3.35.27 \(\int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} (250 x+100 x^2+10 x^3)+e^{5/x} (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5)}{25 x^3+10 x^4+x^5} \, dx\)

Optimal. Leaf size=32 \[ \frac {20}{5+x}-\left (-1-e^{5/x}+x+\frac {3 (5+x)}{x}\right )^2 \]

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Rubi [B]  time = 0.96, antiderivative size = 66, normalized size of antiderivative = 2.06, number of steps used = 25, number of rules used = 9, integrand size = 102, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {1594, 27, 6742, 2209, 44, 43, 2206, 2210, 2212} \begin {gather*} -x^2-\frac {225}{x^2}+2 e^{5/x} x-4 x+4 e^{5/x}-e^{10/x}+\frac {20}{x+5}+\frac {30 e^{5/x}}{x}-\frac {60}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(11250 + 6000*x + 1050*x^2 - 60*x^3 - 90*x^4 - 24*x^5 - 2*x^6 + E^(10/x)*(250*x + 100*x^2 + 10*x^3) + E^(5
/x)*(-3750 - 2750*x - 900*x^2 - 100*x^3 + 10*x^4 + 2*x^5))/(25*x^3 + 10*x^4 + x^5),x]

[Out]

4*E^(5/x) - E^(10/x) - 225/x^2 - 60/x + (30*E^(5/x))/x - 4*x + 2*E^(5/x)*x - x^2 + 20/(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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 2206

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[((c + d*x)*F^(a + b*(c + d*x)^n))/d, x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2210

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Simp[(F^a*ExpIntegralEi[
b*(c + d*x)^n*Log[F]])/(f*n), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

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 {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{x^3 \left (25+10 x+x^2\right )} \, dx\\ &=\int \frac {11250+6000 x+1050 x^2-60 x^3-90 x^4-24 x^5-2 x^6+e^{10/x} \left (250 x+100 x^2+10 x^3\right )+e^{5/x} \left (-3750-2750 x-900 x^2-100 x^3+10 x^4+2 x^5\right )}{x^3 (5+x)^2} \, dx\\ &=\int \left (\frac {10 e^{10/x}}{x^2}-\frac {60}{(5+x)^2}+\frac {11250}{x^3 (5+x)^2}+\frac {6000}{x^2 (5+x)^2}+\frac {1050}{x (5+x)^2}-\frac {90 x}{(5+x)^2}-\frac {24 x^2}{(5+x)^2}-\frac {2 x^3}{(5+x)^2}+\frac {2 e^{5/x} \left (-75-25 x-5 x^2+x^3\right )}{x^3}\right ) \, dx\\ &=\frac {60}{5+x}-2 \int \frac {x^3}{(5+x)^2} \, dx+2 \int \frac {e^{5/x} \left (-75-25 x-5 x^2+x^3\right )}{x^3} \, dx+10 \int \frac {e^{10/x}}{x^2} \, dx-24 \int \frac {x^2}{(5+x)^2} \, dx-90 \int \frac {x}{(5+x)^2} \, dx+1050 \int \frac {1}{x (5+x)^2} \, dx+6000 \int \frac {1}{x^2 (5+x)^2} \, dx+11250 \int \frac {1}{x^3 (5+x)^2} \, dx\\ &=-e^{10/x}+\frac {60}{5+x}+2 \int \left (e^{5/x}-\frac {75 e^{5/x}}{x^3}-\frac {25 e^{5/x}}{x^2}-\frac {5 e^{5/x}}{x}\right ) \, dx-2 \int \left (-10+x-\frac {125}{(5+x)^2}+\frac {75}{5+x}\right ) \, dx-24 \int \left (1+\frac {25}{(5+x)^2}-\frac {10}{5+x}\right ) \, dx-90 \int \left (-\frac {5}{(5+x)^2}+\frac {1}{5+x}\right ) \, dx+1050 \int \left (\frac {1}{25 x}-\frac {1}{5 (5+x)^2}-\frac {1}{25 (5+x)}\right ) \, dx+6000 \int \left (\frac {1}{25 x^2}-\frac {2}{125 x}+\frac {1}{25 (5+x)^2}+\frac {2}{125 (5+x)}\right ) \, dx+11250 \int \left (\frac {1}{25 x^3}-\frac {2}{125 x^2}+\frac {3}{625 x}-\frac {1}{125 (5+x)^2}-\frac {3}{625 (5+x)}\right ) \, dx\\ &=-e^{10/x}-\frac {225}{x^2}-\frac {60}{x}-4 x-x^2+\frac {20}{5+x}+2 \int e^{5/x} \, dx-10 \int \frac {e^{5/x}}{x} \, dx-50 \int \frac {e^{5/x}}{x^2} \, dx-150 \int \frac {e^{5/x}}{x^3} \, dx\\ &=10 e^{5/x}-e^{10/x}-\frac {225}{x^2}-\frac {60}{x}+\frac {30 e^{5/x}}{x}-4 x+2 e^{5/x} x-x^2+\frac {20}{5+x}+10 \text {Ei}\left (\frac {5}{x}\right )+10 \int \frac {e^{5/x}}{x} \, dx+30 \int \frac {e^{5/x}}{x^2} \, dx\\ &=4 e^{5/x}-e^{10/x}-\frac {225}{x^2}-\frac {60}{x}+\frac {30 e^{5/x}}{x}-4 x+2 e^{5/x} x-x^2+\frac {20}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 59, normalized size = 1.84 \begin {gather*} 2 \left (-\frac {e^{10/x}}{2}-\frac {225}{2 x^2}-\frac {30}{x}-2 x-\frac {x^2}{2}+\frac {10}{5+x}+e^{5/x} \left (2+\frac {15}{x}+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(11250 + 6000*x + 1050*x^2 - 60*x^3 - 90*x^4 - 24*x^5 - 2*x^6 + E^(10/x)*(250*x + 100*x^2 + 10*x^3)
+ E^(5/x)*(-3750 - 2750*x - 900*x^2 - 100*x^3 + 10*x^4 + 2*x^5))/(25*x^3 + 10*x^4 + x^5),x]

[Out]

2*(-1/2*E^(10/x) - 225/(2*x^2) - 30/x - 2*x - x^2/2 + 10/(5 + x) + E^(5/x)*(2 + 15/x + x))

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fricas [B]  time = 0.73, size = 77, normalized size = 2.41 \begin {gather*} -\frac {x^{5} + 9 \, x^{4} + 20 \, x^{3} + 40 \, x^{2} + {\left (x^{3} + 5 \, x^{2}\right )} e^{\frac {10}{x}} - 2 \, {\left (x^{4} + 7 \, x^{3} + 25 \, x^{2} + 75 \, x\right )} e^{\frac {5}{x}} + 525 \, x + 1125}{x^{3} + 5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2-2750*x-3750)*exp(5/x)-2*x^6-24*x^5-
90*x^4-60*x^3+1050*x^2+6000*x+11250)/(x^5+10*x^4+25*x^3),x, algorithm="fricas")

[Out]

-(x^5 + 9*x^4 + 20*x^3 + 40*x^2 + (x^3 + 5*x^2)*e^(10/x) - 2*(x^4 + 7*x^3 + 25*x^2 + 75*x)*e^(5/x) + 525*x + 1
125)/(x^3 + 5*x^2)

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giac [B]  time = 0.17, size = 105, normalized size = 3.28 \begin {gather*} \frac {\frac {2 \, e^{\frac {5}{x}}}{x} - \frac {9}{x} - \frac {e^{\frac {10}{x}}}{x^{2}} + \frac {14 \, e^{\frac {5}{x}}}{x^{2}} - \frac {24}{x^{2}} - \frac {5 \, e^{\frac {10}{x}}}{x^{3}} + \frac {50 \, e^{\frac {5}{x}}}{x^{3}} - \frac {60}{x^{3}} + \frac {150 \, e^{\frac {5}{x}}}{x^{4}} - \frac {525}{x^{4}} - \frac {1125}{x^{5}} - 1}{\frac {1}{x^{2}} + \frac {5}{x^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2-2750*x-3750)*exp(5/x)-2*x^6-24*x^5-
90*x^4-60*x^3+1050*x^2+6000*x+11250)/(x^5+10*x^4+25*x^3),x, algorithm="giac")

[Out]

(2*e^(5/x)/x - 9/x - e^(10/x)/x^2 + 14*e^(5/x)/x^2 - 24/x^2 - 5*e^(10/x)/x^3 + 50*e^(5/x)/x^3 - 60/x^3 + 150*e
^(5/x)/x^4 - 525/x^4 - 1125/x^5 - 1)/(1/x^2 + 5/x^3)

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maple [A]  time = 0.12, size = 56, normalized size = 1.75

method	result	size
risch	\(-x^{2}-4 x +\frac {-40 x^{2}-525 x -1125}{x^{2} \left (5+x \right )}-{\mathrm e}^{\frac {10}{x}}+\frac {2 \left (x^{2}+2 x +15\right ) {\mathrm e}^{\frac {5}{x}}}{x}\)	\(56\)
derivativedivides	\(-x^{2}-4 x -\frac {4}{1+\frac {5}{x}}-\frac {60}{x}-\frac {225}{x^{2}}+2 x \,{\mathrm e}^{\frac {5}{x}}+4 \,{\mathrm e}^{\frac {5}{x}}-{\mathrm e}^{\frac {10}{x}}+\frac {30 \,{\mathrm e}^{\frac {5}{x}}}{x}\)	\(69\)
default	\(-x^{2}-4 x -\frac {4}{1+\frac {5}{x}}-\frac {60}{x}-\frac {225}{x^{2}}+2 x \,{\mathrm e}^{\frac {5}{x}}+4 \,{\mathrm e}^{\frac {5}{x}}-{\mathrm e}^{\frac {10}{x}}+\frac {30 \,{\mathrm e}^{\frac {5}{x}}}{x}\)	\(69\)
norman	\(\frac {-1125+60 x^{2}-525 x -9 x^{4}-x^{5}+150 x \,{\mathrm e}^{\frac {5}{x}}+50 x^{2} {\mathrm e}^{\frac {5}{x}}-5 x^{2} {\mathrm e}^{\frac {10}{x}}+14 \,{\mathrm e}^{\frac {5}{x}} x^{3}+2 \,{\mathrm e}^{\frac {5}{x}} x^{4}-x^{3} {\mathrm e}^{\frac {10}{x}}}{x^{2} \left (5+x \right )}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2-2750*x-3750)*exp(5/x)-2*x^6-24*x^5-90*x^4
-60*x^3+1050*x^2+6000*x+11250)/(x^5+10*x^4+25*x^3),x,method=_RETURNVERBOSE)

[Out]

-x^2-4*x+(-40*x^2-525*x-1125)/x^2/(5+x)-exp(10/x)+2*(x^2+2*x+15)/x*exp(5/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -x^{2} + 2 \, x e^{\frac {5}{x}} - 4 \, x + \frac {45 \, {\left (6 \, x^{2} + 15 \, x - 25\right )}}{x^{3} + 5 \, x^{2}} - \frac {240 \, {\left (2 \, x + 5\right )}}{x^{2} + 5 \, x} + \frac {170}{x + 5} - e^{\frac {10}{x}} - 2 \, \int \frac {25 \, {\left (x + 3\right )} e^{\frac {5}{x}}}{x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^3+100*x^2+250*x)*exp(5/x)^2+(2*x^5+10*x^4-100*x^3-900*x^2-2750*x-3750)*exp(5/x)-2*x^6-24*x^5-
90*x^4-60*x^3+1050*x^2+6000*x+11250)/(x^5+10*x^4+25*x^3),x, algorithm="maxima")

[Out]

-x^2 + 2*x*e^(5/x) - 4*x + 45*(6*x^2 + 15*x - 25)/(x^3 + 5*x^2) - 240*(2*x + 5)/(x^2 + 5*x) + 170/(x + 5) - e^
(10/x) - 2*integrate(25*(x + 3)*e^(5/x)/x^3, x)

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mupad [B]  time = 2.20, size = 71, normalized size = 2.22 \begin {gather*} 4\,{\mathrm {e}}^{5/x}-{\mathrm {e}}^{10/x}+x\,\left (2\,{\mathrm {e}}^{5/x}-4\right )-x^2+\frac {x\,\left (150\,{\mathrm {e}}^{5/x}-525\right )+x^2\,\left (30\,{\mathrm {e}}^{5/x}-40\right )-1125}{x^2\,\left (x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5/x)*(2750*x + 900*x^2 + 100*x^3 - 10*x^4 - 2*x^5 + 3750) - 6000*x - exp(10/x)*(250*x + 100*x^2 + 10
*x^3) - 1050*x^2 + 60*x^3 + 90*x^4 + 24*x^5 + 2*x^6 - 11250)/(25*x^3 + 10*x^4 + x^5),x)

[Out]

4*exp(5/x) - exp(10/x) + x*(2*exp(5/x) - 4) - x^2 + (x*(150*exp(5/x) - 525) + x^2*(30*exp(5/x) - 40) - 1125)/(
x^2*(x + 5))

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sympy [B]  time = 0.21, size = 48, normalized size = 1.50 \begin {gather*} - x^{2} - 4 x - \frac {40 x^{2} + 525 x + 1125}{x^{3} + 5 x^{2}} + \frac {- x e^{\frac {10}{x}} + \left (2 x^{2} + 4 x + 30\right ) e^{\frac {5}{x}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**3+100*x**2+250*x)*exp(5/x)**2+(2*x**5+10*x**4-100*x**3-900*x**2-2750*x-3750)*exp(5/x)-2*x**6
-24*x**5-90*x**4-60*x**3+1050*x**2+6000*x+11250)/(x**5+10*x**4+25*x**3),x)

[Out]

-x**2 - 4*x - (40*x**2 + 525*x + 1125)/(x**3 + 5*x**2) + (-x*exp(10/x) + (2*x**2 + 4*x + 30)*exp(5/x))/x

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