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

Optimal. Leaf size=102 \[ \frac {563 \sqrt {-x^4+x^2+2} x}{9800 \left (5 x^2+7\right )}-\frac {17 \sqrt {-x^4+x^2+2} x}{350 \left (5 x^2+7\right )^2}-\frac {1251 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{24500}+\frac {191 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{9800}+\frac {9879 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{343000} \]

[Out]

191/9800*EllipticE(1/2*x*2^(1/2),I*2^(1/2))-1251/24500*EllipticF(1/2*x*2^(1/2),I*2^(1/2))+9879/343000*Elliptic
Pi(1/2*x*2^(1/2),-10/7,I*2^(1/2))-17/350*x*(-x^4+x^2+2)^(1/2)/(5*x^2+7)^2+563/9800*x*(-x^4+x^2+2)^(1/2)/(5*x^2
+7)

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Rubi [A]  time = 0.50, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 27, 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, 1223, 1696, 1716, 1180, 524, 1212, 537} \[ \frac {563 \sqrt {-x^4+x^2+2} x}{9800 \left (5 x^2+7\right )}-\frac {17 \sqrt {-x^4+x^2+2} x}{350 \left (5 x^2+7\right )^2}-\frac {1251 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{24500}+\frac {191 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{9800}+\frac {9879 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{343000} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-17*x*Sqrt[2 + x^2 - x^4])/(350*(7 + 5*x^2)^2) + (563*x*Sqrt[2 + x^2 - x^4])/(9800*(7 + 5*x^2)) + (191*Ellipt
icE[ArcSin[x/Sqrt[2]], -2])/9800 - (1251*EllipticF[ArcSin[x/Sqrt[2]], -2])/24500 + (9879*EllipticPi[-10/7, Arc
Sin[x/Sqrt[2]], -2])/343000

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 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 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 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 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 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 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 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 {\left (2+x^2-x^4\right )^{3/2}}{\left (7+5 x^2\right )^3} \, dx &=\int \left (-\frac {31}{625 \sqrt {2+x^2-x^4}}+\frac {x^2}{125 \sqrt {2+x^2-x^4}}+\frac {1156}{625 \left (7+5 x^2\right )^3 \sqrt {2+x^2-x^4}}-\frac {1292}{625 \left (7+5 x^2\right )^2 \sqrt {2+x^2-x^4}}+\frac {429}{625 \left (7+5 x^2\right ) \sqrt {2+x^2-x^4}}\right ) \, dx\\ &=\frac {1}{125} \int \frac {x^2}{\sqrt {2+x^2-x^4}} \, dx-\frac {31}{625} \int \frac {1}{\sqrt {2+x^2-x^4}} \, dx+\frac {429}{625} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx+\frac {1156}{625} \int \frac {1}{\left (7+5 x^2\right )^3 \sqrt {2+x^2-x^4}} \, dx-\frac {1292}{625} \int \frac {1}{\left (7+5 x^2\right )^2 \sqrt {2+x^2-x^4}} \, dx\\ &=-\frac {17 x \sqrt {2+x^2-x^4}}{350 \left (7+5 x^2\right )^2}+\frac {19 x \sqrt {2+x^2-x^4}}{175 \left (7+5 x^2\right )}+\frac {17 \int \frac {186-190 x^2+25 x^4}{\left (7+5 x^2\right )^2 \sqrt {2+x^2-x^4}} \, dx}{8750}-\frac {19 \int \frac {118-70 x^2-25 x^4}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx}{4375}+\frac {2}{125} \int \frac {x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx-\frac {62}{625} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx+\frac {858}{625} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2} \left (7+5 x^2\right )} \, dx\\ &=-\frac {17 x \sqrt {2+x^2-x^4}}{350 \left (7+5 x^2\right )^2}+\frac {563 x \sqrt {2+x^2-x^4}}{9800 \left (7+5 x^2\right )}-\frac {31}{625} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {429 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{4375}+\frac {\int \frac {37698-32690 x^2-12525 x^4}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx}{245000}+\frac {19 \int \frac {175+125 x^2}{\sqrt {2+x^2-x^4}} \, dx}{109375}+\frac {1}{125} \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx-\frac {2}{125} \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx-\frac {3173 \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx}{4375}\\ &=-\frac {17 x \sqrt {2+x^2-x^4}}{350 \left (7+5 x^2\right )^2}+\frac {563 x \sqrt {2+x^2-x^4}}{9800 \left (7+5 x^2\right )}+\frac {1}{125} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {36}{625} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )+\frac {429 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{4375}-\frac {\int \frac {75775+62625 x^2}{\sqrt {2+x^2-x^4}} \, dx}{6125000}+\frac {38 \int \frac {175+125 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{109375}+\frac {11783 \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+x^2-x^4}} \, dx}{49000}-\frac {6346 \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2} \left (7+5 x^2\right )} \, dx}{4375}\\ &=-\frac {17 x \sqrt {2+x^2-x^4}}{350 \left (7+5 x^2\right )^2}+\frac {563 x \sqrt {2+x^2-x^4}}{9800 \left (7+5 x^2\right )}+\frac {1}{125} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {36}{625} F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {34 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{6125}-\frac {\int \frac {75775+62625 x^2}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{3062500}+\frac {76 \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{4375}+\frac {19}{875} \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx+\frac {11783 \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2} \left (7+5 x^2\right )} \, dx}{24500}\\ &=-\frac {17 x \sqrt {2+x^2-x^4}}{350 \left (7+5 x^2\right )^2}+\frac {563 x \sqrt {2+x^2-x^4}}{9800 \left (7+5 x^2\right )}+\frac {26}{875} E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )-\frac {214 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{4375}+\frac {9879 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{343000}-\frac {263 \int \frac {1}{\sqrt {4-2 x^2} \sqrt {2+2 x^2}} \, dx}{61250}-\frac {501 \int \frac {\sqrt {2+2 x^2}}{\sqrt {4-2 x^2}} \, dx}{49000}\\ &=-\frac {17 x \sqrt {2+x^2-x^4}}{350 \left (7+5 x^2\right )^2}+\frac {563 x \sqrt {2+x^2-x^4}}{9800 \left (7+5 x^2\right )}+\frac {191 E\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{9800}-\frac {1251 F\left (\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{24500}+\frac {9879 \Pi \left (-\frac {10}{7};\left .\sin ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-2\right )}{343000}\\ \end {align*}

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Mathematica [C]  time = 0.41, size = 244, normalized size = 2.39 \[ \frac {-197050 x^7-45500 x^5+636650 x^3-2541 i \sqrt {2} \left (5 x^2+7\right )^2 \sqrt {-x^4+x^2+2} F\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )+13370 i \sqrt {2} \left (5 x^2+7\right )^2 \sqrt {-x^4+x^2+2} E\left (i \sinh ^{-1}(x)|-\frac {1}{2}\right )-246975 i \sqrt {2} \sqrt {-x^4+x^2+2} x^4 \Pi \left (\frac {5}{7};i \sinh ^{-1}(x)|-\frac {1}{2}\right )-691530 i \sqrt {2} \sqrt {-x^4+x^2+2} x^2 \Pi \left (\frac {5}{7};i \sinh ^{-1}(x)|-\frac {1}{2}\right )-484071 i \sqrt {2} \sqrt {-x^4+x^2+2} \Pi \left (\frac {5}{7};i \sinh ^{-1}(x)|-\frac {1}{2}\right )+485100 x}{686000 \left (5 x^2+7\right )^2 \sqrt {-x^4+x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(485100*x + 636650*x^3 - 45500*x^5 - 197050*x^7 + (13370*I)*Sqrt[2]*(7 + 5*x^2)^2*Sqrt[2 + x^2 - x^4]*Elliptic
E[I*ArcSinh[x], -1/2] - (2541*I)*Sqrt[2]*(7 + 5*x^2)^2*Sqrt[2 + x^2 - x^4]*EllipticF[I*ArcSinh[x], -1/2] - (48
4071*I)*Sqrt[2]*Sqrt[2 + x^2 - x^4]*EllipticPi[5/7, I*ArcSinh[x], -1/2] - (691530*I)*Sqrt[2]*x^2*Sqrt[2 + x^2
- x^4]*EllipticPi[5/7, I*ArcSinh[x], -1/2] - (246975*I)*Sqrt[2]*x^4*Sqrt[2 + x^2 - x^4]*EllipticPi[5/7, I*ArcS
inh[x], -1/2])/(686000*(7 + 5*x^2)^2*Sqrt[2 + x^2 - x^4])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}}}{125 \, x^{6} + 525 \, x^{4} + 735 \, x^{2} + 343}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-x^4 + x^2 + 2)^(3/2)/(125*x^6 + 525*x^4 + 735*x^2 + 343), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 189, normalized size = 1.85 \[ -\frac {17 \sqrt {-x^{4}+x^{2}+2}\, x}{350 \left (5 x^{2}+7\right )^{2}}+\frac {563 \sqrt {-x^{4}+x^{2}+2}\, x}{9800 \left (5 x^{2}+7\right )}+\frac {191 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{19600 \sqrt {-x^{4}+x^{2}+2}}-\frac {1251 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {\sqrt {2}\, x}{2}, i \sqrt {2}\right )}{49000 \sqrt {-x^{4}+x^{2}+2}}+\frac {9879 \sqrt {2}\, \sqrt {-\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {\sqrt {2}\, x}{2}, -\frac {10}{7}, i \sqrt {2}\right )}{343000 \sqrt {-x^{4}+x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17/350*(-x^4+x^2+2)^(1/2)/(5*x^2+7)^2*x+563/9800*(-x^4+x^2+2)^(1/2)/(5*x^2+7)*x-1251/49000*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))+191/19600*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))+9879/343000*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 \[ \int \frac {{\left (-x^{4} + x^{2} + 2\right )}^{\frac {3}{2}}}{{\left (5 \, x^{2} + 7\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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