3.60.76 \(\int \frac {-2 x^4+e^{9+x} (-2+x+5 x^2-4 x^3-5 e x^3-4 x^4)+5 e^{9+x} x^3 \log (x)}{x^3} \, dx\)

Optimal. Leaf size=26 \[ -x^2+e^{9+x} \left (\frac {1}{x^2}+x-5 (e+x-\log (x))\right ) \]

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Rubi [A]  time = 0.51, antiderivative size = 51, normalized size of antiderivative = 1.96, number of steps used = 17, number of rules used = 8, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {14, 6742, 2199, 2194, 2177, 2178, 2176, 2554} \begin {gather*} -x^2+\frac {e^{x+9}}{x^2}-4 e^{x+9} x+4 e^{x+9}-(4+5 e) e^{x+9}+5 e^{x+9} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^4 + E^(9 + x)*(-2 + x + 5*x^2 - 4*x^3 - 5*E*x^3 - 4*x^4) + 5*E^(9 + x)*x^3*Log[x])/x^3,x]

[Out]

4*E^(9 + x) - E^(9 + x)*(4 + 5*E) + E^(9 + x)/x^2 - 4*E^(9 + x)*x - x^2 + 5*E^(9 + 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^{9+x} \left (-2+x+5 x^2-4 \left (1+\frac {5 e}{4}\right ) x^3-4 x^4+5 x^3 \log (x)\right )}{x^3}\right ) \, dx\\ &=-x^2+\int \frac {e^{9+x} \left (-2+x+5 x^2-4 \left (1+\frac {5 e}{4}\right ) x^3-4 x^4+5 x^3 \log (x)\right )}{x^3} \, dx\\ &=-x^2+\int \left (\frac {e^{9+x} \left (-2+x+5 x^2-(4+5 e) x^3-4 x^4\right )}{x^3}+5 e^{9+x} \log (x)\right ) \, dx\\ &=-x^2+5 \int e^{9+x} \log (x) \, dx+\int \frac {e^{9+x} \left (-2+x+5 x^2-(4+5 e) x^3-4 x^4\right )}{x^3} \, dx\\ &=-x^2+5 e^{9+x} \log (x)-5 \int \frac {e^{9+x}}{x} \, dx+\int \left (-4 e^{9+x} \left (1+\frac {5 e}{4}\right )-\frac {2 e^{9+x}}{x^3}+\frac {e^{9+x}}{x^2}+\frac {5 e^{9+x}}{x}-4 e^{9+x} x\right ) \, dx\\ &=-x^2-5 e^9 \text {Ei}(x)+5 e^{9+x} \log (x)-2 \int \frac {e^{9+x}}{x^3} \, dx-4 \int e^{9+x} x \, dx+5 \int \frac {e^{9+x}}{x} \, dx-(4+5 e) \int e^{9+x} \, dx+\int \frac {e^{9+x}}{x^2} \, dx\\ &=-e^{9+x} (4+5 e)+\frac {e^{9+x}}{x^2}-\frac {e^{9+x}}{x}-4 e^{9+x} x-x^2+5 e^{9+x} \log (x)+4 \int e^{9+x} \, dx-\int \frac {e^{9+x}}{x^2} \, dx+\int \frac {e^{9+x}}{x} \, dx\\ &=4 e^{9+x}-e^{9+x} (4+5 e)+\frac {e^{9+x}}{x^2}-4 e^{9+x} x-x^2+e^9 \text {Ei}(x)+5 e^{9+x} \log (x)-\int \frac {e^{9+x}}{x} \, dx\\ &=4 e^{9+x}-e^{9+x} (4+5 e)+\frac {e^{9+x}}{x^2}-4 e^{9+x} x-x^2+5 e^{9+x} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 38, normalized size = 1.46 \begin {gather*} -x^2+e^x \left (-5 e^{10}+\frac {e^9}{x^2}-4 e^9 x\right )+5 e^{9+x} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^4 + E^(9 + x)*(-2 + x + 5*x^2 - 4*x^3 - 5*E*x^3 - 4*x^4) + 5*E^(9 + x)*x^3*Log[x])/x^3,x]

[Out]

-x^2 + E^x*(-5*E^10 + E^9/x^2 - 4*E^9*x) + 5*E^(9 + x)*Log[x]

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fricas [A]  time = 0.49, size = 39, normalized size = 1.50 \begin {gather*} -\frac {x^{4} - 5 \, x^{2} e^{\left (x + 9\right )} \log \relax (x) + {\left (4 \, x^{3} + 5 \, x^{2} e - 1\right )} e^{\left (x + 9\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^3*exp(x+9)*log(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp(x+9)-2*x^4)/x^3,x, algorithm="frica
s")

[Out]

-(x^4 - 5*x^2*e^(x + 9)*log(x) + (4*x^3 + 5*x^2*e - 1)*e^(x + 9))/x^2

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giac [A]  time = 0.14, size = 44, normalized size = 1.69 \begin {gather*} -\frac {x^{4} + 4 \, x^{3} e^{\left (x + 9\right )} - 5 \, x^{2} e^{\left (x + 9\right )} \log \relax (x) + 5 \, x^{2} e^{\left (x + 10\right )} - e^{\left (x + 9\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^3*exp(x+9)*log(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp(x+9)-2*x^4)/x^3,x, algorithm="giac"
)

[Out]

-(x^4 + 4*x^3*e^(x + 9) - 5*x^2*e^(x + 9)*log(x) + 5*x^2*e^(x + 10) - e^(x + 9))/x^2

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maple [A]  time = 0.14, size = 43, normalized size = 1.65

method	result	size
risch	\(5 \,{\mathrm e}^{x +9} \ln \relax (x )-\frac {5 x^{2} {\mathrm e}^{x +10}+x^{4}+4 \,{\mathrm e}^{x +9} x^{3}-{\mathrm e}^{x +9}}{x^{2}}\)	\(43\)
norman	\(\frac {-x^{4}-4 \,{\mathrm e}^{x +9} x^{3}-5 x^{2} {\mathrm e} \,{\mathrm e}^{x +9}+5 x^{2} {\mathrm e}^{x +9} \ln \relax (x )+{\mathrm e}^{x +9}}{x^{2}}\)	\(46\)
default	\(\frac {-4 \,{\mathrm e}^{x +9} x^{3}-5 x^{2} {\mathrm e} \,{\mathrm e}^{x +9}+5 x^{2} {\mathrm e}^{x +9} \ln \relax (x )+{\mathrm e}^{x +9}}{x^{2}}-x^{2}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^3*exp(x+9)*ln(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp(x+9)-2*x^4)/x^3,x,method=_RETURNVERBOSE)

[Out]

5*exp(x+9)*ln(x)-(5*x^2*exp(x+10)+x^4+4*exp(x+9)*x^3-exp(x+9))/x^2

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maxima [C]  time = 0.39, size = 56, normalized size = 2.15 \begin {gather*} -x^{2} - 4 \, {\left (x e^{9} - e^{9}\right )} e^{x} + e^{9} \Gamma \left (-1, -x\right ) + 2 \, e^{9} \Gamma \left (-2, -x\right ) + 5 \, e^{\left (x + 9\right )} \log \relax (x) - 5 \, e^{\left (x + 10\right )} - 4 \, e^{\left (x + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^3*exp(x+9)*log(x)+(-5*x^3*exp(1)-4*x^4-4*x^3+5*x^2+x-2)*exp(x+9)-2*x^4)/x^3,x, algorithm="maxim
a")

[Out]

-x^2 - 4*(x*e^9 - e^9)*e^x + e^9*gamma(-1, -x) + 2*e^9*gamma(-2, -x) + 5*e^(x + 9)*log(x) - 5*e^(x + 10) - 4*e
^(x + 9)

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mupad [B]  time = 4.38, size = 35, normalized size = 1.35 \begin {gather*} \frac {{\mathrm {e}}^{x+9}}{x^2}-4\,x\,{\mathrm {e}}^{x+9}-5\,{\mathrm {e}}^{x+10}+5\,{\mathrm {e}}^{x+9}\,\ln \relax (x)-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + 9)*(5*x^3*exp(1) - x - 5*x^2 + 4*x^3 + 4*x^4 + 2) + 2*x^4 - 5*x^3*exp(x + 9)*log(x))/x^3,x)

[Out]

exp(x + 9)/x^2 - 4*x*exp(x + 9) - 5*exp(x + 10) + 5*exp(x + 9)*log(x) - x^2

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sympy [A]  time = 0.29, size = 34, normalized size = 1.31 \begin {gather*} - x^{2} + \frac {\left (- 4 x^{3} + 5 x^{2} \log {\relax (x )} - 5 e x^{2} + 1\right ) e^{x + 9}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**3*exp(x+9)*ln(x)+(-5*x**3*exp(1)-4*x**4-4*x**3+5*x**2+x-2)*exp(x+9)-2*x**4)/x**3,x)

[Out]

-x**2 + (-4*x**3 + 5*x**2*log(x) - 5*E*x**2 + 1)*exp(x + 9)/x**2

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