\(\int \frac {(2+3 x^2+x^4)^{3/2}}{7+5 x^2} \, dx\) [297]

Optimal result

Integrand size = 24, antiderivative size = 207 \[ \int \frac {\left (2+3 x^2+x^4\right )^{3/2}}{7+5 x^2} \, dx=\frac {24 x \left (2+x^2\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \left (11+3 x^2\right ) \sqrt {2+3 x^2+x^4}-\frac {24 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {56 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{375 \sqrt {2+3 x^2+x^4}}-\frac {9 \sqrt {2} \left (2+x^2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {2+x^2}{1+x^2}} \sqrt {2+3 x^2+x^4}} \]

[Out]

24/125*x*(x^2+2)/(x^4+3*x^2+2)^(1/2)-9/875*(x^2+2)*(1/(x^2+1))^(1/2)*(x^2+1)^(1/2)*EllipticPi(x/(x^2+1)^(1/2),
2/7,1/2*2^(1/2))*2^(1/2)/((x^2+2)/(x^2+1))^(1/2)/(x^4+3*x^2+2)^(1/2)-24/125*(x^2+1)^(3/2)*(1/(x^2+1))^(1/2)*El
lipticE(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)+56/375*(x^2+1)^(3/2)*
(1/(x^2+1))^(1/2)*EllipticF(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)+1
/75*x*(3*x^2+11)*(x^4+3*x^2+2)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1222, 1190, 1203, 1113, 1149, 1228, 1470, 553} \[ \int \frac {\left (2+3 x^2+x^4\right )^{3/2}}{7+5 x^2} \, dx=\frac {56 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{375 \sqrt {x^4+3 x^2+2}}-\frac {24 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{125 \sqrt {x^4+3 x^2+2}}-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}+\frac {24 x \left (x^2+2\right )}{125 \sqrt {x^4+3 x^2+2}}+\frac {1}{75} x \left (3 x^2+11\right ) \sqrt {x^4+3 x^2+2} \]

[In]

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

[Out]

(24*x*(2 + x^2))/(125*Sqrt[2 + 3*x^2 + x^4]) + (x*(11 + 3*x^2)*Sqrt[2 + 3*x^2 + x^4])/75 - (24*Sqrt[2]*(1 + x^
2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticE[ArcTan[x], 1/2])/(125*Sqrt[2 + 3*x^2 + x^4]) + (56*Sqrt[2]*(1 + x^2)*Sq
rt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/(375*Sqrt[2 + 3*x^2 + x^4]) - (9*Sqrt[2]*(2 + x^2)*Elliptic
Pi[2/7, ArcTan[x], 1/2])/(875*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])

Rule 553

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

Rule 1113

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)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*Elli
pticF[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], 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 1149

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)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q
/(b + q))], 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 1190

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*(2*b*e*p + c*d*(4*p +
3) + c*e*(4*p + 1)*x^2)*((a + b*x^2 + c*x^4)^p/(c*(4*p + 1)*(4*p + 3))), x] + Dist[2*(p/(c*(4*p + 1)*(4*p + 3
))), Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) - b^2*e*(2*p + 1))*x^2, x]*(a +
b*x^2 + c*x^4)^(p - 1), 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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]

Rule 1203

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 1222

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[-(e^2)^(-1), Int[(c*d -
b*e - c*e*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] + Dist[(c*d^2 - b*d*e + a*e^2)/e^2, Int[(a + b*x^2 + c*x^4
)^(p - 1)/(d + e*x^2), 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] && IGtQ[p + 1/2, 0]

Rule 1228

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 1470

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/e)*x^n)^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]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{25} \int \left (-8-5 x^2\right ) \sqrt {2+3 x^2+x^4} \, dx\right )-\frac {6}{25} \int \frac {\sqrt {2+3 x^2+x^4}}{7+5 x^2} \, dx \\ & = \frac {1}{75} x \left (11+3 x^2\right ) \sqrt {2+3 x^2+x^4}-\frac {1}{375} \int \frac {-130-90 x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {6}{625} \int \frac {-8-5 x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {36}{625} \int \frac {1}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx \\ & = \frac {1}{75} x \left (11+3 x^2\right ) \sqrt {2+3 x^2+x^4}+\frac {18}{625} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {6}{125} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx-\frac {9}{125} \int \frac {2+2 x^2}{\left (7+5 x^2\right ) \sqrt {2+3 x^2+x^4}} \, dx-\frac {48}{625} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {6}{25} \int \frac {x^2}{\sqrt {2+3 x^2+x^4}} \, dx+\frac {26}{75} \int \frac {1}{\sqrt {2+3 x^2+x^4}} \, dx \\ & = \frac {24 x \left (2+x^2\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \left (11+3 x^2\right ) \sqrt {2+3 x^2+x^4}-\frac {24 \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 {56 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{375 \sqrt {2+3 x^2+x^4}}-\frac {\left (9 \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 {24 x \left (2+x^2\right )}{125 \sqrt {2+3 x^2+x^4}}+\frac {1}{75} x \left (11+3 x^2\right ) \sqrt {2+3 x^2+x^4}-\frac {24 \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 {56 \sqrt {2} \left (1+x^2\right ) \sqrt {\frac {2+x^2}{1+x^2}} F\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{375 \sqrt {2+3 x^2+x^4}}-\frac {9 \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] (verified)

Result contains complex when optimal does not.

Time = 10.24 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.71 \[ \int \frac {\left (2+3 x^2+x^4\right )^{3/2}}{7+5 x^2} \, dx=\frac {3850 x+6825 x^3+3500 x^5+525 x^7-2520 i \sqrt {1+x^2} \sqrt {2+x^2} E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-1022 i \sqrt {1+x^2} \sqrt {2+x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )-108 i \sqrt {1+x^2} \sqrt {2+x^2} \operatorname {EllipticPi}\left (\frac {10}{7},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{13125 \sqrt {2+3 x^2+x^4}} \]

[In]

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

[Out]

(3850*x + 6825*x^3 + 3500*x^5 + 525*x^7 - (2520*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticE[I*ArcSinh[x/Sqrt[2]],
2] - (1022*I)*Sqrt[1 + x^2]*Sqrt[2 + x^2]*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] - (108*I)*Sqrt[1 + x^2]*Sqrt[2 +
x^2]*EllipticPi[10/7, I*ArcSinh[x/Sqrt[2]], 2])/(13125*Sqrt[2 + 3*x^2 + x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.69 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {x^{3} \sqrt {x^{4}+3 x^{2}+2}}{25}+\frac {11 x \sqrt {x^{4}+3 x^{2}+2}}{75}-\frac {73 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{1875 \sqrt {x^{4}+3 x^{2}+2}}-\frac {12 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{125 \sqrt {x^{4}+3 x^{2}+2}}-\frac {36 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{4375 \sqrt {x^{4}+3 x^{2}+2}}\)	\(170\)
elliptic	\(\frac {x^{3} \sqrt {x^{4}+3 x^{2}+2}}{25}+\frac {11 x \sqrt {x^{4}+3 x^{2}+2}}{75}-\frac {73 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{1875 \sqrt {x^{4}+3 x^{2}+2}}-\frac {12 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{125 \sqrt {x^{4}+3 x^{2}+2}}-\frac {36 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{4375 \sqrt {x^{4}+3 x^{2}+2}}\)	\(170\)
risch	\(\frac {x \left (3 x^{2}+11\right ) \sqrt {x^{4}+3 x^{2}+2}}{75}-\frac {253 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{1875 \sqrt {x^{4}+3 x^{2}+2}}+\frac {12 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{125 \sqrt {x^{4}+3 x^{2}+2}}-\frac {36 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{4375 \sqrt {x^{4}+3 x^{2}+2}}\)	\(174\)

[In]

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

[Out]

1/25*x^3*(x^4+3*x^2+2)^(1/2)+11/75*x*(x^4+3*x^2+2)^(1/2)-73/1875*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))-12/125*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))-36/4375*I*2^(1/2)*(1+1/2*x^2)^(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))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (2+3 x^2+x^4\right )^{3/2}}{7+5 x^2} \, dx=\int \frac {\left (\left (x^{2} + 1\right ) \left (x^{2} + 2\right )\right )^{\frac {3}{2}}}{5 x^{2} + 7}\, dx \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (2+3 x^2+x^4\right )^{3/2}}{7+5 x^2} \, dx=\int \frac {{\left (x^4+3\,x^2+2\right )}^{3/2}}{5\,x^2+7} \,d x \]

[In]

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

[Out]

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