3.305 \(\int \frac {1}{(7+5 x^2)^2 \sqrt {2+3 x^2+x^4}} \, dx\)

Optimal. Leaf size=209 \[ -\frac {25 \sqrt {x^4+3 x^2+2} x}{84 \left (5 x^2+7\right )}+\frac {5 \left (x^2+2\right ) x}{84 \sqrt {x^4+3 x^2+2}}+\frac {9 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{56 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{42 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {65 \left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{1176 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

[Out]

5/84*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-65/2352*(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)-5/84*(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)+9/112*(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)-25/8
4*x*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)

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Rubi [A]  time = 0.19, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1223, 1716, 1189, 1099, 1135, 1214, 1456, 539} \[ -\frac {25 \sqrt {x^4+3 x^2+2} x}{84 \left (5 x^2+7\right )}+\frac {5 \left (x^2+2\right ) x}{84 \sqrt {x^4+3 x^2+2}}+\frac {9 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{56 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {5 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{42 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {65 \left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{1176 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((7 + 5*x^2)^2*Sqrt[2 + 3*x^2 + x^4]),x]

[Out]

(5*x*(2 + x^2))/(84*Sqrt[2 + 3*x^2 + x^4]) - (25*x*Sqrt[2 + 3*x^2 + x^4])/(84*(7 + 5*x^2)) - (5*(1 + x^2)*Sqrt
[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(42*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) + (9*(1 + x^2)*Sqrt[(2 + x
^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(56*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4]) - (65*(2 + x^2)*EllipticPi[2/7, A
rcTan[x], 1/2])/(1176*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 1223

Int[((d_) + (e_.)*(x_)^2)^(q_)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Simp[(e^2*x*(d + e*x^2)
^(q + 1)*Sqrt[a + b*x^2 + c*x^4])/(2*d*(q + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(2*d*(q + 1)*(c*d^2 - b*d
*e + a*e^2)), Int[((d + e*x^2)^(q + 1)*Simp[a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1) - 2*e*(c*d*(q + 1) - b*e
*(q + 2))*x^2 + c*e^2*(2*q + 5)*x^4, x])/Sqrt[a + b*x^2 + c*x^4], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b
^2 - 4*a*c, 0] && ILtQ[q, -1]

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]

Rule 1716

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[P4x,
x, 0], B = Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Dist[(e^2)^(-1), Int[(C*d - B*e - C*e*x^2)/Sqrt[a + b*x^
2 + c*x^4], x], x] + Dist[(C*d^2 - B*d*e + A*e^2)/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] && PolyQ[P4x, x^2, 2] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && Ne
Q[c*d^2 - a*e^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (7+5 x^2\right )^2 \sqrt {2+3 x^2+x^4}} \, dx &=-\frac {25 x \sqrt {2+3 x^2+x^4}}{84 \left (7+5 x^2\right )}+\frac {1}{84} \int \frac {62+70 x^2+25 x^4}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=-\frac {25 x \sqrt {2+3 x^2+x^4}}{84 \left (7+5 x^2\right )}-\frac {\int \frac {-175-125 x^2}{\sqrt {2+3 x^2+x^4}} \, dx}{2100}+\frac {13}{84} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=-\frac {25 x \sqrt {2+3 x^2+x^4}}{84 \left (7+5 x^2\right )}+\frac {5}{84} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {13}{168} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {1}{12} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {65}{336} \int \frac {2+2 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {5 x \left (2+x^2\right )}{84 \sqrt {2+3 x^2+x^4}}-\frac {25 x \sqrt {2+3 x^2+x^4}}{84 \left (7+5 x^2\right )}-\frac {5 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{42 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {9 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{56 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {\left (65 \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}{336 \sqrt {2+3 x^2+x^4}}\\ &=\frac {5 x \left (2+x^2\right )}{84 \sqrt {2+3 x^2+x^4}}-\frac {25 x \sqrt {2+3 x^2+x^4}}{84 \left (7+5 x^2\right )}-\frac {5 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{42 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {9 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{56 \sqrt {2} \sqrt {2+3 x^2+x^4}}-\frac {65 \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{1176 \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 = 0.26, size = 208, normalized size = 1.00 \[ \frac {-175 x^5-525 x^3-14 i \sqrt {x^2+1} \sqrt {x^2+2} \left (5 x^2+7\right ) F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-35 i \sqrt {x^2+1} \sqrt {x^2+2} \left (5 x^2+7\right ) E\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-65 i \sqrt {x^2+1} \sqrt {x^2+2} x^2 \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-91 i \sqrt {x^2+1} \sqrt {x^2+2} \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-350 x}{588 \left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((7 + 5*x^2)^2*Sqrt[2 + 3*x^2 + x^4]),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
] - (14*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(7 + 5*x^2)*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] - (91*I)*Sqrt[1 + x^2]*S
qrt[2 + x^2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2] - (65*I)*x^2*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticPi[10/
7, I*ArcSinh[x/Sqrt[2]], 2])/(588*(7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4])

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 3 \, x^{2} + 2}}{25 \, x^{8} + 145 \, x^{6} + 309 \, x^{4} + 287 \, x^{2} + 98}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 3*x^2 + 2)/(25*x^8 + 145*x^6 + 309*x^4 + 287*x^2 + 98), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(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 \[ -\frac {25 \sqrt {x^{4}+3 x^{2}+2}\, x}{84 \left (5 x^{2}+7\right )}-\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{168 \sqrt {x^{4}+3 x^{2}+2}}-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{84 \sqrt {x^{4}+3 x^{2}+2}}-\frac {13 i \sqrt {2}\, \sqrt {\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{588 \sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/84*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)*x-1/84*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*Ellipt
icF(1/2*I*2^(1/2)*x,2^(1/2))-5/168*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))-13/588*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 \[ \int \frac {1}{\sqrt {x^{4} + 3 \, x^{2} + 2} {\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(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 \[ \int \frac {1}{{\left (5\,x^2+7\right )}^2\,\sqrt {x^4+3\,x^2+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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