3.54.83 \(\int \frac {36+90 e^{2 x} x+18 e^{2 x} x \log (5 x)}{2000 x-200 e^{2 x} x+5 e^{4 x} x+(400 x-40 e^{2 x} x+e^{4 x} x) \log (5 x)+(-800 x+40 e^{2 x} x+(-160 x+8 e^{2 x} x) \log (5 x)) \log (5+\log (5 x))+(80 x+16 x \log (5 x)) \log ^2(5+\log (5 x))} \, dx\)

Optimal. Leaf size=22 \[ \frac {9}{20-e^{2 x}-4 \log (5+\log (5 x))} \]

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Rubi [A]  time = 0.49, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 127, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6688, 12, 6686} \begin {gather*} \frac {9}{-e^{2 x}-4 \log (\log (5 x)+5)+20} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(36 + 90*E^(2*x)*x + 18*E^(2*x)*x*Log[5*x])/(2000*x - 200*E^(2*x)*x + 5*E^(4*x)*x + (400*x - 40*E^(2*x)*x
+ E^(4*x)*x)*Log[5*x] + (-800*x + 40*E^(2*x)*x + (-160*x + 8*E^(2*x)*x)*Log[5*x])*Log[5 + Log[5*x]] + (80*x +
16*x*Log[5*x])*Log[5 + Log[5*x]]^2),x]

[Out]

9/(20 - E^(2*x) - 4*Log[5 + Log[5*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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {18 \left (2+5 e^{2 x} x+e^{2 x} x \log (5 x)\right )}{x (5+\log (5 x)) \left (20-e^{2 x}-4 \log (5+\log (5 x))\right )^2} \, dx\\ &=18 \int \frac {2+5 e^{2 x} x+e^{2 x} x \log (5 x)}{x (5+\log (5 x)) \left (20-e^{2 x}-4 \log (5+\log (5 x))\right )^2} \, dx\\ &=\frac {9}{20-e^{2 x}-4 \log (5+\log (5 x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 0.91 \begin {gather*} -\frac {9}{-20+e^{2 x}+4 \log (5+\log (5 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(36 + 90*E^(2*x)*x + 18*E^(2*x)*x*Log[5*x])/(2000*x - 200*E^(2*x)*x + 5*E^(4*x)*x + (400*x - 40*E^(2
*x)*x + E^(4*x)*x)*Log[5*x] + (-800*x + 40*E^(2*x)*x + (-160*x + 8*E^(2*x)*x)*Log[5*x])*Log[5 + Log[5*x]] + (8
0*x + 16*x*Log[5*x])*Log[5 + Log[5*x]]^2),x]

[Out]

-9/(-20 + E^(2*x) + 4*Log[5 + Log[5*x]])

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fricas [A]  time = 0.59, size = 19, normalized size = 0.86 \begin {gather*} -\frac {9}{e^{\left (2 \, x\right )} + 4 \, \log \left (\log \left (5 \, x\right ) + 5\right ) - 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(2*x)*log(5*x)+90*x*exp(2*x)+36)/((16*x*log(5*x)+80*x)*log(log(5*x)+5)^2+((8*x*exp(2*x)-160
*x)*log(5*x)+40*x*exp(2*x)-800*x)*log(log(5*x)+5)+(x*exp(2*x)^2-40*x*exp(2*x)+400*x)*log(5*x)+5*x*exp(2*x)^2-2
00*x*exp(2*x)+2000*x),x, algorithm="fricas")

[Out]

-9/(e^(2*x) + 4*log(log(5*x) + 5) - 20)

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giac [A]  time = 0.22, size = 19, normalized size = 0.86 \begin {gather*} -\frac {9}{e^{\left (2 \, x\right )} + 4 \, \log \left (\log \relax (5) + \log \relax (x) + 5\right ) - 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(2*x)*log(5*x)+90*x*exp(2*x)+36)/((16*x*log(5*x)+80*x)*log(log(5*x)+5)^2+((8*x*exp(2*x)-160
*x)*log(5*x)+40*x*exp(2*x)-800*x)*log(log(5*x)+5)+(x*exp(2*x)^2-40*x*exp(2*x)+400*x)*log(5*x)+5*x*exp(2*x)^2-2
00*x*exp(2*x)+2000*x),x, algorithm="giac")

[Out]

-9/(e^(2*x) + 4*log(log(5) + log(x) + 5) - 20)

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maple [A]  time = 0.05, size = 20, normalized size = 0.91

method	result	size
risch	\(-\frac {9}{-20+{\mathrm e}^{2 x}+4 \ln \left (\ln \left (5 x \right )+5\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x*exp(2*x)*ln(5*x)+90*x*exp(2*x)+36)/((16*x*ln(5*x)+80*x)*ln(ln(5*x)+5)^2+((8*x*exp(2*x)-160*x)*ln(5*x
)+40*x*exp(2*x)-800*x)*ln(ln(5*x)+5)+(x*exp(2*x)^2-40*x*exp(2*x)+400*x)*ln(5*x)+5*x*exp(2*x)^2-200*x*exp(2*x)+
2000*x),x,method=_RETURNVERBOSE)

[Out]

-9/(-20+exp(2*x)+4*ln(ln(5*x)+5))

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maxima [A]  time = 0.53, size = 19, normalized size = 0.86 \begin {gather*} -\frac {9}{e^{\left (2 \, x\right )} + 4 \, \log \left (\log \relax (5) + \log \relax (x) + 5\right ) - 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(2*x)*log(5*x)+90*x*exp(2*x)+36)/((16*x*log(5*x)+80*x)*log(log(5*x)+5)^2+((8*x*exp(2*x)-160
*x)*log(5*x)+40*x*exp(2*x)-800*x)*log(log(5*x)+5)+(x*exp(2*x)^2-40*x*exp(2*x)+400*x)*log(5*x)+5*x*exp(2*x)^2-2
00*x*exp(2*x)+2000*x),x, algorithm="maxima")

[Out]

-9/(e^(2*x) + 4*log(log(5) + log(x) + 5) - 20)

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mupad [B]  time = 3.80, size = 19, normalized size = 0.86 \begin {gather*} -\frac {9}{4\,\ln \left (\ln \left (5\,x\right )+5\right )+{\mathrm {e}}^{2\,x}-20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*x*exp(2*x) + 18*x*log(5*x)*exp(2*x) + 36)/(2000*x + log(5*x)*(400*x - 40*x*exp(2*x) + x*exp(4*x)) - 20
0*x*exp(2*x) + 5*x*exp(4*x) + log(log(5*x) + 5)^2*(80*x + 16*x*log(5*x)) - log(log(5*x) + 5)*(800*x - 40*x*exp
(2*x) + log(5*x)*(160*x - 8*x*exp(2*x)))),x)

[Out]

-9/(4*log(log(5*x) + 5) + exp(2*x) - 20)

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sympy [A]  time = 0.37, size = 19, normalized size = 0.86 \begin {gather*} - \frac {9}{e^{2 x} + 4 \log {\left (\log {\left (5 x \right )} + 5 \right )} - 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(2*x)*ln(5*x)+90*x*exp(2*x)+36)/((16*x*ln(5*x)+80*x)*ln(ln(5*x)+5)**2+((8*x*exp(2*x)-160*x)
*ln(5*x)+40*x*exp(2*x)-800*x)*ln(ln(5*x)+5)+(x*exp(2*x)**2-40*x*exp(2*x)+400*x)*ln(5*x)+5*x*exp(2*x)**2-200*x*
exp(2*x)+2000*x),x)

[Out]

-9/(exp(2*x) + 4*log(log(5*x) + 5) - 20)

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