3.37.93 \(\int \frac {2 x^3+e^x (-36+15 x-x^2+e x^2)+e^x (36-36 x+14 x^2-x^3) \log (x)}{x^2} \, dx\)

Optimal. Leaf size=25 \[ x^2-e^x \left (-e+\frac {(-12+x) (-3+x) \log (x)}{x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.64, antiderivative size = 42, normalized size of antiderivative = 1.68, number of steps used = 17, number of rules used = 8, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {14, 6742, 2199, 2194, 2177, 2178, 2176, 2554} \begin {gather*} x^2-(1-e) e^x+e^x-e^x x \log (x)+15 e^x \log (x)-\frac {36 e^x \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^3 + E^x*(-36 + 15*x - x^2 + E*x^2) + E^x*(36 - 36*x + 14*x^2 - x^3)*Log[x])/x^2,x]

[Out]

E^x - (1 - E)*E^x + x^2 + 15*E^x*Log[x] - (36*E^x*Log[x])/x - E^x*x*Log[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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), 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] && LtQ[m, -1] && 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 x+\frac {e^x \left (-36+15 x-(1-e) x^2+36 \log (x)-36 x \log (x)+14 x^2 \log (x)-x^3 \log (x)\right )}{x^2}\right ) \, dx\\ &=x^2+\int \frac {e^x \left (-36+15 x-(1-e) x^2+36 \log (x)-36 x \log (x)+14 x^2 \log (x)-x^3 \log (x)\right )}{x^2} \, dx\\ &=x^2+\int \left (\frac {e^x \left (-36+15 x-(1-e) x^2\right )}{x^2}-\frac {e^x \left (-36+36 x-14 x^2+x^3\right ) \log (x)}{x^2}\right ) \, dx\\ &=x^2+\int \frac {e^x \left (-36+15 x-(1-e) x^2\right )}{x^2} \, dx-\int \frac {e^x \left (-36+36 x-14 x^2+x^3\right ) \log (x)}{x^2} \, dx\\ &=x^2+15 e^x \log (x)-\frac {36 e^x \log (x)}{x}-e^x x \log (x)+\int \left ((-1+e) e^x-\frac {36 e^x}{x^2}+\frac {15 e^x}{x}\right ) \, dx+\int \frac {e^x \left (36-15 x+x^2\right )}{x^2} \, dx\\ &=x^2+15 e^x \log (x)-\frac {36 e^x \log (x)}{x}-e^x x \log (x)+15 \int \frac {e^x}{x} \, dx-36 \int \frac {e^x}{x^2} \, dx+(-1+e) \int e^x \, dx+\int \left (e^x+\frac {36 e^x}{x^2}-\frac {15 e^x}{x}\right ) \, dx\\ &=-\left ((1-e) e^x\right )+\frac {36 e^x}{x}+x^2+15 \text {Ei}(x)+15 e^x \log (x)-\frac {36 e^x \log (x)}{x}-e^x x \log (x)-15 \int \frac {e^x}{x} \, dx+36 \int \frac {e^x}{x^2} \, dx-36 \int \frac {e^x}{x} \, dx+\int e^x \, dx\\ &=e^x-(1-e) e^x+x^2-36 \text {Ei}(x)+15 e^x \log (x)-\frac {36 e^x \log (x)}{x}-e^x x \log (x)+36 \int \frac {e^x}{x} \, dx\\ &=e^x-(1-e) e^x+x^2+15 e^x \log (x)-\frac {36 e^x \log (x)}{x}-e^x x \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 27, normalized size = 1.08 \begin {gather*} e^{1+x}+x^2-\frac {e^x \left (36-15 x+x^2\right ) \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^3 + E^x*(-36 + 15*x - x^2 + E*x^2) + E^x*(36 - 36*x + 14*x^2 - x^3)*Log[x])/x^2,x]

[Out]

E^(1 + x) + x^2 - (E^x*(36 - 15*x + x^2)*Log[x])/x

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 28, normalized size = 1.12 \begin {gather*} \frac {x^{3} - {\left (x^{2} - 15 \, x + 36\right )} e^{x} \log \relax (x) + x e^{\left (x + 1\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3+14*x^2-36*x+36)*exp(x)*log(x)+(x^2*exp(1)-x^2+15*x-36)*exp(x)+2*x^3)/x^2,x, algorithm="fricas
")

[Out]

(x^3 - (x^2 - 15*x + 36)*e^x*log(x) + x*e^(x + 1))/x

________________________________________________________________________________________

giac [A]  time = 0.14, size = 39, normalized size = 1.56 \begin {gather*} -\frac {x^{2} e^{x} \log \relax (x) - x^{3} - 15 \, x e^{x} \log \relax (x) - x e^{\left (x + 1\right )} + 36 \, e^{x} \log \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3+14*x^2-36*x+36)*exp(x)*log(x)+(x^2*exp(1)-x^2+15*x-36)*exp(x)+2*x^3)/x^2,x, algorithm="giac")

[Out]

-(x^2*e^x*log(x) - x^3 - 15*x*e^x*log(x) - x*e^(x + 1) + 36*e^x*log(x))/x

________________________________________________________________________________________

maple [A]  time = 0.08, size = 26, normalized size = 1.04

method	result	size
risch	\(-\frac {\left (x^{2}-15 x +36\right ) {\mathrm e}^{x} \ln \relax (x )}{x}+{\mathrm e}^{x +1}+x^{2}\)	\(26\)
norman	\(\frac {x^{3}+x \,{\mathrm e} \,{\mathrm e}^{x}-36 \,{\mathrm e}^{x} \ln \relax (x )+15 x \,{\mathrm e}^{x} \ln \relax (x )-x^{2} {\mathrm e}^{x} \ln \relax (x )}{x}\)	\(37\)
default	\(\frac {x \,{\mathrm e} \,{\mathrm e}^{x}-36 \,{\mathrm e}^{x} \ln \relax (x )+15 x \,{\mathrm e}^{x} \ln \relax (x )-x^{2} {\mathrm e}^{x} \ln \relax (x )}{x}+x^{2}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3+14*x^2-36*x+36)*exp(x)*ln(x)+(x^2*exp(1)-x^2+15*x-36)*exp(x)+2*x^3)/x^2,x,method=_RETURNVERBOSE)

[Out]

-(x^2-15*x+36)/x*exp(x)*ln(x)+exp(x+1)+x^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} x^{2} - \frac {{\left (x^{2} - x + 36\right )} e^{x} \log \relax (x)}{x} + 14 \, e^{x} \log \relax (x) + {\rm Ei}\relax (x) + e^{\left (x + 1\right )} - e^{x} - 36 \, \Gamma \left (-1, -x\right ) + \int \frac {{\left (x^{2} - x + 36\right )} e^{x}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3+14*x^2-36*x+36)*exp(x)*log(x)+(x^2*exp(1)-x^2+15*x-36)*exp(x)+2*x^3)/x^2,x, algorithm="maxima
")

[Out]

x^2 - (x^2 - x + 36)*e^x*log(x)/x + 14*e^x*log(x) + Ei(x) + e^(x + 1) - e^x - 36*gamma(-1, -x) + integrate((x^
2 - x + 36)*e^x/x^2, x)

________________________________________________________________________________________

mupad [B]  time = 2.20, size = 30, normalized size = 1.20 \begin {gather*} {\mathrm {e}}^{x+1}+15\,{\mathrm {e}}^x\,\ln \relax (x)+x^2-x\,{\mathrm {e}}^x\,\ln \relax (x)-\frac {36\,{\mathrm {e}}^x\,\ln \relax (x)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(15*x + x^2*exp(1) - x^2 - 36) + 2*x^3 - exp(x)*log(x)*(36*x - 14*x^2 + x^3 - 36))/x^2,x)

[Out]

exp(x + 1) + 15*exp(x)*log(x) + x^2 - x*exp(x)*log(x) - (36*exp(x)*log(x))/x

________________________________________________________________________________________

sympy [A]  time = 0.31, size = 31, normalized size = 1.24 \begin {gather*} x^{2} + \frac {\left (- x^{2} \log {\relax (x )} + 15 x \log {\relax (x )} + e x - 36 \log {\relax (x )}\right ) e^{x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3+14*x**2-36*x+36)*exp(x)*ln(x)+(x**2*exp(1)-x**2+15*x-36)*exp(x)+2*x**3)/x**2,x)

[Out]

x**2 + (-x**2*log(x) + 15*x*log(x) + E*x - 36*log(x))*exp(x)/x

________________________________________________________________________________________