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

Optimal. Leaf size=25 \[ (-x+\log (400)) \left (-\frac {2}{5} e^{e^2+x^2}+\log ^2(x)\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 16, number of rules used = 12, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {12, 14, 2226, 2204, 2212, 2209, 43, 2295, 6742, 2346, 2301, 2296} \begin {gather*} \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) \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2204

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 2209

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 2212

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 2226

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 2295

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

Rule 2296

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 2301

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 2346

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 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}{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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 28, normalized size = 1.12 \begin {gather*} \frac {1}{5} (x-\log (400)) \left (2 e^{e^2+x^2}-5 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.54, size = 28, normalized size = 1.12 \begin {gather*} -{\left (x - 2 \, \log \left (20\right )\right )} \log \relax (x)^{2} + \frac {2}{5} \, {\left (x - 2 \, \log \left (20\right )\right )} e^{\left (x^{2} + e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.21, size = 37, normalized size = 1.48 \begin {gather*} -x \log \relax (x)^{2} + 2 \, \log \left (20\right ) \log \relax (x)^{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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.04, size = 38, normalized size = 1.52

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

Verification of antiderivative is not currently implemented for this CAS.

[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)^2-4/5*exp(x^2)*ln(20)*exp(exp(2))+2/5*exp(x^2)*exp(exp(2))*x

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maxima [B]  time = 0.36, size = 51, normalized size = 2.04 \begin {gather*} 2 \, \log \left (20\right ) \log \relax (x)^{2} - {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 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 \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 5.95, size = 24, normalized size = 0.96 \begin {gather*} \frac {\left (2\,{\mathrm {e}}^{x^2+{\mathrm {e}}^2}-5\,{\ln \relax (x)}^2\right )\,\left (x-\ln \left (400\right )\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.35, size = 36, normalized size = 1.44 \begin {gather*} \left (- x + 2 \log {\left (20 \right )}\right ) \log {\relax (x )}^{2} + \frac {\left (2 x e^{e^{2}} - 4 e^{e^{2}} \log {\left (20 \right )}\right ) e^{x^{2}}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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