\(\int \frac {e^{e^2+x^2} (2 x+4 x^3-4 x^2 \log (400))+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx\) [7567]

Optimal result

Integrand size = 52, antiderivative size = 25 \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=(-x+\log (400)) \left (-\frac {2}{5} e^{e^2+x^2}+\log ^2(x)\right ) \]

[Out]

(ln(x)^2-2/5*exp(x^2)*exp(exp(2)))*(2*ln(20)-x)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {12, 14, 2258, 2235, 2243, 2240, 45, 2332, 6874, 2388, 2338, 2333} \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=\frac {2}{5} e^{x^2+e^2} x-\frac {2}{5} e^{x^2+e^2} \log (400)-x \log ^2(x)+\log (400) \log ^2(x) \]

[In]

Int[(E^(E^2 + x^2)*(2*x + 4*x^3 - 4*x^2*Log[400]) + (-10*x + 10*Log[400])*Log[x] - 5*x*Log[x]^2)/(5*x),x]

[Out]

(2*E^(E^2 + x^2)*x)/5 - (2*E^(E^2 + x^2)*Log[400])/5 - x*Log[x]^2 + Log[400]*Log[x]^2

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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2240

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

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2388

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[(d
+ e*x)^(q - 1)*((a + b*Log[c*x^n])^p/x), x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{x} \, dx \\ & = \frac {1}{5} \int \left (2 e^{e^2+x^2} \left (1+2 x^2-2 x \log (400)\right )-\frac {5 \log (x) (2 x-2 \log (400)+x \log (x))}{x}\right ) \, dx \\ & = \frac {2}{5} \int e^{e^2+x^2} \left (1+2 x^2-2 x \log (400)\right ) \, dx-\int \frac {\log (x) (2 x-2 \log (400)+x \log (x))}{x} \, dx \\ & = \frac {2}{5} \int \left (e^{e^2+x^2}+2 e^{e^2+x^2} x^2-2 e^{e^2+x^2} x \log (400)\right ) \, dx-\int \left (\frac {2 (x-\log (400)) \log (x)}{x}+\log ^2(x)\right ) \, dx \\ & = \frac {2}{5} \int e^{e^2+x^2} \, dx+\frac {4}{5} \int e^{e^2+x^2} x^2 \, dx-2 \int \frac {(x-\log (400)) \log (x)}{x} \, dx-\frac {1}{5} (4 \log (400)) \int e^{e^2+x^2} x \, dx-\int \log ^2(x) \, dx \\ & = \frac {2}{5} e^{e^2+x^2} x+\frac {1}{5} e^{e^2} \sqrt {\pi } \text {erfi}(x)-\frac {2}{5} e^{e^2+x^2} \log (400)-x \log ^2(x)-\frac {2}{5} \int e^{e^2+x^2} \, dx+(2 \log (400)) \int \frac {\log (x)}{x} \, dx \\ & = \frac {2}{5} e^{e^2+x^2} x-\frac {2}{5} e^{e^2+x^2} \log (400)-x \log ^2(x)+\log (400) \log ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=\frac {1}{5} (x-\log (400)) \left (2 e^{e^2+x^2}-5 \log ^2(x)\right ) \]

[In]

Integrate[(E^(E^2 + x^2)*(2*x + 4*x^3 - 4*x^2*Log[400]) + (-10*x + 10*Log[400])*Log[x] - 5*x*Log[x]^2)/(5*x),x
]

[Out]

((x - Log[400])*(2*E^(E^2 + x^2) - 5*Log[x]^2))/5

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52

method	result	size
default	\(-x \ln \left (x \right )^{2}+2 \ln \left (20\right ) \ln \left (x \right )^{2}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{2}} \left (-{\mathrm e}^{x^{2}} x +2 \ln \left (20\right ) {\mathrm e}^{x^{2}}\right )}{5}\)	\(38\)
parallelrisch	\(-\frac {4 \ln \left (20\right ) {\mathrm e}^{{\mathrm e}^{2}} {\mathrm e}^{x^{2}}}{5}+2 \ln \left (20\right ) \ln \left (x \right )^{2}+\frac {2 \,{\mathrm e}^{x^{2}} {\mathrm e}^{{\mathrm e}^{2}} x}{5}-x \ln \left (x \right )^{2}\)	\(38\)
parts	\(-x \ln \left (x \right )^{2}+2 \ln \left (20\right ) \ln \left (x \right )^{2}-\frac {2 \,{\mathrm e}^{{\mathrm e}^{2}} \left (-{\mathrm e}^{x^{2}} x +2 \ln \left (20\right ) {\mathrm e}^{x^{2}}\right )}{5}\)	\(38\)
risch	\(\frac {\left (-5 x +20 \ln \left (2\right )+10 \ln \left (5\right )\right ) \ln \left (x \right )^{2}}{5}-\frac {2 \left (4 \ln \left (2\right )+2 \ln \left (5\right )-x \right ) {\mathrm e}^{{\mathrm e}^{2}+x^{2}}}{5}\)	\(41\)

[In]

int(1/5*(-5*x*ln(x)^2+(20*ln(20)-10*x)*ln(x)+(-8*x^2*ln(20)+4*x^3+2*x)*exp(x^2)*exp(exp(2)))/x,x,method=_RETUR
NVERBOSE)

[Out]

-x*ln(x)^2+2*ln(20)*ln(x)^2-2/5*exp(exp(2))*(-exp(x^2)*x+2*ln(20)*exp(x^2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=-{\left (x - 2 \, \log \left (20\right )\right )} \log \left (x\right )^{2} + \frac {2}{5} \, {\left (x - 2 \, \log \left (20\right )\right )} e^{\left (x^{2} + e^{2}\right )} \]

[In]

integrate(1/5*(-5*x*log(x)^2+(20*log(20)-10*x)*log(x)+(-8*x^2*log(20)+4*x^3+2*x)*exp(x^2)*exp(exp(2)))/x,x, al
gorithm="fricas")

[Out]

-(x - 2*log(20))*log(x)^2 + 2/5*(x - 2*log(20))*e^(x^2 + e^2)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=\left (- x + 2 \log {\left (20 \right )}\right ) \log {\left (x \right )}^{2} + \frac {\left (2 x e^{e^{2}} - 4 e^{e^{2}} \log {\left (20 \right )}\right ) e^{x^{2}}}{5} \]

[In]

integrate(1/5*(-5*x*ln(x)**2+(20*ln(20)-10*x)*ln(x)+(-8*x**2*ln(20)+4*x**3+2*x)*exp(x**2)*exp(exp(2)))/x,x)

[Out]

(-x + 2*log(20))*log(x)**2 + (2*x*exp(exp(2)) - 4*exp(exp(2))*log(20))*exp(x**2)/5

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (24) = 48\).

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=2 \, \log \left (20\right ) \log \left (x\right )^{2} - {\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x + \frac {2}{5} \, x e^{\left (x^{2} + e^{2}\right )} - \frac {4}{5} \, e^{\left (x^{2} + e^{2}\right )} \log \left (20\right ) - 2 \, x \log \left (x\right ) + 2 \, x \]

[In]

integrate(1/5*(-5*x*log(x)^2+(20*log(20)-10*x)*log(x)+(-8*x^2*log(20)+4*x^3+2*x)*exp(x^2)*exp(exp(2)))/x,x, al
gorithm="maxima")

[Out]

2*log(20)*log(x)^2 - (log(x)^2 - 2*log(x) + 2)*x + 2/5*x*e^(x^2 + e^2) - 4/5*e^(x^2 + e^2)*log(20) - 2*x*log(x
) + 2*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=-x \log \left (x\right )^{2} + 2 \, \log \left (20\right ) \log \left (x\right )^{2} + \frac {2}{5} \, x e^{\left (x^{2} + e^{2}\right )} - \frac {4}{5} \, e^{\left (x^{2} + e^{2}\right )} \log \left (20\right ) \]

[In]

integrate(1/5*(-5*x*log(x)^2+(20*log(20)-10*x)*log(x)+(-8*x^2*log(20)+4*x^3+2*x)*exp(x^2)*exp(exp(2)))/x,x, al
gorithm="giac")

[Out]

-x*log(x)^2 + 2*log(20)*log(x)^2 + 2/5*x*e^(x^2 + e^2) - 4/5*e^(x^2 + e^2)*log(20)

Mupad [B] (verification not implemented)

Time = 13.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^2+x^2} \left (2 x+4 x^3-4 x^2 \log (400)\right )+(-10 x+10 \log (400)) \log (x)-5 x \log ^2(x)}{5 x} \, dx=\frac {\left (2\,{\mathrm {e}}^{x^2+{\mathrm {e}}^2}-5\,{\ln \left (x\right )}^2\right )\,\left (x-\ln \left (400\right )\right )}{5} \]

[In]

int(-(x*log(x)^2 + (log(x)*(10*x - 20*log(20)))/5 - (exp(x^2)*exp(exp(2))*(2*x - 8*x^2*log(20) + 4*x^3))/5)/x,
x)

[Out]

((2*exp(exp(2) + x^2) - 5*log(x)^2)*(x - log(400)))/5