3.24.29 \(\int \frac {e^{\frac {3}{\log (1+\log (3 x))}} (12+3 x)+(x+12 x^3+3 x^4+(x+12 x^3+3 x^4) \log (3 x)) \log ^2(1+\log (3 x))}{(4 x+x^2+(4 x+x^2) \log (3 x)) \log ^2(1+\log (3 x))} \, dx\)

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

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Rubi [A]  time = 0.56, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6688, 14, 1850, 6706} \begin {gather*} x^3-e^{\frac {3}{\log (\log (3 x)+1)}}+\log (x+4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3/Log[1 + Log[3*x]])*(12 + 3*x) + (x + 12*x^3 + 3*x^4 + (x + 12*x^3 + 3*x^4)*Log[3*x])*Log[1 + Log[3*x
]]^2)/((4*x + x^2 + (4*x + x^2)*Log[3*x])*Log[1 + Log[3*x]]^2),x]

[Out]

-E^(3/Log[1 + Log[3*x]]) + x^3 + Log[4 + 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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {x+12 x^3+3 x^4}{4+x}+\frac {3 e^{\frac {3}{\log (1+\log (3 x))}}}{(1+\log (3 x)) \log ^2(1+\log (3 x))}}{x} \, dx\\ &=\int \left (\frac {1+12 x^2+3 x^3}{4+x}+\frac {3 e^{\frac {3}{\log (1+\log (3 x))}}}{x (1+\log (3 x)) \log ^2(1+\log (3 x))}\right ) \, dx\\ &=3 \int \frac {e^{\frac {3}{\log (1+\log (3 x))}}}{x (1+\log (3 x)) \log ^2(1+\log (3 x))} \, dx+\int \frac {1+12 x^2+3 x^3}{4+x} \, dx\\ &=3 \operatorname {Subst}\left (\int \frac {e^{\frac {3}{\log (1+x)}}}{(1+x) \log ^2(1+x)} \, dx,x,\log (3 x)\right )+\int \left (3 x^2+\frac {1}{4+x}\right ) \, dx\\ &=-e^{\frac {3}{\log (1+\log (3 x))}}+x^3+\log (4+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 23, normalized size = 1.00 \begin {gather*} -e^{\frac {3}{\log (1+\log (3 x))}}+x^3+\log (4+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3/Log[1 + Log[3*x]])*(12 + 3*x) + (x + 12*x^3 + 3*x^4 + (x + 12*x^3 + 3*x^4)*Log[3*x])*Log[1 + L
og[3*x]]^2)/((4*x + x^2 + (4*x + x^2)*Log[3*x])*Log[1 + Log[3*x]]^2),x]

[Out]

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

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fricas [A]  time = 0.83, size = 22, normalized size = 0.96 \begin {gather*} x^{3} - e^{\left (\frac {3}{\log \left (\log \left (3 \, x\right ) + 1\right )}\right )} + \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+12)*exp(3/log(log(3*x)+1))+((3*x^4+12*x^3+x)*log(3*x)+3*x^4+12*x^3+x)*log(log(3*x)+1)^2)/((x^2
+4*x)*log(3*x)+x^2+4*x)/log(log(3*x)+1)^2,x, algorithm="fricas")

[Out]

x^3 - e^(3/log(log(3*x) + 1)) + log(x + 4)

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giac [A]  time = 1.60, size = 22, normalized size = 0.96 \begin {gather*} x^{3} - e^{\left (\frac {3}{\log \left (\log \left (3 \, x\right ) + 1\right )}\right )} + \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+12)*exp(3/log(log(3*x)+1))+((3*x^4+12*x^3+x)*log(3*x)+3*x^4+12*x^3+x)*log(log(3*x)+1)^2)/((x^2
+4*x)*log(3*x)+x^2+4*x)/log(log(3*x)+1)^2,x, algorithm="giac")

[Out]

x^3 - e^(3/log(log(3*x) + 1)) + log(x + 4)

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maple [A]  time = 0.06, size = 23, normalized size = 1.00

method	result	size
risch	\(x^{3}+\ln \left (4+x \right )-{\mathrm e}^{\frac {3}{\ln \left (\ln \left (3 x \right )+1\right )}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x+12)*exp(3/ln(ln(3*x)+1))+((3*x^4+12*x^3+x)*ln(3*x)+3*x^4+12*x^3+x)*ln(ln(3*x)+1)^2)/((x^2+4*x)*ln(3*
x)+x^2+4*x)/ln(ln(3*x)+1)^2,x,method=_RETURNVERBOSE)

[Out]

x^3+ln(4+x)-exp(3/ln(ln(3*x)+1))

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maxima [B]  time = 0.93, size = 47, normalized size = 2.04 \begin {gather*} x^{3} - \frac {x e^{\left (\frac {3}{\log \left (\log \relax (3) + \log \relax (x) + 1\right )}\right )}}{x + 4} - \frac {4 \, e^{\left (\frac {3}{\log \left (\log \relax (3) + \log \relax (x) + 1\right )}\right )}}{x + 4} + \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+12)*exp(3/log(log(3*x)+1))+((3*x^4+12*x^3+x)*log(3*x)+3*x^4+12*x^3+x)*log(log(3*x)+1)^2)/((x^2
+4*x)*log(3*x)+x^2+4*x)/log(log(3*x)+1)^2,x, algorithm="maxima")

[Out]

x^3 - x*e^(3/log(log(3) + log(x) + 1))/(x + 4) - 4*e^(3/log(log(3) + log(x) + 1))/(x + 4) + log(x + 4)

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mupad [B]  time = 1.49, size = 22, normalized size = 0.96 \begin {gather*} \ln \left (x+4\right )-{\mathrm {e}}^{\frac {3}{\ln \left (\ln \left (3\,x\right )+1\right )}}+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3/log(log(3*x) + 1))*(3*x + 12) + log(log(3*x) + 1)^2*(x + 12*x^3 + 3*x^4 + log(3*x)*(x + 12*x^3 + 3*
x^4)))/(log(log(3*x) + 1)^2*(4*x + log(3*x)*(4*x + x^2) + x^2)),x)

[Out]

log(x + 4) - exp(3/log(log(3*x) + 1)) + x^3

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sympy [A]  time = 0.57, size = 19, normalized size = 0.83 \begin {gather*} x^{3} - e^{\frac {3}{\log {\left (\log {\left (3 x \right )} + 1 \right )}}} + \log {\left (x + 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x+12)*exp(3/ln(ln(3*x)+1))+((3*x**4+12*x**3+x)*ln(3*x)+3*x**4+12*x**3+x)*ln(ln(3*x)+1)**2)/((x**
2+4*x)*ln(3*x)+x**2+4*x)/ln(ln(3*x)+1)**2,x)

[Out]

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

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