3.61 \(\int e^{\frac{1}{2} \tanh ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=282 \[ -\frac{x (1-a x)^{3/4} (a x+1)^{5/4}}{3 a^2}-\frac{(1-a x)^{3/4} (a x+1)^{5/4}}{12 a^3}-\frac{3 (1-a x)^{3/4} \sqrt [4]{a x+1}}{8 a^3}-\frac{3 \log \left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}-\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt{2} a^3}+\frac{3 \log \left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}+\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt{2} a^3}+\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{8 \sqrt{2} a^3}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt{2} a^3} \]

[Out]

(-3*(1 - a*x)^(3/4)*(1 + a*x)^(1/4))/(8*a^3) - ((1 - a*x)^(3/4)*(1 + a*x)^(5/4))/(12*a^3) - (x*(1 - a*x)^(3/4)
*(1 + a*x)^(5/4))/(3*a^2) + (3*ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a^3) - (3*Arc
Tan[1 + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a^3) - (3*Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] -
(Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(16*Sqrt[2]*a^3) + (3*Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] + (Sqrt
[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(16*Sqrt[2]*a^3)

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Rubi [A]  time = 0.214073, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {6126, 90, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac{x (1-a x)^{3/4} (a x+1)^{5/4}}{3 a^2}-\frac{(1-a x)^{3/4} (a x+1)^{5/4}}{12 a^3}-\frac{3 (1-a x)^{3/4} \sqrt [4]{a x+1}}{8 a^3}-\frac{3 \log \left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}-\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt{2} a^3}+\frac{3 \log \left (\frac{\sqrt{1-a x}}{\sqrt{a x+1}}+\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt{2} a^3}+\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{8 \sqrt{2} a^3}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt{2} a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 - a*x)^(3/4)*(1 + a*x)^(1/4))/(8*a^3) - ((1 - a*x)^(3/4)*(1 + a*x)^(5/4))/(12*a^3) - (x*(1 - a*x)^(3/4)
*(1 + a*x)^(5/4))/(3*a^2) + (3*ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a^3) - (3*Arc
Tan[1 + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a^3) - (3*Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] -
(Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(16*Sqrt[2]*a^3) + (3*Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] + (Sqrt
[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(16*Sqrt[2]*a^3)

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]

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 297

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

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 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 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 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 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]

Rubi steps

\begin{align*} \int e^{\frac{1}{2} \tanh ^{-1}(a x)} x^2 \, dx &=\int \frac{x^2 \sqrt [4]{1+a x}}{\sqrt [4]{1-a x}} \, dx\\ &=-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}-\frac{\int \frac{\left (-1-\frac{a x}{2}\right ) \sqrt [4]{1+a x}}{\sqrt [4]{1-a x}} \, dx}{3 a^2}\\ &=-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}+\frac{3 \int \frac{\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}} \, dx}{8 a^2}\\ &=-\frac{3 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}+\frac{3 \int \frac{1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx}{16 a^2}\\ &=-\frac{3 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}-\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-a x}\right )}{4 a^3}\\ &=-\frac{3 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}-\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{4 a^3}\\ &=-\frac{3 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}+\frac{3 \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 a^3}-\frac{3 \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 a^3}\\ &=-\frac{3 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}-\frac{3 \operatorname{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 )}{16 a^3}-\frac{3 \operatorname{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 )}{16 a^3}-\frac{3 \operatorname{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 )}{16 \sqrt{2} a^3}-\frac{3 \operatorname{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 )}{16 \sqrt{2} a^3}\\ &=-\frac{3 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}-\frac{3 \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 )}{16 \sqrt{2} a^3}+\frac{3 \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 )}{16 \sqrt{2} a^3}-\frac{3 \operatorname{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 )}{8 \sqrt{2} a^3}+\frac{3 \operatorname{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 )}{8 \sqrt{2} a^3}\\ &=-\frac{3 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}-\frac{(1-a x)^{3/4} (1+a x)^{5/4}}{12 a^3}-\frac{x (1-a x)^{3/4} (1+a x)^{5/4}}{3 a^2}+\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt{2} a^3}-\frac{3 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt{2} a^3}-\frac{3 \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 )}{16 \sqrt{2} a^3}+\frac{3 \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 )}{16 \sqrt{2} a^3}\\ \end{align*}

Mathematica [C]  time = 0.0364549, size = 69, normalized size = 0.24 \[ -\frac{(1-a x)^{3/4} \left (6 \sqrt [4]{2} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{3}{4},\frac{7}{4},\frac{1}{2} (1-a x)\right )+\sqrt [4]{a x+1} \left (4 a^2 x^2+5 a x+1\right )\right )}{12 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((1 - a*x)^(3/4)*((1 + a*x)^(1/4)*(1 + 5*a*x + 4*a^2*x^2) + 6*2^(1/4)*Hypergeometric2F1[-1/4, 3/4, 7/4, (1 -
a*x)/2]))/(12*a^3)

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Maple [F]  time = 0.118, size = 0, normalized size = 0. \begin{align*} \int \sqrt{{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.88041, size = 1350, normalized size = 4.79 \begin{align*} -\frac{36 \, \sqrt{2} a^{3} \frac{1}{a^{12}}^{\frac{1}{4}} \arctan \left (\sqrt{2} a^{9} \sqrt{\frac{\sqrt{2}{\left (a^{4} x - a^{3}\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{1}{4}} +{\left (a^{7} x - a^{6}\right )} \sqrt{\frac{1}{a^{12}}} - \sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{3}{4}} - \sqrt{2} a^{9} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{3}{4}} - 1\right ) + 36 \, \sqrt{2} a^{3} \frac{1}{a^{12}}^{\frac{1}{4}} \arctan \left (\sqrt{2} a^{9} \sqrt{-\frac{\sqrt{2}{\left (a^{4} x - a^{3}\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{1}{4}} -{\left (a^{7} x - a^{6}\right )} \sqrt{\frac{1}{a^{12}}} + \sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{3}{4}} - \sqrt{2} a^{9} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{3}{4}} + 1\right ) - 9 \, \sqrt{2} a^{3} \frac{1}{a^{12}}^{\frac{1}{4}} \log \left (\frac{\sqrt{2}{\left (a^{4} x - a^{3}\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{1}{4}} +{\left (a^{7} x - a^{6}\right )} \sqrt{\frac{1}{a^{12}}} - \sqrt{-a^{2} x^{2} + 1}}{a x - 1}\right ) + 9 \, \sqrt{2} a^{3} \frac{1}{a^{12}}^{\frac{1}{4}} \log \left (-\frac{\sqrt{2}{\left (a^{4} x - a^{3}\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}} \frac{1}{a^{12}}^{\frac{1}{4}} -{\left (a^{7} x - a^{6}\right )} \sqrt{\frac{1}{a^{12}}} + \sqrt{-a^{2} x^{2} + 1}}{a x - 1}\right ) - 4 \,{\left (8 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 11\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}}}{96 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(36*sqrt(2)*a^3*(a^(-12))^(1/4)*arctan(sqrt(2)*a^9*sqrt((sqrt(2)*(a^4*x - a^3)*sqrt(-sqrt(-a^2*x^2 + 1)/
(a*x - 1))*(a^(-12))^(1/4) + (a^7*x - a^6)*sqrt(a^(-12)) - sqrt(-a^2*x^2 + 1))/(a*x - 1))*(a^(-12))^(3/4) - sq
rt(2)*a^9*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-12))^(3/4) - 1) + 36*sqrt(2)*a^3*(a^(-12))^(1/4)*arctan(sqr
t(2)*a^9*sqrt(-(sqrt(2)*(a^4*x - a^3)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-12))^(1/4) - (a^7*x - a^6)*sqrt
(a^(-12)) + sqrt(-a^2*x^2 + 1))/(a*x - 1))*(a^(-12))^(3/4) - sqrt(2)*a^9*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(
a^(-12))^(3/4) + 1) - 9*sqrt(2)*a^3*(a^(-12))^(1/4)*log((sqrt(2)*(a^4*x - a^3)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x -
1))*(a^(-12))^(1/4) + (a^7*x - a^6)*sqrt(a^(-12)) - sqrt(-a^2*x^2 + 1))/(a*x - 1)) + 9*sqrt(2)*a^3*(a^(-12))^
(1/4)*log(-(sqrt(2)*(a^4*x - a^3)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-12))^(1/4) - (a^7*x - a^6)*sqrt(a^(
-12)) + sqrt(-a^2*x^2 + 1))/(a*x - 1)) - 4*(8*a^3*x^3 + 2*a^2*x^2 + a*x - 11)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x -
1)))/a^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt{\frac{a x + 1}{\sqrt{- a^{2} x^{2} + 1}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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