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

Optimal. Leaf size=222 \[ -\frac {3 \sqrt {x^4+3 x^2+2} x}{175 \left (5 x^2+7\right )}+\frac {1}{75} \sqrt {x^4+3 x^2+2} x+\frac {9 \left (x^2+2\right ) x}{175 \sqrt {x^4+3 x^2+2}}+\frac {59 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{2 x^2+2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{1050 \sqrt {x^4+3 x^2+2}}-\frac {9 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{175 \sqrt {x^4+3 x^2+2}}+\frac {9 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{2 x^2+2}} \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{2450 \sqrt {x^4+3 x^2+2}} \]

[Out]

9/175*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-9/175*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(1/2),1/2*2^(1/2
))*2^(1/2)*((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)+59/1050*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*EllipticF(x/(x^
2+1)^(1/2),1/2*2^(1/2))*((x^2+2)/(2*x^2+2))^(1/2)/(x^4+3*x^2+2)^(1/2)+9/2450*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*E
llipticPi(x/(x^2+1)^(1/2),2/7,1/2*2^(1/2))*((x^2+2)/(2*x^2+2))^(1/2)/(x^4+3*x^2+2)^(1/2)+1/75*x*(x^4+3*x^2+2)^
(1/2)-3/175*x*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)

________________________________________________________________________________________

Rubi [A]  time = 0.44, antiderivative size = 333, normalized size of antiderivative = 1.50, number of steps used = 21, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {1228, 1099, 1135, 1122, 1189, 1223, 1716, 1214, 1456, 539} \[ -\frac {3 \sqrt {x^4+3 x^2+2} x}{175 \left (5 x^2+7\right )}+\frac {1}{75} \sqrt {x^4+3 x^2+2} x+\frac {9 \left (x^2+2\right ) x}{175 \sqrt {x^4+3 x^2+2}}+\frac {44 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{1875 \sqrt {x^4+3 x^2+2}}+\frac {81 \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{8750 \sqrt {2} \sqrt {x^4+3 x^2+2}}-\frac {9 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{175 \sqrt {x^4+3 x^2+2}}+\frac {3 \sqrt {2} \left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}-\frac {39 \left (x^2+2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{12250 \sqrt {2} \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*x*(2 + x^2))/(175*Sqrt[2 + 3*x^2 + x^4]) + (x*Sqrt[2 + 3*x^2 + x^4])/75 - (3*x*Sqrt[2 + 3*x^2 + x^4])/(175*
(7 + 5*x^2)) - (9*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(175*Sqrt[2 + 3*x^2 +
x^4]) + (81*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(8750*Sqrt[2]*Sqrt[2 + 3*x^2 + x^4
]) + (44*Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(1875*Sqrt[2 + 3*x^2 + x^4]) -
(39*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(12250*Sqrt[2]*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4]
) + (3*Sqrt[2]*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(875*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4]
)

Rule 539

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(c*Sqrt[e +
f*x^2]*EllipticPi[1 - (b*c)/(a*d), ArcTan[Rt[d/c, 2]*x], 1 - (c*f)/(d*e)])/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sq
rt[(c*(e + f*x^2))/(e*(c + d*x^2))]), x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 1099

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b +
q)*x^2)*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)]
)/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 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 1135

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b +
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], (2*q)/(b + q)])/(2*c*Sqrt[a + b*x^2
+ c*x^4]), x] /; PosQ[(b + q)/a] &&  !(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; Fre
eQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1189

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[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1214

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*c)/(2*c*d - e*(b - q)), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/(2*c*d - e*(b - q)), Int[
(b - q + 2*c*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), 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 1456

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPar
t[p]), Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c*x^n)/e)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p,
q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p]

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+3 x^2+x^4\right )^{3/2}}{\left (7+5 x^2\right )^2} \, dx &=\int \left (\frac {52}{625 \sqrt {2+3 x^2+x^4}}+\frac {16 x^2}{125 \sqrt {2+3 x^2+x^4}}+\frac {x^4}{25 \sqrt {2+3 x^2+x^4}}+\frac {36}{625 \left (7+5 x^2\right )^2 \sqrt {2+3 x^2+x^4}}-\frac {12}{625 \left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}}\right ) \, dx\\ &=-\left (\frac {12}{625} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\right )+\frac {1}{25} \int \frac {x^4}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {36}{625} \int \frac {1}{\left (7+5 x^2\right )^2 \sqrt {2+3 x^2+x^4}} \, dx+\frac {52}{625} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {16}{125} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {16 x \left (2+x^2\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \sqrt {2+3 x^2+x^4}-\frac {3 x \sqrt {2+3 x^2+x^4}}{175 \left (7+5 x^2\right )}-\frac {16 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {26 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{625 \sqrt {2+3 x^2+x^4}}+\frac {3 \int \frac {62+70 x^2+25 x^4}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx}{4375}-\frac {6}{625} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {1}{75} \int \frac {2+6 x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {3}{125} \int \frac {2+2 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx\\ &=\frac {16 x \left (2+x^2\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \sqrt {2+3 x^2+x^4}-\frac {3 x \sqrt {2+3 x^2+x^4}}{175 \left (7+5 x^2\right )}-\frac {16 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {23 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{625 \sqrt {2+3 x^2+x^4}}-\frac {3 \int \frac {-175-125 x^2}{\sqrt {2+3 x^2+x^4}} \, dx}{109375}+\frac {39 \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx}{4375}-\frac {2}{75} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {2}{25} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {\left (3 \sqrt {1+\frac {x^2}{2}} \sqrt {2+2 x^2}\right ) \int \frac {\sqrt {2+2 x^2}}{\sqrt {1+\frac {x^2}{2}} \left (7+5 x^2\right )} \, dx}{125 \sqrt {2+3 x^2+x^4}}\\ &=\frac {6 x \left (2+x^2\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \sqrt {2+3 x^2+x^4}-\frac {3 x \sqrt {2+3 x^2+x^4}}{175 \left (7+5 x^2\right )}-\frac {6 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {44 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{1875 \sqrt {2+3 x^2+x^4}}+\frac {3 \sqrt {2} \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{875 \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}}+\frac {3}{875} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {39 \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx}{8750}+\frac {3}{625} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {39 \int \frac {2+2 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx}{3500}\\ &=\frac {9 x \left (2+x^2\right )}{175 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \sqrt {2+3 x^2+x^4}-\frac {3 x \sqrt {2+3 x^2+x^4}}{175 \left (7+5 x^2\right )}-\frac {9 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{175 \sqrt {2+3 x^2+x^4}}+\frac {81 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{8750 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {44 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{1875 \sqrt {2+3 x^2+x^4}}+\frac {3 \sqrt {2} \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{875 \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}}-\frac {\left (39 \sqrt {1+\frac {x^2}{2}} \sqrt {2+2 x^2}\right ) \int \frac {\sqrt {2+2 x^2}}{\sqrt {1+\frac {x^2}{2}} \left (7+5 x^2\right )} \, dx}{3500 \sqrt {2+3 x^2+x^4}}\\ &=\frac {9 x \left (2+x^2\right )}{175 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \sqrt {2+3 x^2+x^4}-\frac {3 x \sqrt {2+3 x^2+x^4}}{175 \left (7+5 x^2\right )}-\frac {9 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{175 \sqrt {2+3 x^2+x^4}}+\frac {81 \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{8750 \sqrt {2} \sqrt {2+3 x^2+x^4}}+\frac {44 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{1875 \sqrt {2+3 x^2+x^4}}-\frac {39 \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{12250 \sqrt {2} \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}}+\frac {3 \sqrt {2} \left (2+x^2\right ) \Pi \left (\frac {2}{7};\tan ^{-1}(x)|\frac {1}{2}\right )}{875 \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.29, size = 213, normalized size = 0.96 \[ \frac {1225 x^7+5075 x^5+6650 x^3-182 i \sqrt {x^2+1} \sqrt {x^2+2} \left (5 x^2+7\right ) F\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-945 i \sqrt {x^2+1} \sqrt {x^2+2} \left (5 x^2+7\right ) E\left (\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+135 i \sqrt {x^2+1} \sqrt {x^2+2} x^2 \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+189 i \sqrt {x^2+1} \sqrt {x^2+2} \Pi \left (\frac {10}{7};\left .i \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )+2800 x}{18375 \left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2800*x + 6650*x^3 + 5075*x^5 + 1225*x^7 - (945*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(7 + 5*x^2)*EllipticE[I*ArcSinh
[x/Sqrt[2]], 2] - (182*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*(7 + 5*x^2)*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + (189*I)
*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2] + (135*I)*x^2*Sqrt[1 + x^2]*Sqrt[2 + x^
2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2])/(18375*(7 + 5*x^2)*Sqrt[2 + 3*x^2 + x^4])

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (x^{4} + 3 \, x^{2} + 2\right )}^{\frac {3}{2}}}{25 \, x^{4} + 70 \, x^{2} + 49}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.02, size = 177, normalized size = 0.80 \[ -\frac {3 \sqrt {x^{4}+3 x^{2}+2}\, x}{175 \left (5 x^{2}+7\right )}+\frac {\sqrt {x^{4}+3 x^{2}+2}\, x}{75}-\frac {9 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticE \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{350 \sqrt {x^{4}+3 x^{2}+2}}-\frac {13 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \EllipticF \left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{2625 \sqrt {x^{4}+3 x^{2}+2}}+\frac {9 i \sqrt {2}\, \sqrt {\frac {x^{2}}{2}+1}\, \sqrt {x^{2}+1}\, \EllipticPi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{6125 \sqrt {x^{4}+3 x^{2}+2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/175*(x^4+3*x^2+2)^(1/2)/(5*x^2+7)*x+1/75*(x^4+3*x^2+2)^(1/2)*x-13/2625*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1
/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2*I*2^(1/2)*x,2^(1/2))-9/350*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+
3*x^2+2)^(1/2)*EllipticE(1/2*I*2^(1/2)*x,2^(1/2))+9/6125*I*2^(1/2)*(1/2*x^2+1)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+
2)^(1/2)*EllipticPi(1/2*I*2^(1/2)*x,10/7,2^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________