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

Optimal. Leaf size=757 \[ -\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}-\frac {\sqrt {2-\sqrt {2}} a \tan ^{-1}\left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}} \sqrt [8]{b}-\frac {2 \sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}\right ) \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )}{4 b^{15/8}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [8]{b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )}{2 \sqrt {2} b^{15/8}}+\frac {\sqrt {2+\sqrt {2}} a \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )}{4 b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )}{2 \sqrt {2} b^{15/8}}-\frac {\sqrt {2-\sqrt {2}} a \tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )}{4 b^{15/8}}-\frac {\sqrt {2+\sqrt {2}} a \tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2+\sqrt {2}}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )}{4 b^{15/8}}-\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{x \left (2 a^3 x^3-2 a b x\right )+x \sqrt {a^2 x^2-b} \left (2 a^2 x^2-b\right )} \]

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Rubi [A]  time = 0.87, antiderivative size = 772, normalized size of antiderivative = 1.02, number of steps used = 31, number of rules used = 14, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2120, 259, 288, 329, 214, 212, 206, 203, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {a \log \left (\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt {2} \sqrt [8]{b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}+\sqrt [4]{b}\right )}{4 \sqrt {2} b^{15/8}}-\frac {a \log \left (\sqrt {\sqrt {a^2 x^2-b}+a x}+\sqrt {2} \sqrt [8]{b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}+\sqrt [4]{b}\right )}{4 \sqrt {2} b^{15/8}}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}+\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 \sqrt {2} b^{15/8}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}+1\right )}{2 \sqrt {2} b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}+\frac {4 a \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{b^2-\left (\sqrt {a^2 x^2-b}+a x\right )^4}+\frac {a \log \left (\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt {2} \sqrt [8]{-b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}+\sqrt [4]{-b}\right )}{4 \sqrt {2} (-b)^{15/8}}-\frac {a \log \left (\sqrt {\sqrt {a^2 x^2-b}+a x}+\sqrt {2} \sqrt [8]{-b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}+\sqrt [4]{-b}\right )}{4 \sqrt {2} (-b)^{15/8}}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}}+\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )}{2 \sqrt {2} (-b)^{15/8}}-\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}+1\right )}{2 \sqrt {2} (-b)^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(b^2 - (a*x + Sqrt[-b + a^2*x^2])^4) - (a*ArcTan[(a*x + Sqrt[-b + a^2*x
^2])^(1/4)/(-b)^(1/8)])/(2*(-b)^(15/8)) - (a*ArcTan[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)])/(2*b^(15/8)) +
(a*ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)])/(2*Sqrt[2]*(-b)^(15/8)) - (a*ArcTan[1 +
(Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)])/(2*Sqrt[2]*(-b)^(15/8)) + (a*ArcTan[1 - (Sqrt[2]*(a*x
+ Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)])/(2*Sqrt[2]*b^(15/8)) - (a*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2]
)^(1/4))/b^(1/8)])/(2*Sqrt[2]*b^(15/8)) - (a*ArcTanh[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/(-b)^(1/8)])/(2*(-b)^(15
/8)) - (a*ArcTanh[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)])/(2*b^(15/8)) + (a*Log[(-b)^(1/4) - Sqrt[2]*(-b)^(
1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]])/(4*Sqrt[2]*(-b)^(15/8)) - (a*Log[(-b)
^(1/4) + Sqrt[2]*(-b)^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]])/(4*Sqrt[2]*(-b
)^(15/8)) + (a*Log[b^(1/4) - Sqrt[2]*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]
])/(4*Sqrt[2]*b^(15/8)) - (a*Log[b^(1/4) + Sqrt[2]*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[
-b + a^2*x^2]]])/(4*Sqrt[2]*b^(15/8))

Rule 203

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

Rule 204

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

Rule 206

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 214

Int[((a_) + (b_.)*(x_)^(n_))^(-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^(n/2)), x], x] + Dist[r/(2*a), Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a,
b}, x] && IGtQ[n/4, 1] &&  !GtQ[a/b, 0]

Rule 259

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2120

Int[(x_)^(p_.)*((g_) + (i_.)*(x_)^2)^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :>
Dist[(1*(i/c)^m)/(2^(2*m + p + 1)*e^(p + 1)*f^(2*m)), Subst[Int[x^(n - 2*m - p - 2)*(-(a*f^2) + x^2)^p*(a*f^2
+ x^2)^(2*m + 1), x], x, e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0
] && EqQ[c*g - a*i, 0] && IntegersQ[p, 2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2 \left (-b+a^2 x^2\right )^{3/2}} \, dx &=(16 a) \operatorname {Subst}\left (\int \frac {x^{13/4}}{\left (-b+x^2\right )^2 \left (b+x^2\right )^2} \, dx,x,a x+\sqrt {-b+a^2 x^2}\right )\\ &=(16 a) \operatorname {Subst}\left (\int \frac {x^{13/4}}{\left (-b^2+x^4\right )^2} \, dx,x,a x+\sqrt {-b+a^2 x^2}\right )\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}+a \operatorname {Subst}\left (\int \frac {1}{x^{3/4} \left (-b^2+x^4\right )} \, dx,x,a x+\sqrt {-b+a^2 x^2}\right )\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}+(4 a) \operatorname {Subst}\left (\int \frac {1}{-b^2+x^{16}} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{b-x^8} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{b+x^8} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}-x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{(-b)^{3/2}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{(-b)^{3/2}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b^{3/2}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b^{3/2}}\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 (-b)^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 (-b)^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {\sqrt [4]{-b}-x^2}{\sqrt {-b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 (-b)^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {\sqrt [4]{-b}+x^2}{\sqrt {-b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 (-b)^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b}-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 b^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b}+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 b^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {\sqrt [4]{b}-x^2}{\sqrt {b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 b^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {\sqrt [4]{b}+x^2}{\sqrt {b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{2 b^{7/4}}\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}+\frac {a \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [8]{-b}+2 x}{-\sqrt [4]{-b}-\sqrt {2} \sqrt [8]{-b} x-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} (-b)^{15/8}}+\frac {a \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [8]{-b}-2 x}{-\sqrt [4]{-b}+\sqrt {2} \sqrt [8]{-b} x-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} (-b)^{15/8}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}-\sqrt {2} \sqrt [8]{-b} x+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 (-b)^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}+\sqrt {2} \sqrt [8]{-b} x+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 (-b)^{7/4}}+\frac {a \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [8]{b}+2 x}{-\sqrt [4]{b}-\sqrt {2} \sqrt [8]{b} x-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} b^{15/8}}+\frac {a \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt [8]{b}-2 x}{-\sqrt [4]{b}+\sqrt {2} \sqrt [8]{b} x-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} b^{15/8}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b}-\sqrt {2} \sqrt [8]{b} x+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 b^{7/4}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{b}+\sqrt {2} \sqrt [8]{b} x+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{4 b^{7/4}}\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}+\frac {a \log \left (\sqrt [4]{-b}-\sqrt {2} \sqrt [8]{-b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} (-b)^{15/8}}-\frac {a \log \left (\sqrt [4]{-b}+\sqrt {2} \sqrt [8]{-b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} (-b)^{15/8}}+\frac {a \log \left (\sqrt [4]{b}-\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} b^{15/8}}-\frac {a \log \left (\sqrt [4]{b}+\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} b^{15/8}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 \sqrt {2} (-b)^{15/8}}+\frac {a \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 \sqrt {2} (-b)^{15/8}}-\frac {a \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 \sqrt {2} b^{15/8}}+\frac {a \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 \sqrt {2} b^{15/8}}\\ &=\frac {4 a \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{b^2-\left (a x+\sqrt {-b+a^2 x^2}\right )^4}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}+\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 \sqrt {2} (-b)^{15/8}}-\frac {a \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 \sqrt {2} (-b)^{15/8}}+\frac {a \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 \sqrt {2} b^{15/8}}-\frac {a \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 \sqrt {2} b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{2 (-b)^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}+\frac {a \log \left (\sqrt [4]{-b}-\sqrt {2} \sqrt [8]{-b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} (-b)^{15/8}}-\frac {a \log \left (\sqrt [4]{-b}+\sqrt {2} \sqrt [8]{-b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} (-b)^{15/8}}+\frac {a \log \left (\sqrt [4]{b}-\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} b^{15/8}}-\frac {a \log \left (\sqrt [4]{b}+\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{4 \sqrt {2} b^{15/8}}\\ \end {align*}

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Mathematica [B]  time = 2.35, size = 2041, normalized size = 2.70 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/(x^2*(-b + a^2*x^2)^(3/2)),x]

[Out]

(-32*b^(15/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) - 4*(-b^2 + (a*x + Sqrt[-b + a^2*x^2])^4)*ArcTan[(a*x + Sqrt[-b
+ a^2*x^2])^(1/4)/b^(1/8)] + 2*Sqrt[2]*(-b^2 + (a*x + Sqrt[-b + a^2*x^2])^4)*ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[
-b + a^2*x^2])^(1/4))/b^(1/8)] + 2*Sqrt[2]*b^2*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)]
- 2*Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^4*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)] + 4*b^
2*ArcTan[((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) - Tan[Pi/8]]*Cos[Pi/8] - 4*(a*x + Sqrt[-b + a^2*
x^2])^4*ArcTan[((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) - Tan[Pi/8]]*Cos[Pi/8] + 4*b^2*ArcTan[((a*
x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) + Tan[Pi/8]]*Cos[Pi/8] - 4*(a*x + Sqrt[-b + a^2*x^2])^4*ArcTa
n[((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) + Tan[Pi/8]]*Cos[Pi/8] - 2*b^2*Log[b^(1/8) - (a*x + Sqr
t[-b + a^2*x^2])^(1/4)] + 2*(a*x + Sqrt[-b + a^2*x^2])^4*Log[b^(1/8) - (a*x + Sqrt[-b + a^2*x^2])^(1/4)] + 2*b
^2*Log[b^(1/8) + (a*x + Sqrt[-b + a^2*x^2])^(1/4)] - 2*(a*x + Sqrt[-b + a^2*x^2])^4*Log[b^(1/8) + (a*x + Sqrt[
-b + a^2*x^2])^(1/4)] - Sqrt[2]*b^2*Log[b^(1/4) - Sqrt[2]*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + Sqrt[a*x
+ Sqrt[-b + a^2*x^2]]] + Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^4*Log[b^(1/4) - Sqrt[2]*b^(1/8)*(a*x + Sqrt[-b + a
^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]] + Sqrt[2]*b^2*Log[b^(1/4) + Sqrt[2]*b^(1/8)*(a*x + Sqrt[-b +
a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]] - Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^4*Log[b^(1/4) + Sqrt[2]
*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]] - 2*b^2*Cos[Pi/8]*Log[b^(1/4) + Sq
rt[a*x + Sqrt[-b + a^2*x^2]] - 2*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4)*Cos[Pi/8]] + 2*(a*x + Sqrt[-b + a^2*
x^2])^4*Cos[Pi/8]*Log[b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]] - 2*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4)*Co
s[Pi/8]] + 2*b^2*Cos[Pi/8]*Log[b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]] + 2*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])
^(1/4)*Cos[Pi/8]] - 2*(a*x + Sqrt[-b + a^2*x^2])^4*Cos[Pi/8]*Log[b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]] + 2*
b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4)*Cos[Pi/8]] - 4*b^2*ArcTan[Cot[Pi/8] - ((a*x + Sqrt[-b + a^2*x^2])^(1/
4)*Csc[Pi/8])/b^(1/8)]*Sin[Pi/8] + 4*(a*x + Sqrt[-b + a^2*x^2])^4*ArcTan[Cot[Pi/8] - ((a*x + Sqrt[-b + a^2*x^2
])^(1/4)*Csc[Pi/8])/b^(1/8)]*Sin[Pi/8] + 4*b^2*ArcTan[Cot[Pi/8] + ((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Csc[Pi/8])
/b^(1/8)]*Sin[Pi/8] - 4*(a*x + Sqrt[-b + a^2*x^2])^4*ArcTan[Cot[Pi/8] + ((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Csc[
Pi/8])/b^(1/8)]*Sin[Pi/8] - 2*b^2*Log[b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]] - 2*b^(1/8)*(a*x + Sqrt[-b + a^
2*x^2])^(1/4)*Sin[Pi/8]]*Sin[Pi/8] + 2*(a*x + Sqrt[-b + a^2*x^2])^4*Log[b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2
]] - 2*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sin[Pi/8]]*Sin[Pi/8] + 2*b^2*Log[b^(1/4) + Sqrt[a*x + Sqrt[-b
+ a^2*x^2]] + 2*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sin[Pi/8]]*Sin[Pi/8] - 2*(a*x + Sqrt[-b + a^2*x^2])^4
*Log[b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]] + 2*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sin[Pi/8]]*Sin[Pi/8
])/(16*b^(15/8)*x*Sqrt[-4*b + 4*a^2*x^2]*(-b + 2*a*x*(a*x + Sqrt[-b + a^2*x^2])))

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IntegrateAlgebraic [A]  time = 5.71, size = 731, normalized size = 0.97 \begin {gather*} -\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x \sqrt {-b+a^2 x^2} \left (-b+2 a^2 x^2\right )+x \left (-2 a b x+2 a^3 x^3\right )}-\frac {a \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}+\frac {a \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{-\sqrt [4]{b}+\sqrt {a x+\sqrt {-b+a^2 x^2}}}\right )}{2 \sqrt {2} b^{15/8}}+\frac {\sqrt {2-\sqrt {2}} a \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{-\sqrt [4]{b}+\sqrt {a x+\sqrt {-b+a^2 x^2}}}\right )}{4 b^{15/8}}+\frac {\sqrt {2+\sqrt {2}} a \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{-\sqrt [4]{b}+\sqrt {a x+\sqrt {-b+a^2 x^2}}}\right )}{4 b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{15/8}}-\frac {a \tanh ^{-1}\left (\frac {\frac {\sqrt [8]{b}}{\sqrt {2}}+\frac {\sqrt {a x+\sqrt {-b+a^2 x^2}}}{\sqrt {2} \sqrt [8]{b}}}{\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )}{2 \sqrt {2} b^{15/8}}-\frac {\sqrt {2+\sqrt {2}} a \tanh ^{-1}\left (\frac {\sqrt {1-\frac {1}{\sqrt {2}}} \sqrt [8]{b}+\frac {\sqrt {1-\frac {1}{\sqrt {2}}} \sqrt {a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}}{\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )}{4 b^{15/8}}-\frac {\sqrt {2-\sqrt {2}} a \tanh ^{-1}\left (\frac {\sqrt {1+\frac {1}{\sqrt {2}}} \sqrt [8]{b}+\frac {\sqrt {1+\frac {1}{\sqrt {2}}} \sqrt {a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}}{\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )}{4 b^{15/8}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/(x^2*(-b + a^2*x^2)^(3/2)),x]

[Out]

-((a*x + Sqrt[-b + a^2*x^2])^(1/4)/(x*Sqrt[-b + a^2*x^2]*(-b + 2*a^2*x^2) + x*(-2*a*b*x + 2*a^3*x^3))) - (a*Ar
cTan[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)])/(2*b^(15/8)) + (a*ArcTan[(Sqrt[2]*b^(1/8)*(a*x + Sqrt[-b + a^2
*x^2])^(1/4))/(-b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]])])/(2*Sqrt[2]*b^(15/8)) + (Sqrt[2 - Sqrt[2]]*a*ArcTan
[(Sqrt[2 - Sqrt[2]]*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]])])/(4
*b^(15/8)) + (Sqrt[2 + Sqrt[2]]*a*ArcTan[(Sqrt[2 + Sqrt[2]]*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b^(1/4
) + Sqrt[a*x + Sqrt[-b + a^2*x^2]])])/(4*b^(15/8)) - (a*ArcTanh[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)])/(2*
b^(15/8)) - (a*ArcTanh[(b^(1/8)/Sqrt[2] + Sqrt[a*x + Sqrt[-b + a^2*x^2]]/(Sqrt[2]*b^(1/8)))/(a*x + Sqrt[-b + a
^2*x^2])^(1/4)])/(2*Sqrt[2]*b^(15/8)) - (Sqrt[2 + Sqrt[2]]*a*ArcTanh[(Sqrt[1 - 1/Sqrt[2]]*b^(1/8) + (Sqrt[1 -
1/Sqrt[2]]*Sqrt[a*x + Sqrt[-b + a^2*x^2]])/b^(1/8))/(a*x + Sqrt[-b + a^2*x^2])^(1/4)])/(4*b^(15/8)) - (Sqrt[2
- Sqrt[2]]*a*ArcTanh[(Sqrt[1 + 1/Sqrt[2]]*b^(1/8) + (Sqrt[1 + 1/Sqrt[2]]*Sqrt[a*x + Sqrt[-b + a^2*x^2]])/b^(1/
8))/(a*x + Sqrt[-b + a^2*x^2])^(1/4)])/(4*b^(15/8))

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fricas [A]  time = 0.57, size = 80, normalized size = 0.11 \begin {gather*} \frac {{\left (2 \, a^{3} x^{3} - 2 \, a b x - {\left (2 \, a^{2} x^{2} - b\right )} \sqrt {a^{2} x^{2} - b}\right )} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}}}{a^{2} b^{2} x^{3} - b^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}}{x^{2} \left (a^{2} x^{2}-b \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}}}{{\left (a^{2} x^{2} - b\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/4}}{x^2\,{\left (a^2\,x^2-b\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{a x + \sqrt {a^{2} x^{2} - b}}}{x^{2} \left (a^{2} x^{2} - b\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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