[x]^2 + x
*Log[x]^3),x]

[Out]

-Log[x] + Log[4 - Log[x]] - Log[25*x^2 + Log[x]^2]

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fricas [A]  time = 0.68, size = 23, normalized size = 1.00 \begin {gather*} -\log \left (25 \, x^{2} + \log \relax (x)^{2}\right ) - \log \relax (x) + \log \left (\log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^3+3*log(x)^2+(-75*x^2+8)*log(x)+325*x^2)/(x*log(x)^3-4*x*log(x)^2+25*x^3*log(x)-100*x^3),x,
algorithm="fricas")

[Out]

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

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giac [A]  time = 0.82, size = 23, normalized size = 1.00 \begin {gather*} -\log \left (25 \, x^{2} + \log \relax (x)^{2}\right ) - \log \relax (x) + \log \left (\log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^3+3*log(x)^2+(-75*x^2+8)*log(x)+325*x^2)/(x*log(x)^3-4*x*log(x)^2+25*x^3*log(x)-100*x^3),x,
algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 24, normalized size = 1.04

method	result	size
norman	\(-\ln \relax (x )-\ln \left (25 x^{2}+\ln \relax (x )^{2}\right )+\ln \left (\ln \relax (x )-4\right )\)	\(24\)
risch	\(-\ln \relax (x )-\ln \left (25 x^{2}+\ln \relax (x )^{2}\right )+\ln \left (\ln \relax (x )-4\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)^3+3*ln(x)^2+(-75*x^2+8)*ln(x)+325*x^2)/(x*ln(x)^3-4*x*ln(x)^2+25*x^3*ln(x)-100*x^3),x,method=_RETU
RNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 23, normalized size = 1.00 \begin {gather*} -\log \left (25 \, x^{2} + \log \relax (x)^{2}\right ) - \log \relax (x) + \log \left (\log \relax (x) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^3+3*log(x)^2+(-75*x^2+8)*log(x)+325*x^2)/(x*log(x)^3-4*x*log(x)^2+25*x^3*log(x)-100*x^3),x,
algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.44, size = 23, normalized size = 1.00 \begin {gather*} \ln \left (\ln \relax (x)-4\right )-\ln \left (25\,x^2+{\ln \relax (x)}^2\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*log(x)^2 - log(x)^3 + 325*x^2 - log(x)*(75*x^2 - 8))/(4*x*log(x)^2 - x*log(x)^3 - 25*x^3*log(x) + 100*
x^3),x)

[Out]

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

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sympy [A]  time = 0.27, size = 20, normalized size = 0.87 \begin {gather*} - \log {\relax (x )} - \log {\left (25 x^{2} + \log {\relax (x )}^{2} \right )} + \log {\left (\log {\relax (x )} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)**3+3*ln(x)**2+(-75*x**2+8)*ln(x)+325*x**2)/(x*ln(x)**3-4*x*ln(x)**2+25*x**3*ln(x)-100*x**3),
x)

[Out]

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

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