3.24.37 \(\int \frac {4+4 x-8 x^2-8 x^2 \log (x+x^2)}{15 x} \, dx\)

Optimal. Leaf size=20 \[ 7-\frac {4}{15} x \left (-\frac {1}{x}+x\right ) \log \left (x+x^2\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 9, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {12, 14, 2495, 30, 43} \begin {gather*} -\frac {4}{15} x^2 \log (x (x+1))+\frac {4 \log (x)}{15}+\frac {4}{15} \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 4*x - 8*x^2 - 8*x^2*Log[x + x^2])/(15*x),x]

[Out]

(4*Log[x])/15 + (4*Log[1 + x])/15 - (4*x^2*Log[x*(1 + x)])/15

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2495

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{15} \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{x} \, dx\\ &=\frac {1}{15} \int \left (-\frac {4 \left (-1-x+2 x^2\right )}{x}-8 x \log (x (1+x))\right ) \, dx\\ &=-\left (\frac {4}{15} \int \frac {-1-x+2 x^2}{x} \, dx\right )-\frac {8}{15} \int x \log (x (1+x)) \, dx\\ &=-\frac {4}{15} x^2 \log (x (1+x))+\frac {4 \int x \, dx}{15}+\frac {4}{15} \int \frac {x^2}{1+x} \, dx-\frac {4}{15} \int \left (-1-\frac {1}{x}+2 x\right ) \, dx\\ &=\frac {4 x}{15}-\frac {2 x^2}{15}+\frac {4 \log (x)}{15}-\frac {4}{15} x^2 \log (x (1+x))+\frac {4}{15} \int \left (-1+x+\frac {1}{1+x}\right ) \, dx\\ &=\frac {4 \log (x)}{15}+\frac {4}{15} \log (1+x)-\frac {4}{15} x^2 \log (x (1+x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.25 \begin {gather*} -\frac {4}{15} \left (-\log (x)-\log (1+x)+x^2 \log (x (1+x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 4*x - 8*x^2 - 8*x^2*Log[x + x^2])/(15*x),x]

[Out]

(-4*(-Log[x] - Log[1 + x] + x^2*Log[x*(1 + x)]))/15

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fricas [A]  time = 0.89, size = 13, normalized size = 0.65 \begin {gather*} -\frac {4}{15} \, {\left (x^{2} - 1\right )} \log \left (x^{2} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-8*x^2*log(x^2+x)-8*x^2+4*x+4)/x,x, algorithm="fricas")

[Out]

-4/15*(x^2 - 1)*log(x^2 + x)

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giac [A]  time = 0.26, size = 22, normalized size = 1.10 \begin {gather*} -\frac {4}{15} \, x^{2} \log \left (x^{2} + x\right ) + \frac {4}{15} \, \log \left (x + 1\right ) + \frac {4}{15} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-8*x^2*log(x^2+x)-8*x^2+4*x+4)/x,x, algorithm="giac")

[Out]

-4/15*x^2*log(x^2 + x) + 4/15*log(x + 1) + 4/15*log(x)

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maple [A]  time = 0.06, size = 21, normalized size = 1.05

method	result	size
norman	\(-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (x^{2}+x \right )}{15}\)	\(21\)
risch	\(-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (x^{2}+x \right )}{15}\)	\(21\)
default	\(\frac {4 \ln \relax (x )}{15}-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (x +1\right )}{15}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*(-8*x^2*ln(x^2+x)-8*x^2+4*x+4)/x,x,method=_RETURNVERBOSE)

[Out]

-4/15*x^2*ln(x^2+x)+4/15*ln(x^2+x)

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maxima [A]  time = 0.38, size = 22, normalized size = 1.10 \begin {gather*} -\frac {4}{15} \, x^{2} \log \left (x^{2} + x\right ) + \frac {4}{15} \, \log \left (x + 1\right ) + \frac {4}{15} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-8*x^2*log(x^2+x)-8*x^2+4*x+4)/x,x, algorithm="maxima")

[Out]

-4/15*x^2*log(x^2 + x) + 4/15*log(x + 1) + 4/15*log(x)

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mupad [B]  time = 1.47, size = 13, normalized size = 0.65 \begin {gather*} -\frac {4\,\ln \left (x^2+x\right )\,\left (x^2-1\right )}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x)/15 - (8*x^2*log(x + x^2))/15 - (8*x^2)/15 + 4/15)/x,x)

[Out]

-(4*log(x + x^2)*(x^2 - 1))/15

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sympy [A]  time = 0.14, size = 22, normalized size = 1.10 \begin {gather*} - \frac {4 x^{2} \log {\left (x^{2} + x \right )}}{15} + \frac {4 \log {\left (x^{2} + x \right )}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*(-8*x**2*ln(x**2+x)-8*x**2+4*x+4)/x,x)

[Out]

-4*x**2*log(x**2 + x)/15 + 4*log(x**2 + x)/15

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