3.291 \(\int \frac {\sqrt {2+3 x^2+x^4}}{(7+5 x^2)^2} \, dx\)

Optimal. Leaf size=209 \[ \frac {\sqrt {x^4+3 x^2+2} x}{14 \left (5 x^2+7\right )}-\frac {\left (x^2+2\right ) x}{70 \sqrt {x^4+3 x^2+2}}+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{140 \sqrt {2} \sqrt {x^4+3 x^2+2}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {\left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{980 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

[Out]

-1/70*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-1/1960*(x^2+2)*(1/(x^2+1))^(1/2)*(x^2+1)^(1/2)*EllipticPi(x/(x^2+1)^(1/2),
2/7,1/2*2^(1/2))*2^(1/2)/((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+1/70*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*Elli
pticE(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+3/280*(x^2+1)^(3/2)*(1/
(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+1/14
*x*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)

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Rubi [A]  time = 0.12, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1226, 1189, 1099, 1135, 1214, 1456, 539} \[ \frac {\sqrt {x^4+3 x^2+2} x}{14 \left (5 x^2+7\right )}-\frac {\left (x^2+2\right ) x}{70 \sqrt {x^4+3 x^2+2}}+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{140 \sqrt {2} \sqrt {x^4+3 x^2+2}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {\left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{980 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + 3*x^2 + x^4]/(7 + 5*x^2)^2,x]

[Out]

-(x*(2 + x^2))/(70*Sqrt[2 + 3*x^2 + x^4]) + (x*Sqrt[2 + 3*x^2 + x^4])/(14*(7 + 5*x^2)) + ((1 + x^2)*Sqrt[(2 +
x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(35*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (3*(1 + x^2)*Sqrt[(2 + x^2)/(1
+ x^2)]*EllipticF[ArcTan[x], 1/2])/(140*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) - ((2 + x^2)*EllipticPi[2/7, ArcTan[x]
, 1/2])/(980*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 1099

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b +
q)*x^2)*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)]
)/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1189

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1214

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(2*c)/(2*c*d - e*(b - q)), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/(2*c*d - e*(b - q)), Int[
(b - q + 2*c*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a
*c, 0] &&  !LtQ[c, 0]

Rule 1226

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

Rule 1456

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPar
t[p]), Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c*x^n)/e)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p,
q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sqrt {2+3 x^2+x^4}}{\left (7+5 x^2\right )^2} \, dx &=\frac {x \sqrt {2+3 x^2+x^4}}{14 \left (7+5 x^2\right )}+\frac {1}{350} \int \frac {7-5 x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {1}{350} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {x \sqrt {2+3 x^2+x^4}}{14 \left (7+5 x^2\right )}+\frac {1}{700} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {1}{280} \int \frac {2+2 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx-\frac {1}{70} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {1}{50} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=-\frac {x \left (2+x^2\right )}{70 \sqrt {2+3 x^2+x^4}}+\frac {x \sqrt {2+3 x^2+x^4}}{14 \left (7+5 x^2\right )}+\frac {\left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{140 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {\left (\sqrt {1+\frac {x^2}{2}} \sqrt {2+2 x^2}\right ) \int \frac {\sqrt {2+2 x^2}}{\sqrt {1+\frac {x^2}{2}} \left (7+5 x^2\right )} \, dx}{280 \sqrt {2+3 x^2+x^4}}\\ &=-\frac {x \left (2+x^2\right )}{70 \sqrt {2+3 x^2+x^4}}+\frac {x \sqrt {2+3 x^2+x^4}}{14 \left (7+5 x^2\right )}+\frac {\left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{140 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {\left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{980 \sqrt {2} \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.28, size = 208, normalized size = 1.00 \[ \frac {175 x^5+525 x^3-84 i \sqrt {x^2+1} \sqrt {x^2+2} \left (5 x^2+7\right ) F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+35 i \sqrt {x^2+1} \sqrt {x^2+2} \left (5 x^2+7\right ) E\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-5 i \sqrt {x^2+1} \sqrt {x^2+2} x^2 \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-7 i \sqrt {x^2+1} \sqrt {x^2+2} \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+350 x}{2450 \left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + 3*x^2 + x^4]/(7 + 5*x^2)^2,x]

[Out]

(350*x + 525*x^3 + 175*x^5 + (35*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(7 + 5*x^2)*EllipticE[I*ArcSinh[x/Sqrt[2]], 2]
- (84*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(7 + 5*x^2)*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] - (7*I)*Sqrt[1 + x^2]*Sqr
t[2 + x^2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2] - (5*I)*x^2*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticPi[10/7,
I*ArcSinh[x/Sqrt[2]], 2])/(2450*(7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4])

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 3 \, x^{2} + 2}}{25 \, x^{4} + 70 \, x^{2} + 49}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+3*x^2+2)^(1/2)/(5*x^2+7)^2,x, algorithm="fricas")

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)/(25*x^4 + 70*x^2 + 49), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{4} + 3 \, x^{2} + 2}}{{\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+3*x^2+2)^(1/2)/(5*x^2+7)^2,x, algorithm="giac")

[Out]

integrate(sqrt(x^4 + 3*x^2 + 2)/(5*x^2 + 7)^2, x)

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maple [C]  time = 0.02, size = 162, normalized size = 0.78 \[ \frac {\sqrt {x^{4}+3 x^{2}+2}\, x}{70 x^{2}+98}+\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{140 \sqrt {x^{4}+3 x^{2}+2}}-\frac {3 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{175 \sqrt {x^{4}+3 x^{2}+2}}-\frac {i \sqrt {2}\, \sqrt {\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{2450 \sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+3*x^2+2)^(1/2)/(5*x^2+7)^2,x)

[Out]

1/14*x*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)-3/175*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*Ellipti
cF(1/2*I*2^(1/2)*x,2^(1/2))+1/140*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticE(1/2*I*
2^(1/2)*x,2^(1/2))-1/2450*I*2^(1/2)*(1/2*x^2+1)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticPi(1/2*I*2^(1/
2)*x,10/7,2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{4} + 3 \, x^{2} + 2}}{{\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+3*x^2+2)^(1/2)/(5*x^2+7)^2,x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 + 3*x^2 + 2)/(5*x^2 + 7)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {x^4+3\,x^2+2}}{{\left (5\,x^2+7\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2 + x^4 + 2)^(1/2)/(5*x^2 + 7)^2,x)

[Out]

int((3*x^2 + x^4 + 2)^(1/2)/(5*x^2 + 7)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\left (x^{2} + 1\right ) \left (x^{2} + 2\right )}}{\left (5 x^{2} + 7\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+3*x**2+2)**(1/2)/(5*x**2+7)**2,x)

[Out]

Integral(sqrt((x**2 + 1)*(x**2 + 2))/(5*x**2 + 7)**2, x)

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