3.82 \(\int e^{\frac {5}{2} \tanh ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=305 \[ \frac {(1-a x)^{3/4} (a x+1)^{9/4}}{3 a^3}+\frac {2 (a x+1)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {11 (1-a x)^{3/4} (a x+1)^{5/4}}{4 a^3}+\frac {55 (1-a x)^{3/4} \sqrt [4]{a x+1}}{8 a^3}+\frac {55 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{8 \sqrt {2} a^3}+\frac {55 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt {2} a^3} \]

[Out]

55/8*(-a*x+1)^(3/4)*(a*x+1)^(1/4)/a^3+11/4*(-a*x+1)^(3/4)*(a*x+1)^(5/4)/a^3+2*(a*x+1)^(9/4)/a^3/(-a*x+1)^(1/4)
+1/3*(-a*x+1)^(3/4)*(a*x+1)^(9/4)/a^3+55/16*arctan(-1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))/a^3*2^(1/2)+55/16*
arctan(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))/a^3*2^(1/2)+55/32*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))/a^3*2^(1/2)-55/32*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))/a^3*2^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {6126, 89, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {(1-a x)^{3/4} (a x+1)^{9/4}}{3 a^3}+\frac {2 (a x+1)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {11 (1-a x)^{3/4} (a x+1)^{5/4}}{4 a^3}+\frac {55 (1-a x)^{3/4} \sqrt [4]{a x+1}}{8 a^3}+\frac {55 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{16 \sqrt {2} a^3}-\frac {55 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{8 \sqrt {2} a^3}+\frac {55 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt {2} a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(55*(1 - a*x)^(3/4)*(1 + a*x)^(1/4))/(8*a^3) + (11*(1 - a*x)^(3/4)*(1 + a*x)^(5/4))/(4*a^3) + (2*(1 + a*x)^(9/
4))/(a^3*(1 - a*x)^(1/4)) + ((1 - a*x)^(3/4)*(1 + a*x)^(9/4))/(3*a^3) - (55*ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4
))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a^3) + (55*ArcTan[1 + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a
^3) + (55*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) -
(55*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 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 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 89

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

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 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 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 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 e^{\frac {5}{2} \tanh ^{-1}(a x)} x^2 \, dx &=\int \frac {x^2 (1+a x)^{5/4}}{(1-a x)^{5/4}} \, dx\\ &=\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}-\frac {2 \int \frac {(1+a x)^{5/4} \left (\frac {5 a}{2}+\frac {a^2 x}{2}\right )}{\sqrt [4]{1-a x}} \, dx}{a^3}\\ &=\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}-\frac {11 \int \frac {(1+a x)^{5/4}}{\sqrt [4]{1-a x}} \, dx}{2 a^2}\\ &=\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}-\frac {55 \int \frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}} \, dx}{8 a^2}\\ &=\frac {55 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}+\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}-\frac {55 \int \frac {1}{\sqrt [4]{1-a x} (1+a x)^{3/4}} \, dx}{16 a^2}\\ &=\frac {55 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}+\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}+\frac {55 \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 {55 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}+\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}+\frac {55 \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 {55 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}+\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}-\frac {55 \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 {55 \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 {55 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}+\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}+\frac {55 \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 {55 \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 {55 \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 {55 \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 {55 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}+\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}+\frac {55 \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 {55 \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 {55 \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 {55 \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 {55 (1-a x)^{3/4} \sqrt [4]{1+a x}}{8 a^3}+\frac {11 (1-a x)^{3/4} (1+a x)^{5/4}}{4 a^3}+\frac {2 (1+a x)^{9/4}}{a^3 \sqrt [4]{1-a x}}+\frac {(1-a x)^{3/4} (1+a x)^{9/4}}{3 a^3}-\frac {55 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a^3}+\frac {55 \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 {55 \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*}

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Mathematica [C]  time = 0.03, size = 66, normalized size = 0.22 \[ \frac {(7-a x) (a x+1)^{9/4}-44 \sqrt [4]{2} (a x-1) \, _2F_1\left (-\frac {5}{4},\frac {3}{4};\frac {7}{4};\frac {1}{2} (1-a x)\right )}{3 a^3 \sqrt [4]{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((7 - a*x)*(1 + a*x)^(9/4) - 44*2^(1/4)*(-1 + a*x)*Hypergeometric2F1[-5/4, 3/4, 7/4, (1 - a*x)/2])/(3*a^3*(1 -
a*x)^(1/4))

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fricas [B]  time = 0.57, size = 545, normalized size = 1.79 \[ \frac {660 \, \sqrt {2} a^{3} \frac {1}{a^{12}}^{\frac {1}{4}} \arctan \left (\sqrt {2} a^{9} \sqrt {\frac {\sqrt {2} {\left (a^{4} x - a^{3}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {1}{4}} + {\left (a^{7} x - a^{6}\right )} \sqrt {\frac {1}{a^{12}}} - \sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {3}{4}} - \sqrt {2} a^{9} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {3}{4}} - 1\right ) + 660 \, \sqrt {2} a^{3} \frac {1}{a^{12}}^{\frac {1}{4}} \arctan \left (\sqrt {2} a^{9} \sqrt {-\frac {\sqrt {2} {\left (a^{4} x - a^{3}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {1}{4}} - {\left (a^{7} x - a^{6}\right )} \sqrt {\frac {1}{a^{12}}} + \sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {3}{4}} - \sqrt {2} a^{9} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {3}{4}} + 1\right ) - 165 \, \sqrt {2} a^{3} \frac {1}{a^{12}}^{\frac {1}{4}} \log \left (\frac {\sqrt {2} {\left (a^{4} x - a^{3}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {1}{4}} + {\left (a^{7} x - a^{6}\right )} \sqrt {\frac {1}{a^{12}}} - \sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right ) + 165 \, \sqrt {2} a^{3} \frac {1}{a^{12}}^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} {\left (a^{4} x - a^{3}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{12}}^{\frac {1}{4}} - {\left (a^{7} x - a^{6}\right )} \sqrt {\frac {1}{a^{12}}} + \sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right ) - 4 \, {\left (8 \, a^{3} x^{3} + 26 \, a^{2} x^{2} + 61 \, a x - 287\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{96 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(660*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) + 660*sqrt(2)*a^3*(a^(-12))^(1/4)*arctan(sq
rt(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)*sqr
t(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) - 165*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)) + 165*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)*sqr
t(a^(-12)) + sqrt(-a^2*x^2 + 1))/(a*x - 1)) - 4*(8*a^3*x^3 + 26*a^2*x^2 + 61*a*x - 287)*sqrt(-sqrt(-a^2*x^2 +
1)/(a*x - 1)))/a^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(a*x-1)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )^{\frac {5}{2}} x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*((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 \[ \int x^2\,{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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))**(5/2)*x**2,x)

[Out]

Timed out

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