\(\int \frac {e^{10+x} (x^3+x^4+e^{\frac {2 (625+4 x^2)}{x^2}} (-2500+x^3))}{x^3} \, dx\) [4353]

Optimal result

Integrand size = 36, antiderivative size = 17 \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx=e^{10+x} \left (e^{8+\frac {1250}{x^2}}+x\right ) \]

[Out]

exp(x+10)*(x+exp(4+625/x^2)^2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {6820, 6874, 2225, 2207, 6838} \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx=e^{\frac {1250}{x^2}+x+18}+e^{x+10} x \]

[In]

Int[(E^(10 + x)*(x^3 + x^4 + E^((2*(625 + 4*x^2))/x^2)*(-2500 + x^3)))/x^3,x]

[Out]

E^(18 + 1250/x^2 + x) + E^(10 + x)*x

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int e^{10+x} \left (1+e^{8+\frac {1250}{x^2}} \left (1-\frac {2500}{x^3}\right )+x\right ) \, dx \\ & = \int \left (e^{10+x}+e^{10+x} x+\frac {e^{18+\frac {1250}{x^2}+x} \left (-2500+x^3\right )}{x^3}\right ) \, dx \\ & = \int e^{10+x} \, dx+\int e^{10+x} x \, dx+\int \frac {e^{18+\frac {1250}{x^2}+x} \left (-2500+x^3\right )}{x^3} \, dx \\ & = e^{10+x}+e^{18+\frac {1250}{x^2}+x}+e^{10+x} x-\int e^{10+x} \, dx \\ & = e^{18+\frac {1250}{x^2}+x}+e^{10+x} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx=e^{10+x} \left (e^{8+\frac {1250}{x^2}}+x\right ) \]

[In]

Integrate[(E^(10 + x)*(x^3 + x^4 + E^((2*(625 + 4*x^2))/x^2)*(-2500 + x^3)))/x^3,x]

[Out]

E^(10 + x)*(E^(8 + 1250/x^2) + x)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35

method	result	size
risch	\({\mathrm e}^{\frac {x^{3}+18 x^{2}+1250}{x^{2}}}+x \,{\mathrm e}^{x +10}\)	\(23\)
parallelrisch	\({\mathrm e}^{x +10} {\mathrm e}^{\frac {8 x^{2}+1250}{x^{2}}}+x \,{\mathrm e}^{x +10}\)	\(27\)
parts	\({\mathrm e}^{x +10} \left (x +10\right )-10 \,{\mathrm e}^{x +10}+{\mathrm e}^{x +10+\frac {8 x^{2}+1250}{x^{2}}}\)	\(32\)
norman	\(\frac {x^{3} {\mathrm e}^{x +10}+x^{2} {\mathrm e}^{x +10} {\mathrm e}^{\frac {8 x^{2}+1250}{x^{2}}}}{x^{2}}\)	\(36\)

[In]

int(((x^3-2500)*exp((4*x^2+625)/x^2)^2+x^4+x^3)*exp(x+10)/x^3,x,method=_RETURNVERBOSE)

[Out]

exp((x^3+18*x^2+1250)/x^2)+x*exp(x+10)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx={\left (x + e^{\left (\frac {2 \, {\left (4 \, x^{2} + 625\right )}}{x^{2}}\right )}\right )} e^{\left (x + 10\right )} \]

[In]

integrate(((x^3-2500)*exp((4*x^2+625)/x^2)^2+x^4+x^3)*exp(x+10)/x^3,x, algorithm="fricas")

[Out]

(x + e^(2*(4*x^2 + 625)/x^2))*e^(x + 10)

Sympy [A] (verification not implemented)

Time = 2.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx=\left (x + e^{\frac {2 \cdot \left (4 x^{2} + 625\right )}{x^{2}}}\right ) e^{x + 10} \]

[In]

integrate(((x**3-2500)*exp((4*x**2+625)/x**2)**2+x**4+x**3)*exp(x+10)/x**3,x)

[Out]

(x + exp(2*(4*x**2 + 625)/x**2))*exp(x + 10)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx={\left (x e^{10} - e^{10}\right )} e^{x} + e^{\left (x + \frac {1250}{x^{2}} + 18\right )} + e^{\left (x + 10\right )} \]

[In]

integrate(((x^3-2500)*exp((4*x^2+625)/x^2)^2+x^4+x^3)*exp(x+10)/x^3,x, algorithm="maxima")

[Out]

(x*e^10 - e^10)*e^x + e^(x + 1250/x^2 + 18) + e^(x + 10)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx=x e^{\left (x + 10\right )} + e^{\left (\frac {x^{3} + 18 \, x^{2} + 1250}{x^{2}}\right )} \]

[In]

integrate(((x^3-2500)*exp((4*x^2+625)/x^2)^2+x^4+x^3)*exp(x+10)/x^3,x, algorithm="giac")

[Out]

x*e^(x + 10) + e^((x^3 + 18*x^2 + 1250)/x^2)

Mupad [B] (verification not implemented)

Time = 10.86 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^{10+x} \left (x^3+x^4+e^{\frac {2 \left (625+4 x^2\right )}{x^2}} \left (-2500+x^3\right )\right )}{x^3} \, dx={\mathrm {e}}^{10}\,{\mathrm {e}}^x\,\left (x+{\mathrm {e}}^8\,{\mathrm {e}}^{\frac {1250}{x^2}}\right ) \]

[In]

int((exp(x + 10)*(exp((2*(4*x^2 + 625))/x^2)*(x^3 - 2500) + x^3 + x^4))/x^3,x)

[Out]

exp(10)*exp(x)*(x + exp(8)*exp(1250/x^2))