3.83.61 \(\int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+(-10 x^2+2 x^2 \log (3)) \log (x)+(-10 x+2 e^{2 x} x+2 x \log (3)) \log (x) \log (\log (x))}{(-5 x+x \log (3)) \log (x)} \, dx\)

Optimal. Leaf size=25 \[ x^2+\left (5+2 x+\frac {e^{2 x}}{-5+\log (3)}\right ) \log (\log (x)) \]

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Rubi [A]  time = 0.78, antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 16, number of rules used = 9, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 12, 6742, 2288, 2353, 2298, 2302, 29, 2520} \begin {gather*} x^2+2 x \log (\log (x))-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}+5 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 + E^(2*x) - 10*x + (5 + 2*x)*Log[3] + (-10*x^2 + 2*x^2*Log[3])*Log[x] + (-10*x + 2*E^(2*x)*x + 2*x*Lo
g[3])*Log[x]*Log[Log[x]])/((-5*x + x*Log[3])*Log[x]),x]

[Out]

x^2 + 5*Log[Log[x]] + 2*x*Log[Log[x]] - (E^(2*x)*Log[Log[x]])/(5 - Log[3])

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 29

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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 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 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, 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 {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{x (-5+\log (3)) \log (x)} \, dx\\ &=\frac {\int \frac {-25+e^{2 x}-10 x+(5+2 x) \log (3)+\left (-10 x^2+2 x^2 \log (3)\right ) \log (x)+\left (-10 x+2 e^{2 x} x+2 x \log (3)\right ) \log (x) \log (\log (x))}{x \log (x)} \, dx}{-5+\log (3)}\\ &=\frac {\int \left (\frac {e^{2 x} (1+2 x \log (x) \log (\log (x)))}{x \log (x)}+\frac {(-5+\log (3)) \left (5+2 x+2 x^2 \log (x)+2 x \log (x) \log (\log (x))\right )}{x \log (x)}\right ) \, dx}{-5+\log (3)}\\ &=\frac {\int \frac {e^{2 x} (1+2 x \log (x) \log (\log (x)))}{x \log (x)} \, dx}{-5+\log (3)}+\int \frac {5+2 x+2 x^2 \log (x)+2 x \log (x) \log (\log (x))}{x \log (x)} \, dx\\ &=-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}+\int \left (\frac {5+2 x+2 x^2 \log (x)}{x \log (x)}+2 \log (\log (x))\right ) \, dx\\ &=-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}+2 \int \log (\log (x)) \, dx+\int \frac {5+2 x+2 x^2 \log (x)}{x \log (x)} \, dx\\ &=2 x \log (\log (x))-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}-2 \int \frac {1}{\log (x)} \, dx+\int \left (2 x+\frac {5+2 x}{x \log (x)}\right ) \, dx\\ &=x^2+2 x \log (\log (x))-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}-2 \text {li}(x)+\int \frac {5+2 x}{x \log (x)} \, dx\\ &=x^2+2 x \log (\log (x))-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}-2 \text {li}(x)+\int \left (\frac {2}{\log (x)}+\frac {5}{x \log (x)}\right ) \, dx\\ &=x^2+2 x \log (\log (x))-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}-2 \text {li}(x)+2 \int \frac {1}{\log (x)} \, dx+5 \int \frac {1}{x \log (x)} \, dx\\ &=x^2+2 x \log (\log (x))-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}+5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=x^2+5 \log (\log (x))+2 x \log (\log (x))-\frac {e^{2 x} \log (\log (x))}{5-\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 39, normalized size = 1.56 \begin {gather*} \frac {x^2 \left (-5+\frac {\log (9)}{2}\right )+\left (-25+e^{2 x}+x (-10+\log (9))+\log (243)\right ) \log (\log (x))}{-5+\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + E^(2*x) - 10*x + (5 + 2*x)*Log[3] + (-10*x^2 + 2*x^2*Log[3])*Log[x] + (-10*x + 2*E^(2*x)*x +
2*x*Log[3])*Log[x]*Log[Log[x]])/((-5*x + x*Log[3])*Log[x]),x]

[Out]

(x^2*(-5 + Log[9]/2) + (-25 + E^(2*x) + x*(-10 + Log[9]) + Log[243])*Log[Log[x]])/(-5 + Log[3])

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fricas [A]  time = 0.66, size = 40, normalized size = 1.60 \begin {gather*} \frac {x^{2} \log \relax (3) - 5 \, x^{2} + {\left ({\left (2 \, x + 5\right )} \log \relax (3) - 10 \, x + e^{\left (2 \, x\right )} - 25\right )} \log \left (\log \relax (x)\right )}{\log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(2*x)+2*x*log(3)-10*x)*log(x)*log(log(x))+(2*x^2*log(3)-10*x^2)*log(x)+exp(2*x)+(5+2*x)*log
(3)-10*x-25)/(x*log(3)-5*x)/log(x),x, algorithm="fricas")

[Out]

(x^2*log(3) - 5*x^2 + ((2*x + 5)*log(3) - 10*x + e^(2*x) - 25)*log(log(x)))/(log(3) - 5)

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giac [B]  time = 0.27, size = 53, normalized size = 2.12 \begin {gather*} \frac {x^{2} \log \relax (3) + 2 \, x \log \relax (3) \log \left (\log \relax (x)\right ) - 5 \, x^{2} - 10 \, x \log \left (\log \relax (x)\right ) + e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right ) + 5 \, \log \relax (3) \log \left (\log \relax (x)\right ) - 25 \, \log \left (\log \relax (x)\right )}{\log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(2*x)+2*x*log(3)-10*x)*log(x)*log(log(x))+(2*x^2*log(3)-10*x^2)*log(x)+exp(2*x)+(5+2*x)*log
(3)-10*x-25)/(x*log(3)-5*x)/log(x),x, algorithm="giac")

[Out]

(x^2*log(3) + 2*x*log(3)*log(log(x)) - 5*x^2 - 10*x*log(log(x)) + e^(2*x)*log(log(x)) + 5*log(3)*log(log(x)) -
25*log(log(x)))/(log(3) - 5)

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maple [A]  time = 0.05, size = 33, normalized size = 1.32

method	result	size
risch	\(\frac {\left (2 x \ln \relax (3)-10 x +{\mathrm e}^{2 x}\right ) \ln \left (\ln \relax (x )\right )}{\ln \relax (3)-5}+x^{2}+5 \ln \left (\ln \relax (x )\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(2*x)+2*x*ln(3)-10*x)*ln(x)*ln(ln(x))+(2*x^2*ln(3)-10*x^2)*ln(x)+exp(2*x)+(5+2*x)*ln(3)-10*x-25)/
(x*ln(3)-5*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

(2*x*ln(3)-10*x+exp(2*x))/(ln(3)-5)*ln(ln(x))+x^2+5*ln(ln(x))

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maxima [B]  time = 0.48, size = 57, normalized size = 2.28 \begin {gather*} \frac {5 \, \log \relax (3) \log \left (\log \relax (x)\right )}{\log \relax (3) - 5} + \frac {x^{2} {\left (\log \relax (3) - 5\right )} + {\left (2 \, x {\left (\log \relax (3) - 5\right )} + e^{\left (2 \, x\right )}\right )} \log \left (\log \relax (x)\right )}{\log \relax (3) - 5} - \frac {25 \, \log \left (\log \relax (x)\right )}{\log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(2*x)+2*x*log(3)-10*x)*log(x)*log(log(x))+(2*x^2*log(3)-10*x^2)*log(x)+exp(2*x)+(5+2*x)*log
(3)-10*x-25)/(x*log(3)-5*x)/log(x),x, algorithm="maxima")

[Out]

5*log(3)*log(log(x))/(log(3) - 5) + (x^2*(log(3) - 5) + (2*x*(log(3) - 5) + e^(2*x))*log(log(x)))/(log(3) - 5)
- 25*log(log(x))/(log(3) - 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{2\,x}-10\,x+\ln \relax (3)\,\left (2\,x+5\right )+\ln \relax (x)\,\left (2\,x^2\,\ln \relax (3)-10\,x^2\right )+\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (2\,x\,{\mathrm {e}}^{2\,x}-10\,x+2\,x\,\ln \relax (3)\right )-25}{\ln \relax (x)\,\left (5\,x-x\,\ln \relax (3)\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x) - 10*x + log(3)*(2*x + 5) + log(x)*(2*x^2*log(3) - 10*x^2) + log(log(x))*log(x)*(2*x*exp(2*x) -
10*x + 2*x*log(3)) - 25)/(log(x)*(5*x - x*log(3))),x)

[Out]

int(-(exp(2*x) - 10*x + log(3)*(2*x + 5) + log(x)*(2*x^2*log(3) - 10*x^2) + log(log(x))*log(x)*(2*x*exp(2*x) -
10*x + 2*x*log(3)) - 25)/(log(x)*(5*x - x*log(3))), x)

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sympy [A]  time = 2.60, size = 32, normalized size = 1.28 \begin {gather*} x^{2} + 2 x \log {\left (\log {\relax (x )} \right )} + \frac {e^{2 x} \log {\left (\log {\relax (x )} \right )}}{-5 + \log {\relax (3 )}} + 5 \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(2*x)+2*x*ln(3)-10*x)*ln(x)*ln(ln(x))+(2*x**2*ln(3)-10*x**2)*ln(x)+exp(2*x)+(5+2*x)*ln(3)-1
0*x-25)/(x*ln(3)-5*x)/ln(x),x)

[Out]

x**2 + 2*x*log(log(x)) + exp(2*x)*log(log(x))/(-5 + log(3)) + 5*log(log(x))

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