3.29.51 \(\int \frac {10 x+4 x^2-6 \log (5)+(4 x^2-2 \log (5)) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=19 \[ 5+x+\frac {2 \left (-5+\frac {\log (5)}{x}\right )}{2+\log (x)} \]

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Rubi [A]  time = 0.46, antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 8, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {6688, 6742, 2353, 2302, 30, 2306, 2309, 2178} \begin {gather*} x-\frac {10}{\log (x)+2}+\frac {2 \log (5)}{x (\log (x)+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x + 4*x^2 - 6*Log[5] + (4*x^2 - 2*Log[5])*Log[x] + x^2*Log[x]^2)/(4*x^2 + 4*x^2*Log[x] + x^2*Log[x]^2)
,x]

[Out]

x - 10/(2 + Log[x]) + (2*Log[5])/(x*(2 + Log[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 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 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{x^2 (2+\log (x))^2} \, dx\\ &=\int \left (1+\frac {2 (5 x-\log (5))}{x^2 (2+\log (x))^2}-\frac {2 \log (5)}{x^2 (2+\log (x))}\right ) \, dx\\ &=x+2 \int \frac {5 x-\log (5)}{x^2 (2+\log (x))^2} \, dx-(2 \log (5)) \int \frac {1}{x^2 (2+\log (x))} \, dx\\ &=x+2 \int \left (\frac {5}{x (2+\log (x))^2}-\frac {\log (5)}{x^2 (2+\log (x))^2}\right ) \, dx-(2 \log (5)) \operatorname {Subst}\left (\int \frac {e^{-x}}{2+x} \, dx,x,\log (x)\right )\\ &=x-2 e^2 \text {Ei}(-2-\log (x)) \log (5)+10 \int \frac {1}{x (2+\log (x))^2} \, dx-(2 \log (5)) \int \frac {1}{x^2 (2+\log (x))^2} \, dx\\ &=x-2 e^2 \text {Ei}(-2-\log (x)) \log (5)+\frac {2 \log (5)}{x (2+\log (x))}+10 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,2+\log (x)\right )+(2 \log (5)) \int \frac {1}{x^2 (2+\log (x))} \, dx\\ &=x-2 e^2 \text {Ei}(-2-\log (x)) \log (5)-\frac {10}{2+\log (x)}+\frac {2 \log (5)}{x (2+\log (x))}+(2 \log (5)) \operatorname {Subst}\left (\int \frac {e^{-x}}{2+x} \, dx,x,\log (x)\right )\\ &=x-\frac {10}{2+\log (x)}+\frac {2 \log (5)}{x (2+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 18, normalized size = 0.95 \begin {gather*} x+\frac {-10 x+\log (25)}{x (2+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + 4*x^2 - 6*Log[5] + (4*x^2 - 2*Log[5])*Log[x] + x^2*Log[x]^2)/(4*x^2 + 4*x^2*Log[x] + x^2*Log
[x]^2),x]

[Out]

x + (-10*x + Log[25])/(x*(2 + Log[x]))

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fricas [A]  time = 0.57, size = 30, normalized size = 1.58 \begin {gather*} \frac {x^{2} \log \relax (x) + 2 \, x^{2} - 10 \, x + 2 \, \log \relax (5)}{x \log \relax (x) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(-2*log(5)+4*x^2)*log(x)-6*log(5)+4*x^2+10*x)/(x^2*log(x)^2+4*x^2*log(x)+4*x^2),x, alg
orithm="fricas")

[Out]

(x^2*log(x) + 2*x^2 - 10*x + 2*log(5))/(x*log(x) + 2*x)

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giac [A]  time = 0.21, size = 22, normalized size = 1.16 \begin {gather*} x - \frac {2 \, {\left (5 \, x - \log \relax (5)\right )}}{x \log \relax (x) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(-2*log(5)+4*x^2)*log(x)-6*log(5)+4*x^2+10*x)/(x^2*log(x)^2+4*x^2*log(x)+4*x^2),x, alg
orithm="giac")

[Out]

x - 2*(5*x - log(5))/(x*log(x) + 2*x)

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maple [A]  time = 0.07, size = 20, normalized size = 1.05

method	result	size
risch	\(x +\frac {2 \ln \relax (5)-10 x}{\left (\ln \relax (x )+2\right ) x}\)	\(20\)
norman	\(\frac {x^{2} \ln \relax (x )-10 x +2 x^{2}+2 \ln \relax (5)}{\left (\ln \relax (x )+2\right ) x}\)	\(30\)
default	\(\frac {x \ln \relax (x )+2 x -10}{\ln \relax (x )+2}+\frac {2 \ln \relax (5)}{x \left (\ln \relax (x )+2\right )}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(x)^2+(-2*ln(5)+4*x^2)*ln(x)-6*ln(5)+4*x^2+10*x)/(x^2*ln(x)^2+4*x^2*ln(x)+4*x^2),x,method=_RETURNVE
RBOSE)

[Out]

x+2/x*(ln(5)-5*x)/(ln(x)+2)

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maxima [A]  time = 0.84, size = 30, normalized size = 1.58 \begin {gather*} \frac {x^{2} \log \relax (x) + 2 \, x^{2} - 10 \, x + 2 \, \log \relax (5)}{x \log \relax (x) + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(-2*log(5)+4*x^2)*log(x)-6*log(5)+4*x^2+10*x)/(x^2*log(x)^2+4*x^2*log(x)+4*x^2),x, alg
orithm="maxima")

[Out]

(x^2*log(x) + 2*x^2 - 10*x + 2*log(5))/(x*log(x) + 2*x)

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mupad [B]  time = 1.79, size = 31, normalized size = 1.63 \begin {gather*} \frac {x^2+5\,x}{x}-\frac {10\,x-\ln \left (25\right )}{x\,\left (\ln \relax (x)+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x - 6*log(5) + x^2*log(x)^2 - log(x)*(2*log(5) - 4*x^2) + 4*x^2)/(4*x^2*log(x) + x^2*log(x)^2 + 4*x^2)
,x)

[Out]

(5*x + x^2)/x - (10*x - log(25))/(x*(log(x) + 2))

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sympy [A]  time = 0.10, size = 17, normalized size = 0.89 \begin {gather*} x + \frac {- 10 x + 2 \log {\relax (5 )}}{x \log {\relax (x )} + 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(x)**2+(-2*ln(5)+4*x**2)*ln(x)-6*ln(5)+4*x**2+10*x)/(x**2*ln(x)**2+4*x**2*ln(x)+4*x**2),x)

[Out]

x + (-10*x + 2*log(5))/(x*log(x) + 2*x)

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