3.84.90 \(\int \frac {20 x^3-40 x^4-19 x^5-2 x^6+e^{\frac {5+x}{x^2}} (250+125 x+20 x^2+x^3)}{25 x^3+10 x^4+x^5} \, dx\)

Optimal. Leaf size=34 \[ -1-e^{\frac {5+x}{x^2}}+x-x^2-\frac {x}{5+x}+\log \left (\frac {\log (4)}{3}\right ) \]

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Rubi [A]  time = 0.68, antiderivative size = 27, normalized size of antiderivative = 0.79, number of steps used = 11, number of rules used = 5, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {1594, 27, 6742, 43, 6706} \begin {gather*} -x^2-e^{\frac {5}{x^2}+\frac {1}{x}}+x+\frac {5}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20*x^3 - 40*x^4 - 19*x^5 - 2*x^6 + E^((5 + x)/x^2)*(250 + 125*x + 20*x^2 + x^3))/(25*x^3 + 10*x^4 + x^5),
x]

[Out]

-E^(5/x^2 + x^(-1)) + x - x^2 + 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 \frac {20 x^3-40 x^4-19 x^5-2 x^6+e^{\frac {5+x}{x^2}} \left (250+125 x+20 x^2+x^3\right )}{x^3 \left (25+10 x+x^2\right )} \, dx\\ &=\int \frac {20 x^3-40 x^4-19 x^5-2 x^6+e^{\frac {5+x}{x^2}} \left (250+125 x+20 x^2+x^3\right )}{x^3 (5+x)^2} \, dx\\ &=\int \left (\frac {20}{(5+x)^2}-\frac {40 x}{(5+x)^2}-\frac {19 x^2}{(5+x)^2}-\frac {2 x^3}{(5+x)^2}+\frac {e^{\frac {5}{x^2}+\frac {1}{x}} (10+x)}{x^3}\right ) \, dx\\ &=-\frac {20}{5+x}-2 \int \frac {x^3}{(5+x)^2} \, dx-19 \int \frac {x^2}{(5+x)^2} \, dx-40 \int \frac {x}{(5+x)^2} \, dx+\int \frac {e^{\frac {5}{x^2}+\frac {1}{x}} (10+x)}{x^3} \, dx\\ &=-e^{\frac {5}{x^2}+\frac {1}{x}}-\frac {20}{5+x}-2 \int \left (-10+x-\frac {125}{(5+x)^2}+\frac {75}{5+x}\right ) \, dx-19 \int \left (1+\frac {25}{(5+x)^2}-\frac {10}{5+x}\right ) \, dx-40 \int \left (-\frac {5}{(5+x)^2}+\frac {1}{5+x}\right ) \, dx\\ &=-e^{\frac {5}{x^2}+\frac {1}{x}}+x-x^2+\frac {5}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 33, normalized size = 0.97 \begin {gather*} -e^{\frac {5}{x^2}+\frac {1}{x}}+\frac {5}{5+x}+11 (5+x)-(5+x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20*x^3 - 40*x^4 - 19*x^5 - 2*x^6 + E^((5 + x)/x^2)*(250 + 125*x + 20*x^2 + x^3))/(25*x^3 + 10*x^4 +
x^5),x]

[Out]

-E^(5/x^2 + x^(-1)) + 5/(5 + x) + 11*(5 + x) - (5 + x)^2

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fricas [A]  time = 0.66, size = 32, normalized size = 0.94 \begin {gather*} -\frac {x^{3} + 4 \, x^{2} + {\left (x + 5\right )} e^{\left (\frac {x + 5}{x^{2}}\right )} - 5 \, x - 5}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+20*x^2+125*x+250)*exp((5+x)/x^2)-2*x^6-19*x^5-40*x^4+20*x^3)/(x^5+10*x^4+25*x^3),x, algorithm=
"fricas")

[Out]

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

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giac [A]  time = 0.16, size = 26, normalized size = 0.76 \begin {gather*} -x^{2} + x + \frac {5}{x + 5} - e^{\left (\frac {1}{x} + \frac {5}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+20*x^2+125*x+250)*exp((5+x)/x^2)-2*x^6-19*x^5-40*x^4+20*x^3)/(x^5+10*x^4+25*x^3),x, algorithm=
"giac")

[Out]

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

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maple [A]  time = 0.09, size = 25, normalized size = 0.74

method	result	size
risch	\(-x^{2}+x +\frac {5}{5+x}-{\mathrm e}^{\frac {5+x}{x^{2}}}\)	\(25\)
derivativedivides	\(-{\mathrm e}^{\frac {1}{x}+\frac {5}{x^{2}}}+x -x^{2}-\frac {1}{1+\frac {5}{x}}\)	\(31\)
default	\(-{\mathrm e}^{\frac {1}{x}+\frac {5}{x^{2}}}+x -x^{2}-\frac {1}{1+\frac {5}{x}}\)	\(31\)
norman	\(\frac {-20 x^{2}-4 x^{4}-x^{5}-5 \,{\mathrm e}^{\frac {5+x}{x^{2}}} x^{2}-{\mathrm e}^{\frac {5+x}{x^{2}}} x^{3}}{x^{2} \left (5+x \right )}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3+20*x^2+125*x+250)*exp((5+x)/x^2)-2*x^6-19*x^5-40*x^4+20*x^3)/(x^5+10*x^4+25*x^3),x,method=_RETURNVER
BOSE)

[Out]

-x^2+x+5/(5+x)-exp((5+x)/x^2)

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maxima [A]  time = 0.42, size = 26, normalized size = 0.76 \begin {gather*} -x^{2} + x + \frac {5}{x + 5} - e^{\left (\frac {1}{x} + \frac {5}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3+20*x^2+125*x+250)*exp((5+x)/x^2)-2*x^6-19*x^5-40*x^4+20*x^3)/(x^5+10*x^4+25*x^3),x, algorithm=
"maxima")

[Out]

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

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mupad [B]  time = 5.49, size = 26, normalized size = 0.76 \begin {gather*} x-{\mathrm {e}}^{\frac {1}{x}+\frac {5}{x^2}}+\frac {5}{x+5}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(40*x^4 - 20*x^3 - exp((x + 5)/x^2)*(125*x + 20*x^2 + x^3 + 250) + 19*x^5 + 2*x^6)/(25*x^3 + 10*x^4 + x^5
),x)

[Out]

x - exp(1/x + 5/x^2) + 5/(x + 5) - x^2

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sympy [A]  time = 0.19, size = 17, normalized size = 0.50 \begin {gather*} - x^{2} + x - e^{\frac {x + 5}{x^{2}}} + \frac {5}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3+20*x**2+125*x+250)*exp((5+x)/x**2)-2*x**6-19*x**5-40*x**4+20*x**3)/(x**5+10*x**4+25*x**3),x)

[Out]

-x**2 + x - exp((x + 5)/x**2) + 5/(x + 5)

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