3.30.34 \(\int \frac {-5 x^3+e^{4/x} (100+65 x+14 x^2+x^3)}{25 x^3+10 x^4+x^5} \, dx\)

Optimal. Leaf size=21 \[ -\frac {e^{4/x}}{x}-\frac {x}{5+x} \]

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Rubi [A]  time = 0.23, antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1594, 27, 6688, 2288} \begin {gather*} \frac {5}{x+5}-\frac {e^{4/x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*x^3 + E^(4/x)*(100 + 65*x + 14*x^2 + x^3))/(25*x^3 + 10*x^4 + x^5),x]

[Out]

-(E^(4/x)/x) + 5/(5 + x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5 x^3+e^{4/x} \left (100+65 x+14 x^2+x^3\right )}{x^3 \left (25+10 x+x^2\right )} \, dx\\ &=\int \frac {-5 x^3+e^{4/x} \left (100+65 x+14 x^2+x^3\right )}{x^3 (5+x)^2} \, dx\\ &=\int \left (\frac {e^{4/x} (4+x)}{x^3}-\frac {5}{(5+x)^2}\right ) \, dx\\ &=\frac {5}{5+x}+\int \frac {e^{4/x} (4+x)}{x^3} \, dx\\ &=-\frac {e^{4/x}}{x}+\frac {5}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.95 \begin {gather*} -\frac {e^{4/x}}{x}+\frac {5}{5+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x^3 + E^(4/x)*(100 + 65*x + 14*x^2 + x^3))/(25*x^3 + 10*x^4 + x^5),x]

[Out]

-(E^(4/x)/x) + 5/(5 + x)

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fricas [A]  time = 0.54, size = 25, normalized size = 1.19 \begin {gather*} -\frac {{\left (x + 5\right )} e^{\frac {4}{x}} - 5 \, x}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+14*x^2+65*x+100)*exp(2/x)^2-5*x^3)/(x^5+10*x^4+25*x^3),x, algorithm="fricas")

[Out]

-((x + 5)*e^(4/x) - 5*x)/(x^2 + 5*x)

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giac [A]  time = 0.22, size = 34, normalized size = 1.62 \begin {gather*} -\frac {\frac {e^{\frac {4}{x}}}{x} + \frac {5 \, e^{\frac {4}{x}}}{x^{2}} + 1}{\frac {5}{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+14*x^2+65*x+100)*exp(2/x)^2-5*x^3)/(x^5+10*x^4+25*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 20, normalized size = 0.95

method	result	size
risch	\(\frac {5}{5+x}-\frac {{\mathrm e}^{\frac {4}{x}}}{x}\)	\(20\)
derivativedivides	\(-\frac {2}{\frac {10}{x}+2}-\frac {{\mathrm e}^{\frac {4}{x}}}{x}\)	\(26\)
default	\(-\frac {2}{\frac {10}{x}+2}-\frac {{\mathrm e}^{\frac {4}{x}}}{x}\)	\(26\)
norman	\(\frac {5 x^{2}-x^{2} {\mathrm e}^{\frac {4}{x}}-5 x \,{\mathrm e}^{\frac {4}{x}}}{x^{2} \left (5+x \right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+14*x^2+65*x+100)*exp(2/x)^2-5*x^3)/(x^5+10*x^4+25*x^3),x,method=_RETURNVERBOSE)

[Out]

5/(5+x)-exp(4/x)/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {5}{x + 5} + \int \frac {{\left (x + 4\right )} e^{\frac {4}{x}}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+14*x^2+65*x+100)*exp(2/x)^2-5*x^3)/(x^5+10*x^4+25*x^3),x, algorithm="maxima")

[Out]

5/(x + 5) + integrate((x + 4)*e^(4/x)/x^3, x)

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mupad [B]  time = 1.79, size = 29, normalized size = 1.38 \begin {gather*} -\frac {5\,{\mathrm {e}}^{4/x}+x\,\left ({\mathrm {e}}^{4/x}-5\right )}{x\,\left (x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4/x)*(65*x + 14*x^2 + x^3 + 100) - 5*x^3)/(25*x^3 + 10*x^4 + x^5),x)

[Out]

-(5*exp(4/x) + x*(exp(4/x) - 5))/(x*(x + 5))

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sympy [A]  time = 0.13, size = 10, normalized size = 0.48 \begin {gather*} \frac {5}{x + 5} - \frac {e^{\frac {4}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+14*x**2+65*x+100)*exp(2/x)**2-5*x**3)/(x**5+10*x**4+25*x**3),x)

[Out]

5/(x + 5) - exp(4/x)/x

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