\(\int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} (32 x-16 x^4)}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx\) [6844]

Optimal result

Integrand size = 94, antiderivative size = 26 \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=2 x+\frac {4 e^{10} x^2}{\left (4-x \left (5-x^2\right )\right )^2} \]

[Out]

4*x^2*exp(5)^2/(4-(-x^2+5)*x)^2+2*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(96\) vs. \(2(26)=52\).

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 3.69, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2099, 652, 628, 632, 212} \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=\frac {21 e^{10} (2 x+1)}{17 \left (-x^2-x+4\right )}+\frac {e^{10} (43 x+166)}{17 \left (-x^2-x+4\right )}+\frac {e^{10} (7 x+20)}{\left (-x^2-x+4\right )^2}+2 x-\frac {5 e^{10}}{1-x}+\frac {e^{10}}{(1-x)^2} \]

[In]

Int[(128 - 480*x + 600*x^2 - 154*x^3 - 240*x^4 + 150*x^5 + 24*x^6 - 30*x^7 + 2*x^9 + E^10*(32*x - 16*x^4))/(64
- 240*x + 300*x^2 - 77*x^3 - 120*x^4 + 75*x^5 + 12*x^6 - 15*x^7 + x^9),x]

[Out]

E^10/(1 - x)^2 - (5*E^10)/(1 - x) + 2*x + (E^10*(20 + 7*x))/(4 - x - x^2)^2 + (21*E^10*(1 + 2*x))/(17*(4 - x -
x^2)) + (E^10*(166 + 43*x))/(17*(4 - x - x^2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1
)*(b^2 - 4*a*c))), x] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b*x + c*x^2)^(p + 1), x] - Dist[(2*p + 3)*((2*c*d - b*e)/((p + 1)*(b^2 -
4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (2-\frac {2 e^{10}}{(-1+x)^3}-\frac {5 e^{10}}{(-1+x)^2}-\frac {2 e^{10} (76+33 x)}{\left (-4+x+x^2\right )^3}+\frac {e^{10} (30+17 x)}{\left (-4+x+x^2\right )^2}+\frac {5 e^{10}}{-4+x+x^2}\right ) \, dx \\ & = \frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+e^{10} \int \frac {30+17 x}{\left (-4+x+x^2\right )^2} \, dx-\left (2 e^{10}\right ) \int \frac {76+33 x}{\left (-4+x+x^2\right )^3} \, dx+\left (5 e^{10}\right ) \int \frac {1}{-4+x+x^2} \, dx \\ & = \frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )}-\frac {1}{17} \left (43 e^{10}\right ) \int \frac {1}{-4+x+x^2} \, dx-\left (10 e^{10}\right ) \text {Subst}\left (\int \frac {1}{17-x^2} \, dx,x,1+2 x\right )+\left (21 e^{10}\right ) \int \frac {1}{\left (-4+x+x^2\right )^2} \, dx \\ & = \frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {21 e^{10} (1+2 x)}{17 \left (4-x-x^2\right )}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )}-\frac {10 e^{10} \tanh ^{-1}\left (\frac {1+2 x}{\sqrt {17}}\right )}{\sqrt {17}}-\frac {1}{17} \left (42 e^{10}\right ) \int \frac {1}{-4+x+x^2} \, dx+\frac {1}{17} \left (86 e^{10}\right ) \text {Subst}\left (\int \frac {1}{17-x^2} \, dx,x,1+2 x\right ) \\ & = \frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {21 e^{10} (1+2 x)}{17 \left (4-x-x^2\right )}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )}-\frac {84 e^{10} \tanh ^{-1}\left (\frac {1+2 x}{\sqrt {17}}\right )}{17 \sqrt {17}}+\frac {1}{17} \left (84 e^{10}\right ) \text {Subst}\left (\int \frac {1}{17-x^2} \, dx,x,1+2 x\right ) \\ & = \frac {e^{10}}{(1-x)^2}-\frac {5 e^{10}}{1-x}+2 x+\frac {e^{10} (20+7 x)}{\left (4-x-x^2\right )^2}+\frac {21 e^{10} (1+2 x)}{17 \left (4-x-x^2\right )}+\frac {e^{10} (166+43 x)}{17 \left (4-x-x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=2 \left (x+\frac {2 e^{10} x^2}{\left (4-5 x+x^3\right )^2}\right ) \]

[In]

Integrate[(128 - 480*x + 600*x^2 - 154*x^3 - 240*x^4 + 150*x^5 + 24*x^6 - 30*x^7 + 2*x^9 + E^10*(32*x - 16*x^4
))/(64 - 240*x + 300*x^2 - 77*x^3 - 120*x^4 + 75*x^5 + 12*x^6 - 15*x^7 + x^9),x]

[Out]

2*(x + (2*E^10*x^2)/(4 - 5*x + x^3)^2)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42

method	result	size
risch	\(2 x +\frac {4 \,{\mathrm e}^{10} x^{2}}{x^{6}-10 x^{4}+8 x^{3}+25 x^{2}-40 x +16}\)	\(37\)
default	\(2 x +\frac {{\mathrm e}^{10} \left (-5 x^{3}-16 x^{2}+16 x +64\right )}{\left (x^{2}+x -4\right )^{2}}+\frac {{\mathrm e}^{10}}{\left (-1+x \right )^{2}}+\frac {5 \,{\mathrm e}^{10}}{-1+x}\)	\(48\)
norman	\(\frac {16 x^{4}+32 x +50 x^{3}+\left (4 \,{\mathrm e}^{10}-80\right ) x^{2}-20 x^{5}+2 x^{7}}{\left (x^{3}-5 x +4\right )^{2}}\)	\(48\)
gosper	\(\frac {2 x \left (x^{6}-10 x^{4}+8 x^{3}+2 x \,{\mathrm e}^{10}+25 x^{2}-40 x +16\right )}{x^{6}-10 x^{4}+8 x^{3}+25 x^{2}-40 x +16}\)	\(59\)
parallelrisch	\(\frac {2 x^{7}-20 x^{5}+4 x^{2} {\mathrm e}^{10}+16 x^{4}+50 x^{3}-80 x^{2}+32 x}{x^{6}-10 x^{4}+8 x^{3}+25 x^{2}-40 x +16}\)	\(65\)

[In]

int(((-16*x^4+32*x)*exp(5)^2+2*x^9-30*x^7+24*x^6+150*x^5-240*x^4-154*x^3+600*x^2-480*x+128)/(x^9-15*x^7+12*x^6
+75*x^5-120*x^4-77*x^3+300*x^2-240*x+64),x,method=_RETURNVERBOSE)

[Out]

2*x+4*exp(10)*x^2/(x^6-10*x^4+8*x^3+25*x^2-40*x+16)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (22) = 44\).

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=\frac {2 \, {\left (x^{7} - 10 \, x^{5} + 8 \, x^{4} + 25 \, x^{3} + 2 \, x^{2} e^{10} - 40 \, x^{2} + 16 \, x\right )}}{x^{6} - 10 \, x^{4} + 8 \, x^{3} + 25 \, x^{2} - 40 \, x + 16} \]

[In]

integrate(((-16*x^4+32*x)*exp(5)^2+2*x^9-30*x^7+24*x^6+150*x^5-240*x^4-154*x^3+600*x^2-480*x+128)/(x^9-15*x^7+
12*x^6+75*x^5-120*x^4-77*x^3+300*x^2-240*x+64),x, algorithm="fricas")

[Out]

2*(x^7 - 10*x^5 + 8*x^4 + 25*x^3 + 2*x^2*e^10 - 40*x^2 + 16*x)/(x^6 - 10*x^4 + 8*x^3 + 25*x^2 - 40*x + 16)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=\frac {4 x^{2} e^{10}}{x^{6} - 10 x^{4} + 8 x^{3} + 25 x^{2} - 40 x + 16} + 2 x \]

[In]

integrate(((-16*x**4+32*x)*exp(5)**2+2*x**9-30*x**7+24*x**6+150*x**5-240*x**4-154*x**3+600*x**2-480*x+128)/(x*
*9-15*x**7+12*x**6+75*x**5-120*x**4-77*x**3+300*x**2-240*x+64),x)

[Out]

4*x**2*exp(10)/(x**6 - 10*x**4 + 8*x**3 + 25*x**2 - 40*x + 16) + 2*x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=\frac {4 \, x^{2} e^{10}}{x^{6} - 10 \, x^{4} + 8 \, x^{3} + 25 \, x^{2} - 40 \, x + 16} + 2 \, x \]

[In]

integrate(((-16*x^4+32*x)*exp(5)^2+2*x^9-30*x^7+24*x^6+150*x^5-240*x^4-154*x^3+600*x^2-480*x+128)/(x^9-15*x^7+
12*x^6+75*x^5-120*x^4-77*x^3+300*x^2-240*x+64),x, algorithm="maxima")

[Out]

4*x^2*e^10/(x^6 - 10*x^4 + 8*x^3 + 25*x^2 - 40*x + 16) + 2*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=2 \, x + \frac {4 \, x^{2} e^{10}}{{\left (x^{3} - 5 \, x + 4\right )}^{2}} \]

[In]

integrate(((-16*x^4+32*x)*exp(5)^2+2*x^9-30*x^7+24*x^6+150*x^5-240*x^4-154*x^3+600*x^2-480*x+128)/(x^9-15*x^7+
12*x^6+75*x^5-120*x^4-77*x^3+300*x^2-240*x+64),x, algorithm="giac")

[Out]

2*x + 4*x^2*e^10/(x^3 - 5*x + 4)^2

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {128-480 x+600 x^2-154 x^3-240 x^4+150 x^5+24 x^6-30 x^7+2 x^9+e^{10} \left (32 x-16 x^4\right )}{64-240 x+300 x^2-77 x^3-120 x^4+75 x^5+12 x^6-15 x^7+x^9} \, dx=2\,x+\frac {4\,x^2\,{\mathrm {e}}^{10}}{{\left (x^3-5\,x+4\right )}^2} \]

[In]

int((exp(10)*(32*x - 16*x^4) - 480*x + 600*x^2 - 154*x^3 - 240*x^4 + 150*x^5 + 24*x^6 - 30*x^7 + 2*x^9 + 128)/
(300*x^2 - 240*x - 77*x^3 - 120*x^4 + 75*x^5 + 12*x^6 - 15*x^7 + x^9 + 64),x)

[Out]

2*x + (4*x^2*exp(10))/(x^3 - 5*x + 4)^2