3.140 \(\int \frac {e^{\frac {1}{4} \tanh ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=312 \[ \frac {a^2 \log \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}+1\right )}{32 \sqrt {2}}-\frac {a^2 \log \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}+\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}+1\right )}{32 \sqrt {2}}-\frac {1}{16} a^2 \tan ^{-1}\left (\frac {\sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}\right )+\frac {a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}\right )}{16 \sqrt {2}}-\frac {a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}+1\right )}{16 \sqrt {2}}-\frac {1}{16} a^2 \tanh ^{-1}\left (\frac {\sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}\right )-\frac {(1-a x)^{7/8} (a x+1)^{9/8}}{2 x^2}-\frac {a (1-a x)^{7/8} \sqrt [8]{a x+1}}{8 x} \]

[Out]

-1/8*a*(-a*x+1)^(7/8)*(a*x+1)^(1/8)/x-1/2*(-a*x+1)^(7/8)*(a*x+1)^(9/8)/x^2-1/16*a^2*arctan((a*x+1)^(1/8)/(-a*x
+1)^(1/8))-1/16*a^2*arctanh((a*x+1)^(1/8)/(-a*x+1)^(1/8))+1/32*a^2*arctan(1-(a*x+1)^(1/8)*2^(1/2)/(-a*x+1)^(1/
8))*2^(1/2)-1/32*a^2*arctan(1+(a*x+1)^(1/8)*2^(1/2)/(-a*x+1)^(1/8))*2^(1/2)+1/64*a^2*ln(1+(a*x+1)^(1/4)/(-a*x+
1)^(1/4)-(a*x+1)^(1/8)*2^(1/2)/(-a*x+1)^(1/8))*2^(1/2)-1/64*a^2*ln(1+(a*x+1)^(1/4)/(-a*x+1)^(1/4)+(a*x+1)^(1/8
)*2^(1/2)/(-a*x+1)^(1/8))*2^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6126, 96, 94, 93, 214, 212, 206, 203, 211, 1165, 628, 1162, 617, 204} \[ \frac {a^2 \log \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}+1\right )}{32 \sqrt {2}}-\frac {a^2 \log \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}+\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}+1\right )}{32 \sqrt {2}}-\frac {1}{16} a^2 \tan ^{-1}\left (\frac {\sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}\right )+\frac {a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}\right )}{16 \sqrt {2}}-\frac {a^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}+1\right )}{16 \sqrt {2}}-\frac {1}{16} a^2 \tanh ^{-1}\left (\frac {\sqrt [8]{a x+1}}{\sqrt [8]{1-a x}}\right )-\frac {(1-a x)^{7/8} (a x+1)^{9/8}}{2 x^2}-\frac {a (1-a x)^{7/8} \sqrt [8]{a x+1}}{8 x} \]

Antiderivative was successfully verified.

[In]

Int[E^(ArcTanh[a*x]/4)/x^3,x]

[Out]

-(a*(1 - a*x)^(7/8)*(1 + a*x)^(1/8))/(8*x) - ((1 - a*x)^(7/8)*(1 + a*x)^(9/8))/(2*x^2) - (a^2*ArcTan[(1 + a*x)
^(1/8)/(1 - a*x)^(1/8)])/16 + (a^2*ArcTan[1 - (Sqrt[2]*(1 + a*x)^(1/8))/(1 - a*x)^(1/8)])/(16*Sqrt[2]) - (a^2*
ArcTan[1 + (Sqrt[2]*(1 + a*x)^(1/8))/(1 - a*x)^(1/8)])/(16*Sqrt[2]) - (a^2*ArcTanh[(1 + a*x)^(1/8)/(1 - a*x)^(
1/8)])/16 + (a^2*Log[1 - (Sqrt[2]*(1 + a*x)^(1/8))/(1 - a*x)^(1/8) + (1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/(32*Sqr
t[2]) - (a^2*Log[1 + (Sqrt[2]*(1 + a*x)^(1/8))/(1 - a*x)^(1/8) + (1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/(32*Sqrt[2]
)

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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 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 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rubi steps

\begin {align*} \int \frac {e^{\frac {1}{4} \tanh ^{-1}(a x)}}{x^3} \, dx &=\int \frac {\sqrt [8]{1+a x}}{x^3 \sqrt [8]{1-a x}} \, dx\\ &=-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}+\frac {1}{8} a \int \frac {\sqrt [8]{1+a x}}{x^2 \sqrt [8]{1-a x}} \, dx\\ &=-\frac {a (1-a x)^{7/8} \sqrt [8]{1+a x}}{8 x}-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}+\frac {1}{32} a^2 \int \frac {1}{x \sqrt [8]{1-a x} (1+a x)^{7/8}} \, dx\\ &=-\frac {a (1-a x)^{7/8} \sqrt [8]{1+a x}}{8 x}-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}+\frac {1}{4} a^2 \operatorname {Subst}\left (\int \frac {1}{-1+x^8} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )\\ &=-\frac {a (1-a x)^{7/8} \sqrt [8]{1+a x}}{8 x}-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}-\frac {1}{8} a^2 \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{8} a^2 \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )\\ &=-\frac {a (1-a x)^{7/8} \sqrt [8]{1+a x}}{8 x}-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}-\frac {1}{16} a^2 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{16} a^2 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{16} a^2 \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{16} a^2 \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )\\ &=-\frac {a (1-a x)^{7/8} \sqrt [8]{1+a x}}{8 x}-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}-\frac {1}{16} a^2 \tan ^{-1}\left (\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{16} a^2 \tanh ^{-1}\left (\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{32} a^2 \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{32} a^2 \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )+\frac {a^2 \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )}{32 \sqrt {2}}+\frac {a^2 \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )}{32 \sqrt {2}}\\ &=-\frac {a (1-a x)^{7/8} \sqrt [8]{1+a x}}{8 x}-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}-\frac {1}{16} a^2 \tan ^{-1}\left (\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )-\frac {1}{16} a^2 \tanh ^{-1}\left (\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )+\frac {a^2 \log \left (1-\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}+\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{32 \sqrt {2}}-\frac {a^2 \log \left (1+\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}+\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{32 \sqrt {2}}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )}{16 \sqrt {2}}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )}{16 \sqrt {2}}\\ &=-\frac {a (1-a x)^{7/8} \sqrt [8]{1+a x}}{8 x}-\frac {(1-a x)^{7/8} (1+a x)^{9/8}}{2 x^2}-\frac {1}{16} a^2 \tan ^{-1}\left (\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )+\frac {a^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )}{16 \sqrt {2}}-\frac {a^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )}{16 \sqrt {2}}-\frac {1}{16} a^2 \tanh ^{-1}\left (\frac {\sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}\right )+\frac {a^2 \log \left (1-\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}+\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{32 \sqrt {2}}-\frac {a^2 \log \left (1+\frac {\sqrt {2} \sqrt [8]{1+a x}}{\sqrt [8]{1-a x}}+\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )}{32 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 73, normalized size = 0.23 \[ -\frac {(1-a x)^{7/8} \left (2 a^2 x^2 \, _2F_1\left (\frac {7}{8},1;\frac {15}{8};\frac {1-a x}{a x+1}\right )+7 \left (5 a^2 x^2+9 a x+4\right )\right )}{56 x^2 (a x+1)^{7/8}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(ArcTanh[a*x]/4)/x^3,x]

[Out]

-1/56*((1 - a*x)^(7/8)*(7*(4 + 9*a*x + 5*a^2*x^2) + 2*a^2*x^2*Hypergeometric2F1[7/8, 1, 15/8, (1 - a*x)/(1 + a
*x)]))/(x^2*(1 + a*x)^(7/8))

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fricas [B]  time = 0.59, size = 585, normalized size = 1.88 \[ -\frac {4 \, a^{2} x^{2} \arctan \left (\left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}}\right ) + 2 \, a^{2} x^{2} \log \left (\left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} + 1\right ) - 2 \, a^{2} x^{2} \log \left (\left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} - 1\right ) - 4 \, \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} x^{2} \arctan \left (-\frac {a^{8} + \sqrt {2} {\left (a^{8}\right )}^{\frac {3}{4}} a^{2} \left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} - \sqrt {2} {\left (a^{8}\right )}^{\frac {3}{4}} \sqrt {a^{4} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} a^{2} \left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} + \sqrt {a^{8}}}}{a^{8}}\right ) - 4 \, \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} x^{2} \arctan \left (\frac {a^{8} - \sqrt {2} {\left (a^{8}\right )}^{\frac {3}{4}} a^{2} \left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} + \sqrt {2} {\left (a^{8}\right )}^{\frac {3}{4}} \sqrt {a^{4} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} a^{2} \left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} + \sqrt {a^{8}}}}{a^{8}}\right ) + \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} x^{2} \log \left (a^{4} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} a^{2} \left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} + \sqrt {a^{8}}\right ) - \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} x^{2} \log \left (a^{4} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {2} {\left (a^{8}\right )}^{\frac {1}{4}} a^{2} \left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}} + \sqrt {a^{8}}\right ) - 8 \, {\left (5 \, a^{2} x^{2} - a x - 4\right )} \left (-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right )^{\frac {1}{4}}}{64 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(4*a^2*x^2*arctan((-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4)) + 2*a^2*x^2*log((-sqrt(-a^2*x^2 + 1)/(a*x - 1))
^(1/4) + 1) - 2*a^2*x^2*log((-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4) - 1) - 4*sqrt(2)*(a^8)^(1/4)*x^2*arctan(-(a^
8 + sqrt(2)*(a^8)^(3/4)*a^2*(-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4) - sqrt(2)*(a^8)^(3/4)*sqrt(a^4*sqrt(-sqrt(-a
^2*x^2 + 1)/(a*x - 1)) + sqrt(2)*(a^8)^(1/4)*a^2*(-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4) + sqrt(a^8)))/a^8) - 4*
sqrt(2)*(a^8)^(1/4)*x^2*arctan((a^8 - sqrt(2)*(a^8)^(3/4)*a^2*(-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4) + sqrt(2)*
(a^8)^(3/4)*sqrt(a^4*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - sqrt(2)*(a^8)^(1/4)*a^2*(-sqrt(-a^2*x^2 + 1)/(a*x -
1))^(1/4) + sqrt(a^8)))/a^8) + sqrt(2)*(a^8)^(1/4)*x^2*log(a^4*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + sqrt(2)*
(a^8)^(1/4)*a^2*(-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4) + sqrt(a^8)) - sqrt(2)*(a^8)^(1/4)*x^2*log(a^4*sqrt(-sqr
t(-a^2*x^2 + 1)/(a*x - 1)) - sqrt(2)*(a^8)^(1/4)*a^2*(-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4) + sqrt(a^8)) - 8*(5
*a^2*x^2 - a*x - 4)*(-sqrt(-a^2*x^2 + 1)/(a*x - 1))^(1/4))/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )^{\frac {1}{4}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [4]{\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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