3.330 \(\int \frac{(2+x^2-x^4)^{3/2}}{(7+5 x^2)^2} \, dx\)

Optimal. Leaf size=93 \[ \frac{458}{875} \text{EllipticF}\left (\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right ),-2\right )-\frac{17 \sqrt{-x^4+x^2+2} x}{175 \left (5 x^2+7\right )}-\frac{1}{75} \sqrt{-x^4+x^2+2} x-\frac{97}{525} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{1241 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{6125} \]

[Out]

-(x*Sqrt[2 + x^2 - x^4])/75 - (17*x*Sqrt[2 + x^2 - x^4])/(175*(7 + 5*x^2)) - (97*EllipticE[ArcSin[x/Sqrt[2]],
-2])/525 + (458*EllipticF[ArcSin[x/Sqrt[2]], -2])/875 - (1241*EllipticPi[-10/7, ArcSin[x/Sqrt[2]], -2])/6125

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Rubi [A]  time = 0.320221, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 13, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.542, Rules used = {1228, 1095, 419, 1132, 493, 424, 1122, 1180, 1223, 1716, 524, 1212, 537} \[ -\frac{17 \sqrt{-x^4+x^2+2} x}{175 \left (5 x^2+7\right )}-\frac{1}{75} \sqrt{-x^4+x^2+2} x+\frac{458}{875} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{97}{525} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{1241 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{6125} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*Sqrt[2 + x^2 - x^4])/75 - (17*x*Sqrt[2 + x^2 - x^4])/(175*(7 + 5*x^2)) - (97*EllipticE[ArcSin[x/Sqrt[2]],
-2])/525 + (458*EllipticF[ArcSin[x/Sqrt[2]], -2])/875 - (1241*EllipticPi[-10/7, ArcSin[x/Sqrt[2]], -2])/6125

Rule 1228

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{aa, bb, cc}, In
t[ExpandIntegrand[1/Sqrt[aa + bb*x^2 + cc*x^4], (d + e*x^2)^q*(aa + bb*x^2 + cc*x^4)^(p + 1/2), x] /. {aa -> a
, bb -> b, cc -> c}, 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]
&& ILtQ[q, 0] && IntegerQ[p + 1/2]

Rule 1095

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

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 1132

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

Rule 493

Int[(x_)^(n_)/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/b, Int[Sqrt[a +
b*x^n]/Sqrt[c + d*x^n], x], x] - Dist[a/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b, c,
d}, x] && NeQ[b*c - a*d, 0] && (EqQ[n, 2] || EqQ[n, 4]) &&  !(EqQ[n, 2] && 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 1122

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d^3*(d*x)^(m - 3)*(a + b*
x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 1)), x] - Dist[d^4/(c*(m + 4*p + 1)), Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b
*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && Gt
Q[m, 3] && NeQ[m + 4*p + 1, 0] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

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 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 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]

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 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 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)
])

Rubi steps

\begin{align*} \int \frac{\left (2+x^2-x^4\right )^{3/2}}{\left (7+5 x^2\right )^2} \, dx &=\int \left (\frac{212}{625 \sqrt{2+x^2-x^4}}-\frac{24 x^2}{125 \sqrt{2+x^2-x^4}}+\frac{x^4}{25 \sqrt{2+x^2-x^4}}+\frac{1156}{625 \left (7+5 x^2\right )^2 \sqrt{2+x^2-x^4}}-\frac{1292}{625 \left (7+5 x^2\right ) \sqrt{2+x^2-x^4}}\right ) \, dx\\ &=\frac{1}{25} \int \frac{x^4}{\sqrt{2+x^2-x^4}} \, dx-\frac{24}{125} \int \frac{x^2}{\sqrt{2+x^2-x^4}} \, dx+\frac{212}{625} \int \frac{1}{\sqrt{2+x^2-x^4}} \, dx+\frac{1156}{625} \int \frac{1}{\left (7+5 x^2\right )^2 \sqrt{2+x^2-x^4}} \, dx-\frac{1292}{625} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx\\ &=-\frac{1}{75} x \sqrt{2+x^2-x^4}-\frac{17 x \sqrt{2+x^2-x^4}}{175 \left (7+5 x^2\right )}+\frac{17 \int \frac{118-70 x^2-25 x^4}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{4375}+\frac{1}{75} \int \frac{2+2 x^2}{\sqrt{2+x^2-x^4}} \, dx-\frac{48}{125} \int \frac{x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx+\frac{424}{625} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx-\frac{2584}{625} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx\\ &=-\frac{1}{75} x \sqrt{2+x^2-x^4}-\frac{17 x \sqrt{2+x^2-x^4}}{175 \left (7+5 x^2\right )}+\frac{212}{625} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{1292 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{4375}-\frac{17 \int \frac{175+125 x^2}{\sqrt{2+x^2-x^4}} \, dx}{109375}+\frac{2}{75} \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx-\frac{24}{125} \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx+\frac{48}{125} \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx+\frac{2839 \int \frac{1}{\left (7+5 x^2\right ) \sqrt{2+x^2-x^4}} \, dx}{4375}\\ &=-\frac{1}{75} x \sqrt{2+x^2-x^4}-\frac{17 x \sqrt{2+x^2-x^4}}{175 \left (7+5 x^2\right )}-\frac{62}{375} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{332}{625} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{1292 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{4375}-\frac{34 \int \frac{175+125 x^2}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{109375}+\frac{5678 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2} \left (7+5 x^2\right )} \, dx}{4375}\\ &=-\frac{1}{75} x \sqrt{2+x^2-x^4}-\frac{17 x \sqrt{2+x^2-x^4}}{175 \left (7+5 x^2\right )}-\frac{62}{375} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{332}{625} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{1241 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{6125}-\frac{68 \int \frac{1}{\sqrt{4-2 x^2} \sqrt{2+2 x^2}} \, dx}{4375}-\frac{17}{875} \int \frac{\sqrt{2+2 x^2}}{\sqrt{4-2 x^2}} \, dx\\ &=-\frac{1}{75} x \sqrt{2+x^2-x^4}-\frac{17 x \sqrt{2+x^2-x^4}}{175 \left (7+5 x^2\right )}-\frac{97}{525} E\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )+\frac{458}{875} F\left (\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )-\frac{1241 \Pi \left (-\frac{10}{7};\left .\sin ^{-1}\left (\frac{x}{\sqrt{2}}\right )\right |-2\right )}{6125}\\ \end{align*}

Mathematica [C]  time = 0.313982, size = 201, normalized size = 2.16 \[ \frac{567 i \sqrt{2} \left (5 x^2+7\right ) \sqrt{-x^4+x^2+2} \text{EllipticF}\left (i \sinh ^{-1}(x),-\frac{1}{2}\right )+2450 x^7+4550 x^5-11900 x^3-6790 i \sqrt{2} \left (5 x^2+7\right ) \sqrt{-x^4+x^2+2} E\left (i \sinh ^{-1}(x)|-\frac{1}{2}\right )+18615 i \sqrt{2} \sqrt{-x^4+x^2+2} x^2 \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )+26061 i \sqrt{2} \sqrt{-x^4+x^2+2} \Pi \left (\frac{5}{7};i \sinh ^{-1}(x)|-\frac{1}{2}\right )-14000 x}{36750 \left (5 x^2+7\right ) \sqrt{-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14000*x - 11900*x^3 + 4550*x^5 + 2450*x^7 - (6790*I)*Sqrt[2]*(7 + 5*x^2)*Sqrt[2 + x^2 - x^4]*EllipticE[I*Arc
Sinh[x], -1/2] + (567*I)*Sqrt[2]*(7 + 5*x^2)*Sqrt[2 + x^2 - x^4]*EllipticF[I*ArcSinh[x], -1/2] + (26061*I)*Sqr
t[2]*Sqrt[2 + x^2 - x^4]*EllipticPi[5/7, I*ArcSinh[x], -1/2] + (18615*I)*Sqrt[2]*x^2*Sqrt[2 + x^2 - x^4]*Ellip
ticPi[5/7, I*ArcSinh[x], -1/2])/(36750*(7 + 5*x^2)*Sqrt[2 + x^2 - x^4])

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Maple [B]  time = 0.021, size = 180, normalized size = 1.9 \begin{align*} -{\frac{17\,x}{875\,{x}^{2}+1225}\sqrt{-{x}^{4}+{x}^{2}+2}}-{\frac{x}{75}\sqrt{-{x}^{4}+{x}^{2}+2}}+{\frac{229\,\sqrt{2}}{875}\sqrt{-2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\it EllipticF} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}}-{\frac{97\,\sqrt{2}}{1050}\sqrt{-2\,{x}^{2}+4}\sqrt{{x}^{2}+1}{\it EllipticE} \left ({\frac{x\sqrt{2}}{2}},i\sqrt{2} \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}}-{\frac{1241\,\sqrt{2}}{6125}\sqrt{1-{\frac{{x}^{2}}{2}}}\sqrt{{x}^{2}+1}{\it EllipticPi} \left ({\frac{x\sqrt{2}}{2}},-{\frac{10}{7}},i\sqrt{2} \right ){\frac{1}{\sqrt{-{x}^{4}+{x}^{2}+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17/175*x*(-x^4+x^2+2)^(1/2)/(5*x^2+7)-1/75*x*(-x^4+x^2+2)^(1/2)+229/875*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2
)/(-x^4+x^2+2)^(1/2)*EllipticF(1/2*x*2^(1/2),I*2^(1/2))-97/1050*2^(1/2)*(-2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(-x^4+x
^2+2)^(1/2)*EllipticE(1/2*x*2^(1/2),I*2^(1/2))-1241/6125*2^(1/2)*(1-1/2*x^2)^(1/2)*(x^2+1)^(1/2)/(-x^4+x^2+2)^
(1/2)*EllipticPi(1/2*x*2^(1/2),-10/7,I*2^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-x^{4} + x^{2} + 2\right )}^{\frac{3}{2}}}{{\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-x^{4} + x^{2} + 2\right )}^{\frac{3}{2}}}{25 \, x^{4} + 70 \, x^{2} + 49}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (x^{2} - 2\right ) \left (x^{2} + 1\right )\right )^{\frac{3}{2}}}{\left (5 x^{2} + 7\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-x^{4} + x^{2} + 2\right )}^{\frac{3}{2}}}{{\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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