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

Optimal. Leaf size=526 \[ \sqrt {2+\sqrt {2}} b^{13/8} \tan ^{-1}\left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}} \sqrt [8]{b}-\frac {2 \sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}\right ) \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )+\sqrt {2-\sqrt {2}} b^{13/8} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt [4]{b}}\right )+\sqrt {2+\sqrt {2}} b^{13/8} \tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )-\sqrt {2-\sqrt {2}} b^{13/8} \tanh ^{-1}\left (\frac {\frac {\sqrt {\sqrt {a^2 x^2-b}+a x}}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b}}{\sqrt {2+\sqrt {2}}}}{\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}\right )+\frac {4 \left (132 a^6 x^6-627 a^4 b x^4+682 a^2 b^2 x^2-152 b^3\right )+4 \sqrt {a^2 x^2-b} \left (132 a^5 x^5-561 a^3 b x^3+418 a b^2 x\right )}{429 \left (\sqrt {a^2 x^2-b}+a x\right )^{11/4}} \]

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Rubi [A]  time = 0.66, antiderivative size = 468, normalized size of antiderivative = 0.89, number of steps used = 17, number of rules used = 13, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2120, 466, 461, 301, 211, 1165, 628, 1162, 617, 204, 212, 206, 203} \begin {gather*} -\frac {b^3}{22 \left (\sqrt {a^2 x^2-b}+a x\right )^{11/4}}+\frac {5 b^2}{6 \left (\sqrt {a^2 x^2-b}+a x\right )^{3/4}}+\frac {1}{26} \left (\sqrt {a^2 x^2-b}+a x\right )^{13/4}-\frac {1}{2} b \left (\sqrt {a^2 x^2-b}+a x\right )^{5/4}-\frac {(-b)^{13/8} \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}}+\frac {(-b)^{13/8} \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}}-2 (-b)^{13/8} \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )-\sqrt {2} (-b)^{13/8} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )+\sqrt {2} (-b)^{13/8} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}+1\right )-2 (-b)^{13/8} \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/22*b^3/(a*x + Sqrt[-b + a^2*x^2])^(11/4) + (5*b^2)/(6*(a*x + Sqrt[-b + a^2*x^2])^(3/4)) - (b*(a*x + Sqrt[-b
+ a^2*x^2])^(5/4))/2 + (a*x + Sqrt[-b + a^2*x^2])^(13/4)/26 - 2*(-b)^(13/8)*ArcTan[(a*x + Sqrt[-b + a^2*x^2])
^(1/4)/(-b)^(1/8)] - Sqrt[2]*(-b)^(13/8)*ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)] + S
qrt[2]*(-b)^(13/8)*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)] - 2*(-b)^(13/8)*ArcTanh[(
a*x + Sqrt[-b + a^2*x^2])^(1/4)/(-b)^(1/8)] - ((-b)^(13/8)*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)^(13/8)*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]

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 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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 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 {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x} \, dx &=\frac {1}{8} \operatorname {Subst}\left (\int \frac {\left (-b+x^2\right )^4}{x^{15/4} \left (b+x^2\right )} \, dx,x,a x+\sqrt {-b+a^2 x^2}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (-b+x^8\right )^4}{x^{12} \left (b+x^8\right )} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {b^3}{x^{12}}-\frac {5 b^2}{x^4}-5 b x^4+x^{12}+\frac {16 b^2 x^4}{b+x^8}\right ) \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )\\ &=-\frac {b^3}{22 \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+\frac {5 b^2}{6 \left (a x+\sqrt {-b+a^2 x^2}\right )^{3/4}}-\frac {1}{2} b \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}+\frac {1}{26} \left (a x+\sqrt {-b+a^2 x^2}\right )^{13/4}+\left (8 b^2\right ) \operatorname {Subst}\left (\int \frac {x^4}{b+x^8} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )\\ &=-\frac {b^3}{22 \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+\frac {5 b^2}{6 \left (a x+\sqrt {-b+a^2 x^2}\right )^{3/4}}-\frac {1}{2} b \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}+\frac {1}{26} \left (a x+\sqrt {-b+a^2 x^2}\right )^{13/4}-\left (4 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}-x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )+\left (4 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-b}+x^4} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )\\ &=-\frac {b^3}{22 \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+\frac {5 b^2}{6 \left (a x+\sqrt {-b+a^2 x^2}\right )^{3/4}}-\frac {1}{2} b \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}+\frac {1}{26} \left (a x+\sqrt {-b+a^2 x^2}\right )^{13/4}-\left (2 (-b)^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}-x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )-\left (2 (-b)^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-b}+x^2} \, dx,x,\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}\right )+\left (2 (-b)^{7/4}\right ) \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 )+\left (2 (-b)^{7/4}\right ) \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 )\\ &=-\frac {b^3}{22 \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+\frac {5 b^2}{6 \left (a x+\sqrt {-b+a^2 x^2}\right )^{3/4}}-\frac {1}{2} b \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}+\frac {1}{26} \left (a x+\sqrt {-b+a^2 x^2}\right )^{13/4}-2 (-b)^{13/8} \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )-2 (-b)^{13/8} \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )-\frac {(-b)^{13/8} \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}}-\frac {(-b)^{13/8} \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)^{7/4} \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)^{7/4} \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 )\\ &=-\frac {b^3}{22 \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+\frac {5 b^2}{6 \left (a x+\sqrt {-b+a^2 x^2}\right )^{3/4}}-\frac {1}{2} b \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}+\frac {1}{26} \left (a x+\sqrt {-b+a^2 x^2}\right )^{13/4}-2 (-b)^{13/8} \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )-2 (-b)^{13/8} \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )-\frac {(-b)^{13/8} \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}}+\frac {(-b)^{13/8} \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}}+\left (\sqrt {2} (-b)^{13/8}\right ) \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 )-\left (\sqrt {2} (-b)^{13/8}\right ) \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 )\\ &=-\frac {b^3}{22 \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+\frac {5 b^2}{6 \left (a x+\sqrt {-b+a^2 x^2}\right )^{3/4}}-\frac {1}{2} b \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}+\frac {1}{26} \left (a x+\sqrt {-b+a^2 x^2}\right )^{13/4}-2 (-b)^{13/8} \tan ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )-\sqrt {2} (-b)^{13/8} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )+\sqrt {2} (-b)^{13/8} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )-2 (-b)^{13/8} \tanh ^{-1}\left (\frac {\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [8]{-b}}\right )-\frac {(-b)^{13/8} \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}}+\frac {(-b)^{13/8} \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}}\\ \end {align*}

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Mathematica [B]  time = 23.49, size = 14841, normalized size = 28.21 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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IntegrateAlgebraic [A]  time = 1.78, size = 501, normalized size = 0.95 \begin {gather*} \frac {4 \sqrt {-b+a^2 x^2} \left (418 a b^2 x-561 a^3 b x^3+132 a^5 x^5\right )+4 \left (-152 b^3+682 a^2 b^2 x^2-627 a^4 b x^4+132 a^6 x^6\right )}{429 \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}-\sqrt {2+\sqrt {2}} b^{13/8} \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 )+\sqrt {2-\sqrt {2}} b^{13/8} \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 )-\sqrt {2-\sqrt {2}} b^{13/8} \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 )+\sqrt {2+\sqrt {2}} b^{13/8} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[-b + a^2*x^2]*(418*a*b^2*x - 561*a^3*b*x^3 + 132*a^5*x^5) + 4*(-152*b^3 + 682*a^2*b^2*x^2 - 627*a^4*b*
x^4 + 132*a^6*x^6))/(429*(a*x + Sqrt[-b + a^2*x^2])^(11/4)) - Sqrt[2 + Sqrt[2]]*b^(13/8)*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]])] + Sqrt[2 - Sqrt[2]
]*b^(13/8)*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]])] - Sqrt[2 - Sqrt[2]]*b^(13/8)*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)] + Sqrt[2 + Sqrt[2]]*b^(13/8)*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)]

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fricas [A]  time = 0.66, size = 683, normalized size = 1.30 \begin {gather*} 2 \, \sqrt {2} \left (-b^{13}\right )^{\frac {1}{8}} \arctan \left (-\frac {b^{13} + \sqrt {2} \left (-b^{13}\right )^{\frac {3}{8}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8} - \sqrt {2} \sqrt {\sqrt {a x + \sqrt {a^{2} x^{2} - b}} b^{16} - \left (-b^{13}\right )^{\frac {1}{4}} b^{13} - \sqrt {2} \left (-b^{13}\right )^{\frac {5}{8}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8}} \left (-b^{13}\right )^{\frac {3}{8}}}{b^{13}}\right ) + 2 \, \sqrt {2} \left (-b^{13}\right )^{\frac {1}{8}} \arctan \left (\frac {b^{13} - \sqrt {2} \left (-b^{13}\right )^{\frac {3}{8}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8} + \sqrt {2} \sqrt {\sqrt {a x + \sqrt {a^{2} x^{2} - b}} b^{16} - \left (-b^{13}\right )^{\frac {1}{4}} b^{13} + \sqrt {2} \left (-b^{13}\right )^{\frac {5}{8}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8}} \left (-b^{13}\right )^{\frac {3}{8}}}{b^{13}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-b^{13}\right )^{\frac {1}{8}} \log \left (4 \, \sqrt {a x + \sqrt {a^{2} x^{2} - b}} b^{16} - 4 \, \left (-b^{13}\right )^{\frac {1}{4}} b^{13} + 4 \, \sqrt {2} \left (-b^{13}\right )^{\frac {5}{8}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8}\right ) - \frac {1}{2} \, \sqrt {2} \left (-b^{13}\right )^{\frac {1}{8}} \log \left (4 \, \sqrt {a x + \sqrt {a^{2} x^{2} - b}} b^{16} - 4 \, \left (-b^{13}\right )^{\frac {1}{4}} b^{13} - 4 \, \sqrt {2} \left (-b^{13}\right )^{\frac {5}{8}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8}\right ) - \frac {4}{429} \, {\left (3 \, a^{3} x^{3} - 38 \, a b x - 4 \, {\left (9 \, a^{2} x^{2} - 38 \, b\right )} \sqrt {a^{2} x^{2} - b}\right )} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} - 4 \, \left (-b^{13}\right )^{\frac {1}{8}} \arctan \left (-\frac {\left (-b^{13}\right )^{\frac {3}{8}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8} - \sqrt {\sqrt {a x + \sqrt {a^{2} x^{2} - b}} b^{16} - \left (-b^{13}\right )^{\frac {1}{4}} b^{13}} \left (-b^{13}\right )^{\frac {3}{8}}}{b^{13}}\right ) - \left (-b^{13}\right )^{\frac {1}{8}} \log \left ({\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8} + \left (-b^{13}\right )^{\frac {5}{8}}\right ) + \left (-b^{13}\right )^{\frac {1}{8}} \log \left ({\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} b^{8} - \left (-b^{13}\right )^{\frac {5}{8}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(2)*(-b^13)^(1/8)*arctan(-(b^13 + sqrt(2)*(-b^13)^(3/8)*(a*x + sqrt(a^2*x^2 - b))^(1/4)*b^8 - sqrt(2)*sq
rt(sqrt(a*x + sqrt(a^2*x^2 - b))*b^16 - (-b^13)^(1/4)*b^13 - sqrt(2)*(-b^13)^(5/8)*(a*x + sqrt(a^2*x^2 - b))^(
1/4)*b^8)*(-b^13)^(3/8))/b^13) + 2*sqrt(2)*(-b^13)^(1/8)*arctan((b^13 - sqrt(2)*(-b^13)^(3/8)*(a*x + sqrt(a^2*
x^2 - b))^(1/4)*b^8 + sqrt(2)*sqrt(sqrt(a*x + sqrt(a^2*x^2 - b))*b^16 - (-b^13)^(1/4)*b^13 + sqrt(2)*(-b^13)^(
5/8)*(a*x + sqrt(a^2*x^2 - b))^(1/4)*b^8)*(-b^13)^(3/8))/b^13) + 1/2*sqrt(2)*(-b^13)^(1/8)*log(4*sqrt(a*x + sq
rt(a^2*x^2 - b))*b^16 - 4*(-b^13)^(1/4)*b^13 + 4*sqrt(2)*(-b^13)^(5/8)*(a*x + sqrt(a^2*x^2 - b))^(1/4)*b^8) -
1/2*sqrt(2)*(-b^13)^(1/8)*log(4*sqrt(a*x + sqrt(a^2*x^2 - b))*b^16 - 4*(-b^13)^(1/4)*b^13 - 4*sqrt(2)*(-b^13)^
(5/8)*(a*x + sqrt(a^2*x^2 - b))^(1/4)*b^8) - 4/429*(3*a^3*x^3 - 38*a*b*x - 4*(9*a^2*x^2 - 38*b)*sqrt(a^2*x^2 -
b))*(a*x + sqrt(a^2*x^2 - b))^(1/4) - 4*(-b^13)^(1/8)*arctan(-((-b^13)^(3/8)*(a*x + sqrt(a^2*x^2 - b))^(1/4)*
b^8 - sqrt(sqrt(a*x + sqrt(a^2*x^2 - b))*b^16 - (-b^13)^(1/4)*b^13)*(-b^13)^(3/8))/b^13) - (-b^13)^(1/8)*log((
a*x + sqrt(a^2*x^2 - b))^(1/4)*b^8 + (-b^13)^(5/8)) + (-b^13)^(1/8)*log((a*x + sqrt(a^2*x^2 - b))^(1/4)*b^8 -
(-b^13)^(5/8))

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a^2*x^2 - b)^(3/2)*(a*x + sqrt(a^2*x^2 - b))^(1/4)/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}\,{\left (a^2\,x^2-b\right )}^{3/2}}{x} \,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)*(a^2*x^2 - b)^(3/2))/x,x)

[Out]

int(((a*x + (a^2*x^2 - b)^(1/2))^(1/4)*(a^2*x^2 - b)^(3/2))/x, 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}} \left (a^{2} x^{2} - b\right )^{\frac {3}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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