3.2.15 \(\int \frac {e^{-\frac {5}{2} \tanh ^{-1}(a x)}}{x} \, dx\) [115]

Optimal. Leaf size=248 \[ \frac {8 \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}+2 \text {ArcTan}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )+\sqrt {2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )-\sqrt {2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )-2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )+\frac {\log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{\sqrt {2}}-\frac {\log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{\sqrt {2}} \]

[Out]

8*(-a*x+1)^(1/4)/(a*x+1)^(1/4)+2*arctan((a*x+1)^(1/4)/(-a*x+1)^(1/4))-2*arctanh((a*x+1)^(1/4)/(-a*x+1)^(1/4))+
1/2*ln(1-(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4)+(-a*x+1)^(1/2)/(a*x+1)^(1/2))*2^(1/2)-1/2*ln(1+(-a*x+1)^(1/4)*2^
(1/2)/(a*x+1)^(1/4)+(-a*x+1)^(1/2)/(a*x+1)^(1/2))*2^(1/2)-arctan(-1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))*2^(1
/2)-arctan(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))*2^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 16, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {6261, 100, 21, 132, 65, 246, 217, 1179, 642, 1176, 631, 210, 95, 304, 209, 212} \begin {gather*} 2 \text {ArcTan}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )+\sqrt {2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )+\frac {8 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^((5*ArcTanh[a*x])/2)*x),x]

[Out]

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

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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

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 100

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

Rule 132

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

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 217

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 304

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

Rule 631

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 642

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 1176

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 1179

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 6261

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.04, size = 90, normalized size = 0.36 \begin {gather*} \frac {\sqrt [4]{1-a x} \left (20-20 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {1-a x}{1+a x}\right )+2^{3/4} (1-a x) \sqrt [4]{1+a x} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2} (1-a x)\right )\right )}{5 \sqrt [4]{1+a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((5*ArcTanh[a*x])/2)*x),x]

[Out]

((1 - a*x)^(1/4)*(20 - 20*Hypergeometric2F1[1/4, 1, 5/4, (1 - a*x)/(1 + a*x)] + 2^(3/4)*(1 - a*x)*(1 + a*x)^(1
/4)*Hypergeometric2F1[5/4, 5/4, 9/4, (1 - a*x)/2]))/(5*(1 + a*x)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )^{\frac {5}{2}} x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x,x)

[Out]

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x,x)

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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(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x,x, algorithm="maxima")

[Out]

integrate(1/(x*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (197) = 394\).
time = 0.39, size = 499, normalized size = 2.01 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (a x + 1\right )} \arctan \left (\sqrt {2} \sqrt {\frac {a x + \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1}{a x - 1}} - \sqrt {2} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) + 4 \, \sqrt {2} {\left (a x + 1\right )} \arctan \left (\sqrt {2} \sqrt {\frac {a x - \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1}{a x - 1}} - \sqrt {2} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + \sqrt {2} {\left (a x + 1\right )} \log \left (\frac {4 \, {\left (a x + \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1\right )}}{a x - 1}\right ) - \sqrt {2} {\left (a x + 1\right )} \log \left (\frac {4 \, {\left (a x - \sqrt {2} {\left (a x - 1\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - \sqrt {-a^{2} x^{2} + 1} - 1\right )}}{a x - 1}\right ) - 4 \, {\left (a x + 1\right )} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) + 2 \, {\left (a x + 1\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) - 2 \, {\left (a x + 1\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) - 16 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{2 \, {\left (a x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x,x, algorithm="fricas")

[Out]

-1/2*(4*sqrt(2)*(a*x + 1)*arctan(sqrt(2)*sqrt((a*x + sqrt(2)*(a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - s
qrt(-a^2*x^2 + 1) - 1)/(a*x - 1)) - sqrt(2)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1) + 4*sqrt(2)*(a*x + 1)*arc
tan(sqrt(2)*sqrt((a*x - sqrt(2)*(a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - sqrt(-a^2*x^2 + 1) - 1)/(a*x -
1)) - sqrt(2)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) + sqrt(2)*(a*x + 1)*log(4*(a*x + sqrt(2)*(a*x - 1)*sqr
t(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - sqrt(-a^2*x^2 + 1) - 1)/(a*x - 1)) - sqrt(2)*(a*x + 1)*log(4*(a*x - sqrt(2)
*(a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - sqrt(-a^2*x^2 + 1) - 1)/(a*x - 1)) - 4*(a*x + 1)*arctan(sqrt(
-sqrt(-a^2*x^2 + 1)/(a*x - 1))) + 2*(a*x + 1)*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) - 2*(a*x + 1)*log(s
qrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1) - 16*sqrt(-a^2*x^2 + 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)))/(a*x + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x+1)/(-a**2*x**2+1)**(1/2))**(5/2)/x,x)

[Out]

Integral(1/(x*((a*x + 1)/sqrt(-a**2*x**2 + 1))**(5/2)), 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(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x,x, algorithm="giac")

[Out]

integrate(1/(x*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x\,{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2)),x)

[Out]

int(1/(x*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2)), x)

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