3.86.69 \(\int \frac {-2 e^{25-x} x^2-20 \log (x) \log (x^2)+(-5+5 \log (x)) \log ^2(x^2)}{10 x^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {e^{25-x}}{5}-\frac {x+\frac {1}{2} \log (x) \log ^2\left (x^2\right )}{x} \]

________________________________________________________________________________________

Rubi [B]  time = 0.27, antiderivative size = 104, normalized size of antiderivative = 3.35, number of steps used = 14, number of rules used = 8, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {12, 14, 2194, 2304, 2303, 2366, 6742, 2305} \begin {gather*} \frac {(1-\log (x)) \log ^2\left (x^2\right )}{2 x}-\frac {\log ^2\left (x^2\right )}{2 x}+\frac {2 (1-\log (x)) \log \left (x^2\right )}{x}+\frac {2 \log (x) \log \left (x^2\right )}{x}-\frac {2 \log \left (x^2\right )}{x}+\frac {e^{25-x}}{5}-\frac {8}{x}+\frac {4 (1-\log (x))}{x}+\frac {4 (\log (x)+1)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^(25 - x)*x^2 - 20*Log[x]*Log[x^2] + (-5 + 5*Log[x])*Log[x^2]^2)/(10*x^2),x]

[Out]

E^(25 - x)/5 - 8/x + (4*(1 - Log[x]))/x + (4*(1 + Log[x]))/x - (2*Log[x^2])/x + (2*(1 - Log[x])*Log[x^2])/x +
(2*Log[x]*Log[x^2])/x - Log[x^2]^2/(2*x) + ((1 - Log[x])*Log[x^2]^2)/(2*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 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 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 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 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] && GtQ[p, 0]

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{x^2} \, dx\\ &=\frac {1}{10} \int \left (-2 e^{25-x}+\frac {5 \log \left (x^2\right ) \left (-4 \log (x)-\log \left (x^2\right )+\log (x) \log \left (x^2\right )\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int e^{25-x} \, dx\right )+\frac {1}{2} \int \frac {\log \left (x^2\right ) \left (-4 \log (x)-\log \left (x^2\right )+\log (x) \log \left (x^2\right )\right )}{x^2} \, dx\\ &=\frac {e^{25-x}}{5}+\frac {1}{2} \int \left (-\frac {4 \log (x) \log \left (x^2\right )}{x^2}+\frac {(-1+\log (x)) \log ^2\left (x^2\right )}{x^2}\right ) \, dx\\ &=\frac {e^{25-x}}{5}+\frac {1}{2} \int \frac {(-1+\log (x)) \log ^2\left (x^2\right )}{x^2} \, dx-2 \int \frac {\log (x) \log \left (x^2\right )}{x^2} \, dx\\ &=\frac {e^{25-x}}{5}+\frac {4 (1-\log (x))}{x}+\frac {2 \log \left (x^2\right )}{x}+\frac {2 (1-\log (x)) \log \left (x^2\right )}{x}+\frac {2 \log (x) \log \left (x^2\right )}{x}+\frac {(1-\log (x)) \log ^2\left (x^2\right )}{2 x}-\frac {1}{2} \int \frac {-8-4 \log \left (x^2\right )-\log ^2\left (x^2\right )}{x^2} \, dx+4 \int \frac {-1-\log (x)}{x^2} \, dx\\ &=\frac {e^{25-x}}{5}+\frac {4}{x}+\frac {4 (1-\log (x))}{x}+\frac {4 (1+\log (x))}{x}+\frac {2 \log \left (x^2\right )}{x}+\frac {2 (1-\log (x)) \log \left (x^2\right )}{x}+\frac {2 \log (x) \log \left (x^2\right )}{x}+\frac {(1-\log (x)) \log ^2\left (x^2\right )}{2 x}-\frac {1}{2} \int \left (-\frac {8}{x^2}-\frac {4 \log \left (x^2\right )}{x^2}-\frac {\log ^2\left (x^2\right )}{x^2}\right ) \, dx\\ &=\frac {e^{25-x}}{5}+\frac {4 (1-\log (x))}{x}+\frac {4 (1+\log (x))}{x}+\frac {2 \log \left (x^2\right )}{x}+\frac {2 (1-\log (x)) \log \left (x^2\right )}{x}+\frac {2 \log (x) \log \left (x^2\right )}{x}+\frac {(1-\log (x)) \log ^2\left (x^2\right )}{2 x}+\frac {1}{2} \int \frac {\log ^2\left (x^2\right )}{x^2} \, dx+2 \int \frac {\log \left (x^2\right )}{x^2} \, dx\\ &=\frac {e^{25-x}}{5}-\frac {4}{x}+\frac {4 (1-\log (x))}{x}+\frac {4 (1+\log (x))}{x}+\frac {2 (1-\log (x)) \log \left (x^2\right )}{x}+\frac {2 \log (x) \log \left (x^2\right )}{x}-\frac {\log ^2\left (x^2\right )}{2 x}+\frac {(1-\log (x)) \log ^2\left (x^2\right )}{2 x}+2 \int \frac {\log \left (x^2\right )}{x^2} \, dx\\ &=\frac {e^{25-x}}{5}-\frac {8}{x}+\frac {4 (1-\log (x))}{x}+\frac {4 (1+\log (x))}{x}-\frac {2 \log \left (x^2\right )}{x}+\frac {2 (1-\log (x)) \log \left (x^2\right )}{x}+\frac {2 \log (x) \log \left (x^2\right )}{x}-\frac {\log ^2\left (x^2\right )}{2 x}+\frac {(1-\log (x)) \log ^2\left (x^2\right )}{2 x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 27, normalized size = 0.87 \begin {gather*} \frac {e^{25-x}}{5}-\frac {\log (x) \log ^2\left (x^2\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(25 - x)*x^2 - 20*Log[x]*Log[x^2] + (-5 + 5*Log[x])*Log[x^2]^2)/(10*x^2),x]

[Out]

E^(25 - x)/5 - (Log[x]*Log[x^2]^2)/(2*x)

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 21, normalized size = 0.68 \begin {gather*} -\frac {10 \, \log \relax (x)^{3} - x e^{\left (-x + 25\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((5*log(x)-5)*log(x^2)^2-20*log(x)*log(x^2)-2*x^2*exp(-x+25))/x^2,x, algorithm="fricas")

[Out]

-1/5*(10*log(x)^3 - x*e^(-x + 25))/x

________________________________________________________________________________________

giac [A]  time = 0.15, size = 21, normalized size = 0.68 \begin {gather*} -\frac {10 \, \log \relax (x)^{3} - x e^{\left (-x + 25\right )}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((5*log(x)-5)*log(x^2)^2-20*log(x)*log(x^2)-2*x^2*exp(-x+25))/x^2,x, algorithm="giac")

[Out]

-1/5*(10*log(x)^3 - x*e^(-x + 25))/x

________________________________________________________________________________________

maple [B]  time = 0.06, size = 55, normalized size = 1.77

method	result	size
default	\(-\frac {2 \ln \relax (x )^{3}}{x}-\frac {2 \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right ) \ln \relax (x )^{2}}{x}-\frac {\left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right )^{2} \ln \relax (x )}{2 x}+\frac {{\mathrm e}^{-x +25}}{5}\)	\(55\)
risch	\(-\frac {2 \ln \relax (x )^{3}}{x}+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) \ln \relax (x )^{2}}{x}+\frac {\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{2} \left (\mathrm {csgn}\left (i x \right )^{4}-4 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{3}+6 \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )^{2}-4 \mathrm {csgn}\left (i x^{2}\right )^{3} \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{4}\right ) \ln \relax (x )}{8 x}+\frac {{\mathrm e}^{-x +25}}{5}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((5*ln(x)-5)*ln(x^2)^2-20*ln(x)*ln(x^2)-2*x^2*exp(-x+25))/x^2,x,method=_RETURNVERBOSE)

[Out]

-2/x*ln(x)^3-2*(ln(x^2)-2*ln(x))/x*ln(x)^2-1/2*(ln(x^2)-2*ln(x))^2/x*ln(x)+1/5*exp(-x+25)

________________________________________________________________________________________

maxima [B]  time = 0.35, size = 105, normalized size = 3.39 \begin {gather*} -\frac {1}{2} \, {\left (\frac {\log \left (x^{2}\right )^{2}}{x} + \frac {4 \, \log \left (x^{2}\right )}{x} + \frac {8}{x}\right )} \log \relax (x) + 2 \, {\left (\frac {\log \left (x^{2}\right )}{x} + \frac {2}{x}\right )} \log \relax (x) + \frac {\log \left (x^{2}\right )^{2}}{2 \, x} - \frac {2 \, {\left (\log \relax (x)^{2} + 4 \, \log \relax (x) + 6\right )}}{x} + \frac {4 \, {\left (\log \relax (x) + 2\right )}}{x} + \frac {2 \, \log \left (x^{2}\right )}{x} + \frac {4}{x} + \frac {1}{5} \, e^{\left (-x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((5*log(x)-5)*log(x^2)^2-20*log(x)*log(x^2)-2*x^2*exp(-x+25))/x^2,x, algorithm="maxima")

[Out]

-1/2*(log(x^2)^2/x + 4*log(x^2)/x + 8/x)*log(x) + 2*(log(x^2)/x + 2/x)*log(x) + 1/2*log(x^2)^2/x - 2*(log(x)^2
+ 4*log(x) + 6)/x + 4*(log(x) + 2)/x + 2*log(x^2)/x + 4/x + 1/5*e^(-x + 25)

________________________________________________________________________________________

mupad [B]  time = 5.27, size = 22, normalized size = 0.71 \begin {gather*} \frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{25}}{5}-\frac {{\ln \left (x^2\right )}^2\,\ln \relax (x)}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(-x)*exp(25))/5 - (log(x^2)^2*log(x))/(2*x)

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 14, normalized size = 0.45 \begin {gather*} \frac {e^{25 - x}}{5} - \frac {2 \log {\relax (x )}^{3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((5*ln(x)-5)*ln(x**2)**2-20*ln(x)*ln(x**2)-2*x**2*exp(-x+25))/x**2,x)

[Out]

exp(25 - x)/5 - 2*log(x)**3/x

________________________________________________________________________________________