3.57.99 \(\int \frac {10+e^2 (10+x)-x \log (x)+(10+x) \log (48 x)}{e^4 x^2+2 e^2 x^2 \log (48 x)+x^2 \log ^2(48 x)} \, dx\)

Optimal. Leaf size=21 \[ 5+\frac {-\frac {10}{x}+\log (x)}{e^2+\log (48 x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.67, antiderivative size = 29, normalized size of antiderivative = 1.38, number of steps used = 17, number of rules used = 10, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6688, 6742, 2306, 2309, 2178, 2302, 30, 2366, 29, 2353} \begin {gather*} \frac {\log (x)}{\log (48 x)+e^2}-\frac {10}{x \left (\log (48 x)+e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + E^2*(10 + x) - x*Log[x] + (10 + x)*Log[48*x])/(E^4*x^2 + 2*E^2*x^2*Log[48*x] + x^2*Log[48*x]^2),x]

[Out]

-10/(x*(E^2 + Log[48*x])) + Log[x]/(E^2 + Log[48*x])

Rule 29

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

Rule 30

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 6688

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

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 {10 \left (1+e^2\right )+e^2 x-x \log (x)+(10+x) \log (48 x)}{x^2 \left (e^2+\log (48 x)\right )^2} \, dx\\ &=\int \left (\frac {10-x \log (x)}{x^2 \left (e^2+\log (48 x)\right )^2}+\frac {10+x}{x^2 \left (e^2+\log (48 x)\right )}\right ) \, dx\\ &=\int \frac {10-x \log (x)}{x^2 \left (e^2+\log (48 x)\right )^2} \, dx+\int \frac {10+x}{x^2 \left (e^2+\log (48 x)\right )} \, dx\\ &=\int \left (\frac {10}{x^2 \left (e^2+\log (48 x)\right )^2}-\frac {\log (x)}{x \left (e^2+\log (48 x)\right )^2}\right ) \, dx+\int \left (\frac {10}{x^2 \left (e^2+\log (48 x)\right )}+\frac {1}{x \left (e^2+\log (48 x)\right )}\right ) \, dx\\ &=10 \int \frac {1}{x^2 \left (e^2+\log (48 x)\right )^2} \, dx+10 \int \frac {1}{x^2 \left (e^2+\log (48 x)\right )} \, dx-\int \frac {\log (x)}{x \left (e^2+\log (48 x)\right )^2} \, dx+\int \frac {1}{x \left (e^2+\log (48 x)\right )} \, dx\\ &=-\frac {10}{x \left (e^2+\log (48 x)\right )}+\frac {\log (x)}{e^2+\log (48 x)}-10 \int \frac {1}{x^2 \left (e^2+\log (48 x)\right )} \, dx+480 \operatorname {Subst}\left (\int \frac {e^{-x}}{e^2+x} \, dx,x,\log (48 x)\right )-\int \frac {1}{x \left (e^2+\log (48 x)\right )} \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^2+\log (48 x)\right )\\ &=480 e^{e^2} \text {Ei}\left (-e^2-\log (48 x)\right )-\frac {10}{x \left (e^2+\log (48 x)\right )}+\frac {\log (x)}{e^2+\log (48 x)}+\log \left (e^2+\log (48 x)\right )-480 \operatorname {Subst}\left (\int \frac {e^{-x}}{e^2+x} \, dx,x,\log (48 x)\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^2+\log (48 x)\right )\\ &=-\frac {10}{x \left (e^2+\log (48 x)\right )}+\frac {\log (x)}{e^2+\log (48 x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 27, normalized size = 1.29 \begin {gather*} -\frac {10+e^2 x+x \log (48)}{e^2 x+x \log (48 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + E^2*(10 + x) - x*Log[x] + (10 + x)*Log[48*x])/(E^4*x^2 + 2*E^2*x^2*Log[48*x] + x^2*Log[48*x]^2
),x]

[Out]

-((10 + E^2*x + x*Log[48])/(E^2*x + x*Log[48*x]))

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 27, normalized size = 1.29 \begin {gather*} -\frac {x e^{2} + x \log \left (48\right ) + 10}{x e^{2} + x \log \left (48\right ) + x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*log(48*x)-x*log(x)+(x+10)*exp(2)+10)/(x^2*log(48*x)^2+2*x^2*exp(2)*log(48*x)+x^2*exp(2)^2),x
, algorithm="fricas")

[Out]

-(x*e^2 + x*log(48) + 10)/(x*e^2 + x*log(48) + x*log(x))

________________________________________________________________________________________

giac [A]  time = 0.22, size = 27, normalized size = 1.29 \begin {gather*} -\frac {x e^{2} + x \log \left (48\right ) + 10}{x e^{2} + x \log \left (48\right ) + x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*log(48*x)-x*log(x)+(x+10)*exp(2)+10)/(x^2*log(48*x)^2+2*x^2*exp(2)*log(48*x)+x^2*exp(2)^2),x
, algorithm="giac")

[Out]

-(x*e^2 + x*log(48) + 10)/(x*e^2 + x*log(48) + x*log(x))

________________________________________________________________________________________

maple [B]  time = 0.11, size = 42, normalized size = 2.00

method	result	size
risch	\(-\frac {20+2 \,{\mathrm e}^{2} x +2 x \ln \relax (3)+8 x \ln \relax (2)}{x \left (2 \,{\mathrm e}^{2}+2 \ln \relax (3)+8 \ln \relax (2)+2 \ln \relax (x )\right )}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+10)*ln(48*x)-x*ln(x)+(x+10)*exp(2)+10)/(x^2*ln(48*x)^2+2*x^2*exp(2)*ln(48*x)+x^2*exp(2)^2),x,method=_R
ETURNVERBOSE)

[Out]

-1/x*(20+2*exp(2)*x+2*x*ln(3)+8*x*ln(2))/(2*exp(2)+2*ln(3)+8*ln(2)+2*ln(x))

________________________________________________________________________________________

maxima [A]  time = 0.48, size = 33, normalized size = 1.57 \begin {gather*} -\frac {x {\left (e^{2} + \log \relax (3) + 4 \, \log \relax (2)\right )} + 10}{x {\left (e^{2} + \log \relax (3) + 4 \, \log \relax (2)\right )} + x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*log(48*x)-x*log(x)+(x+10)*exp(2)+10)/(x^2*log(48*x)^2+2*x^2*exp(2)*log(48*x)+x^2*exp(2)^2),x
, algorithm="maxima")

[Out]

-(x*(e^2 + log(3) + 4*log(2)) + 10)/(x*(e^2 + log(3) + 4*log(2)) + x*log(x))

________________________________________________________________________________________

mupad [B]  time = 3.82, size = 19, normalized size = 0.90 \begin {gather*} \frac {x\,\ln \relax (x)-10}{x\,\left (\ln \left (48\,x\right )+{\mathrm {e}}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(48*x)*(x + 10) + exp(2)*(x + 10) - x*log(x) + 10)/(x^2*exp(4) + x^2*log(48*x)^2 + 2*x^2*log(48*x)*exp
(2)),x)

[Out]

(x*log(x) - 10)/(x*(log(48*x) + exp(2)))

________________________________________________________________________________________

sympy [A]  time = 0.26, size = 27, normalized size = 1.29 \begin {gather*} \frac {- x e^{2} - x \log {\left (48 \right )} - 10}{x \log {\relax (x )} + x \log {\left (48 \right )} + x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+10)*ln(48*x)-x*ln(x)+(x+10)*exp(2)+10)/(x**2*ln(48*x)**2+2*x**2*exp(2)*ln(48*x)+x**2*exp(2)**2),
x)

[Out]

(-x*exp(2) - x*log(48) - 10)/(x*log(x) + x*log(48) + x*exp(2))

________________________________________________________________________________________