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

3.3.97.1 Optimal result

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

output

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)*Elli 
pticE(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)
 

3.3.97.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

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

input

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

output

(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 
])
 

3.3.97.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {1529, 27, 1786, 414, 2207, 27, 2207, 27, 1503, 1412, 1455}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1529

\(\displaystyle \frac {1}{500} \int \frac {4 \left (25 x^6+115 x^4+164 x^2+74\right )}{\sqrt {x^4+3 x^2+2}}dx-\frac {9}{125} \int \frac {2 \left (x^2+1\right )}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{125} \int \frac {25 x^6+115 x^4+164 x^2+74}{\sqrt {x^4+3 x^2+2}}dx-\frac {18}{125} \int \frac {x^2+1}{\left (5 x^2+7\right ) \sqrt {x^4+3 x^2+2}}dx\)

\(\Big \downarrow \) 1786

\(\displaystyle \frac {1}{125} \int \frac {25 x^6+115 x^4+164 x^2+74}{\sqrt {x^4+3 x^2+2}}dx-\frac {18 \sqrt {x^2+1} \sqrt {x^2+2} \int \frac {\sqrt {x^2+1}}{\sqrt {x^2+2} \left (5 x^2+7\right )}dx}{125 \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 414

\(\displaystyle \frac {1}{125} \int \frac {25 x^6+115 x^4+164 x^2+74}{\sqrt {x^4+3 x^2+2}}dx-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{125} \left (\frac {1}{5} \int \frac {5 \left (55 x^4+134 x^2+74\right )}{\sqrt {x^4+3 x^2+2}}dx+5 \sqrt {x^4+3 x^2+2} x^3\right )-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{125} \left (\int \frac {55 x^4+134 x^2+74}{\sqrt {x^4+3 x^2+2}}dx+5 \sqrt {x^4+3 x^2+2} x^3\right )-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 2207

\(\displaystyle \frac {1}{125} \left (\frac {1}{3} \int \frac {8 \left (9 x^2+14\right )}{\sqrt {x^4+3 x^2+2}}dx+\frac {55}{3} \sqrt {x^4+3 x^2+2} x+5 \sqrt {x^4+3 x^2+2} x^3\right )-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{125} \left (\frac {8}{3} \int \frac {9 x^2+14}{\sqrt {x^4+3 x^2+2}}dx+\frac {55}{3} \sqrt {x^4+3 x^2+2} x+5 \sqrt {x^4+3 x^2+2} x^3\right )-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{125} \left (\frac {8}{3} \left (14 \int \frac {1}{\sqrt {x^4+3 x^2+2}}dx+9 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx\right )+\frac {55}{3} \sqrt {x^4+3 x^2+2} x+5 \sqrt {x^4+3 x^2+2} x^3\right )-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1412

\(\displaystyle \frac {1}{125} \left (\frac {8}{3} \left (9 \int \frac {x^2}{\sqrt {x^4+3 x^2+2}}dx+\frac {7 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}}\right )+\frac {55}{3} \sqrt {x^4+3 x^2+2} x+5 \sqrt {x^4+3 x^2+2} x^3\right )-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1455

\(\displaystyle \frac {1}{125} \left (\frac {8}{3} \left (\frac {7 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^4+3 x^2+2}}+9 \left (\frac {x \left (x^2+2\right )}{\sqrt {x^4+3 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {x^2+2}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+3 x^2+2}}\right )\right )+\frac {55}{3} \sqrt {x^4+3 x^2+2} x+5 \sqrt {x^4+3 x^2+2} x^3\right )-\frac {9 \sqrt {2} \left (x^2+2\right ) \operatorname {EllipticPi}\left (\frac {2}{7},\arctan (x),\frac {1}{2}\right )}{875 \sqrt {\frac {x^2+2}{x^2+1}} \sqrt {x^4+3 x^2+2}}\)

input

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

output

((55*x*Sqrt[2 + 3*x^2 + x^4])/3 + 5*x^3*Sqrt[2 + 3*x^2 + x^4] + (8*(9*((x* 
(2 + x^2))/Sqrt[2 + 3*x^2 + x^4] - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + x^2)/(1 + 
x^2)]*EllipticE[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + x^4]) + (7*Sqrt[2]*(1 + 
x^2)*Sqrt[(2 + x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/2])/Sqrt[2 + 3*x^2 + 
 x^4]))/3)/125 - (9*Sqrt[2]*(2 + x^2)*EllipticPi[2/7, ArcTan[x], 1/2])/(87 
5*Sqrt[(2 + x^2)/(1 + x^2)]*Sqrt[2 + 3*x^2 + x^4])
 

3.3.97.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 414

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)), ArcTan[ 
Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ 
[d/c]
 

rule 1412

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]))*EllipticF 
[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)])] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1455

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)])] /; FreeQ[{a, b, 
c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1503

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d   Int[1/Sqrt[a + b*x^2 + c*x^4] 
, x], x] + Simp[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 1529

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symb 
ol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(2*c*d - e*(b + q)))*((c*d^2 - 
 b*d*e + a*e^2)^(p - 1/2)/(4*c*e^(2*p)))   Int[(b - q + 2*c*x^2)/((d + e*x^ 
2)*Sqrt[a + b*x^2 + c*x^4]), x], x] + Simp[1/(4*c*e^(2*p))   Int[(1/Sqrt[a 
+ b*x^2 + c*x^4])*ExpandToSum[(4*c*e^(2*p)*(a + b*x^2 + c*x^4)^(p + 1/2) + 
(2*c*d - e*(b + q))*(c*d^2 - b*d*e + a*e^2)^(p - 1/2)*(b - q + 2*c*x^2))/(d 
 + e*x^2), x], 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] && PosQ[b^2 - 4*a*c] & 
& PosQ[c/a]
 

rule 1786

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + ( 
b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[(a + b*x^n + c*x 
^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPart[p]) 
   Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c/e)*x^n)^p, x], x] /; Fre 
eQ[{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 2207

Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{n = 
 Expon[Px, x^2], e = Coeff[Px, x^2, Expon[Px, x^2]]}, Simp[e*x^(2*n - 3)*(( 
a + b*x^2 + c*x^4)^(p + 1)/(c*(2*n + 4*p + 1))), x] + Simp[1/(c*(2*n + 4*p 
+ 1))   Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*n + 4*p + 1)*Px - a*e*(2 
*n - 3)*x^(2*n - 4) - b*e*(2*n + 2*p - 1)*x^(2*n - 2) - c*e*(2*n + 4*p + 1) 
*x^(2*n), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Px, x^2] && Expon[ 
Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]
 

3.3.97.4 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	\(\frac {x \left (3 x^{2}+11\right ) \sqrt {x^{4}+3 x^{2}+2}}{75}-\frac {253 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )}{1875 \sqrt {x^{4}+3 x^{2}+2}}+\frac {12 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \left (F\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )-E\left (\frac {i \sqrt {2}\, x}{2}, \sqrt {2}\right )\right )}{125 \sqrt {x^{4}+3 x^{2}+2}}-\frac {36 i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i \sqrt {2}\, x}{2}, \frac {10}{7}, \sqrt {2}\right )}{4375 \sqrt {x^{4}+3 x^{2}+2}}\)	\(174\)

input

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

output

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 
))
 

3.3.97.5 Fricas [F]

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

input

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

output

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

3.3.97.6 Sympy [F]

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

input

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

output

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

3.3.97.7 Maxima [F]

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

input

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

output

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

3.3.97.8 Giac [F]

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

input

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

output

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

3.3.97.9 Mupad [F(-1)]

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

input

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

output

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