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

Optimal. Leaf size=305 \[ \frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right ),\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}+\frac{4 \sqrt{x^4+3 x^2+4} x}{175 \left (x^2+2\right )}+\frac{22 \sqrt{x^4+3 x^2+4} x}{175 \left (5 x^2+7\right )}+\frac{1}{75} \sqrt{x^4+3 x^2+4} x+\frac{13}{350} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )-\frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}+\frac{2431 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{36750 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/75 + (4*x*Sqrt[4 + 3*x^2 + x^4])/(175*(2 + x^2)) + (22*x*Sqrt[4 + 3*x^2 + x^4])/(175
*(7 + 5*x^2)) + (13*Sqrt[11/35]*ArcTan[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/350 - (4*Sqrt[2]*(2 + x^2)*Sq
rt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4]) + (4*Sqrt[2
]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4
]) + (2431*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(36750*
Sqrt[2]*Sqrt[4 + 3*x^2 + x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.534379, antiderivative size = 372, normalized size of antiderivative = 1.22, number of steps used = 19, number of rules used = 11, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.458, Rules used = {1228, 1103, 1139, 1195, 1122, 1197, 1223, 1714, 1708, 1706, 1216} \[ \frac{4 \sqrt{x^4+3 x^2+4} x}{175 \left (x^2+2\right )}+\frac{22 \sqrt{x^4+3 x^2+4} x}{175 \left (5 x^2+7\right )}+\frac{1}{75} \sqrt{x^4+3 x^2+4} x+\frac{13}{350} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{x^4+3 x^2+4}}\right )+\frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}-\frac{4 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{175 \sqrt{x^4+3 x^2+4}}+\frac{187 \sqrt{2} \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{13125 \sqrt{x^4+3 x^2+4}}+\frac{6919 \left (x^2+2\right ) \sqrt{\frac{x^4+3 x^2+4}{\left (x^2+2\right )^2}} \Pi \left (-\frac{9}{280};2 \tan ^{-1}\left (\frac{x}{\sqrt{2}}\right )|\frac{1}{8}\right )}{183750 \sqrt{2} \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[4 + 3*x^2 + x^4])/75 + (4*x*Sqrt[4 + 3*x^2 + x^4])/(175*(2 + x^2)) + (22*x*Sqrt[4 + 3*x^2 + x^4])/(175
*(7 + 5*x^2)) + (13*Sqrt[11/35]*ArcTan[(2*Sqrt[11/35]*x)/Sqrt[4 + 3*x^2 + x^4]])/350 - (4*Sqrt[2]*(2 + x^2)*Sq
rt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticE[2*ArcTan[x/Sqrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4]) + (4*Sqrt[2
]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticF[2*ArcTan[x/Sqrt[2]], 1/8])/(175*Sqrt[4 + 3*x^2 + x^4
]) + (6919*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticPi[-9/280, 2*ArcTan[x/Sqrt[2]], 1/8])/(183750
*Sqrt[2]*Sqrt[4 + 3*x^2 + x^4]) + (187*Sqrt[2]*(2 + x^2)*Sqrt[(4 + 3*x^2 + x^4)/(2 + x^2)^2]*EllipticPi[-9/280
, 2*ArcTan[x/Sqrt[2]], 1/8])/(13125*Sqrt[4 + 3*x^2 + x^4])

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 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 1139

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (4+3 x^2+x^4\right )^{3/2}}{\left (7+5 x^2\right )^2} \, dx &=\int \left (\frac{152}{625 \sqrt{4+3 x^2+x^4}}+\frac{16 x^2}{125 \sqrt{4+3 x^2+x^4}}+\frac{x^4}{25 \sqrt{4+3 x^2+x^4}}+\frac{1936}{625 \left (7+5 x^2\right )^2 \sqrt{4+3 x^2+x^4}}+\frac{88}{625 \left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}}\right ) \, dx\\ &=\frac{1}{25} \int \frac{x^4}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{16}{125} \int \frac{x^2}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{88}{625} \int \frac{1}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx+\frac{152}{625} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{1936}{625} \int \frac{1}{\left (7+5 x^2\right )^2 \sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{1}{75} x \sqrt{4+3 x^2+x^4}+\frac{22 x \sqrt{4+3 x^2+x^4}}{175 \left (7+5 x^2\right )}+\frac{38 \sqrt{2} \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 )}{625 \sqrt{4+3 x^2+x^4}}-\frac{22 \int \frac{12+70 x^2+25 x^4}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx}{4375}-\frac{1}{75} \int \frac{4+6 x^2}{\sqrt{4+3 x^2+x^4}} \, dx-\frac{88 \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx}{1875}+\frac{32}{125} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx-\frac{32}{125} \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{176}{375} \int \frac{1+\frac{x^2}{2}}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{1}{75} x \sqrt{4+3 x^2+x^4}+\frac{16 x \sqrt{4+3 x^2+x^4}}{125 \left (2+x^2\right )}+\frac{22 x \sqrt{4+3 x^2+x^4}}{175 \left (7+5 x^2\right )}+\frac{2}{125} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{4+3 x^2+x^4}}\right )-\frac{16 \sqrt{2} \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 )}{125 \sqrt{4+3 x^2+x^4}}+\frac{212 \sqrt{2} \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 )}{1875 \sqrt{4+3 x^2+x^4}}+\frac{187 \sqrt{2} \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 )}{13125 \sqrt{4+3 x^2+x^4}}-\frac{22 \int \frac{410+425 x^2}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx}{21875}+\frac{44}{875} \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx+\frac{4}{25} \int \frac{1-\frac{x^2}{2}}{\sqrt{4+3 x^2+x^4}} \, dx-\frac{16}{75} \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx\\ &=\frac{1}{75} x \sqrt{4+3 x^2+x^4}+\frac{4 x \sqrt{4+3 x^2+x^4}}{175 \left (2+x^2\right )}+\frac{22 x \sqrt{4+3 x^2+x^4}}{175 \left (7+5 x^2\right )}+\frac{2}{125} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{4+3 x^2+x^4}}\right )-\frac{4 \sqrt{2} \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 )}{175 \sqrt{4+3 x^2+x^4}}+\frac{112 \sqrt{2} \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 )}{1875 \sqrt{4+3 x^2+x^4}}+\frac{187 \sqrt{2} \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 )}{13125 \sqrt{4+3 x^2+x^4}}-\frac{1936 \int \frac{1}{\sqrt{4+3 x^2+x^4}} \, dx}{13125}+\frac{1628 \int \frac{1+\frac{x^2}{2}}{\left (7+5 x^2\right ) \sqrt{4+3 x^2+x^4}} \, dx}{2625}\\ &=\frac{1}{75} x \sqrt{4+3 x^2+x^4}+\frac{4 x \sqrt{4+3 x^2+x^4}}{175 \left (2+x^2\right )}+\frac{22 x \sqrt{4+3 x^2+x^4}}{175 \left (7+5 x^2\right )}+\frac{13}{350} \sqrt{\frac{11}{35}} \tan ^{-1}\left (\frac{2 \sqrt{\frac{11}{35}} x}{\sqrt{4+3 x^2+x^4}}\right )-\frac{4 \sqrt{2} \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 )}{175 \sqrt{4+3 x^2+x^4}}+\frac{4 \sqrt{2} \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 )}{175 \sqrt{4+3 x^2+x^4}}+\frac{6919 \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 )}{183750 \sqrt{2} \sqrt{4+3 x^2+x^4}}+\frac{187 \sqrt{2} \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 )}{13125 \sqrt{4+3 x^2+x^4}}\\ \end{align*}

Mathematica [C]  time = 0.585088, size = 309, normalized size = 1.01 \[ \frac{\frac{175 x \left (7 x^2+23\right ) \left (x^4+3 x^2+4\right )}{5 x^2+7}-i \sqrt{6+2 i \sqrt{7}} \sqrt{1-\frac{2 i x^2}{\sqrt{7}-3 i}} \sqrt{1+\frac{2 i x^2}{\sqrt{7}+3 i}} \left (7 \left (158+15 i \sqrt{7}\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{\sqrt{7}-3 i}} x\right ),\frac{-\sqrt{7}+3 i}{\sqrt{7}+3 i}\right )+105 \left (3-i \sqrt{7}\right ) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )+429 \Pi \left (\frac{5}{14} \left (3+i \sqrt{7}\right );i \sinh ^{-1}\left (\sqrt{-\frac{2 i}{-3 i+\sqrt{7}}} x\right )|\frac{3 i-\sqrt{7}}{3 i+\sqrt{7}}\right )\right )}{18375 \sqrt{x^4+3 x^2+4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((175*x*(23 + 7*x^2)*(4 + 3*x^2 + x^4))/(7 + 5*x^2) - I*Sqrt[6 + (2*I)*Sqrt[7]]*Sqrt[1 - ((2*I)*x^2)/(-3*I + S
qrt[7])]*Sqrt[1 + ((2*I)*x^2)/(3*I + Sqrt[7])]*(105*(3 - I*Sqrt[7])*EllipticE[I*ArcSinh[Sqrt[(-2*I)/(-3*I + Sq
rt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + 7*(158 + (15*I)*Sqrt[7])*EllipticF[I*ArcSinh[Sqrt[(-2*I)/(-3*I
+ Sqrt[7])]*x], (3*I - Sqrt[7])/(3*I + Sqrt[7])] + 429*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])]))/(18375*Sqrt[4 + 3*x^2 + x^4])

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Maple [C]  time = 0.023, size = 425, normalized size = 1.4 \begin{align*}{\frac{22\,x}{875\,{x}^{2}+1225}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{x}{75}\sqrt{{x}^{4}+3\,{x}^{2}+4}}+{\frac{232}{375\,\sqrt{-6+2\,i\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}-{\frac{128}{175\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticF} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}+{\frac{128}{175\,\sqrt{-6+2\,i\sqrt{7}} \left ( i\sqrt{7}+3 \right ) }\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticE} \left ({\frac{x\sqrt{-6+2\,i\sqrt{7}}}{4}},{\frac{\sqrt{2+6\,i\sqrt{7}}}{4}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}}+{\frac{286}{6125\,\sqrt{-3/8+i/8\sqrt{7}}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}-{\frac{i}{8}}{x}^{2}\sqrt{7}}\sqrt{1+{\frac{3\,{x}^{2}}{8}}+{\frac{i}{8}}{x}^{2}\sqrt{7}}{\it EllipticPi} \left ( \sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}x,-{\frac{5}{-{\frac{21}{8}}+{\frac{7\,i}{8}}\sqrt{7}}},{\frac{\sqrt{-{\frac{3}{8}}-{\frac{i}{8}}\sqrt{7}}}{\sqrt{-{\frac{3}{8}}+{\frac{i}{8}}\sqrt{7}}}} \right ){\frac{1}{\sqrt{{x}^{4}+3\,{x}^{2}+4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

22/175*x*(x^4+3*x^2+4)^(1/2)/(5*x^2+7)+1/75*x*(x^4+3*x^2+4)^(1/2)+232/375/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/
8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)*EllipticF(1/4*x*(-6+2*I*7^(1/2)
)^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))-128/175/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x
^2+1/8*I*x^2*7^(1/2))^(1/2)/(x^4+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*EllipticF(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*
I*7^(1/2))^(1/2))+128/175/(-6+2*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1
/2))^(1/2)/(x^4+3*x^2+4)^(1/2)/(I*7^(1/2)+3)*EllipticE(1/4*x*(-6+2*I*7^(1/2))^(1/2),1/4*(2+6*I*7^(1/2))^(1/2))
+286/6125/(-3/8+1/8*I*7^(1/2))^(1/2)*(1+3/8*x^2-1/8*I*x^2*7^(1/2))^(1/2)*(1+3/8*x^2+1/8*I*x^2*7^(1/2))^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{4} + 3 \, x^{2} + 4\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+3*x^2+4)^(3/2)/(5*x^2+7)^2,x, algorithm="maxima")

[Out]

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

[Out]

integral((x^4 + 3*x^2 + 4)^(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} - x + 2\right ) \left (x^{2} + x + 2\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+3*x**2+4)**(3/2)/(5*x**2+7)**2,x)

[Out]

Integral(((x**2 - x + 2)*(x**2 + x + 2))**(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} + 3 \, x^{2} + 4\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+3*x^2+4)^(3/2)/(5*x^2+7)^2,x, algorithm="giac")

[Out]

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