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

Optimal. Leaf size=23 \[ 4 \log \left ((1+x) \left (-x+\frac {x \log (3 x)}{\log (1+x)}\right )\right ) \]

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Rubi [A]  time = 0.89, antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 9, number of rules used = 7, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.092, Rules used = {6741, 6742, 72, 6684, 2390, 2302, 29} \begin {gather*} 4 \log (x)+4 \log (x+1)+4 \log (\log (3 x)-\log (x+1))-4 \log (\log (x+1)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*Log[x] + 4*Log[1 + x] + 4*Log[Log[3*x] - Log[1 + x]] - 4*Log[Log[1 + x]]

Rule 29

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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 2390

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

Rule 6684

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

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

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Mathematica [A]  time = 0.22, size = 28, normalized size = 1.22 \begin {gather*} 4 (\log (x)+\log (1+x)+\log (\log (3 x)-\log (1+x))-\log (\log (1+x))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*(Log[x] + Log[1 + x] + Log[Log[3*x] - Log[1 + x]] - Log[Log[1 + x]])

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fricas [A]  time = 0.80, size = 30, normalized size = 1.30 \begin {gather*} 4 \, \log \left (x^{2} + x\right ) + 4 \, \log \left (\log \left (3 \, x\right ) - \log \left (x + 1\right )\right ) - 4 \, \log \left (\log \left (x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(x^2 + x) + 4*log(log(3*x) - log(x + 1)) - 4*log(log(x + 1))

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giac [A]  time = 0.17, size = 32, normalized size = 1.39 \begin {gather*} 4 \, \log \left (x + 1\right ) + 4 \, \log \relax (x) + 4 \, \log \left (-\log \left (3 \, x\right ) + \log \left (x + 1\right )\right ) - 4 \, \log \left (\log \left (x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(x + 1) + 4*log(x) + 4*log(-log(3*x) + log(x + 1)) - 4*log(log(x + 1))

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maple [A]  time = 0.04, size = 31, normalized size = 1.35

method	result	size
risch	\(4 \ln \left (x^{2}+x \right )-4 \ln \left (\ln \left (x +1\right )\right )+4 \ln \left (\ln \left (x +1\right )-\ln \left (3 x \right )\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*ln(x^2+x)-4*ln(ln(x+1))+4*ln(ln(x+1)-ln(3*x))

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maxima [A]  time = 0.71, size = 34, normalized size = 1.48 \begin {gather*} 4 \, \log \left (x + 1\right ) + 4 \, \log \relax (x) + 4 \, \log \left (-\log \relax (3) + \log \left (x + 1\right ) - \log \relax (x)\right ) - 4 \, \log \left (\log \left (x + 1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(x + 1) + 4*log(x) + 4*log(-log(3) + log(x + 1) - log(x)) - 4*log(log(x + 1))

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mupad [B]  time = 2.45, size = 30, normalized size = 1.30 \begin {gather*} 4\,\ln \left (x\,\left (x+1\right )\right )-4\,\ln \left (\ln \left (x+1\right )\right )+4\,\ln \left (\ln \left (3\,x\right )-\ln \left (x+1\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(x*(x + 1)) - 4*log(log(x + 1)) + 4*log(log(3*x) - log(x + 1))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: PolynomialError

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