3.31.39 \(\int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 (6 x^2+32 x^4+32 x^6)}{1+12 x^2+48 x^4+64 x^6} \, dx\)

Optimal. Leaf size=28 \[ -4 x+\left (x+\frac {e^5 x^2}{1+4 x^2}\right )^2-\log (3) \]

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Rubi [A]  time = 0.13, antiderivative size = 55, normalized size of antiderivative = 1.96, number of steps used = 7, number of rules used = 5, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 2073, 261, 639, 203} \begin {gather*} x^2-\frac {e^5 \left (4 x+e^5\right )}{8 \left (4 x^2+1\right )}+\frac {e^{10}}{16 \left (4 x^2+1\right )^2}-\frac {1}{2} \left (8-e^5\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 2*x - 48*x^2 + 24*x^3 + 4*E^10*x^3 - 192*x^4 + 96*x^5 - 256*x^6 + 128*x^7 + E^5*(6*x^2 + 32*x^4 + 32
*x^6))/(1 + 12*x^2 + 48*x^4 + 64*x^6),x]

[Out]

-1/2*((8 - E^5)*x) + x^2 + E^10/(16*(1 + 4*x^2)^2) - (E^5*(E^5 + 4*x))/(8*(1 + 4*x^2))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 203

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 639

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

Rule 2073

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4+2 x-48 x^2+\left (24+4 e^{10}\right ) x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx\\ &=\int \left (\frac {1}{2} \left (-8+e^5\right )+2 x-\frac {e^{10} x}{\left (1+4 x^2\right )^3}+\frac {e^5 \left (-1+e^5 x\right )}{\left (1+4 x^2\right )^2}+\frac {e^5}{2 \left (1+4 x^2\right )}\right ) \, dx\\ &=-\frac {1}{2} \left (8-e^5\right ) x+x^2+\frac {1}{2} e^5 \int \frac {1}{1+4 x^2} \, dx+e^5 \int \frac {-1+e^5 x}{\left (1+4 x^2\right )^2} \, dx-e^{10} \int \frac {x}{\left (1+4 x^2\right )^3} \, dx\\ &=-\frac {1}{2} \left (8-e^5\right ) x+x^2+\frac {e^{10}}{16 \left (1+4 x^2\right )^2}-\frac {e^5 \left (e^5+4 x\right )}{8 \left (1+4 x^2\right )}+\frac {1}{4} e^5 \tan ^{-1}(2 x)-\frac {1}{2} e^5 \int \frac {1}{1+4 x^2} \, dx\\ &=-\frac {1}{2} \left (8-e^5\right ) x+x^2+\frac {e^{10}}{16 \left (1+4 x^2\right )^2}-\frac {e^5 \left (e^5+4 x\right )}{8 \left (1+4 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 50, normalized size = 1.79 \begin {gather*} 2 \left (\frac {1}{2} (-4+x) x+\frac {e^5 x^3}{1+4 x^2}-\frac {e^{10} \left (1+8 x^2\right )}{32 \left (1+4 x^2\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 2*x - 48*x^2 + 24*x^3 + 4*E^10*x^3 - 192*x^4 + 96*x^5 - 256*x^6 + 128*x^7 + E^5*(6*x^2 + 32*x^
4 + 32*x^6))/(1 + 12*x^2 + 48*x^4 + 64*x^6),x]

[Out]

2*(((-4 + x)*x)/2 + (E^5*x^3)/(1 + 4*x^2) - (E^10*(1 + 8*x^2))/(32*(1 + 4*x^2)^2))

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fricas [B]  time = 0.49, size = 69, normalized size = 2.46 \begin {gather*} \frac {256 \, x^{6} - 1024 \, x^{5} + 128 \, x^{4} - 512 \, x^{3} + 16 \, x^{2} - {\left (8 \, x^{2} + 1\right )} e^{10} + 32 \, {\left (4 \, x^{5} + x^{3}\right )} e^{5} - 64 \, x}{16 \, {\left (16 \, x^{4} + 8 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96*x^5-192*x^4+24*x^3-48*x^2+2*x-4)/(64
*x^6+48*x^4+12*x^2+1),x, algorithm="fricas")

[Out]

1/16*(256*x^6 - 1024*x^5 + 128*x^4 - 512*x^3 + 16*x^2 - (8*x^2 + 1)*e^10 + 32*(4*x^5 + x^3)*e^5 - 64*x)/(16*x^
4 + 8*x^2 + 1)

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giac [A]  time = 0.16, size = 45, normalized size = 1.61 \begin {gather*} x^{2} + \frac {1}{2} \, x e^{5} - 4 \, x - \frac {32 \, x^{3} e^{5} + 8 \, x^{2} e^{10} + 8 \, x e^{5} + e^{10}}{16 \, {\left (4 \, x^{2} + 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96*x^5-192*x^4+24*x^3-48*x^2+2*x-4)/(64
*x^6+48*x^4+12*x^2+1),x, algorithm="giac")

[Out]

x^2 + 1/2*x*e^5 - 4*x - 1/16*(32*x^3*e^5 + 8*x^2*e^10 + 8*x*e^5 + e^10)/(4*x^2 + 1)^2

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maple [A]  time = 0.07, size = 48, normalized size = 1.71

method	result	size
default	\(x^{2}+\frac {x \,{\mathrm e}^{5}}{2}-4 x +\frac {-2 x^{3} {\mathrm e}^{5}-\frac {x^{2} {\mathrm e}^{10}}{2}-\frac {x \,{\mathrm e}^{5}}{2}-\frac {{\mathrm e}^{10}}{16}}{\left (4 x^{2}+1\right )^{2}}\)	\(48\)
risch	\(\frac {x \,{\mathrm e}^{5}}{2}+x^{2}-4 x +\frac {-\frac {x^{2} {\mathrm e}^{10}}{32}-\frac {x^{3} {\mathrm e}^{5}}{8}-\frac {{\mathrm e}^{10}}{256}-\frac {x \,{\mathrm e}^{5}}{32}}{x^{4}+\frac {1}{2} x^{2}+\frac {1}{16}}\)	\(50\)
norman	\(\frac {\left (-64+8 \,{\mathrm e}^{5}\right ) x^{5}+\left (-32+2 \,{\mathrm e}^{5}\right ) x^{3}+x^{2}+\left ({\mathrm e}^{10}+8\right ) x^{4}-4 x +16 x^{6}}{\left (4 x^{2}+1\right )^{2}}\)	\(53\)
gosper	\(\frac {x \left (x^{3} {\mathrm e}^{10}+8 x^{4} {\mathrm e}^{5}+16 x^{5}-64 x^{4}+2 x^{2} {\mathrm e}^{5}+8 x^{3}-32 x^{2}+x -4\right )}{16 x^{4}+8 x^{2}+1}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96*x^5-192*x^4+24*x^3-48*x^2+2*x-4)/(64*x^6+4
8*x^4+12*x^2+1),x,method=_RETURNVERBOSE)

[Out]

x^2+1/2*x*exp(5)-4*x+8*(-1/4*x^3*exp(5)-1/16*x^2*exp(10)-1/16*x*exp(5)-1/128*exp(10))/(4*x^2+1)^2

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maxima [A]  time = 0.38, size = 49, normalized size = 1.75 \begin {gather*} x^{2} + \frac {1}{2} \, x {\left (e^{5} - 8\right )} - \frac {32 \, x^{3} e^{5} + 8 \, x^{2} e^{10} + 8 \, x e^{5} + e^{10}}{16 \, {\left (16 \, x^{4} + 8 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96*x^5-192*x^4+24*x^3-48*x^2+2*x-4)/(64
*x^6+48*x^4+12*x^2+1),x, algorithm="maxima")

[Out]

x^2 + 1/2*x*(e^5 - 8) - 1/16*(32*x^3*e^5 + 8*x^2*e^10 + 8*x*e^5 + e^10)/(16*x^4 + 8*x^2 + 1)

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mupad [B]  time = 0.10, size = 50, normalized size = 1.79 \begin {gather*} x^2-\frac {4\,{\mathrm {e}}^5\,x^3+{\mathrm {e}}^{10}\,x^2+{\mathrm {e}}^5\,x+\frac {{\mathrm {e}}^{10}}{8}}{32\,x^4+16\,x^2+2}+x\,\left (\frac {{\mathrm {e}}^5}{2}-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 4*x^3*exp(10) + exp(5)*(6*x^2 + 32*x^4 + 32*x^6) - 48*x^2 + 24*x^3 - 192*x^4 + 96*x^5 - 256*x^6 + 1
28*x^7 - 4)/(12*x^2 + 48*x^4 + 64*x^6 + 1),x)

[Out]

x^2 - (exp(10)/8 + x*exp(5) + 4*x^3*exp(5) + x^2*exp(10))/(16*x^2 + 32*x^4 + 2) + x*(exp(5)/2 - 4)

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sympy [B]  time = 0.50, size = 51, normalized size = 1.82 \begin {gather*} x^{2} + x \left (-4 + \frac {e^{5}}{2}\right ) + \frac {- 32 x^{3} e^{5} - 8 x^{2} e^{10} - 8 x e^{5} - e^{10}}{256 x^{4} + 128 x^{2} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**3*exp(5)**2+(32*x**6+32*x**4+6*x**2)*exp(5)+128*x**7-256*x**6+96*x**5-192*x**4+24*x**3-48*x**2
+2*x-4)/(64*x**6+48*x**4+12*x**2+1),x)

[Out]

x**2 + x*(-4 + exp(5)/2) + (-32*x**3*exp(5) - 8*x**2*exp(10) - 8*x*exp(5) - exp(10))/(256*x**4 + 128*x**2 + 16
)

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