3.13.67 \(\int \frac {4+8 x+x^2+2 x^3+(2 x^2+2 x^3) \log (-1-x)+(2 x^2+2 x^3) \log (x)}{x+x^2} \, dx\)

Optimal. Leaf size=17 \[ 3+\left (4+x^2\right ) (\log (-1-x)+\log (x)) \]

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Rubi [A]  time = 0.46, antiderivative size = 27, normalized size of antiderivative = 1.59, number of steps used = 11, number of rules used = 6, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.113, Rules used = {1593, 6742, 1620, 2395, 43, 2304} \begin {gather*} x^2 \log (-x-1)+x^2 \log (x)+4 \log (x)+4 \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2*Log[-1 - x] + 4*Log[x] + x^2*Log[x] + 4*Log[1 + 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 1593

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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 2395

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

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 {4+8 x+x^2+2 x^3+\left (2 x^2+2 x^3\right ) \log (-1-x)+\left (2 x^2+2 x^3\right ) \log (x)}{x (1+x)} \, dx\\ &=\int \left (\frac {4+8 x+x^2+2 x^3+2 x^2 \log (-1-x)+2 x^3 \log (-1-x)}{x (1+x)}+2 x \log (x)\right ) \, dx\\ &=2 \int x \log (x) \, dx+\int \frac {4+8 x+x^2+2 x^3+2 x^2 \log (-1-x)+2 x^3 \log (-1-x)}{x (1+x)} \, dx\\ &=-\frac {x^2}{2}+x^2 \log (x)+\int \left (\frac {4+8 x+x^2+2 x^3}{x (1+x)}+2 x \log (-1-x)\right ) \, dx\\ &=-\frac {x^2}{2}+x^2 \log (x)+2 \int x \log (-1-x) \, dx+\int \frac {4+8 x+x^2+2 x^3}{x (1+x)} \, dx\\ &=-\frac {x^2}{2}+x^2 \log (-1-x)+x^2 \log (x)+\int \frac {x^2}{-1-x} \, dx+\int \left (-1+\frac {4}{x}+2 x+\frac {5}{1+x}\right ) \, dx\\ &=-x+\frac {x^2}{2}+x^2 \log (-1-x)+4 \log (x)+x^2 \log (x)+5 \log (1+x)+\int \left (1+\frac {1}{-1-x}-x\right ) \, dx\\ &=x^2 \log (-1-x)+4 \log (x)+x^2 \log (x)+4 \log (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 27, normalized size = 1.59 \begin {gather*} x^2 \log (-1-x)+4 \log (x)+x^2 \log (x)+4 \log (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 8*x + x^2 + 2*x^3 + (2*x^2 + 2*x^3)*Log[-1 - x] + (2*x^2 + 2*x^3)*Log[x])/(x + x^2),x]

[Out]

x^2*Log[-1 - x] + 4*Log[x] + x^2*Log[x] + 4*Log[1 + x]

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fricas [A]  time = 0.55, size = 21, normalized size = 1.24 \begin {gather*} {\left (x^{2} + 4\right )} \log \relax (x) + {\left (x^{2} + 4\right )} \log \left (-x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.64, size = 27, normalized size = 1.59 \begin {gather*} x^{2} \log \relax (x) + x^{2} \log \left (-x - 1\right ) + 4 \, \log \left (x + 1\right ) + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.21, size = 26, normalized size = 1.53

method	result	size
risch	\(x^{2} \ln \left (-x -1\right )+x^{2} \ln \relax (x )+4 \ln \left (x^{2}+x \right )\)	\(26\)
default	\(5 \ln \left (x +1\right )+4 \ln \relax (x )+x^{2} \ln \relax (x )+\left (-x -1\right )^{2} \ln \left (-x -1\right )+\frac {3}{2}+2 \left (-x -1\right ) \ln \left (-x -1\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.53, size = 53, normalized size = 3.12 \begin {gather*} x^{2} \log \relax (x) + {\left (x^{2} - 2 \, x + 2 \, \log \left (x + 1\right )\right )} \log \left (-x - 1\right ) + 2 \, {\left (x - \log \left (x + 1\right )\right )} \log \left (-x - 1\right ) + 4 \, \log \left (x + 1\right ) + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.10, size = 25, normalized size = 1.47 \begin {gather*} 4\,\ln \left (x\,\left (x+1\right )\right )+x^2\,\ln \relax (x)+x^2\,\ln \left (-x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.51, size = 32, normalized size = 1.88 \begin {gather*} x^{2} \log {\relax (x )} + \left (x^{2} - \frac {1}{3}\right ) \log {\left (- x - 1 \right )} + 4 \log {\relax (x )} + \frac {13 \log {\left (x + 1 \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+2*x**2)*ln(x)+(2*x**3+2*x**2)*ln(-x-1)+2*x**3+x**2+8*x+4)/(x**2+x),x)

[Out]

x**2*log(x) + (x**2 - 1/3)*log(-x - 1) + 4*log(x) + 13*log(x + 1)/3

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