3.45.20 \(\int \frac {-10-2 x+2 x \log (x)+32 e^{e^{16 x^2}+16 x^2} x^2 \log ^2(x)+(-2+6 x+2 x^2) \log ^2(x)}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ e^{e^{16 x^2}}+x+(5+x) \left (x+\frac {2}{\log (x)}\right )-\log \left (x^2\right ) \]

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Rubi [A]  time = 0.68, antiderivative size = 33, normalized size of antiderivative = 1.14, number of steps used = 17, number of rules used = 11, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.193, Rules used = {6742, 6715, 2282, 2194, 6688, 14, 2353, 2297, 2298, 2302, 30} \begin {gather*} x^2+e^{e^{16 x^2}}+6 x+\frac {2 x}{\log (x)}-2 \log (x)+\frac {10}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 - 2*x + 2*x*Log[x] + 32*E^(E^(16*x^2) + 16*x^2)*x^2*Log[x]^2 + (-2 + 6*x + 2*x^2)*Log[x]^2)/(x*Log[x]
^2),x]

[Out]

E^E^(16*x^2) + 6*x + x^2 + 10/Log[x] + (2*x)/Log[x] - 2*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

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 6688

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

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), 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 \left (32 e^{e^{16 x^2}+16 x^2} x+\frac {2 \left (-5-x+x \log (x)-\log ^2(x)+3 x \log ^2(x)+x^2 \log ^2(x)\right )}{x \log ^2(x)}\right ) \, dx\\ &=2 \int \frac {-5-x+x \log (x)-\log ^2(x)+3 x \log ^2(x)+x^2 \log ^2(x)}{x \log ^2(x)} \, dx+32 \int e^{e^{16 x^2}+16 x^2} x \, dx\\ &=2 \int \frac {-5-x+x \log (x)+\left (-1+3 x+x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx+16 \operatorname {Subst}\left (\int e^{e^{16 x}+16 x} \, dx,x,x^2\right )\\ &=2 \int \left (\frac {-1+3 x+x^2}{x}+\frac {-5-x}{x \log ^2(x)}+\frac {1}{\log (x)}\right ) \, dx+\operatorname {Subst}\left (\int e^x \, dx,x,e^{16 x^2}\right )\\ &=e^{e^{16 x^2}}+2 \int \frac {-1+3 x+x^2}{x} \, dx+2 \int \frac {-5-x}{x \log ^2(x)} \, dx+2 \int \frac {1}{\log (x)} \, dx\\ &=e^{e^{16 x^2}}+2 \text {li}(x)+2 \int \left (3-\frac {1}{x}+x\right ) \, dx+2 \int \left (-\frac {1}{\log ^2(x)}-\frac {5}{x \log ^2(x)}\right ) \, dx\\ &=e^{e^{16 x^2}}+6 x+x^2-2 \log (x)+2 \text {li}(x)-2 \int \frac {1}{\log ^2(x)} \, dx-10 \int \frac {1}{x \log ^2(x)} \, dx\\ &=e^{e^{16 x^2}}+6 x+x^2+\frac {2 x}{\log (x)}-2 \log (x)+2 \text {li}(x)-2 \int \frac {1}{\log (x)} \, dx-10 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=e^{e^{16 x^2}}+6 x+x^2+\frac {10}{\log (x)}+\frac {2 x}{\log (x)}-2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 28, normalized size = 0.97 \begin {gather*} e^{e^{16 x^2}}+x (6+x)+\frac {2 (5+x)}{\log (x)}-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 2*x + 2*x*Log[x] + 32*E^(E^(16*x^2) + 16*x^2)*x^2*Log[x]^2 + (-2 + 6*x + 2*x^2)*Log[x]^2)/(x*
Log[x]^2),x]

[Out]

E^E^(16*x^2) + x*(6 + x) + (2*(5 + x))/Log[x] - 2*Log[x]

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fricas [B]  time = 0.49, size = 67, normalized size = 2.31 \begin {gather*} \frac {{\left ({\left (x^{2} + 6 \, x\right )} e^{\left (16 \, x^{2}\right )} \log \relax (x) - 2 \, e^{\left (16 \, x^{2}\right )} \log \relax (x)^{2} + 2 \, {\left (x + 5\right )} e^{\left (16 \, x^{2}\right )} + e^{\left (16 \, x^{2} + e^{\left (16 \, x^{2}\right )}\right )} \log \relax (x)\right )} e^{\left (-16 \, x^{2}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2*exp(16*x^2)*log(x)^2*exp(exp(16*x^2))+(2*x^2+6*x-2)*log(x)^2+2*x*log(x)-2*x-10)/x/log(x)^2,x
, algorithm="fricas")

[Out]

((x^2 + 6*x)*e^(16*x^2)*log(x) - 2*e^(16*x^2)*log(x)^2 + 2*(x + 5)*e^(16*x^2) + e^(16*x^2 + e^(16*x^2))*log(x)
)*e^(-16*x^2)/log(x)

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giac [B]  time = 0.18, size = 80, normalized size = 2.76 \begin {gather*} \frac {{\left (x^{2} e^{\left (16 \, x^{2}\right )} \log \relax (x) + 6 \, x e^{\left (16 \, x^{2}\right )} \log \relax (x) - 2 \, e^{\left (16 \, x^{2}\right )} \log \relax (x)^{2} + 2 \, x e^{\left (16 \, x^{2}\right )} + e^{\left (16 \, x^{2} + e^{\left (16 \, x^{2}\right )}\right )} \log \relax (x) + 10 \, e^{\left (16 \, x^{2}\right )}\right )} e^{\left (-16 \, x^{2}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2*exp(16*x^2)*log(x)^2*exp(exp(16*x^2))+(2*x^2+6*x-2)*log(x)^2+2*x*log(x)-2*x-10)/x/log(x)^2,x
, algorithm="giac")

[Out]

(x^2*e^(16*x^2)*log(x) + 6*x*e^(16*x^2)*log(x) - 2*e^(16*x^2)*log(x)^2 + 2*x*e^(16*x^2) + e^(16*x^2 + e^(16*x^
2))*log(x) + 10*e^(16*x^2))*e^(-16*x^2)/log(x)

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maple [A]  time = 0.03, size = 28, normalized size = 0.97

method	result	size
risch	\(6 x +x^{2}-2 \ln \relax (x )+\frac {2 x +10}{\ln \relax (x )}+{\mathrm e}^{{\mathrm e}^{16 x^{2}}}\)	\(28\)
default	\(6 x +x^{2}-2 \ln \relax (x )+\frac {2 x}{\ln \relax (x )}+\frac {10}{\ln \relax (x )}+{\mathrm e}^{{\mathrm e}^{16 x^{2}}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x^2*exp(16*x^2)*ln(x)^2*exp(exp(16*x^2))+(2*x^2+6*x-2)*ln(x)^2+2*x*ln(x)-2*x-10)/x/ln(x)^2,x,method=_R
ETURNVERBOSE)

[Out]

6*x+x^2-2*ln(x)+2*(5+x)/ln(x)+exp(exp(16*x^2))

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maxima [C]  time = 0.40, size = 37, normalized size = 1.28 \begin {gather*} x^{2} + 6 \, x + \frac {10}{\log \relax (x)} + 2 \, {\rm Ei}\left (\log \relax (x)\right ) + e^{\left (e^{\left (16 \, x^{2}\right )}\right )} - 2 \, \Gamma \left (-1, -\log \relax (x)\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2*exp(16*x^2)*log(x)^2*exp(exp(16*x^2))+(2*x^2+6*x-2)*log(x)^2+2*x*log(x)-2*x-10)/x/log(x)^2,x
, algorithm="maxima")

[Out]

x^2 + 6*x + 10/log(x) + 2*Ei(log(x)) + e^(e^(16*x^2)) - 2*gamma(-1, -log(x)) - 2*log(x)

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mupad [B]  time = 3.30, size = 33, normalized size = 1.14 \begin {gather*} 8\,x+{\mathrm {e}}^{{\mathrm {e}}^{16\,x^2}}-2\,\ln \relax (x)+\frac {2\,x-2\,x\,\ln \relax (x)+10}{\ln \relax (x)}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(6*x + 2*x^2 - 2) - 2*x + 2*x*log(x) + 32*x^2*exp(exp(16*x^2))*exp(16*x^2)*log(x)^2 - 10)/(x*log
(x)^2),x)

[Out]

8*x + exp(exp(16*x^2)) - 2*log(x) + (2*x - 2*x*log(x) + 10)/log(x) + x^2

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sympy [A]  time = 0.39, size = 27, normalized size = 0.93 \begin {gather*} x^{2} + 6 x + \frac {2 x + 10}{\log {\relax (x )}} + e^{e^{16 x^{2}}} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x**2*exp(16*x**2)*ln(x)**2*exp(exp(16*x**2))+(2*x**2+6*x-2)*ln(x)**2+2*x*ln(x)-2*x-10)/x/ln(x)**
2,x)

[Out]

x**2 + 6*x + (2*x + 10)/log(x) + exp(exp(16*x**2)) - 2*log(x)

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