\(\int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx\) [105]

Optimal result

Integrand size = 14, antiderivative size = 250 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=-\frac {2467 \sqrt [4]{1-\frac {1}{a x}}}{192 a^4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {521 \sqrt [4]{1-\frac {1}{a x}} x}{192 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {475 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{64 a^4}+\frac {475 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{64 a^4} \]

[Out]

-2467/192*(1-1/a/x)^(1/4)/a^4/(1+1/a/x)^(1/4)-521/192*(1-1/a/x)^(1/4)*x/a^3/(1+1/a/x)^(1/4)+113/96*(1-1/a/x)^(
1/4)*x^2/a^2/(1+1/a/x)^(1/4)-17/24*(1-1/a/x)^(1/4)*x^3/a/(1+1/a/x)^(1/4)+1/4*(1-1/a/x)^(1/4)*x^4/(1+1/a/x)^(1/
4)-475/64*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^4+475/64*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^4

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6306, 100, 156, 160, 12, 95, 304, 209, 212} \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=-\frac {475 \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{64 a^4}+\frac {475 \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{64 a^4}-\frac {2467 \sqrt [4]{1-\frac {1}{a x}}}{192 a^4 \sqrt [4]{\frac {1}{a x}+1}}-\frac {521 x \sqrt [4]{1-\frac {1}{a x}}}{192 a^3 \sqrt [4]{\frac {1}{a x}+1}}+\frac {113 x^2 \sqrt [4]{1-\frac {1}{a x}}}{96 a^2 \sqrt [4]{\frac {1}{a x}+1}}+\frac {x^4 \sqrt [4]{1-\frac {1}{a x}}}{4 \sqrt [4]{\frac {1}{a x}+1}}-\frac {17 x^3 \sqrt [4]{1-\frac {1}{a x}}}{24 a \sqrt [4]{\frac {1}{a x}+1}} \]

[In]

Int[x^3/E^((5*ArcCoth[a*x])/2),x]

[Out]

(-2467*(1 - 1/(a*x))^(1/4))/(192*a^4*(1 + 1/(a*x))^(1/4)) - (521*(1 - 1/(a*x))^(1/4)*x)/(192*a^3*(1 + 1/(a*x))
^(1/4)) + (113*(1 - 1/(a*x))^(1/4)*x^2)/(96*a^2*(1 + 1/(a*x))^(1/4)) - (17*(1 - 1/(a*x))^(1/4)*x^3)/(24*a*(1 +
1/(a*x))^(1/4)) + ((1 - 1/(a*x))^(1/4)*x^4)/(4*(1 + 1/(a*x))^(1/4)) - (475*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/
(a*x))^(1/4)])/(64*a^4) + (475*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(64*a^4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 156

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, -1]

Rule 160

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 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 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 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 6306

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]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{5/4}}{x^5 \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}+\frac {1}{4} \text {Subst}\left (\int \frac {\frac {17}{2 a}-\frac {8 x}{a^2}}{x^4 \left (1-\frac {x}{a}\right )^{3/4} \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {1}{12} \text {Subst}\left (\int \frac {\frac {113}{4 a^2}-\frac {51 x}{2 a^3}}{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 {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}+\frac {1}{24} \text {Subst}\left (\int \frac {\frac {521}{8 a^3}-\frac {113 x}{2 a^4}}{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 {521 \sqrt [4]{1-\frac {1}{a x}} x}{192 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {1}{24} \text {Subst}\left (\int \frac {\frac {1425}{16 a^4}-\frac {521 x}{8 a^5}}{x \left (1-\frac {x}{a}\right )^{3/4} \left (1+\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {2467 \sqrt [4]{1-\frac {1}{a x}}}{192 a^4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {521 \sqrt [4]{1-\frac {1}{a x}} x}{192 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {1}{12} a \text {Subst}\left (\int \frac {1425}{32 a^5 x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {2467 \sqrt [4]{1-\frac {1}{a x}}}{192 a^4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {521 \sqrt [4]{1-\frac {1}{a x}} x}{192 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {475 \text {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 )}{128 a^4} \\ & = -\frac {2467 \sqrt [4]{1-\frac {1}{a x}}}{192 a^4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {521 \sqrt [4]{1-\frac {1}{a x}} x}{192 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {475 \text {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 )}{32 a^4} \\ & = -\frac {2467 \sqrt [4]{1-\frac {1}{a x}}}{192 a^4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {521 \sqrt [4]{1-\frac {1}{a x}} x}{192 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}+\frac {475 \text {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 )}{64 a^4}-\frac {475 \text {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 )}{64 a^4} \\ & = -\frac {2467 \sqrt [4]{1-\frac {1}{a x}}}{192 a^4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {521 \sqrt [4]{1-\frac {1}{a x}} x}{192 a^3 \sqrt [4]{1+\frac {1}{a x}}}+\frac {113 \sqrt [4]{1-\frac {1}{a x}} x^2}{96 a^2 \sqrt [4]{1+\frac {1}{a x}}}-\frac {17 \sqrt [4]{1-\frac {1}{a x}} x^3}{24 a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\sqrt [4]{1-\frac {1}{a x}} x^4}{4 \sqrt [4]{1+\frac {1}{a x}}}-\frac {475 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{64 a^4}+\frac {475 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{64 a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.64 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=\frac {-3072 e^{-\frac {1}{2} \coth ^{-1}(a x)}+\frac {1536 e^{\frac {15}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^4}-\frac {5248 e^{\frac {11}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^3}+\frac {7376 e^{\frac {7}{2} \coth ^{-1}(a x)}}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}-\frac {6292 e^{\frac {3}{2} \coth ^{-1}(a x)}}{-1+e^{2 \coth ^{-1}(a x)}}+2850 \arctan \left (e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )-1425 \log \left (1-e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )+1425 \log \left (1+e^{-\frac {1}{2} \coth ^{-1}(a x)}\right )}{384 a^4} \]

[In]

Integrate[x^3/E^((5*ArcCoth[a*x])/2),x]

[Out]

(-3072/E^(ArcCoth[a*x]/2) + (1536*E^((15*ArcCoth[a*x])/2))/(-1 + E^(2*ArcCoth[a*x]))^4 - (5248*E^((11*ArcCoth[
a*x])/2))/(-1 + E^(2*ArcCoth[a*x]))^3 + (7376*E^((7*ArcCoth[a*x])/2))/(-1 + E^(2*ArcCoth[a*x]))^2 - (6292*E^((
3*ArcCoth[a*x])/2))/(-1 + E^(2*ArcCoth[a*x])) + 2850*ArcTan[E^(-1/2*ArcCoth[a*x])] - 1425*Log[1 - E^(-1/2*ArcC
oth[a*x])] + 1425*Log[1 + E^(-1/2*ArcCoth[a*x])])/(384*a^4)

Maple [F]

\[\int x^{3} \left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}d x\]

[In]

int(x^3*((a*x-1)/(a*x+1))^(5/4),x)

[Out]

int(x^3*((a*x-1)/(a*x+1))^(5/4),x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.44 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=\frac {2 \, {\left (48 \, a^{4} x^{4} - 136 \, a^{3} x^{3} + 226 \, a^{2} x^{2} - 521 \, a x - 2467\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 2850 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + 1425 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - 1425 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{384 \, a^{4}} \]

[In]

integrate(x^3*((a*x-1)/(a*x+1))^(5/4),x, algorithm="fricas")

[Out]

1/384*(2*(48*a^4*x^4 - 136*a^3*x^3 + 226*a^2*x^2 - 521*a*x - 2467)*((a*x - 1)/(a*x + 1))^(1/4) + 2850*arctan((
(a*x - 1)/(a*x + 1))^(1/4)) + 1425*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) - 1425*log(((a*x - 1)/(a*x + 1))^(1/4)
- 1))/a^4

Sympy [F]

\[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=\int x^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}}\, dx \]

[In]

integrate(x**3*((a*x-1)/(a*x+1))**(5/4),x)

[Out]

Integral(x**3*((a*x - 1)/(a*x + 1))**(5/4), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.98 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=-\frac {1}{384} \, a {\left (\frac {4 \, {\left (1573 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {13}{4}} - 2875 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} + 2343 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - 657 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {4 \, {\left (a x - 1\right )} a^{5}}{a x + 1} - \frac {6 \, {\left (a x - 1\right )}^{2} a^{5}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3} a^{5}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{5}}{{\left (a x + 1\right )}^{4}} - a^{5}} - \frac {2850 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{5}} - \frac {1425 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{5}} + \frac {1425 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{5}} + \frac {3072 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{5}}\right )} \]

[In]

integrate(x^3*((a*x-1)/(a*x+1))^(5/4),x, algorithm="maxima")

[Out]

-1/384*a*(4*(1573*((a*x - 1)/(a*x + 1))^(13/4) - 2875*((a*x - 1)/(a*x + 1))^(9/4) + 2343*((a*x - 1)/(a*x + 1))
^(5/4) - 657*((a*x - 1)/(a*x + 1))^(1/4))/(4*(a*x - 1)*a^5/(a*x + 1) - 6*(a*x - 1)^2*a^5/(a*x + 1)^2 + 4*(a*x
- 1)^3*a^5/(a*x + 1)^3 - (a*x - 1)^4*a^5/(a*x + 1)^4 - a^5) - 2850*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^5 - 1
425*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^5 + 1425*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^5 + 3072*((a*x - 1)
/(a*x + 1))^(1/4)/a^5)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.89 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=\frac {1}{384} \, a {\left (\frac {2850 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{5}} + \frac {1425 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{5}} - \frac {1425 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{5}} - \frac {3072 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{5}} + \frac {4 \, {\left (\frac {2343 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \frac {2875 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{{\left (a x + 1\right )}^{2}} + \frac {1573 \, {\left (a x - 1\right )}^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{{\left (a x + 1\right )}^{3}} - 657 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{5} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{4}}\right )} \]

[In]

integrate(x^3*((a*x-1)/(a*x+1))^(5/4),x, algorithm="giac")

[Out]

1/384*a*(2850*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^5 + 1425*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^5 - 1425*l
og(abs(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^5 - 3072*((a*x - 1)/(a*x + 1))^(1/4)/a^5 + 4*(2343*(a*x - 1)*((a*x
- 1)/(a*x + 1))^(1/4)/(a*x + 1) - 2875*(a*x - 1)^2*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1)^2 + 1573*(a*x - 1)^3*
((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1)^3 - 657*((a*x - 1)/(a*x + 1))^(1/4))/(a^5*((a*x - 1)/(a*x + 1) - 1)^4))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.87 \[ \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x^3 \, dx=\frac {475\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{64\,a^4}-\frac {\frac {219\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{32}-\frac {781\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{32}+\frac {2875\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}}{96}-\frac {1573\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{13/4}}{96}}{a^4+\frac {6\,a^4\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {4\,a^4\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {a^4\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}-\frac {4\,a^4\,\left (a\,x-1\right )}{a\,x+1}}-\frac {8\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a^4}-\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )\,475{}\mathrm {i}}{64\,a^4} \]

[In]

int(x^3*((a*x - 1)/(a*x + 1))^(5/4),x)

[Out]

(475*atan(((a*x - 1)/(a*x + 1))^(1/4)))/(64*a^4) - (8*((a*x - 1)/(a*x + 1))^(1/4))/a^4 - ((219*((a*x - 1)/(a*x
+ 1))^(1/4))/32 - (781*((a*x - 1)/(a*x + 1))^(5/4))/32 + (2875*((a*x - 1)/(a*x + 1))^(9/4))/96 - (1573*((a*x
- 1)/(a*x + 1))^(13/4))/96)/(a^4 + (6*a^4*(a*x - 1)^2)/(a*x + 1)^2 - (4*a^4*(a*x - 1)^3)/(a*x + 1)^3 + (a^4*(a
*x - 1)^4)/(a*x + 1)^4 - (4*a^4*(a*x - 1))/(a*x + 1)) - (atan(((a*x - 1)/(a*x + 1))^(1/4)*1i)*475i)/(64*a^4)