\(\int \frac {-2 e^{25-x} x^2-20 \log (x) \log (x^2)+(-5+5 \log (x)) \log ^2(x^2)}{10 x^2} \, dx\) [8569]

Optimal result

Integrand size = 41, antiderivative size = 31 \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=\frac {e^{25-x}}{5}-\frac {x+\frac {1}{2} \log (x) \log ^2\left (x^2\right )}{x} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(104\) vs. \(2(31)=62\).

Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.35, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {12, 14, 2225, 2341, 2340, 2413, 6874, 2342} \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=\frac {(1-\log (x)) \log ^2\left (x^2\right )}{2 x}-\frac {\log ^2\left (x^2\right )}{2 x}+\frac {2 (1-\log (x)) \log \left (x^2\right )}{x}+\frac {2 \log (x) \log \left (x^2\right )}{x}-\frac {2 \log \left (x^2\right )}{x}+\frac {e^{25-x}}{5}-\frac {8}{x}+\frac {4 (1-\log (x))}{x}+\frac {4 (\log (x)+1)}{x} \]

[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 2225

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 2340

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 2341

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 2342

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 2413

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 6874

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=\frac {e^{25-x}}{5}-\frac {\log (x) \log ^2\left (x^2\right )}{2 x} \]

[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)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\(\frac {-20 \ln \left (x \right ) \ln \left (x^{2}\right )^{2}+8 x \,{\mathrm e}^{-x +25}}{40 x}\)	\(26\)
parts	\(-\frac {\ln \left (x \right ) {\left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}^{2}}{2 x}-\frac {{\left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}^{2}}{2 x}-\frac {2 \ln \left (x \right )^{2}}{x}-\frac {4 \ln \left (x \right )}{x}-\frac {4}{x}-\frac {2 \ln \left (x \right )^{3}}{x}-\frac {2 \ln \left (x \right ) \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}{x}-\frac {2 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}{x}-\frac {2 \ln \left (x \right )^{2} \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}{x}+\frac {4+\frac {\ln \left (x^{2}\right )^{2}}{2}+2 \ln \left (x^{2}\right )}{x}+\frac {{\mathrm e}^{-x +25}}{5}\)	\(142\)
default	\(-\frac {\ln \left (x \right ) {\left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}^{2}}{2 x}-\frac {{\left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}^{2}}{2 x}-\frac {2 \ln \left (x \right )^{2}}{x}-\frac {4 \ln \left (x \right )}{x}-\frac {4}{x}-\frac {2 \ln \left (x \right )^{3}}{x}-\frac {2 \ln \left (x \right ) \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}{x}-\frac {2 \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}{x}-\frac {2 \ln \left (x \right )^{2} \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right )}{x}+\frac {40+5 \ln \left (x^{2}\right )^{2}+20 \ln \left (x^{2}\right )}{10 x}+\frac {{\mathrm e}^{-x +25}}{5}\)	\(143\)
risch	\(-\frac {2 \ln \left (x \right )^{3}}{x}+\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right ) \ln \left (x \right )^{2}}{x}+\frac {\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{2} \left (\operatorname {csgn}\left (i x \right )^{4}-4 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{3}+6 \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )^{2}-4 \operatorname {csgn}\left (i x^{2}\right )^{3} \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{4}\right ) \ln \left (x \right )}{8 x}+\frac {{\mathrm e}^{-x +25}}{5}\)	\(154\)

[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]

1/40/x*(-20*ln(x)*ln(x^2)^2+8*x*exp(-x+25))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=-\frac {10 \, \log \left (x\right )^{3} - x e^{\left (-x + 25\right )}}{5 \, x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.45 \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=\frac {e^{25 - x}}{5} - \frac {2 \log {\left (x \right )}^{3}}{x} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (27) = 54\).

Time = 0.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.39 \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=-\frac {1}{2} \, {\left (\frac {\log \left (x^{2}\right )^{2}}{x} + \frac {4 \, \log \left (x^{2}\right )}{x} + \frac {8}{x}\right )} \log \left (x\right ) + 2 \, {\left (\frac {\log \left (x^{2}\right )}{x} + \frac {2}{x}\right )} \log \left (x\right ) + \frac {\log \left (x^{2}\right )^{2}}{2 \, x} - \frac {2 \, {\left (\log \left (x\right )^{2} + 4 \, \log \left (x\right ) + 6\right )}}{x} + \frac {4 \, {\left (\log \left (x\right ) + 2\right )}}{x} + \frac {2 \, \log \left (x^{2}\right )}{x} + \frac {4}{x} + \frac {1}{5} \, e^{\left (-x + 25\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=-\frac {10 \, \log \left (x\right )^{3} - x e^{\left (-x + 25\right )}}{5 \, x} \]

[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

Mupad [B] (verification not implemented)

Time = 14.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {-2 e^{25-x} x^2-20 \log (x) \log \left (x^2\right )+(-5+5 \log (x)) \log ^2\left (x^2\right )}{10 x^2} \, dx=\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{25}}{5}-\frac {{\ln \left (x^2\right )}^2\,\ln \left (x\right )}{2\,x} \]

[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)