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

Optimal. Leaf size=102 \[ -\frac {12525 \sqrt {-x^4+x^2+2} x}{453152 \left (5 x^2+7\right )}-\frac {25 \sqrt {-x^4+x^2+2} x}{952 \left (5 x^2+7\right )^2}-\frac {263 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{226576}-\frac {2505 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{453152}+\frac {58915 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{3172064} \]

[Out]

-2505/453152*EllipticE(1/2*x*2^(1/2),I*2^(1/2))-263/226576*EllipticF(1/2*x*2^(1/2),I*2^(1/2))+58915/3172064*El
lipticPi(1/2*x*2^(1/2),-10/7,I*2^(1/2))-25/952*x*(-x^4+x^2+2)^(1/2)/(5*x^2+7)^2-12525/453152*x*(-x^4+x^2+2)^(1
/2)/(5*x^2+7)

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Rubi [A]  time = 0.19, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1223, 1696, 1716, 1180, 524, 424, 419, 1212, 537} \[ -\frac {12525 \sqrt {-x^4+x^2+2} x}{453152 \left (5 x^2+7\right )}-\frac {25 \sqrt {-x^4+x^2+2} x}{952 \left (5 x^2+7\right )^2}-\frac {263 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{226576}-\frac {2505 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{453152}+\frac {58915 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{3172064} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*x*Sqrt[2 + x^2 - x^4])/(952*(7 + 5*x^2)^2) - (12525*x*Sqrt[2 + x^2 - x^4])/(453152*(7 + 5*x^2)) - (2505*E
llipticE[ArcSin[x/Sqrt[2]], -2])/453152 - (263*EllipticF[ArcSin[x/Sqrt[2]], -2])/226576 + (58915*EllipticPi[-1
0/7, ArcSin[x/Sqrt[2]], -2])/3172064

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 1180

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[2*Sqrt[-c], Int[(d + e*x^2)/(Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c,
d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[c, 0]

Rule 1212

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*Sqrt[-c], Int[1/((d + e*x^2)*Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), 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 1696

Int[((P4x_)*((d_) + (e_.)*(x_)^2)^(q_))/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]}, -Simp[((C*d^2 - B*d*e + A*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*d*(C*d - B*e) + A*(a*e^2*(2*q + 3) + 2*d*(c*d - b*e)*(q + 1)) - 2*((B*d
- A*e)*(b*e*(q + 2) - c*d*(q + 1)) - C*d*(b*d + a*e*(q + 1)))*x^2 + c*(C*d^2 - B*d*e + A*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] && PolyQ[P4x, x^2] && LeQ[Expon[P4x, x], 4] &
& NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[q, -1]

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 )^3 \sqrt {2+x^2-x^4}} \, dx &=-\frac {25 x \sqrt {2+x^2-x^4}}{952 \left (7+5 x^2\right )^2}+\frac {1}{952} \int \frac {186-190 x^2+25 x^4}{\left (7+5 x^2\right )^2 \sqrt {2+x^2-x^4}} \, dx\\ &=-\frac {25 x \sqrt {2+x^2-x^4}}{952 \left (7+5 x^2\right )^2}-\frac {12525 x \sqrt {2+x^2-x^4}}{453152 \left (7+5 x^2\right )}+\frac {\int \frac {37698-32690 x^2-12525 x^4}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx}{453152}\\ &=-\frac {25 x \sqrt {2+x^2-x^4}}{952 \left (7+5 x^2\right )^2}-\frac {12525 x \sqrt {2+x^2-x^4}}{453152 \left (7+5 x^2\right )}-\frac {\int \frac {75775+62625 x^2}{\sqrt {2+x^2-x^4}} \, dx}{11328800}+\frac {58915 \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx}{453152}\\ &=-\frac {25 x \sqrt {2+x^2-x^4}}{952 \left (7+5 x^2\right )^2}-\frac {12525 x \sqrt {2+x^2-x^4}}{453152 \left (7+5 x^2\right )}-\frac {\int \frac {75775+62625 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{5664400}+\frac {58915 \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2} \left (7+5 x^2\right )} \, dx}{226576}\\ &=-\frac {25 x \sqrt {2+x^2-x^4}}{952 \left (7+5 x^2\right )^2}-\frac {12525 x \sqrt {2+x^2-x^4}}{453152 \left (7+5 x^2\right )}+\frac {58915 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{3172064}-\frac {263 \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{113288}-\frac {2505 \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx}{453152}\\ &=-\frac {25 x \sqrt {2+x^2-x^4}}{952 \left (7+5 x^2\right )^2}-\frac {12525 x \sqrt {2+x^2-x^4}}{453152 \left (7+5 x^2\right )}-\frac {2505 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{453152}-\frac {263 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{226576}+\frac {58915 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{3172064}\\ \end {align*}

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Mathematica [C]  time = 0.42, size = 108, normalized size = 1.06 \[ \frac {\frac {350 x \left (2505 x^6+1478 x^4-8993 x^2-7966\right )}{\left (5 x^2+7\right )^2 \sqrt {-x^4+x^2+2}}+56287 i \sqrt {2} F\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )-35070 i \sqrt {2} E\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )-58915 i \sqrt {2} \Pi \left (\frac {5}{7};i \sinh ^{-1}(x)|-\frac {1}{2}\right )}{6344128} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((350*x*(-7966 - 8993*x^2 + 1478*x^4 + 2505*x^6))/((7 + 5*x^2)^2*Sqrt[2 + x^2 - x^4]) - (35070*I)*Sqrt[2]*Elli
pticE[I*ArcSinh[x], -1/2] + (56287*I)*Sqrt[2]*EllipticF[I*ArcSinh[x], -1/2] - (58915*I)*Sqrt[2]*EllipticPi[5/7
, I*ArcSinh[x], -1/2])/6344128

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-x^{4} + x^{2} + 2}}{125 \, x^{10} + 400 \, x^{8} - 40 \, x^{6} - 1442 \, x^{4} - 1813 \, x^{2} - 686}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-x^4 + x^2 + 2)/(125*x^10 + 400*x^8 - 40*x^6 - 1442*x^4 - 1813*x^2 - 686), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 189, normalized size = 1.85 \[ -\frac {25 \sqrt {-x^{4}+x^{2}+2}\, x}{952 \left (5 x^{2}+7\right )^{2}}-\frac {12525 \sqrt {-x^{4}+x^{2}+2}\, x}{453152 \left (5 x^{2}+7\right )}-\frac {2505 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{906304 \sqrt {-x^{4}+x^{2}+2}}-\frac {263 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{453152 \sqrt {-x^{4}+x^{2}+2}}+\frac {58915 \sqrt {2}\, \sqrt {-\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {\sqrt {2}\, x}{2}, -\frac {10}{7}, i \sqrt {2}\right )}{3172064 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/952*(-x^4+x^2+2)^(1/2)/(5*x^2+7)^2*x-12525/453152*(-x^4+x^2+2)^(1/2)/(5*x^2+7)*x-263/453152*2^(1/2)*(-2*x^
2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticF(1/2*2^(1/2)*x,I*2^(1/2))-2505/906304*2^(1/2)*(-2*x^2+4)^
(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticE(1/2*2^(1/2)*x,I*2^(1/2))+58915/3172064*2^(1/2)*(-1/2*x^2+1)^(
1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^(1/2)*EllipticPi(1/2*2^(1/2)*x,-10/7,I*2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(-(x**2 - 2)*(x**2 + 1))*(5*x**2 + 7)**3), x)

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