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

Optimal. Leaf size=725 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {\sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}} \sqrt [8]{b}-\frac {2 \sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}\right ) \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )}{b^{11/8}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [8]{b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )}{2 \sqrt {2} b^{11/8}}-\frac {\sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )}{b^{11/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}+\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )}{2 \sqrt {2} b^{11/8}}-\frac {\sqrt {2+\sqrt {2}} \tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )}{b^{11/8}}+\frac {\sqrt {2-\sqrt {2}} \tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2+\sqrt {2}}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )}{b^{11/8}}+\frac {\left (\sqrt {a^2 x^2-b}+a x\right )^{5/4}}{b \left (-a x \sqrt {a^2 x^2-b}-a^2 x^2+b\right )} \]

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Rubi [A]  time = 0.91, antiderivative size = 746, normalized size of antiderivative = 1.03, 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, 466, 471, 584, 301, 211, 1165, 628, 1162, 617, 204, 212, 206, 203} \begin {gather*} -\frac {\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^{11/8}}+\frac {\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^{11/8}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {\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^{11/8}}+\frac {\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^{11/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}+\frac {2 \left (\sqrt {a^2 x^2-b}+a x\right )^{5/4}}{b \left (b-\left (\sqrt {a^2 x^2-b}+a x\right )^2\right )}-\frac {\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 )}{\sqrt {2} (-b)^{11/8}}+\frac {\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 )}{\sqrt {2} (-b)^{11/8}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}+\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}+1\right )}{(-b)^{11/8}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*x + Sqrt[-b + a^2*x^2])^(5/4))/(b*(b - (a*x + Sqrt[-b + a^2*x^2])^2)) - (2*ArcTan[(a*x + Sqrt[-b + a^2*x
^2])^(1/4)/(-b)^(1/8)])/(-b)^(11/8) - ArcTan[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)]/(2*b^(11/8)) - (Sqrt[2]
*ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)])/(-b)^(11/8) + (Sqrt[2]*ArcTan[1 + (Sqrt[2]
*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)])/(-b)^(11/8) - ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(
1/4))/b^(1/8)]/(2*Sqrt[2]*b^(11/8)) + ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)]/(2*Sqrt[2
]*b^(11/8)) - (2*ArcTanh[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/(-b)^(1/8)])/(-b)^(11/8) - ArcTanh[(a*x + Sqrt[-b +
a^2*x^2])^(1/4)/b^(1/8)]/(2*b^(11/8)) - 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)^(11/8)) + 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)^(11/8)) - 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^(11/8)) + 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^(11/8))

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 301

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(
a/b), 2]]}, Dist[s/(2*b), Int[x^(m - n/2)/(r + s*x^(n/2)), x], x] - Dist[s/(2*b), Int[x^(m - n/2)/(r - s*x^(n/
2)), x], x]] /; FreeQ[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LeQ[n/2, m] && LtQ[m, n] &&  !GtQ[a/b, 0]

Rule 466

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 471

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 584

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

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 \left (-b+a^2 x^2\right )^{3/2}} \, dx &=8 \operatorname {Subst}\left (\int \frac {x^{9/4}}{\left (-b+x^2\right )^2 \left (b+x^2\right )} \, dx,x,a x+\sqrt {-b+a^2 x^2}\right )\\ &=32 \operatorname {Subst}\left (\int \frac {x^{12}}{\left (-b+x^8\right )^2 \left (b+x^8\right )} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {x^4 \left (5 b-3 x^8\right )}{\left (-b+x^8\right ) \left (b+x^8\right )} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}+\frac {2 \operatorname {Subst}\left (\int \left (\frac {x^4}{-b+x^8}-\frac {4 x^4}{b+x^8}\right ) \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {x^4}{-b+x^8} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}-\frac {8 \operatorname {Subst}\left (\int \frac {x^4}{b+x^8} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}+\frac {4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}-x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{b}\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{(-b)^{5/4}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )}{(-b)^{5/4}}+\frac {2 \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 )}{(-b)^{5/4}}+\frac {2 \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 )}{(-b)^{5/4}}-\frac {\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^{5/4}}-\frac {\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^{5/4}}+\frac {\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^{5/4}}+\frac {\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^{5/4}}\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {\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 )}{\sqrt {2} (-b)^{11/8}}-\frac {\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 )}{\sqrt {2} (-b)^{11/8}}+\frac {\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 )}{(-b)^{5/4}}+\frac {\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 )}{(-b)^{5/4}}-\frac {\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^{11/8}}-\frac {\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^{11/8}}+\frac {\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^{5/4}}+\frac {\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^{5/4}}\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {\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 )}{\sqrt {2} (-b)^{11/8}}+\frac {\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 )}{\sqrt {2} (-b)^{11/8}}-\frac {\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^{11/8}}+\frac {\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^{11/8}}+\frac {\sqrt {2} \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 )}{(-b)^{11/8}}-\frac {\sqrt {2} \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 )}{(-b)^{11/8}}+\frac {\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^{11/8}}-\frac {\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^{11/8}}\\ &=\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-\left (a x+\sqrt {-b+a^2 x^2}\right )^2\right )}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}+\frac {\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\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^{11/8}}+\frac {\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^{11/8}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )}{(-b)^{11/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {\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 )}{\sqrt {2} (-b)^{11/8}}+\frac {\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 )}{\sqrt {2} (-b)^{11/8}}-\frac {\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^{11/8}}+\frac {\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^{11/8}}\\ \end {align*}

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Mathematica [B]  time = 3.95, size = 2000, normalized size = 2.76

result too large to display

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-16*b^(3/8)*(a*x + Sqrt[-b + a^2*x^2])^(5/4) - 4*(-b + (a*x + Sqrt[-b + a^2*x^2])^2)*ArcTan[(a*x + Sqrt[-b +
a^2*x^2])^(1/4)/b^(1/8)] - 2*Sqrt[2]*(-b + (a*x + Sqrt[-b + a^2*x^2])^2)*ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[-b +
a^2*x^2])^(1/4))/b^(1/8)] - 2*Sqrt[2]*b*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)] + 2*Sqr
t[2]*(a*x + Sqrt[-b + a^2*x^2])^2*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/b^(1/8)] - 16*b*ArcTan
[Cot[Pi/8] - ((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Csc[Pi/8])/b^(1/8)]*Cos[Pi/8] + 16*(a*x + Sqrt[-b + a^2*x^2])^2
*ArcTan[Cot[Pi/8] - ((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Csc[Pi/8])/b^(1/8)]*Cos[Pi/8] + 16*b*ArcTan[Cot[Pi/8] +
((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Csc[Pi/8])/b^(1/8)]*Cos[Pi/8] - 16*(a*x + Sqrt[-b + a^2*x^2])^2*ArcTan[Cot[P
i/8] + ((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Csc[Pi/8])/b^(1/8)]*Cos[Pi/8] - 2*b*Log[b^(1/8) - (a*x + Sqrt[-b + a^
2*x^2])^(1/4)] + 2*(a*x + Sqrt[-b + a^2*x^2])^2*Log[b^(1/8) - (a*x + Sqrt[-b + a^2*x^2])^(1/4)] + 2*b*Log[b^(1
/8) + (a*x + Sqrt[-b + a^2*x^2])^(1/4)] - 2*(a*x + Sqrt[-b + a^2*x^2])^2*Log[b^(1/8) + (a*x + Sqrt[-b + a^2*x^
2])^(1/4)] + Sqrt[2]*b*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])^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]*b*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])^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]]] - 8*b*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)*Sin[Pi/8]] + 8*(a*x + Sqrt[-b + a^2*x^2])^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)*Sin[Pi/8]] + 8*b*Co
s[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)*Sin[Pi/8]] -
8*(a*x + Sqrt[-b + a^2*x^2])^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)*Sin[Pi/8]] - 16*b*ArcTan[((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) - Tan[Pi/8
]]*Sin[Pi/8] + 16*(a*x + Sqrt[-b + a^2*x^2])^2*ArcTan[((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) - T
an[Pi/8]]*Sin[Pi/8] - 16*b*ArcTan[((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) + Tan[Pi/8]]*Sin[Pi/8]
+ 16*(a*x + Sqrt[-b + a^2*x^2])^2*ArcTan[((a*x + Sqrt[-b + a^2*x^2])^(1/4)*Sec[Pi/8])/b^(1/8) + Tan[Pi/8]]*Sin
[Pi/8] + 8*b*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]]*Sin[Pi/8] - 8*(a*x + Sqrt[-b + a^2*x^2])^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)*Cos[Pi/8]]*Sin[Pi/8] - 8*b*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]]*Sin[Pi/8] + 8*(a*x + Sqrt[-b + a^2*x^2])^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)*Cos[Pi/8]]*Sin[Pi/8])/(8*b^(11/8)*Sqrt[-4*
b + 4*a^2*x^2]*(a*x + Sqrt[-b + a^2*x^2]))

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IntegrateAlgebraic [A]  time = 3.25, size = 698, normalized size = 0.96 \begin {gather*} \frac {\left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}{b \left (b-a^2 x^2-a x \sqrt {-b+a^2 x^2}\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}-\frac {\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^{11/8}}+\frac {\sqrt {2+\sqrt {2}} \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 )}{b^{11/8}}-\frac {\sqrt {2-\sqrt {2}} \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 )}{b^{11/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{b}}\right )}{2 b^{11/8}}+\frac {\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^{11/8}}+\frac {\sqrt {2-\sqrt {2}} \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 )}{b^{11/8}}-\frac {\sqrt {2+\sqrt {2}} \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 )}{b^{11/8}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x + Sqrt[-b + a^2*x^2])^(5/4)/(b*(b - a^2*x^2 - a*x*Sqrt[-b + a^2*x^2])) - ArcTan[(a*x + Sqrt[-b + a^2*x^2]
)^(1/4)/b^(1/8)]/(2*b^(11/8)) - 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^(11/8)) + (Sqrt[2 + Sqrt[2]]*ArcTan[(Sqrt[2 - Sqrt[2]]*b^(1/8)*(a*x + Sq
rt[-b + a^2*x^2])^(1/4))/(-b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]])])/b^(11/8) - (Sqrt[2 - Sqrt[2]]*ArcTan[(S
qrt[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]])])/b^(11
/8) - ArcTanh[(a*x + Sqrt[-b + a^2*x^2])^(1/4)/b^(1/8)]/(2*b^(11/8)) + ArcTanh[(b^(1/8)/Sqrt[2] + Sqrt[a*x + S
qrt[-b + a^2*x^2]]/(Sqrt[2]*b^(1/8)))/(a*x + Sqrt[-b + a^2*x^2])^(1/4)]/(2*Sqrt[2]*b^(11/8)) + (Sqrt[2 - Sqrt[
2]]*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)])/b^(11/8) - (Sqrt[2 + Sqrt[2]]*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)])/b^(11/8)

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fricas [A]  time = 0.62, size = 50, normalized size = 0.07 \begin {gather*} -\frac {\sqrt {a^{2} x^{2} - b} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}}}{a^{2} b x^{2} - b^{2}} \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/(a^2*x^2-b)^(3/2),x, algorithm="fricas")

[Out]

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

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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/(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 \[\int \frac {\left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}}{x \left (a^{2} x^{2}-b \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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