3.73.12 \(\int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log (x^2)}{5 x^2+10 x^3+5 x^4} \, dx\)

Optimal. Leaf size=23 \[ \log (2 x)-\frac {1296 \left (x+\log \left (x^2\right )\right )}{5 x (1+x)} \]

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Rubi [A]  time = 0.33, antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 16, number of rules used = 10, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1594, 27, 12, 6742, 44, 43, 2357, 2304, 2314, 31} \begin {gather*} -\frac {1296 \log \left (x^2\right )}{5 x}-\frac {1296 x \log \left (x^2\right )}{5 (x+1)}-\frac {1296}{5 (x+1)}+\frac {2597 \log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2592 - 2587*x + 1306*x^2 + 5*x^3 + (1296 + 2592*x)*Log[x^2])/(5*x^2 + 10*x^3 + 5*x^4),x]

[Out]

-1296/(5*(1 + x)) + (2597*Log[x])/5 - (1296*Log[x^2])/(5*x) - (1296*x*Log[x^2])/(5*(1 + 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 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2314

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

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 {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{x^2 \left (5+10 x+5 x^2\right )} \, dx\\ &=\int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2 (1+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{x^2 (1+x)^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {1306}{(1+x)^2}-\frac {2592}{x^2 (1+x)^2}-\frac {2587}{x (1+x)^2}+\frac {5 x}{(1+x)^2}+\frac {1296 (1+2 x) \log \left (x^2\right )}{x^2 (1+x)^2}\right ) \, dx\\ &=-\frac {1306}{5 (1+x)}+\frac {1296}{5} \int \frac {(1+2 x) \log \left (x^2\right )}{x^2 (1+x)^2} \, dx-\frac {2587}{5} \int \frac {1}{x (1+x)^2} \, dx-\frac {2592}{5} \int \frac {1}{x^2 (1+x)^2} \, dx+\int \frac {x}{(1+x)^2} \, dx\\ &=-\frac {1306}{5 (1+x)}+\frac {1296}{5} \int \left (\frac {\log \left (x^2\right )}{x^2}-\frac {\log \left (x^2\right )}{(1+x)^2}\right ) \, dx-\frac {2587}{5} \int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^2}\right ) \, dx-\frac {2592}{5} \int \left (\frac {1}{x^2}-\frac {2}{x}+\frac {1}{(1+x)^2}+\frac {2}{1+x}\right ) \, dx+\int \left (-\frac {1}{(1+x)^2}+\frac {1}{1+x}\right ) \, dx\\ &=\frac {2592}{5 x}-\frac {1296}{5 (1+x)}+\frac {2597 \log (x)}{5}-\frac {2592}{5} \log (1+x)+\frac {1296}{5} \int \frac {\log \left (x^2\right )}{x^2} \, dx-\frac {1296}{5} \int \frac {\log \left (x^2\right )}{(1+x)^2} \, dx\\ &=-\frac {1296}{5 (1+x)}+\frac {2597 \log (x)}{5}-\frac {1296 \log \left (x^2\right )}{5 x}-\frac {1296 x \log \left (x^2\right )}{5 (1+x)}-\frac {2592}{5} \log (1+x)+\frac {2592}{5} \int \frac {1}{1+x} \, dx\\ &=-\frac {1296}{5 (1+x)}+\frac {2597 \log (x)}{5}-\frac {1296 \log \left (x^2\right )}{5 x}-\frac {1296 x \log \left (x^2\right )}{5 (1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 1.09 \begin {gather*} \frac {1}{5} \left (5 \log (x)-\frac {1296 \left (x+\log \left (x^2\right )\right )}{x (1+x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2592 - 2587*x + 1306*x^2 + 5*x^3 + (1296 + 2592*x)*Log[x^2])/(5*x^2 + 10*x^3 + 5*x^4),x]

[Out]

(5*Log[x] - (1296*(x + Log[x^2]))/(x*(1 + x)))/5

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fricas [A]  time = 1.04, size = 28, normalized size = 1.22 \begin {gather*} \frac {{\left (5 \, x^{2} + 5 \, x - 2592\right )} \log \left (x^{2}\right ) - 2592 \, x}{10 \, {\left (x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2592*x+1296)*log(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^3+5*x^2),x, algorithm="fricas")

[Out]

1/10*((5*x^2 + 5*x - 2592)*log(x^2) - 2592*x)/(x^2 + x)

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giac [A]  time = 0.15, size = 27, normalized size = 1.17 \begin {gather*} \frac {1296}{5} \, {\left (\frac {1}{x + 1} - \frac {1}{x}\right )} \log \left (x^{2}\right ) - \frac {1296}{5 \, {\left (x + 1\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2592*x+1296)*log(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^3+5*x^2),x, algorithm="giac")

[Out]

1296/5*(1/(x + 1) - 1/x)*log(x^2) - 1296/5/(x + 1) + log(x)

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maple [A]  time = 0.05, size = 23, normalized size = 1.00

method	result	size
norman	\(\frac {-\frac {1296 x}{5}-\frac {1296 \ln \left (x^{2}\right )}{5}}{\left (x +1\right ) x}+\ln \relax (x )\)	\(23\)
risch	\(-\frac {1296 \ln \left (x^{2}\right )}{5 x \left (x +1\right )}+\frac {5 x \ln \relax (x )+5 \ln \relax (x )-1296}{5 x +5}\)	\(34\)
default	\(\frac {2592}{5 x}+\ln \relax (x )-\frac {1296}{5 \left (x +1\right )}+\frac {-2592-2592 x -1296 \ln \left (x^{2}\right )}{5 x \left (x +1\right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2592*x+1296)*ln(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^3+5*x^2),x,method=_RETURNVERBOSE)

[Out]

(-1296/5*x-1296/5*ln(x^2))/(x+1)/x+ln(x)

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maxima [A]  time = 0.42, size = 38, normalized size = 1.65 \begin {gather*} \frac {2592 \, {\left (2 \, x + 1\right )}}{5 \, {\left (x^{2} + x\right )}} - \frac {2592 \, {\left (x + \log \relax (x) + 1\right )}}{5 \, {\left (x^{2} + x\right )}} - \frac {3888}{5 \, {\left (x + 1\right )}} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2592*x+1296)*log(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^3+5*x^2),x, algorithm="maxima")

[Out]

2592/5*(2*x + 1)/(x^2 + x) - 2592/5*(x + log(x) + 1)/(x^2 + x) - 3888/5/(x + 1) + log(x)

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mupad [B]  time = 4.38, size = 27, normalized size = 1.17 \begin {gather*} \frac {\ln \left (x^2\right )}{2}-\frac {\frac {1296\,x}{5}+\frac {1296\,\ln \left (x^2\right )}{5}}{x\,\left (x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1306*x^2 - 2587*x + 5*x^3 + log(x^2)*(2592*x + 1296) - 2592)/(5*x^2 + 10*x^3 + 5*x^4),x)

[Out]

log(x^2)/2 - ((1296*x)/5 + (1296*log(x^2))/5)/(x*(x + 1))

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sympy [A]  time = 0.21, size = 24, normalized size = 1.04 \begin {gather*} \log {\relax (x )} - \frac {1296 \log {\left (x^{2} \right )}}{5 x^{2} + 5 x} - \frac {1296}{5 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2592*x+1296)*ln(x**2)+5*x**3+1306*x**2-2587*x-2592)/(5*x**4+10*x**3+5*x**2),x)

[Out]

log(x) - 1296*log(x**2)/(5*x**2 + 5*x) - 1296/(5*x + 5)

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