3.97 \(\int e^{-\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=179 \[ -\frac {17 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {17 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {23 x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{24 a^2}+\frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {7 x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{12 a} \]

[Out]

23/24*(1-1/a/x)^(3/4)*(1+1/a/x)^(1/4)*x/a^2-7/12*(1-1/a/x)^(3/4)*(1+1/a/x)^(1/4)*x^2/a+1/3*(1-1/a/x)^(3/4)*(1+
1/a/x)^(1/4)*x^3-17/8*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^3-17/8*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4)
)/a^3

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Rubi [A]  time = 0.09, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6171, 99, 151, 12, 93, 212, 206, 203} \[ \frac {23 x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{24 a^2}-\frac {17 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {17 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}+\frac {1}{3} x^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {7 x^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}}{12 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4)*x)/(24*a^2) - (7*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4)*x^2)/(12*
a) + ((1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4)*x^3)/3 - (17*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(8
*a^3) - (17*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(8*a^3)

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 99

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

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 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]

Rubi steps

\begin {align*} \int e^{-\frac {3}{2} \coth ^{-1}(a x)} x^2 \, dx &=-\operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{3/4}}{x^4 \left (1+\frac {x}{a}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3-\frac {1}{3} \operatorname {Subst}\left (\int \frac {-\frac {7}{2 a}+\frac {2 x}{a^2}}{x^3 \sqrt [4]{1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {7 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}+\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3+\frac {1}{6} \operatorname {Subst}\left (\int \frac {-\frac {23}{4 a^2}+\frac {7 x}{2 a^3}}{x^2 \sqrt [4]{1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {23 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2}-\frac {7 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}+\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3-\frac {1}{6} \operatorname {Subst}\left (\int -\frac {51}{8 a^3 x \sqrt [4]{1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {23 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2}-\frac {7 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}+\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3+\frac {17 \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 {23 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2}-\frac {7 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}+\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3+\frac {17 \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 {23 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2}-\frac {7 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}+\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3-\frac {17 \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 {17 \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 {23 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x}{24 a^2}-\frac {7 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^2}{12 a}+\frac {1}{3} \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x^3-\frac {17 \tan ^{-1}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}-\frac {17 \tanh ^{-1}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{8 a^3}\\ \end {align*}

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Mathematica [C]  time = 8.65, size = 389, normalized size = 2.17 \[ -\frac {e^{-\frac {7}{2} \coth ^{-1}(a x)} \left (256 e^{6 \coth ^{-1}(a x)} \left (850 e^{2 \coth ^{-1}(a x)}+325 e^{4 \coth ^{-1}(a x)}+557\right ) \, _4F_3\left (\frac {5}{4},2,2,2;1,1,\frac {17}{4};e^{2 \coth ^{-1}(a x)}\right )+2048 e^{6 \coth ^{-1}(a x)} \left (34 e^{2 \coth ^{-1}(a x)}+15 e^{4 \coth ^{-1}(a x)}+19\right ) \, _5F_4\left (\frac {5}{4},2,2,2,2;1,1,1,\frac {17}{4};e^{2 \coth ^{-1}(a x)}\right )+4096 e^{6 \coth ^{-1}(a x)} \, _6F_5\left (\frac {5}{4},2,2,2,2,2;1,1,1,1,\frac {17}{4};e^{2 \coth ^{-1}(a x)}\right )+8192 e^{8 \coth ^{-1}(a x)} \, _6F_5\left (\frac {5}{4},2,2,2,2,2;1,1,1,1,\frac {17}{4};e^{2 \coth ^{-1}(a x)}\right )+4096 e^{10 \coth ^{-1}(a x)} \, _6F_5\left (\frac {5}{4},2,2,2,2,2;1,1,1,1,\frac {17}{4};e^{2 \coth ^{-1}(a x)}\right )[phone] e^{2 \coth ^{-1}(a x)} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{2 \coth ^{-1}(a x)}\right )-2472210 e^{4 \coth ^{-1}(a x)} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{2 \coth ^{-1}(a x)}\right )-3598920 e^{6 \coth ^{-1}(a x)} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{2 \coth ^{-1}(a x)}\right )+21645 e^{8 \coth ^{-1}(a x)} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{2 \coth ^{-1}(a x)}\right )[phone] \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{2 \coth ^{-1}(a x)}\right )-17312841 e^{2 \coth ^{-1}(a x)}-1213875 e^{4 \coth ^{-1}(a x)}+2199249 e^{6 \coth ^{-1}(a x)}-15779205\right )}{112320 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/112320*(-15779205 - 17312841*E^(2*ArcCoth[a*x]) - 1213875*E^(4*ArcCoth[a*x]) + 2199249*E^(6*ArcCoth[a*x]) +
15779205*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])] + 14157000*E^(2*ArcCoth[a*x])*Hypergeometric2F1[1
/4, 1, 5/4, E^(2*ArcCoth[a*x])] - 2472210*E^(4*ArcCoth[a*x])*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])
] - 3598920*E^(6*ArcCoth[a*x])*Hypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])] + 21645*E^(8*ArcCoth[a*x])*H
ypergeometric2F1[1/4, 1, 5/4, E^(2*ArcCoth[a*x])] + 256*E^(6*ArcCoth[a*x])*(557 + 850*E^(2*ArcCoth[a*x]) + 325
*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{5/4, 2, 2, 2}, {1, 1, 17/4}, E^(2*ArcCoth[a*x])] + 2048*E^(6*ArcCoth[a
*x])*(19 + 34*E^(2*ArcCoth[a*x]) + 15*E^(4*ArcCoth[a*x]))*HypergeometricPFQ[{5/4, 2, 2, 2, 2}, {1, 1, 1, 17/4}
, E^(2*ArcCoth[a*x])] + 4096*E^(6*ArcCoth[a*x])*HypergeometricPFQ[{5/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 17/4}, E^
(2*ArcCoth[a*x])] + 8192*E^(8*ArcCoth[a*x])*HypergeometricPFQ[{5/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 17/4}, E^(2*A
rcCoth[a*x])] + 4096*E^(10*ArcCoth[a*x])*HypergeometricPFQ[{5/4, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 17/4}, E^(2*ArcC
oth[a*x])])/(a^3*E^((7*ArcCoth[a*x])/2))

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fricas [A]  time = 0.50, size = 103, normalized size = 0.58 \[ \frac {2 \, {\left (8 \, a^{3} x^{3} - 6 \, a^{2} x^{2} + 9 \, a x + 23\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}} + 102 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{48 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*((a*x-1)/(a*x+1))^(3/4),x, algorithm="fricas")

[Out]

1/48*(2*(8*a^3*x^3 - 6*a^2*x^2 + 9*a*x + 23)*((a*x - 1)/(a*x + 1))^(3/4) + 102*arctan(((a*x - 1)/(a*x + 1))^(1
/4)) - 51*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) + 51*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3

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giac [A]  time = 0.22, size = 172, normalized size = 0.96 \[ \frac {1}{48} \, a {\left (\frac {102 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} - \frac {51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} + \frac {51 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{4}} + \frac {4 \, {\left (\frac {30 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - \frac {45 \, {\left (a x - 1\right )}^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{{\left (a x + 1\right )}^{2}} - 17 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{4} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*((a*x-1)/(a*x+1))^(3/4),x, algorithm="giac")

[Out]

1/48*a*(102*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^4 - 51*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 + 51*log(abs
(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^4 + 4*(30*(a*x - 1)*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1) - 45*(a*x - 1)^
2*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1)^2 - 17*((a*x - 1)/(a*x + 1))^(3/4))/(a^4*((a*x - 1)/(a*x + 1) - 1)^3))

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*((a*x-1)/(a*x+1))^(3/4),x)

[Out]

int(x^2*((a*x-1)/(a*x+1))^(3/4),x)

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maxima [A]  time = 0.41, size = 187, normalized size = 1.04 \[ -\frac {1}{48} \, a {\left (\frac {4 \, {\left (45 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {11}{4}} - 30 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{4}} + 17 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{4}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{4}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{4}}{{\left (a x + 1\right )}^{3}} - a^{4}} - \frac {102 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{4}} + \frac {51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{4}} - \frac {51 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*((a*x-1)/(a*x+1))^(3/4),x, algorithm="maxima")

[Out]

-1/48*a*(4*(45*((a*x - 1)/(a*x + 1))^(11/4) - 30*((a*x - 1)/(a*x + 1))^(7/4) + 17*((a*x - 1)/(a*x + 1))^(3/4))
/(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) - 102*arctan(
((a*x - 1)/(a*x + 1))^(1/4))/a^4 + 51*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^4 - 51*log(((a*x - 1)/(a*x + 1))^
(1/4) - 1)/a^4)

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mupad [B]  time = 0.06, size = 157, normalized size = 0.88 \[ \frac {\frac {17\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/4}}{12}-\frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/4}}{2}+\frac {15\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{11/4}}{4}}{a^3+\frac {3\,a^3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a^3\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,a^3\,\left (a\,x-1\right )}{a\,x+1}}+\frac {17\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3}-\frac {17\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*((a*x - 1)/(a*x + 1))^(3/4),x)

[Out]

((17*((a*x - 1)/(a*x + 1))^(3/4))/12 - (5*((a*x - 1)/(a*x + 1))^(7/4))/2 + (15*((a*x - 1)/(a*x + 1))^(11/4))/4
)/(a^3 + (3*a^3*(a*x - 1)^2)/(a*x + 1)^2 - (a^3*(a*x - 1)^3)/(a*x + 1)^3 - (3*a^3*(a*x - 1))/(a*x + 1)) + (17*
atan(((a*x - 1)/(a*x + 1))^(1/4)))/(8*a^3) - (17*atanh(((a*x - 1)/(a*x + 1))^(1/4)))/(8*a^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*((a*x-1)/(a*x+1))**(3/4),x)

[Out]

Integral(x**2*((a*x - 1)/(a*x + 1))**(3/4), x)

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