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

Optimal. Leaf size=284 \[ -\frac {\sqrt {x^4+3 x^2+4} x}{70 \left (x^2+2\right )}+\frac {\sqrt {x^4+3 x^2+4} x}{14 \left (5 x^2+7\right )}+\frac {51 \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {x^4+3 x^2+4}}\right )}{280 \sqrt {385}}-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{35 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{35 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {289 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{9800 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

[Out]

51/107800*arctan(2/35*x*385^(1/2)/(x^4+3*x^2+4)^(1/2))*385^(1/2)-1/70*x*(x^4+3*x^2+4)^(1/2)/(x^2+2)+1/14*x*(x^
4+3*x^2+4)^(1/2)/(5*x^2+7)+1/70*(x^2+2)*(cos(2*arctan(1/2*x*2^(1/2)))^2)^(1/2)/cos(2*arctan(1/2*x*2^(1/2)))*El
lipticE(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)-
1/70*(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)+289/19600*(x^2+2)*(cos(2*ar
ctan(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 = 0.15, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1226, 1197, 1103, 1195, 1216, 1706} \[ -\frac {\sqrt {x^4+3 x^2+4} x}{70 \left (x^2+2\right )}+\frac {\sqrt {x^4+3 x^2+4} x}{14 \left (5 x^2+7\right )}+\frac {51 \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {x^4+3 x^2+4}}\right )}{280 \sqrt {385}}-\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{35 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {\left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{35 \sqrt {2} \sqrt {x^4+3 x^2+4}}+\frac {289 \left (x^2+2\right ) \sqrt {\frac {x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac {9}{280};2 \tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )|\frac {1}{8}\right )}{9800 \sqrt {2} \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*Sqrt[4 + 3*x^2 + x^4])/(70*(2 + x^2)) + (x*Sqrt[4 + 3*x^2 + x^4])/(14*(7 + 5*x^2)) + (51*ArcTan[(2*Sqrt[11
/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/(280*Sqrt[385]) + ((2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*A
rcTan[x/Sqrt[2]], 1/8])/(35*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) - ((2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*El
lipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(35*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (289*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/
(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(9800*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 1197

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

Rule 1216

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Di
st[(c*d + a*e*q)/(c*d^2 - a*e^2), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[(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}, 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]

Rule 1226

Int[Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*Sqrt[a + b*x^2 + c*
x^4])/(2*d*(d + e*x^2)), x] + (Dist[c/(2*d*e^2), Int[(d - e*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[(c*d^2
- a*e^2)/(2*d*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] && NeQ[
b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {4+3 x^2+x^4}}{\left (7+5 x^2\right )^2} \, dx &=\frac {x \sqrt {4+3 x^2+x^4}}{14 \left (7+5 x^2\right )}+\frac {1}{350} \int \frac {7-5 x^2}{\sqrt {4+3 x^2+x^4}} \, dx+\frac {51}{350} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx\\ &=\frac {x \sqrt {4+3 x^2+x^4}}{14 \left (7+5 x^2\right )}-\frac {3}{350} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx+\frac {1}{35} \int \frac {1-\frac {x^2}{2}}{\sqrt {4+3 x^2+x^4}} \, dx-\frac {17}{350} \int \frac {1}{\sqrt {4+3 x^2+x^4}} \, dx+\frac {17}{35} \int \frac {1+\frac {x^2}{2}}{\left (7+5 x^2\right ) \sqrt {4+3 x^2+x^4}} \, dx\\ &=-\frac {x \sqrt {4+3 x^2+x^4}}{70 \left (2+x^2\right )}+\frac {x \sqrt {4+3 x^2+x^4}}{14 \left (7+5 x^2\right )}+\frac {51 \tan ^{-1}\left (\frac {2 \sqrt {\frac {11}{35}} x}{\sqrt {4+3 x^2+x^4}}\right )}{280 \sqrt {385}}+\frac {\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 )}{35 \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 )}{35 \sqrt {2} \sqrt {4+3 x^2+x^4}}+\frac {289 \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 )}{9800 \sqrt {2} \sqrt {4+3 x^2+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.77, size = 481, normalized size = 1.69 \[ \frac {-98 i \left (5 x^2+7\right ) \sqrt {2-\frac {4 i x^2}{\sqrt {7}-3 i}} \sqrt {1+\frac {2 i x^2}{\sqrt {7}+3 i}} F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )-102 i \left (5 x^2+7\right ) \sqrt {2-\frac {4 i x^2}{\sqrt {7}-3 i}} \sqrt {1+\frac {2 i x^2}{\sqrt {7}+3 i}} \Pi \left (\frac {5}{14} \left (3+i \sqrt {7}\right );i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )+35 \left (\sqrt {7}+3 i\right ) \left (5 x^2+7\right ) \sqrt {2-\frac {4 i x^2}{\sqrt {7}-3 i}} \sqrt {1+\frac {2 i x^2}{\sqrt {7}+3 i}} \left (E\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )-F\left (i \sinh ^{-1}\left (\sqrt {-\frac {2 i}{-3 i+\sqrt {7}}} x\right )|\frac {3 i-\sqrt {7}}{3 i+\sqrt {7}}\right )\right )+700 \sqrt {-\frac {i}{\sqrt {7}-3 i}} x \left (x^4+3 x^2+4\right )}{9800 \sqrt {-\frac {i}{\sqrt {7}-3 i}} \left (5 x^2+7\right ) \sqrt {x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[4 + 3*x^2 + x^4]/(7 + 5*x^2)^2,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])] - (102*I)*(7 + 5*x^2)*Sq
rt[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])])/(9800*Sqrt[(-I)/(-3*I + Sqrt[7])
]*(7 + 5*x^2)*Sqrt[4 + 3*x^2 + x^4])

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{4} + 3 \, x^{2} + 4}}{25 \, x^{4} + 70 \, x^{2} + 49}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(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 = 1.44 \[ \frac {\sqrt {x^{4}+3 x^{2}+4}\, x}{70 x^{2}+98}-\frac {16 \sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticE \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{35 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )}+\frac {2 \sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{25 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}}+\frac {16 \sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticF \left (\frac {\sqrt {-6+2 i \sqrt {7}}\, x}{4}, \frac {\sqrt {2+6 i \sqrt {7}}}{4}\right )}{35 \sqrt {-6+2 i \sqrt {7}}\, \sqrt {x^{4}+3 x^{2}+4}\, \left (i \sqrt {7}+3\right )}+\frac {51 \sqrt {\frac {3 x^{2}}{8}-\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \sqrt {\frac {3 x^{2}}{8}+\frac {i \sqrt {7}\, x^{2}}{8}+1}\, \EllipticPi \left (\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, x , -\frac {5}{7 \left (-\frac {3}{8}+\frac {i \sqrt {7}}{8}\right )}, \frac {\sqrt {-\frac {3}{8}-\frac {i \sqrt {7}}{8}}}{\sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}}\right )}{2450 \sqrt {-\frac {3}{8}+\frac {i \sqrt {7}}{8}}\, \sqrt {x^{4}+3 x^{2}+4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*x*(x^4+3*x^2+4)^(1/2)/(5*x^2+7)+2/25/(-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
))+16/35/(-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))-16/35/(-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))+51/2450/(-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 \[ \int \frac {\sqrt {x^{4} + 3 \, x^{2} + 4}}{{\left (5 \, x^{2} + 7\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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