3.69.44 $\int \frac{128-480x+600x^2-154x^3-240x^4+150x^5+24x^6-30x^7+2x^9+e^{10}(32x-16x^4)}{64-240x+300x^2-77x^3-120x^4+75x^5+12x^6-15x^7+x^9}dx$

Optimal. Leaf size=26 $$2x+\frac{4e^{10}{x}^2}{{(4-x(5-x^2))}^2}$$

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Rubi [B]  time = 0.31, antiderivative size = 96, normalized size of antiderivative = 3.69, number of steps used = 11, number of rules used = 5, integrand size = 94, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.053, Rules used = {2074, 638, 614, 618, 206} $$\begin{array}{c}\frac{21e^{10}(2x+1)}{17(-x^2-x+4)}+\frac{e^{10}(43x+166)}{17(-x^2-x+4)}+\frac{e^{10}(7x+20)}{{(-x^2-x+4)}^2}+2x-\frac{5e^{10}}{1-x}+\frac{e^{10}}{(1-x{)}^2}\end{array}$$

Antiderivative was successfully verified.

[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 206

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

Rule 614

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 618

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 638

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)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), 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 2074

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

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

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Mathematica [A]  time = 0.02, size = 22, normalized size = 0.85 $$\begin{array}{c}2(x+\frac{2e^{10}{x}^2}{{(4-5x+x^3)}^2})\end{array}$$

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 0.61, size = 61, normalized size = 2.35 $$\begin{array}{c}\frac{2(x^7-10x^5+8x^4+25x^3+2x^2{e}^{10}-40x^2+16x)}{x^6-10x^4+8x^3+25x^2-40x+16}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.15, size = 21, normalized size = 0.81 $$\begin{array}{c}2x+\frac{4x^2{e}^{10}}{{(x^3-5x+4)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.08, size = 37, normalized size = 1.42

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.35, size = 36, normalized size = 1.38 $$\begin{array}{c}\frac{4x^2{e}^{10}}{x^6-10x^4+8x^3+25x^2-40x+16}+2x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.09, size = 21, normalized size = 0.81 $$\begin{array}{c}2x+\frac{4x^2{\mathrm{e}}^{10}}{{(x^3-5x+4)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.46, size = 34, normalized size = 1.31 $$\begin{array}{c}\frac{4x^2{e}^{10}}{x^6-10x^4+8x^3+25x^2-40x+16}+2x\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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