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

Optimal. Leaf size=17 \[ e^{10+x} \left (e^{8+\frac {1250}{x^2}}+x\right ) \]

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Rubi [A]  time = 0.36, antiderivative size = 18, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {6688, 6742, 2194, 2176, 6706} \begin {gather*} e^{\frac {1250}{x^2}+x+18}+e^{x+10} x \end {gather*}

Antiderivative was successfully verified.

[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 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 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 6688

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

Rule 6706

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 17, normalized size = 1.00 \begin {gather*} e^{10+x} \left (e^{8+\frac {1250}{x^2}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.57, size = 20, normalized size = 1.18 \begin {gather*} {\left (x + e^{\left (\frac {2 \, {\left (4 \, x^{2} + 625\right )}}{x^{2}}\right )}\right )} e^{\left (x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.15, size = 22, normalized size = 1.29 \begin {gather*} x e^{\left (x + 10\right )} + e^{\left (\frac {x^{3} + 18 \, x^{2} + 1250}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.07, size = 23, normalized size = 1.35

method	result	size
risch	\({\mathrm e}^{\frac {x^{3}+18 x^{2}+1250}{x^{2}}}+x \,{\mathrm e}^{x +10}\)	\(23\)
norman	\(\frac {x^{3} {\mathrm e}^{x +10}+x^{2} {\mathrm e}^{x +10} {\mathrm e}^{\frac {8 x^{2}+1250}{x^{2}}}}{x^{2}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.42, size = 26, normalized size = 1.53 \begin {gather*} {\left (x e^{10} - e^{10}\right )} e^{x} + e^{\left (x + \frac {1250}{x^{2}} + 18\right )} + e^{\left (x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.29, size = 16, normalized size = 0.94 \begin {gather*} {\mathrm {e}}^{10}\,{\mathrm {e}}^x\,\left (x+{\mathrm {e}}^8\,{\mathrm {e}}^{\frac {1250}{x^2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 4.50, size = 17, normalized size = 1.00 \begin {gather*} \left (x + e^{\frac {2 \left (4 x^{2} + 625\right )}{x^{2}}}\right ) e^{x + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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