3.99.24 \(\int \frac {x+2 e^{2 x} x+2 x^2+12 x^3+4 x^4+e^x (2+2 x+2 x^2)+(2+2 x+2 e^x x) \log (x)}{x} \, dx\)

Optimal. Leaf size=29 \[ -x+\frac {1}{3} x^3 (-x+4 (3+x))+\left (e^x+x+\log (x)\right )^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 45, normalized size of antiderivative = 1.55, number of steps used = 9, number of rules used = 6, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2194, 2288, 2346, 2301, 2295} \begin {gather*} x^4+4 x^3+x^2+\frac {2 e^x \left (x^2+x \log (x)\right )}{x}-x+e^{2 x}+\log ^2(x)+2 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + 2*E^(2*x)*x + 2*x^2 + 12*x^3 + 4*x^4 + E^x*(2 + 2*x + 2*x^2) + (2 + 2*x + 2*E^x*x)*Log[x])/x,x]

[Out]

E^(2*x) - x + x^2 + 4*x^3 + x^4 + 2*x*Log[x] + Log[x]^2 + (2*E^x*(x^2 + x*Log[x]))/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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, 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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^{2 x}+\frac {2 e^x \left (1+x+x^2+x \log (x)\right )}{x}+\frac {x+2 x^2+12 x^3+4 x^4+2 \log (x)+2 x \log (x)}{x}\right ) \, dx\\ &=2 \int e^{2 x} \, dx+2 \int \frac {e^x \left (1+x+x^2+x \log (x)\right )}{x} \, dx+\int \frac {x+2 x^2+12 x^3+4 x^4+2 \log (x)+2 x \log (x)}{x} \, dx\\ &=e^{2 x}+\frac {2 e^x \left (x^2+x \log (x)\right )}{x}+\int \left (1+2 x+12 x^2+4 x^3+\frac {2 (1+x) \log (x)}{x}\right ) \, dx\\ &=e^{2 x}+x+x^2+4 x^3+x^4+\frac {2 e^x \left (x^2+x \log (x)\right )}{x}+2 \int \frac {(1+x) \log (x)}{x} \, dx\\ &=e^{2 x}+x+x^2+4 x^3+x^4+\frac {2 e^x \left (x^2+x \log (x)\right )}{x}+2 \int \log (x) \, dx+2 \int \frac {\log (x)}{x} \, dx\\ &=e^{2 x}-x+x^2+4 x^3+x^4+2 x \log (x)+\log ^2(x)+\frac {2 e^x \left (x^2+x \log (x)\right )}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 39, normalized size = 1.34 \begin {gather*} e^{2 x}-x+2 e^x x+x^2+4 x^3+x^4+2 \left (e^x+x\right ) \log (x)+\log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + 2*E^(2*x)*x + 2*x^2 + 12*x^3 + 4*x^4 + E^x*(2 + 2*x + 2*x^2) + (2 + 2*x + 2*E^x*x)*Log[x])/x,x]

[Out]

E^(2*x) - x + 2*E^x*x + x^2 + 4*x^3 + x^4 + 2*(E^x + x)*Log[x] + Log[x]^2

________________________________________________________________________________________

fricas [A]  time = 0.83, size = 36, normalized size = 1.24 \begin {gather*} x^{4} + 4 \, x^{3} + x^{2} + 2 \, x e^{x} + 2 \, {\left (x + e^{x}\right )} \log \relax (x) + \log \relax (x)^{2} - x + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x+2)*log(x)+2*x*exp(x)^2+(2*x^2+2*x+2)*exp(x)+4*x^4+12*x^3+2*x^2+x)/x,x, algorithm="f
ricas")

[Out]

x^4 + 4*x^3 + x^2 + 2*x*e^x + 2*(x + e^x)*log(x) + log(x)^2 - x + e^(2*x)

________________________________________________________________________________________

giac [B]  time = 0.15, size = 39, normalized size = 1.34 \begin {gather*} x^{4} + 4 \, x^{3} + x^{2} + 2 \, x e^{x} + 2 \, x \log \relax (x) + 2 \, e^{x} \log \relax (x) + \log \relax (x)^{2} - x + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x+2)*log(x)+2*x*exp(x)^2+(2*x^2+2*x+2)*exp(x)+4*x^4+12*x^3+2*x^2+x)/x,x, algorithm="g
iac")

[Out]

x^4 + 4*x^3 + x^2 + 2*x*e^x + 2*x*log(x) + 2*e^x*log(x) + log(x)^2 - x + e^(2*x)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 40, normalized size = 1.38

method	result	size
default	\(-x +2 \,{\mathrm e}^{x} x +2 \,{\mathrm e}^{x} \ln \relax (x )+x^{2}+4 x^{3}+x^{4}+{\mathrm e}^{2 x}+2 x \ln \relax (x )+\ln \relax (x )^{2}\)	\(40\)
risch	\(\ln \relax (x )^{2}+\left (2 \,{\mathrm e}^{x}+2 x \right ) \ln \relax (x )+x^{4}+4 x^{3}+x^{2}+2 \,{\mathrm e}^{x} x +{\mathrm e}^{2 x}-x\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x+2*x+2)*ln(x)+2*x*exp(x)^2+(2*x^2+2*x+2)*exp(x)+4*x^4+12*x^3+2*x^2+x)/x,x,method=_RETURNVERBOS
E)

[Out]

-x+2*exp(x)*x+2*exp(x)*ln(x)+x^2+4*x^3+x^4+exp(x)^2+2*x*ln(x)+ln(x)^2

________________________________________________________________________________________

maxima [B]  time = 0.40, size = 45, normalized size = 1.55 \begin {gather*} x^{4} + 4 \, x^{3} + x^{2} + 2 \, {\left (x - 1\right )} e^{x} + 2 \, x \log \relax (x) + 2 \, e^{x} \log \relax (x) + \log \relax (x)^{2} - x + e^{\left (2 \, x\right )} + 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x+2)*log(x)+2*x*exp(x)^2+(2*x^2+2*x+2)*exp(x)+4*x^4+12*x^3+2*x^2+x)/x,x, algorithm="m
axima")

[Out]

x^4 + 4*x^3 + x^2 + 2*(x - 1)*e^x + 2*x*log(x) + 2*e^x*log(x) + log(x)^2 - x + e^(2*x) + 2*e^x

________________________________________________________________________________________

mupad [B]  time = 5.87, size = 39, normalized size = 1.34 \begin {gather*} {\mathrm {e}}^{2\,x}-x+2\,{\mathrm {e}}^x\,\ln \relax (x)+{\ln \relax (x)}^2+2\,x\,{\mathrm {e}}^x+2\,x\,\ln \relax (x)+x^2+4\,x^3+x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2*x*exp(2*x) + log(x)*(2*x + 2*x*exp(x) + 2) + exp(x)*(2*x + 2*x^2 + 2) + 2*x^2 + 12*x^3 + 4*x^4)/x,x
)

[Out]

exp(2*x) - x + 2*exp(x)*log(x) + log(x)^2 + 2*x*exp(x) + 2*x*log(x) + x^2 + 4*x^3 + x^4

________________________________________________________________________________________

sympy [B]  time = 0.40, size = 41, normalized size = 1.41 \begin {gather*} x^{4} + 4 x^{3} + x^{2} + 2 x \log {\relax (x )} - x + \left (2 x + 2 \log {\relax (x )}\right ) e^{x} + e^{2 x} + \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+2*x+2)*ln(x)+2*x*exp(x)**2+(2*x**2+2*x+2)*exp(x)+4*x**4+12*x**3+2*x**2+x)/x,x)

[Out]

x**4 + 4*x**3 + x**2 + 2*x*log(x) - x + (2*x + 2*log(x))*exp(x) + exp(2*x) + log(x)**2

________________________________________________________________________________________