3.99.73 \(\int \frac {e^3 (10+x^2)+e^{3+x} (-5 x^2-5 x^3)}{x^2} \, dx\)

Optimal. Leaf size=21 \[ 4+e^3 \left (x-5 \left (\frac {2}{x}+e^x x\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.48, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {14, 2176, 2194} \begin {gather*} e^3 x+5 e^{x+3}-5 e^{x+3} (x+1)-\frac {10 e^3}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(10 + x^2) + E^(3 + x)*(-5*x^2 - 5*x^3))/x^2,x]

[Out]

5*E^(3 + x) - (10*E^3)/x + E^3*x - 5*E^(3 + x)*(1 + 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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-5 e^{3+x} (1+x)+\frac {e^3 \left (10+x^2\right )}{x^2}\right ) \, dx\\ &=-\left (5 \int e^{3+x} (1+x) \, dx\right )+e^3 \int \frac {10+x^2}{x^2} \, dx\\ &=-5 e^{3+x} (1+x)+5 \int e^{3+x} \, dx+e^3 \int \left (1+\frac {10}{x^2}\right ) \, dx\\ &=5 e^{3+x}-\frac {10 e^3}{x}+e^3 x-5 e^{3+x} (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.81 \begin {gather*} e^3 \left (-\frac {10}{x}+x-5 e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(10 + x^2) + E^(3 + x)*(-5*x^2 - 5*x^3))/x^2,x]

[Out]

E^3*(-10/x + x - 5*E^x*x)

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fricas [A]  time = 0.63, size = 24, normalized size = 1.14 \begin {gather*} -\frac {5 \, x^{2} e^{\left (x + 3\right )} - {\left (x^{2} - 10\right )} e^{3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-5*x^2)*exp(3)*exp(x)+(x^2+10)*exp(3))/x^2,x, algorithm="fricas")

[Out]

-(5*x^2*e^(x + 3) - (x^2 - 10)*e^3)/x

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giac [A]  time = 0.13, size = 24, normalized size = 1.14 \begin {gather*} \frac {x^{2} e^{3} - 5 \, x^{2} e^{\left (x + 3\right )} - 10 \, e^{3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-5*x^2)*exp(3)*exp(x)+(x^2+10)*exp(3))/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 20, normalized size = 0.95

method	result	size
risch	\(x \,{\mathrm e}^{3}-\frac {10 \,{\mathrm e}^{3}}{x}-5 \,{\mathrm e}^{3+x} x\)	\(20\)
norman	\(\frac {x^{2} {\mathrm e}^{3}-5 x^{2} {\mathrm e}^{3} {\mathrm e}^{x}-10 \,{\mathrm e}^{3}}{x}\)	\(25\)
default	\(-\frac {10 \,{\mathrm e}^{3}}{x}-5 \,{\mathrm e}^{x} {\mathrm e}^{3}-5 \,{\mathrm e}^{3} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )+x \,{\mathrm e}^{3}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^3-5*x^2)*exp(3)*exp(x)+(x^2+10)*exp(3))/x^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(3)-10*exp(3)/x-5*exp(3+x)*x

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maxima [A]  time = 0.36, size = 31, normalized size = 1.48 \begin {gather*} x e^{3} - 5 \, {\left (x e^{3} - e^{3}\right )} e^{x} - \frac {10 \, e^{3}}{x} - 5 \, e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-5*x^2)*exp(3)*exp(x)+(x^2+10)*exp(3))/x^2,x, algorithm="maxima")

[Out]

x*e^3 - 5*(x*e^3 - e^3)*e^x - 10*e^3/x - 5*e^(x + 3)

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mupad [B]  time = 0.07, size = 19, normalized size = 0.90 \begin {gather*} -\frac {10\,{\mathrm {e}}^3}{x}-x\,{\mathrm {e}}^3\,\left (5\,{\mathrm {e}}^x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(x^2 + 10) - exp(3)*exp(x)*(5*x^2 + 5*x^3))/x^2,x)

[Out]

- (10*exp(3))/x - x*exp(3)*(5*exp(x) - 1)

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sympy [A]  time = 0.14, size = 20, normalized size = 0.95 \begin {gather*} - 5 x e^{3} e^{x} + x e^{3} - \frac {10 e^{3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**3-5*x**2)*exp(3)*exp(x)+(x**2+10)*exp(3))/x**2,x)

[Out]

-5*x*exp(3)*exp(x) + x*exp(3) - 10*exp(3)/x

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