3.21.65 \(\int \frac {x^2}{\sqrt {b x+a x^3} (-b^2+a^2 x^4)} \, dx\) [2065]

3.21.65.1 Optimal result
3.21.65.2 Mathematica [A] (verified)
3.21.65.3 Rubi [A] (verified)
3.21.65.4 Maple [A] (verified)
3.21.65.5 Fricas [C] (verification not implemented)
3.21.65.6 Sympy [F]
3.21.65.7 Maxima [F]
3.21.65.8 Giac [F]
3.21.65.9 Mupad [F(-1)]

3.21.65.1 Optimal result

Integrand size = 32, antiderivative size = 149 \[ \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx=\frac {\sqrt {b x+a x^3}}{2 a b \left (b+a x^2\right )}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {b x+a x^3}}{b+a x^2}\right )}{4 \sqrt {2} a^{5/4} b^{5/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {b x+a x^3}}{b+a x^2}\right )}{4 \sqrt {2} a^{5/4} b^{5/4}} \]

output
1/2*(a*x^3+b*x)^(1/2)/a/b/(a*x^2+b)-1/8*arctan(2^(1/2)*a^(1/4)*b^(1/4)*(a* 
x^3+b*x)^(1/2)/(a*x^2+b))*2^(1/2)/a^(5/4)/b^(5/4)-1/8*arctanh(2^(1/2)*a^(1 
/4)*b^(1/4)*(a*x^3+b*x)^(1/2)/(a*x^2+b))*2^(1/2)/a^(5/4)/b^(5/4)
 
3.21.65.2 Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.02 \[ \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx=\frac {\sqrt {x} \left (4 \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}-\sqrt {2} \sqrt {b+a x^2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {b+a x^2}}\right )-\sqrt {2} \sqrt {b+a x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {b+a x^2}}\right )\right )}{8 a^{5/4} b^{5/4} \sqrt {x \left (b+a x^2\right )}} \]

input
Integrate[x^2/(Sqrt[b*x + a*x^3]*(-b^2 + a^2*x^4)),x]
 
output
(Sqrt[x]*(4*a^(1/4)*b^(1/4)*Sqrt[x] - Sqrt[2]*Sqrt[b + a*x^2]*ArcTan[(Sqrt 
[2]*a^(1/4)*b^(1/4)*Sqrt[x])/Sqrt[b + a*x^2]] - Sqrt[2]*Sqrt[b + a*x^2]*Ar 
cTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/Sqrt[b + a*x^2]]))/(8*a^(5/4)*b^(5 
/4)*Sqrt[x*(b + a*x^2)])
 
3.21.65.3 Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {2467, 25, 1388, 368, 971, 920, 756, 216, 219}

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.

\(\displaystyle \int \frac {x^2}{\sqrt {a x^3+b x} \left (a^2 x^4-b^2\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt {x} \sqrt {a x^2+b} \int -\frac {x^{3/2}}{\sqrt {a x^2+b} \left (b^2-a^2 x^4\right )}dx}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {x} \sqrt {a x^2+b} \int \frac {x^{3/2}}{\sqrt {a x^2+b} \left (b^2-a^2 x^4\right )}dx}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 1388

\(\displaystyle -\frac {\sqrt {x} \sqrt {a x^2+b} \int \frac {x^{3/2}}{\left (b-a x^2\right ) \left (a x^2+b\right )^{3/2}}dx}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 368

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {a x^2+b} \int \frac {x^2}{\left (b-a x^2\right ) \left (a x^2+b\right )^{3/2}}d\sqrt {x}}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 971

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {a x^2+b} \left (\frac {\int \frac {\sqrt {a x^2+b}}{b-a x^2}d\sqrt {x}}{4 a b}-\frac {\sqrt {x}}{4 a b \sqrt {a x^2+b}}\right )}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 920

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {a x^2+b} \left (\frac {\int \frac {1}{1-4 a b x^2}d\frac {\sqrt {x}}{\sqrt {a x^2+b}}}{4 a b}-\frac {\sqrt {x}}{4 a b \sqrt {a x^2+b}}\right )}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 756

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {a x^2+b} \left (\frac {\frac {1}{2} \int \frac {1}{1-2 \sqrt {a} \sqrt {b} x}d\frac {\sqrt {x}}{\sqrt {a x^2+b}}+\frac {1}{2} \int \frac {1}{2 \sqrt {a} \sqrt {b} x+1}d\frac {\sqrt {x}}{\sqrt {a x^2+b}}}{4 a b}-\frac {\sqrt {x}}{4 a b \sqrt {a x^2+b}}\right )}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {a x^2+b} \left (\frac {\frac {1}{2} \int \frac {1}{1-2 \sqrt {a} \sqrt {b} x}d\frac {\sqrt {x}}{\sqrt {a x^2+b}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a x^2+b}}\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{4 a b}-\frac {\sqrt {x}}{4 a b \sqrt {a x^2+b}}\right )}{\sqrt {a x^3+b x}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 \sqrt {x} \sqrt {a x^2+b} \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a x^2+b}}\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a x^2+b}}\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{4 a b}-\frac {\sqrt {x}}{4 a b \sqrt {a x^2+b}}\right )}{\sqrt {a x^3+b x}}\)

input
Int[x^2/(Sqrt[b*x + a*x^3]*(-b^2 + a^2*x^4)),x]
 
output
(-2*Sqrt[x]*Sqrt[b + a*x^2]*(-1/4*Sqrt[x]/(a*b*Sqrt[b + a*x^2]) + (ArcTan[ 
(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/Sqrt[b + a*x^2]]/(2*Sqrt[2]*a^(1/4)*b^(1 
/4)) + ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/Sqrt[b + a*x^2]]/(2*Sqrt[ 
2]*a^(1/4)*b^(1/4)))/(4*a*b)))/Sqrt[b*x + a*x^3]
 

3.21.65.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 368
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], 
 x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m 
] && IntegerQ[p]
 

rule 756
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 920
Int[Sqrt[(a_) + (b_.)*(x_)^4]/((c_) + (d_.)*(x_)^4), x_Symbol] :> Simp[a/c 
  Subst[Int[1/(1 - 4*a*b*x^4), x], x, x/Sqrt[a + b*x^4]], x] /; FreeQ[{a, b 
, c, d}, x] && EqQ[b*c + a*d, 0] && PosQ[a*b]
 

rule 971
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* 
((c + d*x^n)^(q + 1)/(n*(b*c - a*d)*(p + 1))), x] - Simp[e^n/(n*(b*c - a*d) 
*(p + 1))   Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - 
 n + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e 
, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n, m - n + 
 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
 

rule 1388
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

rule 2467
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 
3.21.65.4 Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.96

method result size
default \(-\frac {\ln \left (\frac {-\sqrt {2}\, \left (a b \right )^{\frac {1}{4}} x -\sqrt {\left (a \,x^{2}+b \right ) x}}{\sqrt {2}\, \left (a b \right )^{\frac {1}{4}} x -\sqrt {\left (a \,x^{2}+b \right ) x}}\right ) \sqrt {2}\, \sqrt {\left (a \,x^{2}+b \right ) x}-2 \arctan \left (\frac {\sqrt {\left (a \,x^{2}+b \right ) x}\, \sqrt {2}}{2 x \left (a b \right )^{\frac {1}{4}}}\right ) \sqrt {2}\, \sqrt {\left (a \,x^{2}+b \right ) x}-8 x \left (a b \right )^{\frac {1}{4}}}{16 \left (a b \right )^{\frac {1}{4}} \sqrt {\left (a \,x^{2}+b \right ) x}\, b a}\) \(143\)
pseudoelliptic \(-\frac {\ln \left (\frac {-\sqrt {2}\, \left (a b \right )^{\frac {1}{4}} x -\sqrt {\left (a \,x^{2}+b \right ) x}}{\sqrt {2}\, \left (a b \right )^{\frac {1}{4}} x -\sqrt {\left (a \,x^{2}+b \right ) x}}\right ) \sqrt {2}\, \sqrt {\left (a \,x^{2}+b \right ) x}-2 \arctan \left (\frac {\sqrt {\left (a \,x^{2}+b \right ) x}\, \sqrt {2}}{2 x \left (a b \right )^{\frac {1}{4}}}\right ) \sqrt {2}\, \sqrt {\left (a \,x^{2}+b \right ) x}-8 x \left (a b \right )^{\frac {1}{4}}}{16 \left (a b \right )^{\frac {1}{4}} \sqrt {\left (a \,x^{2}+b \right ) x}\, b a}\) \(143\)
elliptic \(\frac {x}{2 a b \sqrt {\left (x^{2}+\frac {b}{a}\right ) a x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{4 a^{2} b \sqrt {a \,x^{3}+b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{4 a^{2} \sqrt {a b}\, \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}-\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{4 a^{2} \sqrt {a b}\, \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}\) \(421\)

input
int(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x,method=_RETURNVERBOSE)
 
output
-1/16*(ln((-2^(1/2)*(a*b)^(1/4)*x-((a*x^2+b)*x)^(1/2))/(2^(1/2)*(a*b)^(1/4 
)*x-((a*x^2+b)*x)^(1/2)))*2^(1/2)*((a*x^2+b)*x)^(1/2)-2*arctan(1/2*((a*x^2 
+b)*x)^(1/2)/x*2^(1/2)/(a*b)^(1/4))*2^(1/2)*((a*x^2+b)*x)^(1/2)-8*x*(a*b)^ 
(1/4))/(a*b)^(1/4)/((a*x^2+b)*x)^(1/2)/b/a
 
3.21.65.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 693, normalized size of antiderivative = 4.65 \[ \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx=-\frac {\left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (a^{2} b x^{2} + a b^{2}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {a^{2} x^{4} + 6 \, a b x^{2} + b^{2} + 8 \, {\left (\left (\frac {1}{4}\right )^{\frac {1}{4}} a^{2} b^{2} x \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} + \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (a^{5} b^{4} x^{2} + a^{4} b^{5}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {3}{4}}\right )} \sqrt {a x^{3} + b x} + 4 \, {\left (a^{4} b^{3} x^{3} + a^{3} b^{4} x\right )} \sqrt {\frac {1}{a^{5} b^{5}}}}{a^{2} x^{4} - 2 \, a b x^{2} + b^{2}}\right ) - \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (a^{2} b x^{2} + a b^{2}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {a^{2} x^{4} + 6 \, a b x^{2} + b^{2} - 8 \, {\left (\left (\frac {1}{4}\right )^{\frac {1}{4}} a^{2} b^{2} x \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} + \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (a^{5} b^{4} x^{2} + a^{4} b^{5}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {3}{4}}\right )} \sqrt {a x^{3} + b x} + 4 \, {\left (a^{4} b^{3} x^{3} + a^{3} b^{4} x\right )} \sqrt {\frac {1}{a^{5} b^{5}}}}{a^{2} x^{4} - 2 \, a b x^{2} + b^{2}}\right ) - \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (i \, a^{2} b x^{2} + i \, a b^{2}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {a^{2} x^{4} + 6 \, a b x^{2} + b^{2} - 8 \, {\left (i \, \left (\frac {1}{4}\right )^{\frac {1}{4}} a^{2} b^{2} x \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} + \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (-i \, a^{5} b^{4} x^{2} - i \, a^{4} b^{5}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {3}{4}}\right )} \sqrt {a x^{3} + b x} - 4 \, {\left (a^{4} b^{3} x^{3} + a^{3} b^{4} x\right )} \sqrt {\frac {1}{a^{5} b^{5}}}}{a^{2} x^{4} - 2 \, a b x^{2} + b^{2}}\right ) - \left (\frac {1}{4}\right )^{\frac {1}{4}} {\left (-i \, a^{2} b x^{2} - i \, a b^{2}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {a^{2} x^{4} + 6 \, a b x^{2} + b^{2} - 8 \, {\left (-i \, \left (\frac {1}{4}\right )^{\frac {1}{4}} a^{2} b^{2} x \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {1}{4}} + \left (\frac {1}{4}\right )^{\frac {3}{4}} {\left (i \, a^{5} b^{4} x^{2} + i \, a^{4} b^{5}\right )} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {3}{4}}\right )} \sqrt {a x^{3} + b x} - 4 \, {\left (a^{4} b^{3} x^{3} + a^{3} b^{4} x\right )} \sqrt {\frac {1}{a^{5} b^{5}}}}{a^{2} x^{4} - 2 \, a b x^{2} + b^{2}}\right ) - 8 \, \sqrt {a x^{3} + b x}}{16 \, {\left (a^{2} b x^{2} + a b^{2}\right )}} \]

input
integrate(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x, algorithm="fricas")
 
output
-1/16*((1/4)^(1/4)*(a^2*b*x^2 + a*b^2)*(1/(a^5*b^5))^(1/4)*log((a^2*x^4 + 
6*a*b*x^2 + b^2 + 8*((1/4)^(1/4)*a^2*b^2*x*(1/(a^5*b^5))^(1/4) + (1/4)^(3/ 
4)*(a^5*b^4*x^2 + a^4*b^5)*(1/(a^5*b^5))^(3/4))*sqrt(a*x^3 + b*x) + 4*(a^4 
*b^3*x^3 + a^3*b^4*x)*sqrt(1/(a^5*b^5)))/(a^2*x^4 - 2*a*b*x^2 + b^2)) - (1 
/4)^(1/4)*(a^2*b*x^2 + a*b^2)*(1/(a^5*b^5))^(1/4)*log((a^2*x^4 + 6*a*b*x^2 
 + b^2 - 8*((1/4)^(1/4)*a^2*b^2*x*(1/(a^5*b^5))^(1/4) + (1/4)^(3/4)*(a^5*b 
^4*x^2 + a^4*b^5)*(1/(a^5*b^5))^(3/4))*sqrt(a*x^3 + b*x) + 4*(a^4*b^3*x^3 
+ a^3*b^4*x)*sqrt(1/(a^5*b^5)))/(a^2*x^4 - 2*a*b*x^2 + b^2)) - (1/4)^(1/4) 
*(I*a^2*b*x^2 + I*a*b^2)*(1/(a^5*b^5))^(1/4)*log((a^2*x^4 + 6*a*b*x^2 + b^ 
2 - 8*(I*(1/4)^(1/4)*a^2*b^2*x*(1/(a^5*b^5))^(1/4) + (1/4)^(3/4)*(-I*a^5*b 
^4*x^2 - I*a^4*b^5)*(1/(a^5*b^5))^(3/4))*sqrt(a*x^3 + b*x) - 4*(a^4*b^3*x^ 
3 + a^3*b^4*x)*sqrt(1/(a^5*b^5)))/(a^2*x^4 - 2*a*b*x^2 + b^2)) - (1/4)^(1/ 
4)*(-I*a^2*b*x^2 - I*a*b^2)*(1/(a^5*b^5))^(1/4)*log((a^2*x^4 + 6*a*b*x^2 + 
 b^2 - 8*(-I*(1/4)^(1/4)*a^2*b^2*x*(1/(a^5*b^5))^(1/4) + (1/4)^(3/4)*(I*a^ 
5*b^4*x^2 + I*a^4*b^5)*(1/(a^5*b^5))^(3/4))*sqrt(a*x^3 + b*x) - 4*(a^4*b^3 
*x^3 + a^3*b^4*x)*sqrt(1/(a^5*b^5)))/(a^2*x^4 - 2*a*b*x^2 + b^2)) - 8*sqrt 
(a*x^3 + b*x))/(a^2*b*x^2 + a*b^2)
 
3.21.65.6 Sympy [F]

\[ \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx=\int \frac {x^{2}}{\sqrt {x \left (a x^{2} + b\right )} \left (a x^{2} - b\right ) \left (a x^{2} + b\right )}\, dx \]

input
integrate(x**2/(a*x**3+b*x)**(1/2)/(a**2*x**4-b**2),x)
 
output
Integral(x**2/(sqrt(x*(a*x**2 + b))*(a*x**2 - b)*(a*x**2 + b)), x)
 
3.21.65.7 Maxima [F]

\[ \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (a^{2} x^{4} - b^{2}\right )} \sqrt {a x^{3} + b x}} \,d x } \]

input
integrate(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x, algorithm="maxima")
 
output
integrate(x^2/((a^2*x^4 - b^2)*sqrt(a*x^3 + b*x)), x)
 
3.21.65.8 Giac [F]

\[ \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (a^{2} x^{4} - b^{2}\right )} \sqrt {a x^{3} + b x}} \,d x } \]

input
integrate(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x, algorithm="giac")
 
output
integrate(x^2/((a^2*x^4 - b^2)*sqrt(a*x^3 + b*x)), x)
 
3.21.65.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx=\text {Hanged} \]

input
int(-x^2/((b^2 - a^2*x^4)*(b*x + a*x^3)^(1/2)),x)
 
output
\text{Hanged}