3.82.44 \(\int \frac {45 e^5+x^2+(-45 e^5+2 x^2) \log (2)+x^2 \log ^2(2)+(90 e^5-4 x^2-4 x^2 \log (2)) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+(-36 x^2-36 x^2 \log (2)) \log (x)+36 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{9} (x-\log (3))+\frac {5 e^5}{x (1+\log (2)-2 \log (x))} \]

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Rubi [A]  time = 0.57, antiderivative size = 24, normalized size of antiderivative = 0.83, number of steps used = 12, number of rules used = 7, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 6688, 12, 6742, 2306, 2309, 2178} \begin {gather*} \frac {x}{9}+\frac {5 e^5}{x (-2 \log (x)+1+\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(45*E^5 + x^2 + (-45*E^5 + 2*x^2)*Log[2] + x^2*Log[2]^2 + (90*E^5 - 4*x^2 - 4*x^2*Log[2])*Log[x] + 4*x^2*L
og[x]^2)/(9*x^2 + 18*x^2*Log[2] + 9*x^2*Log[2]^2 + (-36*x^2 - 36*x^2*Log[2])*Log[x] + 36*x^2*Log[x]^2),x]

[Out]

x/9 + (5*E^5)/(x*(1 + Log[2] - 2*Log[x]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 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 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 {45 e^5+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \left (1+\log ^2(2)\right )+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx\\ &=\int \frac {45 e^5+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \left (1+\log ^2(2)\right )+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2 \log ^2(2)+x^2 (9+18 \log (2))+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx\\ &=\int \frac {45 e^5+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \left (1+\log ^2(2)\right )+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{x^2 \left (9+18 \log (2)+9 \log ^2(2)\right )+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx\\ &=\int \frac {-45 e^5 (-1+\log (2))+x^2 \left (1+\log ^2(2)+\log (4)\right )+\left (90 e^5-4 x^2 (1+\log (2))\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2 (1+\log (2)-2 \log (x))^2} \, dx\\ &=\frac {1}{9} \int \frac {-45 e^5 (-1+\log (2))+x^2 \left (1+\log ^2(2)+\log (4)\right )+\left (90 e^5-4 x^2 (1+\log (2))\right ) \log (x)+4 x^2 \log ^2(x)}{x^2 (1+\log (2)-2 \log (x))^2} \, dx\\ &=\frac {1}{9} \int \left (1+\frac {90 e^5}{x^2 (1+\log (2)-2 \log (x))^2}-\frac {45 e^5}{x^2 (1+\log (2)-2 \log (x))}\right ) \, dx\\ &=\frac {x}{9}-\left (5 e^5\right ) \int \frac {1}{x^2 (1+\log (2)-2 \log (x))} \, dx+\left (10 e^5\right ) \int \frac {1}{x^2 (1+\log (2)-2 \log (x))^2} \, dx\\ &=\frac {x}{9}+\frac {5 e^5}{x (1+\log (2)-2 \log (x))}+\left (5 e^5\right ) \int \frac {1}{x^2 (1+\log (2)-2 \log (x))} \, dx-\left (5 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{1-2 x+\log (2)} \, dx,x,\log (x)\right )\\ &=\frac {x}{9}+\frac {5 e^{9/2} \text {Ei}\left (\frac {1}{2} (1+\log (2)-2 \log (x))\right )}{2 \sqrt {2}}+\frac {5 e^5}{x (1+\log (2)-2 \log (x))}+\left (5 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{1-2 x+\log (2)} \, dx,x,\log (x)\right )\\ &=\frac {x}{9}+\frac {5 e^5}{x (1+\log (2)-2 \log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 38, normalized size = 1.31 \begin {gather*} \frac {90 e^5+x^2 (2+\log (4))-4 x^2 \log (x)}{18 x (1+\log (2)-2 \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(45*E^5 + x^2 + (-45*E^5 + 2*x^2)*Log[2] + x^2*Log[2]^2 + (90*E^5 - 4*x^2 - 4*x^2*Log[2])*Log[x] + 4
*x^2*Log[x]^2)/(9*x^2 + 18*x^2*Log[2] + 9*x^2*Log[2]^2 + (-36*x^2 - 36*x^2*Log[2])*Log[x] + 36*x^2*Log[x]^2),x
]

[Out]

(90*E^5 + x^2*(2 + Log[4]) - 4*x^2*Log[x])/(18*x*(1 + Log[2] - 2*Log[x]))

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fricas [A]  time = 0.84, size = 36, normalized size = 1.24 \begin {gather*} \frac {x^{2} \log \relax (2) - 2 \, x^{2} \log \relax (x) + x^{2} + 45 \, e^{5}}{9 \, {\left (x \log \relax (2) - 2 \, x \log \relax (x) + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(x)^2+(-4*x^2*log(2)+90*exp(5)-4*x^2)*log(x)+x^2*log(2)^2+(-45*exp(5)+2*x^2)*log(2)+45*exp
(5)+x^2)/(36*x^2*log(x)^2+(-36*x^2*log(2)-36*x^2)*log(x)+9*x^2*log(2)^2+18*x^2*log(2)+9*x^2),x, algorithm="fri
cas")

[Out]

1/9*(x^2*log(2) - 2*x^2*log(x) + x^2 + 45*e^5)/(x*log(2) - 2*x*log(x) + x)

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giac [A]  time = 0.15, size = 36, normalized size = 1.24 \begin {gather*} \frac {x^{2} \log \relax (2) - 2 \, x^{2} \log \relax (x) + x^{2} + 45 \, e^{5}}{9 \, {\left (x \log \relax (2) - 2 \, x \log \relax (x) + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(x)^2+(-4*x^2*log(2)+90*exp(5)-4*x^2)*log(x)+x^2*log(2)^2+(-45*exp(5)+2*x^2)*log(2)+45*exp
(5)+x^2)/(36*x^2*log(x)^2+(-36*x^2*log(2)-36*x^2)*log(x)+9*x^2*log(2)^2+18*x^2*log(2)+9*x^2),x, algorithm="gia
c")

[Out]

1/9*(x^2*log(2) - 2*x^2*log(x) + x^2 + 45*e^5)/(x*log(2) - 2*x*log(x) + x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.76

method	result	size
risch	\(\frac {x}{9}+\frac {5 \,{\mathrm e}^{5}}{\left (1+\ln \relax (2)-2 \ln \relax (x )\right ) x}\)	\(22\)
norman	\(\frac {\left (\frac {\ln \relax (2)}{9}+\frac {1}{9}\right ) x^{2}-\frac {2 x^{2} \ln \relax (x )}{9}+5 \,{\mathrm e}^{5}}{x \left (1+\ln \relax (2)-2 \ln \relax (x )\right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2*ln(x)^2+(-4*x^2*ln(2)+90*exp(5)-4*x^2)*ln(x)+x^2*ln(2)^2+(-45*exp(5)+2*x^2)*ln(2)+45*exp(5)+x^2)/(3
6*x^2*ln(x)^2+(-36*x^2*ln(2)-36*x^2)*ln(x)+9*x^2*ln(2)^2+18*x^2*ln(2)+9*x^2),x,method=_RETURNVERBOSE)

[Out]

1/9*x+5*exp(5)/(1+ln(2)-2*ln(x))/x

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maxima [A]  time = 0.48, size = 36, normalized size = 1.24 \begin {gather*} \frac {x^{2} {\left (\log \relax (2) + 1\right )} - 2 \, x^{2} \log \relax (x) + 45 \, e^{5}}{9 \, {\left (x {\left (\log \relax (2) + 1\right )} - 2 \, x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2*log(x)^2+(-4*x^2*log(2)+90*exp(5)-4*x^2)*log(x)+x^2*log(2)^2+(-45*exp(5)+2*x^2)*log(2)+45*exp
(5)+x^2)/(36*x^2*log(x)^2+(-36*x^2*log(2)-36*x^2)*log(x)+9*x^2*log(2)^2+18*x^2*log(2)+9*x^2),x, algorithm="max
ima")

[Out]

1/9*(x^2*(log(2) + 1) - 2*x^2*log(x) + 45*e^5)/(x*(log(2) + 1) - 2*x*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {45\,{\mathrm {e}}^5+x^2\,{\ln \relax (2)}^2-\ln \relax (2)\,\left (45\,{\mathrm {e}}^5-2\,x^2\right )-\ln \relax (x)\,\left (4\,x^2\,\ln \relax (2)-90\,{\mathrm {e}}^5+4\,x^2\right )+4\,x^2\,{\ln \relax (x)}^2+x^2}{9\,x^2\,{\ln \relax (2)}^2+36\,x^2\,{\ln \relax (x)}^2-\ln \relax (x)\,\left (36\,x^2\,\ln \relax (2)+36\,x^2\right )+18\,x^2\,\ln \relax (2)+9\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*exp(5) + x^2*log(2)^2 - log(2)*(45*exp(5) - 2*x^2) - log(x)*(4*x^2*log(2) - 90*exp(5) + 4*x^2) + 4*x^2
*log(x)^2 + x^2)/(9*x^2*log(2)^2 + 36*x^2*log(x)^2 - log(x)*(36*x^2*log(2) + 36*x^2) + 18*x^2*log(2) + 9*x^2),
x)

[Out]

int((45*exp(5) + x^2*log(2)^2 - log(2)*(45*exp(5) - 2*x^2) - log(x)*(4*x^2*log(2) - 90*exp(5) + 4*x^2) + 4*x^2
*log(x)^2 + x^2)/(9*x^2*log(2)^2 + 36*x^2*log(x)^2 - log(x)*(36*x^2*log(2) + 36*x^2) + 18*x^2*log(2) + 9*x^2),
x)

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sympy [A]  time = 0.13, size = 20, normalized size = 0.69 \begin {gather*} \frac {x}{9} - \frac {5 e^{5}}{2 x \log {\relax (x )} - x - x \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2*ln(x)**2+(-4*x**2*ln(2)+90*exp(5)-4*x**2)*ln(x)+x**2*ln(2)**2+(-45*exp(5)+2*x**2)*ln(2)+45*e
xp(5)+x**2)/(36*x**2*ln(x)**2+(-36*x**2*ln(2)-36*x**2)*ln(x)+9*x**2*ln(2)**2+18*x**2*ln(2)+9*x**2),x)

[Out]

x/9 - 5*exp(5)/(2*x*log(x) - x - x*log(2))

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