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

Optimal. Leaf size=178 \[ \frac {x \left (x^2+2\right )}{5 \sqrt {x^4+3 x^2+2}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^2+2}{2 x^2+2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {x^4+3 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {x^4+3 x^2+2}}+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {2} \sqrt {x^4+3 x^2+2}} \]

[Out]

1/5*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)+3/70*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticPi(x/(x^2+1)^(1/2),2/7,1/2*2^(1
/2))*((x^2+2)/(x^2+1))^(1/2)*2^(1/2)/(x^4+3*x^2+2)^(1/2)-1/5*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticE(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)+1/5*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*
EllipticF(x/(x^2+1)^(1/2),1/2*2^(1/2))*((x^2+2)/(2*x^2+2))^(1/2)/(x^4+3*x^2+2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 232, normalized size of antiderivative = 1.30, 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 = {1208, 1189, 1099, 1135, 1214, 1456, 539} \[ \frac {x \left (x^2+2\right )}{5 \sqrt {x^4+3 x^2+2}}+\frac {4 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{25 \sqrt {x^4+3 x^2+2}}-\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{25 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{5 \sqrt {x^4+3 x^2+2}}+\frac {3 \left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2 + x^2))/(5*Sqrt[2 + 3*x^2 + x^4]) - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/
2])/(5*Sqrt[2 + 3*x^2 + x^4]) - (3*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(25*Sqrt[2]*
Sqrt[2 + 3*x^2 + x^4]) + (4*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(25*Sqrt[2
+ 3*x^2 + x^4]) + (3*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(35*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 1208

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d -
b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/e^2, Int[(a + b*x^2 + c*x^4
)^(p - 1)/(d + e*x^2), 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] && IGtQ[p + 1/2, 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 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}}{7+5 x^2} \, dx &=-\left (\frac {1}{25} \int \frac {-8-5 x^2}{\sqrt {2+3 x^2+x^4}} \, dx\right )-\frac {6}{25} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=-\left (\frac {3}{25} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx\right )+\frac {1}{5} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {3}{10} \int \frac {2+2 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx+\frac {8}{25} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {x \left (2+x^2\right )}{5 \sqrt {2+3 x^2+x^4}}-\frac {\sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{5 \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 )}{25 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {4 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{25 \sqrt {2+3 x^2+x^4}}+\frac {\left (3 \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}{10 \sqrt {2+3 x^2+x^4}}\\ &=\frac {x \left (2+x^2\right )}{5 \sqrt {2+3 x^2+x^4}}-\frac {\sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{5 \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 )}{25 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {4 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{25 \sqrt {2+3 x^2+x^4}}+\frac {3 \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{35 \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 = 0.15, size = 90, normalized size = 0.51 \[ -\frac {i \sqrt {x^2+1} \sqrt {x^2+2} \left (21 F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+35 E\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-6 \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )\right )}{175 \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/175*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(35*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] + 21*EllipticF[I*ArcSinh[x/Sqrt
[2]], 2] - 6*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2]))/Sqrt[2 + 3*x^2 + x^4]

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 3 \, x^{2} + 2}}{5 \, x^{2} + 7}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.04, size = 138, normalized size = 0.78 \[ -\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{10 \sqrt {x^{4}+3 x^{2}+2}}-\frac {3 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{50 \sqrt {x^{4}+3 x^{2}+2}}+\frac {6 i \sqrt {2}\, \sqrt {\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{175 \sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/50*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*2^(1/2)*x,2^(1/2))-1/10*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))+6/175*I*2^(1/2)*(1+
1/2*x^2)^(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 \[ \int \frac {\sqrt {x^{4} + 3 \, x^{2} + 2}}{5 \, x^{2} + 7}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {x^4+3\,x^2+2}}{5\,x^2+7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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