### 3.79 $$\int e^{\frac{5}{2} \coth ^{-1}(a x)} x^2 \, dx$$

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

[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)

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Rubi [A]  time = 0.112202, 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, 212, 206, 203} $\frac{61 x \sqrt [4]{\frac{1}{a x}+1}}{24 a^2 \sqrt [4]{1-\frac{1}{a x}}}-\frac{287 \sqrt [4]{\frac{1}{a x}+1}}{24 a^3 \sqrt [4]{1-\frac{1}{a x}}}+\frac{55 \tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}+\frac{55 \tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{8 a^3}+\frac{x^3 \sqrt [4]{\frac{1}{a x}+1}}{3 \sqrt [4]{1-\frac{1}{a x}}}+\frac{13 x^2 \sqrt [4]{\frac{1}{a x}+1}}{12 a \sqrt [4]{1-\frac{1}{a x}}}$

Antiderivative was successfully veriﬁed.

[In]

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

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 )^{5/4} \left (1+\frac{x}{a}\right )^{3/4}} \, 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 )^{5/4} \left (1+\frac{x}{a}\right )^{3/4}} \, 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 )^{5/4} \left (1+\frac{x}{a}\right )^{3/4}} \, 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 \sqrt [4]{1-\frac{x}{a}} \left (1+\frac{x}{a}\right )^{3/4}} \, 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{55 \operatorname{Subst}\left (\int \frac{1}{x \sqrt [4]{1-\frac{x}{a}} \left (1+\frac{x}{a}\right )^{3/4}} \, 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{1}{-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 = 9.0558, size = 441, normalized size = 2.07 $-\frac{8 e^{\frac{9}{2} \coth ^{-1}(a x)} \left (\frac{e^{2 \coth ^{-1}(a x)} \left (1906 e^{2 \coth ^{-1}(a x)}+821 e^{4 \coth ^{-1}(a x)}+1117\right ) \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{13}{4}\right \},\left \{1,1,\frac{25}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )}{3094}+\frac{4 e^{2 \coth ^{-1}(a x)} \left (50 e^{2 \coth ^{-1}(a x)}+23 e^{4 \coth ^{-1}(a x)}+27\right ) \text{HypergeometricPFQ}\left (\left \{2,2,2,2,\frac{13}{4}\right \},\left \{1,1,1,\frac{25}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac{8 e^{2 \coth ^{-1}(a x)} \text{HypergeometricPFQ}\left (\left \{2,2,2,2,2,\frac{13}{4}\right \},\left \{1,1,1,1,\frac{25}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac{16 e^{4 \coth ^{-1}(a x)} \text{HypergeometricPFQ}\left (\left \{2,2,2,2,2,\frac{13}{4}\right \},\left \{1,1,1,1,\frac{25}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac{8 e^{6 \coth ^{-1}(a x)} \text{HypergeometricPFQ}\left (\left \{2,2,2,2,2,\frac{13}{4}\right \},\left \{1,1,1,1,\frac{25}{4}\right \},e^{2 \coth ^{-1}(a x)}\right )}{1547}+\frac{899079}{512} e^{-8 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac{60267}{64} e^{-6 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},e^{2 \coth ^{-1}(a x)}\right )-\frac{382227}{256} e^{-4 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},e^{2 \coth ^{-1}(a x)}\right )-\frac{40827}{64} e^{-2 \coth ^{-1}(a x)} \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},e^{2 \coth ^{-1}(a x)}\right )+\frac{133407}{512} \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},e^{2 \coth ^{-1}(a x)}\right )-\frac{899079}{512} e^{-8 \coth ^{-1}(a x)}-\frac{3309759 e^{-6 \coth ^{-1}(a x)}}{2560}+\frac{8521937 e^{-4 \coth ^{-1}(a x)}}{7680}+\frac{69571361 e^{-2 \coth ^{-1}(a x)}}{99840}-\frac{653}{390} e^{2 \coth ^{-1}(a x)}-\frac{27653}{195}\right )}{9 a^3}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*E^((9*ArcCoth[a*x])/2)*(-27653/195 - 899079/(512*E^(8*ArcCoth[a*x])) - 3309759/(2560*E^(6*ArcCoth[a*x])) +
8521937/(7680*E^(4*ArcCoth[a*x])) + 69571361/(99840*E^(2*ArcCoth[a*x])) - (653*E^(2*ArcCoth[a*x]))/390 + (133
407*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/512 + (899079*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcC
oth[a*x])])/(512*E^(8*ArcCoth[a*x])) + (60267*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(64*E^(6*Arc
Coth[a*x])) - (382227*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(256*E^(4*ArcCoth[a*x])) - (40827*Hy
pergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])])/(64*E^(2*ArcCoth[a*x])) + (E^(2*ArcCoth[a*x])*(1117 + 1906*E
^(2*ArcCoth[a*x]) + 821*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{2, 2, 2, 13/4}, {1, 1, 25/4}, E^(2*ArcCoth[a*x]
)])/3094 + (4*E^(2*ArcCoth[a*x])*(27 + 50*E^(2*ArcCoth[a*x]) + 23*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{2, 2,
2, 2, 13/4}, {1, 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/1547 + (8*E^(2*ArcCoth[a*x])*HypergeometricPFQ[{2, 2, 2, 2
, 2, 13/4}, {1, 1, 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/1547 + (16*E^(4*ArcCoth[a*x])*HypergeometricPFQ[{2, 2, 2,
2, 2, 13/4}, {1, 1, 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/1547 + (8*E^(6*ArcCoth[a*x])*HypergeometricPFQ[{2, 2, 2
, 2, 2, 13/4}, {1, 1, 1, 1, 25/4}, E^(2*ArcCoth[a*x])])/1547))/(9*a^3)

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Maple [F]  time = 0.322, 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(1/((a*x-1)/(a*x+1))^(5/4)*x^2,x)

[Out]

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

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Maxima [A]  time = 1.55226, size = 274, normalized size = 1.29 \begin{align*} -\frac{1}{48} \, a{\left (\frac{4 \,{\left (\frac{425 \,{\left (a x - 1\right )}}{a x + 1} - \frac{462 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac{165 \,{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 96\right )}}{a^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{13}{4}} - 3 \, a^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{9}{4}} + 3 \, a^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{4}} - a^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{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}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*a*(4*(425*(a*x - 1)/(a*x + 1) - 462*(a*x - 1)^2/(a*x + 1)^2 + 165*(a*x - 1)^3/(a*x + 1)^3 - 96)/(a^4*((a
*x - 1)/(a*x + 1))^(13/4) - 3*a^4*((a*x - 1)/(a*x + 1))^(9/4) + 3*a^4*((a*x - 1)/(a*x + 1))^(5/4) - a^4*((a*x
- 1)/(a*x + 1))^(1/4)) + 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)

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Fricas [A]  time = 1.58515, size = 358, normalized size = 1.68 \begin{align*} -\frac{330 \,{\left (a x - 1\right )} \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right ) - 165 \,{\left (a x - 1\right )} \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right ) + 165 \,{\left (a x - 1\right )} \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1\right ) - 2 \,{\left (8 \, a^{4} x^{4} + 34 \, a^{3} x^{3} + 87 \, a^{2} x^{2} - 226 \, a x - 287\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}{48 \,{\left (a^{4} x - a^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(330*(a*x - 1)*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 165*(a*x - 1)*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)
+ 165*(a*x - 1)*log(((a*x - 1)/(a*x + 1))^(1/4) - 1) - 2*(8*a^4*x^4 + 34*a^3*x^3 + 87*a^2*x^2 - 226*a*x - 287)
*((a*x - 1)/(a*x + 1))^(3/4))/(a^4*x - a^3)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18243, 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}{a^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}} - \frac{4 \,{\left (\frac{174 \,{\left (a x - 1\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}{a x + 1} - \frac{69 \,{\left (a x - 1\right )}^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}{{\left (a x + 1\right )}^{2}} - 137 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{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(1/((a*x-1)/(a*x+1))^(5/4)*x^2,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^4*((a*x - 1)/(a*x + 1))^(1/4)) - 4*(174*(a*x - 1)*((a*x - 1
)/(a*x + 1))^(3/4)/(a*x + 1) - 69*(a*x - 1)^2*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1)^2 - 137*((a*x - 1)/(a*x +
1))^(3/4))/(a^4*((a*x - 1)/(a*x + 1) - 1)^3))