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

Optimal. Leaf size=149 \[ \frac {\sqrt {b x+a x^3}}{2 a b \left (b+a x^2\right )}-\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {b x+a x^3}}{b+a x^2}\right )}{4 \sqrt {2} a^{5/4} b^{5/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {b x+a x^3}}{b+a x^2}\right )}{4 \sqrt {2} a^{5/4} b^{5/4}} \]

[Out]

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)

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Rubi [A]
time = 0.22, antiderivative size = 187, normalized size of antiderivative = 1.26, number of steps used = 9, number of rules used = 9, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {2081, 1268, 477, 482, 21, 413, 218, 212, 209} \begin {gather*} -\frac {\sqrt {x} \sqrt {a x^2+b} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a x^2+b}}\right )}{4 \sqrt {2} a^{5/4} b^{5/4} \sqrt {a x^3+b x}}-\frac {\sqrt {x} \sqrt {a x^2+b} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a x^2+b}}\right )}{4 \sqrt {2} a^{5/4} b^{5/4} \sqrt {a x^3+b x}}+\frac {x}{2 a b \sqrt {a x^3+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(Sqrt[b*x + a*x^3]*(-b^2 + a^2*x^4)),x]

[Out]

x/(2*a*b*Sqrt[b*x + a*x^3]) - (Sqrt[x]*Sqrt[b + a*x^2]*ArcTan[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/Sqrt[b + a*x^2
]])/(4*Sqrt[2]*a^(5/4)*b^(5/4)*Sqrt[b*x + a*x^3]) - (Sqrt[x]*Sqrt[b + a*x^2]*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*
Sqrt[x])/Sqrt[b + a*x^2]])/(4*Sqrt[2]*a^(5/4)*b^(5/4)*Sqrt[b*x + a*x^3])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 209

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

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

Rule 413

Int[Sqrt[(a_) + (b_.)*(x_)^4]/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[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 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 482

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] - Dist[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 1268

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(f*x)^m*(d +
e*x^2)^(q + p)*(a/d + (c/e)*x^2)^p, x] /; FreeQ[{a, c, d, e, f, q, m, q}, x] && EqQ[c*d^2 + a*e^2, 0] && Integ
erQ[p]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {b x+a x^3} \left (-b^2+a^2 x^4\right )} \, dx &=\frac {\left (\sqrt {x} \sqrt {b+a x^2}\right ) \int \frac {x^{3/2}}{\sqrt {b+a x^2} \left (-b^2+a^2 x^4\right )} \, dx}{\sqrt {b x+a x^3}}\\ &=\frac {\left (\sqrt {x} \sqrt {b+a x^2}\right ) \int \frac {x^{3/2}}{\left (-b+a x^2\right ) \left (b+a x^2\right )^{3/2}} \, dx}{\sqrt {b x+a x^3}}\\ &=\frac {\left (2 \sqrt {x} \sqrt {b+a x^2}\right ) \text {Subst}\left (\int \frac {x^4}{\left (-b+a x^4\right ) \left (b+a x^4\right )^{3/2}} \, dx,x,\sqrt {x}\right )}{\sqrt {b x+a x^3}}\\ &=\frac {x}{2 a b \sqrt {b x+a x^3}}-\frac {\left (\sqrt {x} \sqrt {b+a x^2}\right ) \text {Subst}\left (\int \frac {-b-a x^4}{\left (-b+a x^4\right ) \sqrt {b+a x^4}} \, dx,x,\sqrt {x}\right )}{2 a b \sqrt {b x+a x^3}}\\ &=\frac {x}{2 a b \sqrt {b x+a x^3}}+\frac {\left (\sqrt {x} \sqrt {b+a x^2}\right ) \text {Subst}\left (\int \frac {\sqrt {b+a x^4}}{-b+a x^4} \, dx,x,\sqrt {x}\right )}{2 a b \sqrt {b x+a x^3}}\\ &=\frac {x}{2 a b \sqrt {b x+a x^3}}-\frac {\left (\sqrt {x} \sqrt {b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1-4 a b x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt {b+a x^2}}\right )}{2 a b \sqrt {b x+a x^3}}\\ &=\frac {x}{2 a b \sqrt {b x+a x^3}}-\frac {\left (\sqrt {x} \sqrt {b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1-2 \sqrt {a} \sqrt {b} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b+a x^2}}\right )}{4 a b \sqrt {b x+a x^3}}-\frac {\left (\sqrt {x} \sqrt {b+a x^2}\right ) \text {Subst}\left (\int \frac {1}{1+2 \sqrt {a} \sqrt {b} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b+a x^2}}\right )}{4 a b \sqrt {b x+a x^3}}\\ &=\frac {x}{2 a b \sqrt {b x+a x^3}}-\frac {\sqrt {x} \sqrt {b+a x^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {b+a x^2}}\right )}{4 \sqrt {2} a^{5/4} b^{5/4} \sqrt {b x+a x^3}}-\frac {\sqrt {x} \sqrt {b+a x^2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {b+a x^2}}\right )}{4 \sqrt {2} a^{5/4} b^{5/4} \sqrt {b x+a x^3}}\\ \end {align*}

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Mathematica [A]
time = 0.74, size = 152, normalized size = 1.02 \begin {gather*} \frac {\sqrt {x} \left (4 \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}-\sqrt {2} \sqrt {b+a x^2} \text {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} \tanh ^{-1}\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 )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(Sqrt[b*x + a*x^3]*(-b^2 + a^2*x^4)),x]

[Out]

(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]*ArcTanh[(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)])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.10, size = 448, normalized size = 3.01

method	result	size
elliptic	\(\frac {x}{2 a b \sqrt {\left (x^{2}+\frac {b}{a}\right ) a x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{4 a^{2} b \sqrt {a \,x^{3}+b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{4 a^{2} \sqrt {a b}\, \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}-\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{4 a^{2} \sqrt {a b}\, \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}\)	\(421\)
default	\(\frac {\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {a b}\, a \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}-\frac {\sqrt {a b}}{a}\right )}-\frac {\sqrt {-a b}\, \sqrt {\frac {x a}{\sqrt {-a b}}+1}\, \sqrt {-\frac {2 x a}{\sqrt {-a b}}+2}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticPi \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, -\frac {\sqrt {-a b}}{a \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {a b}\, a \sqrt {a \,x^{3}+b x}\, \left (-\frac {\sqrt {-a b}}{a}+\frac {\sqrt {a b}}{a}\right )}}{2 a}+\frac {\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{a}\right ) a x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 b a \sqrt {a \,x^{3}+b x}}}{2 a}\)	\(448\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x,method=_RETURNVERBOSE)

[Out]

1/2/a*(1/2/(a*b)^(1/2)/a*(-a*b)^(1/2)*(x*a/(-a*b)^(1/2)+1)^(1/2)*(-2*x*a/(-a*b)^(1/2)+2)^(1/2)*(-x*a/(-a*b)^(1
/2))^(1/2)/(a*x^3+b*x)^(1/2)/(-1/a*(-a*b)^(1/2)-1/a*(a*b)^(1/2))*EllipticPi(((x+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/
2))^(1/2),-1/a*(-a*b)^(1/2)/(-1/a*(-a*b)^(1/2)-1/a*(a*b)^(1/2)),1/2*2^(1/2))-1/2/(a*b)^(1/2)/a*(-a*b)^(1/2)*(x
*a/(-a*b)^(1/2)+1)^(1/2)*(-2*x*a/(-a*b)^(1/2)+2)^(1/2)*(-x*a/(-a*b)^(1/2))^(1/2)/(a*x^3+b*x)^(1/2)/(-1/a*(-a*b
)^(1/2)+1/a*(a*b)^(1/2))*EllipticPi(((x+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),-1/a*(-a*b)^(1/2)/(-1/a*(-a*b)
^(1/2)+1/a*(a*b)^(1/2)),1/2*2^(1/2)))+1/2/a*(x/b/((x^2+b/a)*a*x)^(1/2)+1/2/b/a*(-a*b)^(1/2)*((x+1/a*(-a*b)^(1/
2))*a/(-a*b)^(1/2))^(1/2)*(-2*(x-1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2)*(-x*a/(-a*b)^(1/2))^(1/2)/(a*x^3+b*x)
^(1/2)*EllipticF(((x+1/a*(-a*b)^(1/2))*a/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x, algorithm="maxima")

[Out]

integrate(x^2/((a^2*x^4 - b^2)*sqrt(a*x^3 + b*x)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (113) = 226\).
time = 0.45, size = 426, normalized size = 2.86 \begin {gather*} -\frac {4 \, \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}} \arctan \left (\frac {4 \, \left (\frac {1}{4}\right )^{\frac {3}{4}} \sqrt {a x^{3} + b x} a^{4} b^{4} \left (\frac {1}{a^{5} b^{5}}\right )^{\frac {3}{4}}}{a x^{2} + b}\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 (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 ) - 8 \, \sqrt {a x^{3} + b x}}{16 \, {\left (a^{2} b x^{2} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x, algorithm="fricas")

[Out]

-1/16*(4*(1/4)^(1/4)*(a^2*b*x^2 + a*b^2)*(1/(a^5*b^5))^(1/4)*arctan(4*(1/4)^(3/4)*sqrt(a*x^3 + b*x)*a^4*b^4*(1
/(a^5*b^5))^(3/4)/(a*x^2 + b)) + (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)) - 8*sqrt(a*x^3 + b*x))/(a^2*b*x^2 + a*b^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a*x**3+b*x)**(1/2)/(a**2*x**4-b**2),x)

[Out]

Integral(x**2/(sqrt(x*(a*x**2 + b))*(a*x**2 - b)*(a*x**2 + b)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a*x^3+b*x)^(1/2)/(a^2*x^4-b^2),x, algorithm="giac")

[Out]

integrate(x^2/((a^2*x^4 - b^2)*sqrt(a*x^3 + b*x)), x)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^2/((b^2 - a^2*x^4)*(b*x + a*x^3)^(1/2)),x)

[Out]

\text{Hanged}

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