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

Optimal. Leaf size=213 \[ \frac{61 x \sqrt [4]{1-\frac{1}{a x}}}{24 a^2 \sqrt [4]{\frac{1}{a x}+1}}+\frac{287 \sqrt [4]{1-\frac{1}{a x}}}{24 a^3 \sqrt [4]{\frac{1}{a x}+1}}+\frac{55 \tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}-\frac{55 \tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}+\frac{x^3 \sqrt [4]{1-\frac{1}{a x}}}{3 \sqrt [4]{\frac{1}{a x}+1}}-\frac{13 x^2 \sqrt [4]{1-\frac{1}{a x}}}{12 a \sqrt [4]{\frac{1}{a x}+1}} \]

[Out]

(287*(1 - 1/(a*x))^(1/4))/(24*a^3*(1 + 1/(a*x))^(1/4)) + (61*(1 - 1/(a*x))^(1/4)*x)/(24*a^2*(1 + 1/(a*x))^(1/4
)) - (13*(1 - 1/(a*x))^(1/4)*x^2)/(12*a*(1 + 1/(a*x))^(1/4)) + ((1 - 1/(a*x))^(1/4)*x^3)/(3*(1 + 1/(a*x))^(1/4
)) + (55*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(8*a^3) - (55*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*
x))^(1/4)])/(8*a^3)

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Rubi [A]  time = 0.110327, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {6171, 98, 151, 155, 12, 93, 298, 203, 206} \[ \frac{61 x \sqrt [4]{1-\frac{1}{a x}}}{24 a^2 \sqrt [4]{\frac{1}{a x}+1}}+\frac{287 \sqrt [4]{1-\frac{1}{a x}}}{24 a^3 \sqrt [4]{\frac{1}{a x}+1}}+\frac{55 \tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}-\frac{55 \tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}+\frac{x^3 \sqrt [4]{1-\frac{1}{a x}}}{3 \sqrt [4]{\frac{1}{a x}+1}}-\frac{13 x^2 \sqrt [4]{1-\frac{1}{a x}}}{12 a \sqrt [4]{\frac{1}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(287*(1 - 1/(a*x))^(1/4))/(24*a^3*(1 + 1/(a*x))^(1/4)) + (61*(1 - 1/(a*x))^(1/4)*x)/(24*a^2*(1 + 1/(a*x))^(1/4
)) - (13*(1 - 1/(a*x))^(1/4)*x^2)/(12*a*(1 + 1/(a*x))^(1/4)) + ((1 - 1/(a*x))^(1/4)*x^3)/(3*(1 + 1/(a*x))^(1/4
)) + (55*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(8*a^3) - (55*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*
x))^(1/4)])/(8*a^3)

Rule 6171

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

Rule 98

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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 298

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 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 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])

Rubi steps

\begin{align*} \int e^{-\frac{5}{2} \coth ^{-1}(a x)} x^2 \, dx &=-\operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^{5/4}}{x^4 \left (1+\frac{x}{a}\right )^{5/4}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{\frac{13}{2 a}-\frac{6 x}{a^2}}{x^3 \left (1-\frac{x}{a}\right )^{3/4} \left (1+\frac{x}{a}\right )^{5/4}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{13 \sqrt [4]{1-\frac{1}{a x}} x^2}{12 a \sqrt [4]{1+\frac{1}{a x}}}+\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}-\frac{1}{6} \operatorname{Subst}\left (\int \frac{\frac{61}{4 a^2}-\frac{13 x}{a^3}}{x^2 \left (1-\frac{x}{a}\right )^{3/4} \left (1+\frac{x}{a}\right )^{5/4}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{61 \sqrt [4]{1-\frac{1}{a x}} x}{24 a^2 \sqrt [4]{1+\frac{1}{a x}}}-\frac{13 \sqrt [4]{1-\frac{1}{a x}} x^2}{12 a \sqrt [4]{1+\frac{1}{a x}}}+\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{\frac{165}{8 a^3}-\frac{61 x}{4 a^4}}{x \left (1-\frac{x}{a}\right )^{3/4} \left (1+\frac{x}{a}\right )^{5/4}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{287 \sqrt [4]{1-\frac{1}{a x}}}{24 a^3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{61 \sqrt [4]{1-\frac{1}{a x}} x}{24 a^2 \sqrt [4]{1+\frac{1}{a x}}}-\frac{13 \sqrt [4]{1-\frac{1}{a x}} x^2}{12 a \sqrt [4]{1+\frac{1}{a x}}}+\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{1}{3} a \operatorname{Subst}\left (\int \frac{165}{16 a^4 x \left (1-\frac{x}{a}\right )^{3/4} \sqrt [4]{1+\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{287 \sqrt [4]{1-\frac{1}{a x}}}{24 a^3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{61 \sqrt [4]{1-\frac{1}{a x}} x}{24 a^2 \sqrt [4]{1+\frac{1}{a x}}}-\frac{13 \sqrt [4]{1-\frac{1}{a x}} x^2}{12 a \sqrt [4]{1+\frac{1}{a x}}}+\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{55 \operatorname{Subst}\left (\int \frac{1}{x \left (1-\frac{x}{a}\right )^{3/4} \sqrt [4]{1+\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )}{16 a^3}\\ &=\frac{287 \sqrt [4]{1-\frac{1}{a x}}}{24 a^3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{61 \sqrt [4]{1-\frac{1}{a x}} x}{24 a^2 \sqrt [4]{1+\frac{1}{a x}}}-\frac{13 \sqrt [4]{1-\frac{1}{a x}} x^2}{12 a \sqrt [4]{1+\frac{1}{a x}}}+\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{55 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^4} \, dx,x,\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^3}\\ &=\frac{287 \sqrt [4]{1-\frac{1}{a x}}}{24 a^3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{61 \sqrt [4]{1-\frac{1}{a x}} x}{24 a^2 \sqrt [4]{1+\frac{1}{a x}}}-\frac{13 \sqrt [4]{1-\frac{1}{a x}} x^2}{12 a \sqrt [4]{1+\frac{1}{a x}}}+\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}-\frac{55 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}+\frac{55 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}\\ &=\frac{287 \sqrt [4]{1-\frac{1}{a x}}}{24 a^3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{61 \sqrt [4]{1-\frac{1}{a x}} x}{24 a^2 \sqrt [4]{1+\frac{1}{a x}}}-\frac{13 \sqrt [4]{1-\frac{1}{a x}} x^2}{12 a \sqrt [4]{1+\frac{1}{a x}}}+\frac{\sqrt [4]{1-\frac{1}{a x}} x^3}{3 \sqrt [4]{1+\frac{1}{a x}}}+\frac{55 \tan ^{-1}\left (\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}-\frac{55 \tanh ^{-1}\left (\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}\\ \end{align*}

Mathematica [C]  time = 8.43072, size = 389, normalized size = 1.83 \[ -\frac{e^{-\frac{5}{2} \coth ^{-1}(a x)} \left (256 e^{4 \coth ^{-1}(a x)} \left (626 e^{2 \coth ^{-1}(a x)}+221 e^{4 \coth ^{-1}(a x)}+437\right ) \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},2,2,2\right \},\left \{1,1,\frac{15}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )+2048 e^{4 \coth ^{-1}(a x)} \left (30 e^{2 \coth ^{-1}(a x)}+13 e^{4 \coth ^{-1}(a x)}+17\right ) \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},2,2,2,2\right \},\left \{1,1,1,\frac{15}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )+4096 e^{4 \coth ^{-1}(a x)} \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{15}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )+8192 e^{6 \coth ^{-1}(a x)} \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{15}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )+4096 e^{8 \coth ^{-1}(a x)} \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{15}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )+824824 e^{2 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},e^{2 \coth ^{-1}(a x)}\right )+248094 e^{4 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},e^{2 \coth ^{-1}(a x)}\right )-85624 e^{6 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},e^{2 \coth ^{-1}(a x)}\right )-2387 e^{8 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},e^{2 \coth ^{-1}(a x)}\right )+818741 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},e^{2 \coth ^{-1}(a x)}\right )-1530529 e^{2 \coth ^{-1}(a x)}-266035 e^{4 \coth ^{-1}(a x)}+7161 e^{6 \coth ^{-1}(a x)}-818741\right )}{44352 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(-818741 - 1530529*E^(2*ArcCoth[a*x]) - 266035*E^(4*ArcCoth[a*x]) + 7161*E^(6*ArcCoth[a*x]) + 818741*Hypergeo
metric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] + 824824*E^(2*ArcCoth[a*x])*Hypergeometric2F1[3/4, 1, 7/4, E^(2*Arc
Coth[a*x])] + 248094*E^(4*ArcCoth[a*x])*Hypergeometric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] - 85624*E^(6*ArcCot
h[a*x])*Hypergeometric2F1[3/4, 1, 7/4, E^(2*ArcCoth[a*x])] - 2387*E^(8*ArcCoth[a*x])*Hypergeometric2F1[3/4, 1,
 7/4, E^(2*ArcCoth[a*x])] + 256*E^(4*ArcCoth[a*x])*(437 + 626*E^(2*ArcCoth[a*x]) + 221*E^(4*ArcCoth[a*x]))*Hyp
ergeometricPFQ[{3/4, 2, 2, 2}, {1, 1, 15/4}, E^(2*ArcCoth[a*x])] + 2048*E^(4*ArcCoth[a*x])*(17 + 30*E^(2*ArcCo
th[a*x]) + 13*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{3/4, 2, 2, 2, 2}, {1, 1, 1, 15/4}, E^(2*ArcCoth[a*x])] +
4096*E^(4*ArcCoth[a*x])*HypergeometricPFQ[{3/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 15/4}, E^(2*ArcCoth[a*x])] + 8192
*E^(6*ArcCoth[a*x])*HypergeometricPFQ[{3/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 15/4}, E^(2*ArcCoth[a*x])] + 4096*E^(
8*ArcCoth[a*x])*HypergeometricPFQ[{3/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 15/4}, E^(2*ArcCoth[a*x])])/(44352*a^3*E^
((5*ArcCoth[a*x])/2))

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Maple [F]  time = 0.331, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{5}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*((a*x-1)/(a*x+1))^(5/4),x)

[Out]

int(x^2*((a*x-1)/(a*x+1))^(5/4),x)

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Maxima [A]  time = 1.54513, size = 279, normalized size = 1.31 \begin{align*} -\frac{1}{48} \, a{\left (\frac{4 \,{\left (137 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{9}{4}} - 174 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{4}} + 69 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}}{\frac{3 \,{\left (a x - 1\right )} a^{4}}{a x + 1} - \frac{3 \,{\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac{{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} + \frac{330 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}{a^{4}} + \frac{165 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right )}{a^{4}} - \frac{165 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1\right )}{a^{4}} - \frac{384 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}{a^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*((a*x-1)/(a*x+1))^(5/4),x, algorithm="maxima")

[Out]

-1/48*a*(4*(137*((a*x - 1)/(a*x + 1))^(9/4) - 174*((a*x - 1)/(a*x + 1))^(5/4) + 69*((a*x - 1)/(a*x + 1))^(1/4)
)/(3*(a*x - 1)*a^4/(a*x + 1) - 3*(a*x - 1)^2*a^4/(a*x + 1)^2 + (a*x - 1)^3*a^4/(a*x + 1)^3 - a^4) + 330*arctan
(((a*x - 1)/(a*x + 1))^(1/4))/a^4 + 165*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 - 165*log(((a*x - 1)/(a*x + 1
))^(1/4) - 1)/a^4 - 384*((a*x - 1)/(a*x + 1))^(1/4)/a^4)

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Fricas [A]  time = 1.59335, size = 284, normalized size = 1.33 \begin{align*} \frac{2 \,{\left (8 \, a^{3} x^{3} - 26 \, a^{2} x^{2} + 61 \, a x + 287\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 330 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right ) - 165 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right ) + 165 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1\right )}{48 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*((a*x-1)/(a*x+1))^(5/4),x, algorithm="fricas")

[Out]

1/48*(2*(8*a^3*x^3 - 26*a^2*x^2 + 61*a*x + 287)*((a*x - 1)/(a*x + 1))^(1/4) - 330*arctan(((a*x - 1)/(a*x + 1))
^(1/4)) - 165*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) + 165*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*((a*x-1)/(a*x+1))**(5/4),x)

[Out]

Timed out

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Giac [A]  time = 1.18864, size = 259, normalized size = 1.22 \begin{align*} -\frac{1}{48} \, a{\left (\frac{330 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}{a^{4}} + \frac{165 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right )}{a^{4}} - \frac{165 \, \log \left ({\left | \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1 \right |}\right )}{a^{4}} - \frac{384 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}{a^{4}} - \frac{4 \,{\left (\frac{174 \,{\left (a x - 1\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}{a x + 1} - \frac{137 \,{\left (a x - 1\right )}^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}{{\left (a x + 1\right )}^{2}} - 69 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}}{a^{4}{\left (\frac{a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*((a*x-1)/(a*x+1))^(5/4),x, algorithm="giac")

[Out]

-1/48*a*(330*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^4 + 165*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 - 165*log(
abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^4 - 384*((a*x - 1)/(a*x + 1))^(1/4)/a^4 - 4*(174*(a*x - 1)*((a*x - 1)/
(a*x + 1))^(1/4)/(a*x + 1) - 137*(a*x - 1)^2*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1)^2 - 69*((a*x - 1)/(a*x + 1)
)^(1/4))/(a^4*((a*x - 1)/(a*x + 1) - 1)^3))