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

Optimal. Leaf size=286 \[ -\frac {5 \sqrt {x^4+3 x^2+4} x}{616 \left (x^2+2\right )}+\frac {25 \sqrt {x^4+3 x^2+4} x}{616 \left (5 x^2+7\right )}+\frac {37 \sqrt {\frac {5}{77}} \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {x^4+3 x^2+4}}\right )}{2464}-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{42 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {5 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{308 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {629 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{51744 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

[Out]

37/189728*arctan(2/35*x*385^(1/2)/(x^4+3*x^2+4)^(1/2))*385^(1/2)-5/616*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+25/616*x*
(x^4+3*x^2+4)^(1/2)/(5*x^2+7)+5/616*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2))
)*EllipticE(sin(2*arctan(1/2*x*2^(1/2))),1/4*2^(1/2))*2^(1/2)*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)/(x^4+3*x^2+4)^(1
/2)-1/84*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2)))*EllipticF(sin(2*arctan(1/
2*x*2^(1/2))),1/4*2^(1/2))*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)*2^(1/2)/(x^4+3*x^2+4)^(1/2)+629/103488*(x^2+2)*(cos
(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2)))*EllipticPi(sin(2*arctan(1/2*x*2^(1/2))),-9/280
,1/4*2^(1/2))*((x^4+3*x^2+4)/(x^2+2)^2)^(1/2)*2^(1/2)/(x^4+3*x^2+4)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1223, 1714, 1195, 1708, 1103, 1706} \[ -\frac {5 \sqrt {x^4+3 x^2+4} x}{616 \left (x^2+2\right )}+\frac {25 \sqrt {x^4+3 x^2+4} x}{616 \left (5 x^2+7\right )}+\frac {37 \sqrt {\frac {5}{77}} \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {x^4+3 x^2+4}}\right )}{2464}-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{42 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {5 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{308 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {629 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{51744 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x*Sqrt[4 + 3*x^2 + x^4])/(616*(2 + x^2)) + (25*x*Sqrt[4 + 3*x^2 + x^4])/(616*(7 + 5*x^2)) + (37*Sqrt[5/77]
*ArcTan[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/2464 + (5*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*Elli
pticE[2*ArcTan[x/Sqrt[2]], 1/8])/(308*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) - ((2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 +
x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(42*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (629*(2 + x^2)*Sqrt[(4 + 3*x
^2 + x^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(51744*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[
(d*x*Sqrt[a + b*x^2 + c*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q
^2*x^2)^2)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), x] /; EqQ[e + d*q^2, 0
]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

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 1706

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[
{q = Rt[B/A, 2]}, -Simp[((B*d - A*e)*ArcTan[(Rt[-b + (c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + b*x^2 + c*x^4]])/(2*d*e
*Rt[-b + (c*d)/e + (a*e)/d, 2]), x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + b*x^2 + c*x^4))/(a*(A + B*x
^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2 - (b*A)/(4*a*B)])/(4*d*e*A*q*Sqrt[
a + b*x^2 + c*x^4]), x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^
2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rule 1708

Int[((A_.) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With
[{q = Rt[c/a, 2]}, Dist[(A*(c*d + a*e*q) - a*B*(e + d*q))/(c*d^2 - a*e^2), Int[1/Sqrt[a + b*x^2 + c*x^4], x],
x] + Dist[(a*(B*d - A*e)*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x]
, x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^2
- a*e^2, 0] && PosQ[c/a] && NeQ[c*A^2 - a*B^2, 0]

Rule 1714

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

Rubi steps

\begin {align*} \int \frac {1}{\left (7+5 x^2\right )^2 \sqrt {4+3 x^2+x^4}} \, dx &=\frac {25 x \sqrt {4+3 x^2+x^4}}{616 \left (7+5 x^2\right )}-\frac {1}{616} \int \frac {12+70 x^2+25 x^4}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {25 x \sqrt {4+3 x^2+x^4}}{616 \left (7+5 x^2\right )}-\frac {\int \frac {410+425 x^2}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx}{3080}+\frac {5}{308} \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx\\ &=-\frac {5 x \sqrt {4+3 x^2+x^4}}{616 \left (2+x^2\right )}+\frac {25 x \sqrt {4+3 x^2+x^4}}{616 \left (7+5 x^2\right )}+\frac {5 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{308 \sqrt {2} \sqrt {4+3 x^2+x^4}}-\frac {1}{21} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx+\frac {185}{924} \int \frac {1+\frac {x^2}{2}}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx\\ &=-\frac {5 x \sqrt {4+3 x^2+x^4}}{616 \left (2+x^2\right )}+\frac {25 x \sqrt {4+3 x^2+x^4}}{616 \left (7+5 x^2\right )}+\frac {37 \sqrt {\frac {5}{77}} \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {4+3 x^2+x^4}}\right )}{2464}+\frac {5 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{308 \sqrt {2} \sqrt {4+3 x^2+x^4}}-\frac {\left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{42 \sqrt {2} \sqrt {4+3 x^2+x^4}}+\frac {629 \left (2+x^2\right ) \sqrt {\frac {4+3 x^2+x^4}{\left (2+x^2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{51744 \sqrt {2} \sqrt {4+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.78, size = 481, normalized size = 1.68 \[ \frac {98 i \left (5 x^2+7\right ) \sqrt {2-\frac {4 i x^2}{\sqrt {7}-3 i}} \sqrt {1+\frac {2 i x^2}{\sqrt {7}+3 i}} F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )-74 i \left (5 x^2+7\right ) \sqrt {2-\frac {4 i x^2}{\sqrt {7}-3 i}} \sqrt {1+\frac {2 i x^2}{\sqrt {7}+3 i}} \Pi \left (\frac {5}{14} \left (3+i \sqrt {7}\right );i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+35 \left (\sqrt {7}+3 i\right ) \left (5 x^2+7\right ) \sqrt {2-\frac {4 i x^2}{\sqrt {7}-3 i}} \sqrt {1+\frac {2 i x^2}{\sqrt {7}+3 i}} \left (E\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )-F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )\right )+700 \sqrt {-\frac {i}{\sqrt {7}-3 i}} x \left (x^4+3 x^2+4\right )}{17248 \sqrt {-\frac {i}{\sqrt {7}-3 i}} \left (5 x^2+7\right ) \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(700*Sqrt[(-I)/(-3*I + Sqrt[7])]*x*(4 + 3*x^2 + x^4) + 35*(3*I + Sqrt[7])*(7 + 5*x^2)*Sqrt[2 - ((4*I)*x^2)/(-3
*I + Sqrt[7])]*Sqrt[1 + ((2*I)*x^2)/(3*I + Sqrt[7])]*(EllipticE[I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3
*I - Sqrt[7])/(3*I + Sqrt[7])] - EllipticF[I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I +
Sqrt[7])]) + (98*I)*(7 + 5*x^2)*Sqrt[2 - ((4*I)*x^2)/(-3*I + Sqrt[7])]*Sqrt[1 + ((2*I)*x^2)/(3*I + Sqrt[7])]*E
llipticF[I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] - (74*I)*(7 + 5*x^2)*Sqr
t[2 - ((4*I)*x^2)/(-3*I + Sqrt[7])]*Sqrt[1 + ((2*I)*x^2)/(3*I + Sqrt[7])]*EllipticPi[(5*(3 + I*Sqrt[7]))/14, I
*ArcSinh[Sqrt[(-2*I)/(-3*I + Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])])/(17248*Sqrt[(-I)/(-3*I + Sqrt[7])
]*(7 + 5*x^2)*Sqrt[4 + 3*x^2 + x^4])

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 3 \, x^{2} + 4}}{25 \, x^{8} + 145 \, x^{6} + 359 \, x^{4} + 427 \, x^{2} + 196}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 3*x^2 + 4)/(25*x^8 + 145*x^6 + 359*x^4 + 427*x^2 + 196), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 410, normalized size = 1.43 \[ \frac {25 \sqrt {x^{4}+3 x^{2}+4}\, x}{616 \left (5 x^{2}+7\right )}-\frac {20 \sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{77 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )}-\frac {\sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{22 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}+\frac {20 \sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{77 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )}+\frac {37 \sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticPi \left (\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, x , -\frac {5}{7 \left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right )}, \frac {\sqrt {-\frac {3}{8}-\frac {i \sqrt {7}}{8}}}{\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}}\right )}{4312 \sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, \sqrt {x^{4}+3 x^{2}+4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/616*(x^4+3*x^2+4)^(1/2)/(5*x^2+7)*x-1/22/(-6+2*I*7^(1/2))^(1/2)*(3/8*x^2-1/8*I*7^(1/2)*x^2+1)^(1/2)*(3/8*x^
2+1/8*I*7^(1/2)*x^2+1)^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*(-6+2*I*7^(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1
/2))+20/77/(-6+2*I*7^(1/2))^(1/2)*(3/8*x^2-1/8*I*7^(1/2)*x^2+1)^(1/2)*(3/8*x^2+1/8*I*7^(1/2)*x^2+1)^(1/2)/(x^4
+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*EllipticF(1/4*(-6+2*I*7^(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2))-20/77/(-6+2*I*
7^(1/2))^(1/2)*(3/8*x^2-1/8*I*7^(1/2)*x^2+1)^(1/2)*(3/8*x^2+1/8*I*7^(1/2)*x^2+1)^(1/2)/(x^4+3*x^2+4)^(1/2)/(I*
7^(1/2)+3)*EllipticE(1/4*(-6+2*I*7^(1/2))^(1/2)*x,1/4*(2+6*I*7^(1/2))^(1/2))+37/4312/(-3/8+1/8*I*7^(1/2))^(1/2
)*(3/8*x^2-1/8*I*7^(1/2)*x^2+1)^(1/2)*(3/8*x^2+1/8*I*7^(1/2)*x^2+1)^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticPi((-3/8
+1/8*I*7^(1/2))^(1/2)*x,-5/7/(-3/8+1/8*I*7^(1/2)),(-3/8-1/8*I*7^(1/2))^(1/2)/(-3/8+1/8*I*7^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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