3.21.10 \(\int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} (4-1596 x-4 x^2)+(16-16 x+4 e^{5+x} x) \log (x)}{8 x} \, dx\)

Optimal. Leaf size=18 \[ \left (398-\frac {e^{5+x}}{4}+x-\log (x)\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 0.32, antiderivative size = 49, normalized size of antiderivative = 2.72, number of steps used = 15, number of rules used = 9, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.155, Rules used = {12, 14, 2194, 6686, 6742, 2199, 2178, 2176, 2554} \begin {gather*} -199 e^{x+5}+\frac {1}{16} e^{2 x+10}-\frac {1}{2} e^{x+5} x+(x-\log (x)+398)^2+\frac {1}{2} e^{x+5} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6368 + 6352*x + E^(10 + 2*x)*x + 16*x^2 + E^(5 + x)*(4 - 1596*x - 4*x^2) + (16 - 16*x + 4*E^(5 + x)*x)*L
og[x])/(8*x),x]

[Out]

-199*E^(5 + x) + E^(10 + 2*x)/16 - (E^(5 + x)*x)/2 + (398 + x - Log[x])^2 + (E^(5 + x)*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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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}{8} \int \frac {-6368+6352 x+e^{10+2 x} x+16 x^2+e^{5+x} \left (4-1596 x-4 x^2\right )+\left (16-16 x+4 e^{5+x} x\right ) \log (x)}{x} \, dx\\ &=\frac {1}{8} \int \left (e^{10+2 x}+\frac {16 (-1+x) (398+x-\log (x))}{x}-\frac {4 e^{5+x} \left (-1+399 x+x^2-x \log (x)\right )}{x}\right ) \, dx\\ &=\frac {1}{8} \int e^{10+2 x} \, dx-\frac {1}{2} \int \frac {e^{5+x} \left (-1+399 x+x^2-x \log (x)\right )}{x} \, dx+2 \int \frac {(-1+x) (398+x-\log (x))}{x} \, dx\\ &=\frac {1}{16} e^{10+2 x}+(398+x-\log (x))^2-\frac {1}{2} \int \left (\frac {e^{5+x} \left (-1+399 x+x^2\right )}{x}-e^{5+x} \log (x)\right ) \, dx\\ &=\frac {1}{16} e^{10+2 x}+(398+x-\log (x))^2-\frac {1}{2} \int \frac {e^{5+x} \left (-1+399 x+x^2\right )}{x} \, dx+\frac {1}{2} \int e^{5+x} \log (x) \, dx\\ &=\frac {1}{16} e^{10+2 x}+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x)-\frac {1}{2} \int \frac {e^{5+x}}{x} \, dx-\frac {1}{2} \int \left (399 e^{5+x}-\frac {e^{5+x}}{x}+e^{5+x} x\right ) \, dx\\ &=\frac {1}{16} e^{10+2 x}-\frac {e^5 \text {Ei}(x)}{2}+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x)+\frac {1}{2} \int \frac {e^{5+x}}{x} \, dx-\frac {1}{2} \int e^{5+x} x \, dx-\frac {399}{2} \int e^{5+x} \, dx\\ &=-\frac {399 e^{5+x}}{2}+\frac {1}{16} e^{10+2 x}-\frac {1}{2} e^{5+x} x+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x)+\frac {1}{2} \int e^{5+x} \, dx\\ &=-199 e^{5+x}+\frac {1}{16} e^{10+2 x}-\frac {1}{2} e^{5+x} x+(398+x-\log (x))^2+\frac {1}{2} e^{5+x} \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 23, normalized size = 1.28 \begin {gather*} \frac {1}{16} \left (-e^{5+x}+4 (398+x)-4 \log (x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6368 + 6352*x + E^(10 + 2*x)*x + 16*x^2 + E^(5 + x)*(4 - 1596*x - 4*x^2) + (16 - 16*x + 4*E^(5 + x
)*x)*Log[x])/(8*x),x]

[Out]

(-E^(5 + x) + 4*(398 + x) - 4*Log[x])^2/16

________________________________________________________________________________________

fricas [B]  time = 0.97, size = 43, normalized size = 2.39 \begin {gather*} x^{2} - \frac {1}{2} \, {\left (x + 398\right )} e^{\left (x + 5\right )} - \frac {1}{2} \, {\left (4 \, x - e^{\left (x + 5\right )} + 1592\right )} \log \relax (x) + \log \relax (x)^{2} + 796 \, x + \frac {1}{16} \, e^{\left (2 \, x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((4*x*exp(5+x)-16*x+16)*log(x)+x*exp(5+x)^2+(-4*x^2-1596*x+4)*exp(5+x)+16*x^2+6352*x-6368)/x,x,
algorithm="fricas")

[Out]

x^2 - 1/2*(x + 398)*e^(x + 5) - 1/2*(4*x - e^(x + 5) + 1592)*log(x) + log(x)^2 + 796*x + 1/16*e^(2*x + 10)

________________________________________________________________________________________

giac [B]  time = 0.24, size = 49, normalized size = 2.72 \begin {gather*} x^{2} - \frac {1}{2} \, x e^{\left (x + 5\right )} - 2 \, x \log \relax (x) + \frac {1}{2} \, e^{\left (x + 5\right )} \log \relax (x) + \log \relax (x)^{2} + 796 \, x + \frac {1}{16} \, e^{\left (2 \, x + 10\right )} - 199 \, e^{\left (x + 5\right )} - 796 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((4*x*exp(5+x)-16*x+16)*log(x)+x*exp(5+x)^2+(-4*x^2-1596*x+4)*exp(5+x)+16*x^2+6352*x-6368)/x,x,
algorithm="giac")

[Out]

x^2 - 1/2*x*e^(x + 5) - 2*x*log(x) + 1/2*e^(x + 5)*log(x) + log(x)^2 + 796*x + 1/16*e^(2*x + 10) - 199*e^(x +
5) - 796*log(x)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 50, normalized size = 2.78

method	result	size
default	\(\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}+\frac {\ln \relax (x ) {\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}+x^{2}+796 x -796 \ln \relax (x )-2 x \ln \relax (x )+\ln \relax (x )^{2}\)	\(50\)
norman	\(\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}+\frac {\ln \relax (x ) {\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}+x^{2}+796 x -796 \ln \relax (x )-2 x \ln \relax (x )+\ln \relax (x )^{2}\)	\(50\)
risch	\(\ln \relax (x )^{2}+\frac {\left (-16 x +4 \,{\mathrm e}^{5+x}\right ) \ln \relax (x )}{8}+x^{2}+796 x -796 \ln \relax (x )+\frac {{\mathrm e}^{2 x +10}}{16}-\frac {x \,{\mathrm e}^{5+x}}{2}-199 \,{\mathrm e}^{5+x}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*((4*x*exp(5+x)-16*x+16)*ln(x)+x*exp(5+x)^2+(-4*x^2-1596*x+4)*exp(5+x)+16*x^2+6352*x-6368)/x,x,method=_
RETURNVERBOSE)

[Out]

1/16*exp(5+x)^2-1/2*x*exp(5+x)+1/2*ln(x)*exp(5+x)-199*exp(5+x)+x^2+796*x-796*ln(x)-2*x*ln(x)+ln(x)^2

________________________________________________________________________________________

maxima [B]  time = 0.57, size = 55, normalized size = 3.06 \begin {gather*} x^{2} - \frac {1}{2} \, {\left (x e^{5} - e^{5}\right )} e^{x} - 2 \, x \log \relax (x) + \frac {1}{2} \, e^{\left (x + 5\right )} \log \relax (x) + \log \relax (x)^{2} + 796 \, x + \frac {1}{16} \, e^{\left (2 \, x + 10\right )} - \frac {399}{2} \, e^{\left (x + 5\right )} - 796 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((4*x*exp(5+x)-16*x+16)*log(x)+x*exp(5+x)^2+(-4*x^2-1596*x+4)*exp(5+x)+16*x^2+6352*x-6368)/x,x,
algorithm="maxima")

[Out]

x^2 - 1/2*(x*e^5 - e^5)*e^x - 2*x*log(x) + 1/2*e^(x + 5)*log(x) + log(x)^2 + 796*x + 1/16*e^(2*x + 10) - 399/2
*e^(x + 5) - 796*log(x)

________________________________________________________________________________________

mupad [B]  time = 1.22, size = 49, normalized size = 2.72 \begin {gather*} 796\,x-199\,{\mathrm {e}}^{x+5}+\frac {{\mathrm {e}}^{2\,x+10}}{16}-796\,\ln \relax (x)-\frac {x\,{\mathrm {e}}^{x+5}}{2}+{\ln \relax (x)}^2+\frac {{\mathrm {e}}^{x+5}\,\ln \relax (x)}{2}-2\,x\,\ln \relax (x)+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((794*x - (exp(x + 5)*(1596*x + 4*x^2 - 4))/8 + (x*exp(2*x + 10))/8 + (log(x)*(4*x*exp(x + 5) - 16*x + 16))
/8 + 2*x^2 - 796)/x,x)

[Out]

796*x - 199*exp(x + 5) + exp(2*x + 10)/16 - 796*log(x) - (x*exp(x + 5))/2 + log(x)^2 + (exp(x + 5)*log(x))/2 -
2*x*log(x) + x^2

________________________________________________________________________________________

sympy [B]  time = 0.42, size = 48, normalized size = 2.67 \begin {gather*} x^{2} - 2 x \log {\relax (x )} + 796 x + \frac {\left (- 16 x + 16 \log {\relax (x )} - 6368\right ) e^{x + 5}}{32} + \frac {e^{2 x + 10}}{16} + \log {\relax (x )}^{2} - 796 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((4*x*exp(5+x)-16*x+16)*ln(x)+x*exp(5+x)**2+(-4*x**2-1596*x+4)*exp(5+x)+16*x**2+6352*x-6368)/x,x
)

[Out]

x**2 - 2*x*log(x) + 796*x + (-16*x + 16*log(x) - 6368)*exp(x + 5)/32 + exp(2*x + 10)/16 + log(x)**2 - 796*log(
x)

________________________________________________________________________________________