3.111 \(\int e^{-\frac {5}{2} \tanh ^{-1}(a x)} x^3 \, dx\)

Optimal. Leaf size=317 \[ \frac {(521-452 a x) (1-a x)^{5/4} (a x+1)^{3/4}}{96 a^4}+\frac {475 \sqrt [4]{1-a x} (a x+1)^{3/4}}{64 a^4}+\frac {475 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{128 \sqrt {2} a^4}-\frac {475 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{128 \sqrt {2} a^4}+\frac {475 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{64 \sqrt {2} a^4}-\frac {475 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{64 \sqrt {2} a^4}+\frac {17 x^2 (1-a x)^{5/4} (a x+1)^{3/4}}{4 a^2}-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}} \]

[Out]

-4*x^3*(-a*x+1)^(5/4)/a/(a*x+1)^(1/4)+475/64*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/a^4+17/4*x^2*(-a*x+1)^(5/4)*(a*x+1)^
(3/4)/a^2+1/96*(-452*a*x+521)*(-a*x+1)^(5/4)*(a*x+1)^(3/4)/a^4-475/128*arctan(-1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1
)^(1/4))/a^4*2^(1/2)-475/128*arctan(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))/a^4*2^(1/2)+475/256*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^4*2^(1/2)-475/256*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^4*2^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {6126, 97, 153, 147, 50, 63, 240, 211, 1165, 628, 1162, 617, 204} \[ \frac {17 x^2 (1-a x)^{5/4} (a x+1)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (a x+1)^{3/4}}{96 a^4}+\frac {475 \sqrt [4]{1-a x} (a x+1)^{3/4}}{64 a^4}+\frac {475 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{128 \sqrt {2} a^4}-\frac {475 \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{128 \sqrt {2} a^4}+\frac {475 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{64 \sqrt {2} a^4}-\frac {475 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{64 \sqrt {2} a^4}-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{a x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*x^3*(1 - a*x)^(5/4))/(a*(1 + a*x)^(1/4)) + (475*(1 - a*x)^(1/4)*(1 + a*x)^(3/4))/(64*a^4) + (17*x^2*(1 - a
*x)^(5/4)*(1 + a*x)^(3/4))/(4*a^2) + ((521 - 452*a*x)*(1 - a*x)^(5/4)*(1 + a*x)^(3/4))/(96*a^4) + (475*ArcTan[
1 - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(64*Sqrt[2]*a^4) - (475*ArcTan[1 + (Sqrt[2]*(1 - a*x)^(1/4))/(
1 + a*x)^(1/4)])/(64*Sqrt[2]*a^4) + (475*Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] - (Sqrt[2]*(1 - a*x)^(1/4))/(1 +
a*x)^(1/4)])/(128*Sqrt[2]*a^4) - (475*Log[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x
)^(1/4)])/(128*Sqrt[2]*a^4)

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 97

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)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 153

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

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 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^3 \, dx &=\int \frac {x^3 (1-a x)^{5/4}}{(1+a x)^{5/4}} \, dx\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {4 \int \frac {x^2 \left (3-\frac {17 a x}{4}\right ) \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}} \, dx}{a}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}-\frac {\int \frac {x \sqrt [4]{1-a x} \left (\frac {17 a}{2}-\frac {113 a^2 x}{8}\right )}{\sqrt [4]{1+a x}} \, dx}{a^3}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}+\frac {475 \int \frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}} \, dx}{64 a^3}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {475 \sqrt [4]{1-a x} (1+a x)^{3/4}}{64 a^4}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}+\frac {475 \int \frac {1}{(1-a x)^{3/4} \sqrt [4]{1+a x}} \, dx}{128 a^3}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {475 \sqrt [4]{1-a x} (1+a x)^{3/4}}{64 a^4}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}-\frac {475 \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-a x}\right )}{32 a^4}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {475 \sqrt [4]{1-a x} (1+a x)^{3/4}}{64 a^4}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}-\frac {475 \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{32 a^4}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {475 \sqrt [4]{1-a x} (1+a x)^{3/4}}{64 a^4}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}-\frac {475 \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 )}{64 a^4}-\frac {475 \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 )}{64 a^4}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {475 \sqrt [4]{1-a x} (1+a x)^{3/4}}{64 a^4}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}-\frac {475 \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 )}{128 a^4}-\frac {475 \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 )}{128 a^4}+\frac {475 \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 )}{128 \sqrt {2} a^4}+\frac {475 \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 )}{128 \sqrt {2} a^4}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {475 \sqrt [4]{1-a x} (1+a x)^{3/4}}{64 a^4}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}+\frac {475 \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 )}{128 \sqrt {2} a^4}-\frac {475 \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 )}{128 \sqrt {2} a^4}-\frac {475 \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 )}{64 \sqrt {2} a^4}+\frac {475 \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 )}{64 \sqrt {2} a^4}\\ &=-\frac {4 x^3 (1-a x)^{5/4}}{a \sqrt [4]{1+a x}}+\frac {475 \sqrt [4]{1-a x} (1+a x)^{3/4}}{64 a^4}+\frac {17 x^2 (1-a x)^{5/4} (1+a x)^{3/4}}{4 a^2}+\frac {(521-452 a x) (1-a x)^{5/4} (1+a x)^{3/4}}{96 a^4}+\frac {475 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{64 \sqrt {2} a^4}-\frac {475 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{64 \sqrt {2} a^4}+\frac {475 \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 )}{128 \sqrt {2} a^4}-\frac {475 \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 )}{128 \sqrt {2} a^4}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 79, normalized size = 0.25 \[ -\frac {(1-a x)^{9/4} \left (3 \left (6 a^2 x^2-5 a x-59\right )+95\ 2^{3/4} \sqrt [4]{a x+1} \, _2F_1\left (\frac {1}{4},\frac {9}{4};\frac {13}{4};\frac {1}{2} (1-a x)\right )\right )}{72 a^4 \sqrt [4]{a x+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/72*((1 - a*x)^(9/4)*(3*(-59 - 5*a*x + 6*a^2*x^2) + 95*2^(3/4)*(1 + a*x)^(1/4)*Hypergeometric2F1[1/4, 9/4, 1
3/4, (1 - a*x)/2]))/(a^4*(1 + a*x)^(1/4))

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fricas [B]  time = 0.54, size = 597, normalized size = 1.88 \[ -\frac {5700 \, \sqrt {2} {\left (a^{5} x + a^{4}\right )} \frac {1}{a^{16}}^{\frac {1}{4}} \arctan \left (\sqrt {2} a^{4} \sqrt {\frac {\sqrt {2} {\left (a^{13} x - a^{12}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {3}{4}} + {\left (a^{9} x - a^{8}\right )} \sqrt {\frac {1}{a^{16}}} - \sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {1}{4}} - \sqrt {2} a^{4} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {1}{4}} - 1\right ) + 5700 \, \sqrt {2} {\left (a^{5} x + a^{4}\right )} \frac {1}{a^{16}}^{\frac {1}{4}} \arctan \left (\sqrt {2} a^{4} \sqrt {-\frac {\sqrt {2} {\left (a^{13} x - a^{12}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {3}{4}} - {\left (a^{9} x - a^{8}\right )} \sqrt {\frac {1}{a^{16}}} + \sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {1}{4}} - \sqrt {2} a^{4} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {1}{4}} + 1\right ) + 1425 \, \sqrt {2} {\left (a^{5} x + a^{4}\right )} \frac {1}{a^{16}}^{\frac {1}{4}} \log \left (\frac {\sqrt {2} {\left (a^{13} x - a^{12}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {3}{4}} + {\left (a^{9} x - a^{8}\right )} \sqrt {\frac {1}{a^{16}}} - \sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right ) - 1425 \, \sqrt {2} {\left (a^{5} x + a^{4}\right )} \frac {1}{a^{16}}^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} {\left (a^{13} x - a^{12}\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} \frac {1}{a^{16}}^{\frac {3}{4}} - {\left (a^{9} x - a^{8}\right )} \sqrt {\frac {1}{a^{16}}} + \sqrt {-a^{2} x^{2} + 1}}{a x - 1}\right ) + 4 \, {\left (48 \, a^{4} x^{4} - 136 \, a^{3} x^{3} + 226 \, a^{2} x^{2} - 521 \, a x - 2467\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{768 \, {\left (a^{5} x + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="fricas")

[Out]

-1/768*(5700*sqrt(2)*(a^5*x + a^4)*(a^(-16))^(1/4)*arctan(sqrt(2)*a^4*sqrt((sqrt(2)*(a^13*x - a^12)*sqrt(-sqrt
(-a^2*x^2 + 1)/(a*x - 1))*(a^(-16))^(3/4) + (a^9*x - a^8)*sqrt(a^(-16)) - sqrt(-a^2*x^2 + 1))/(a*x - 1))*(a^(-
16))^(1/4) - sqrt(2)*a^4*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-16))^(1/4) - 1) + 5700*sqrt(2)*(a^5*x + a^4)
*(a^(-16))^(1/4)*arctan(sqrt(2)*a^4*sqrt(-(sqrt(2)*(a^13*x - a^12)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-16
))^(3/4) - (a^9*x - a^8)*sqrt(a^(-16)) + sqrt(-a^2*x^2 + 1))/(a*x - 1))*(a^(-16))^(1/4) - sqrt(2)*a^4*sqrt(-sq
rt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-16))^(1/4) + 1) + 1425*sqrt(2)*(a^5*x + a^4)*(a^(-16))^(1/4)*log((sqrt(2)*(a^
13*x - a^12)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))*(a^(-16))^(3/4) + (a^9*x - a^8)*sqrt(a^(-16)) - sqrt(-a^2*x^2
+ 1))/(a*x - 1)) - 1425*sqrt(2)*(a^5*x + a^4)*(a^(-16))^(1/4)*log(-(sqrt(2)*(a^13*x - a^12)*sqrt(-sqrt(-a^2*x
^2 + 1)/(a*x - 1))*(a^(-16))^(3/4) - (a^9*x - a^8)*sqrt(a^(-16)) + sqrt(-a^2*x^2 + 1))/(a*x - 1)) + 4*(48*a^4*
x^4 - 136*a^3*x^3 + 226*a^2*x^2 - 521*a*x - 2467)*sqrt(-a^2*x^2 + 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)))/(a^5
*x + a^4)

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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(x^3/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)

[Out]

int(x^3/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3/((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 \[ \int \frac {x^3}{{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2),x)

[Out]

int(x^3/((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/((a*x+1)/(-a**2*x**2+1)**(1/2))**(5/2),x)

[Out]

Integral(x**3/((a*x + 1)/sqrt(-a**2*x**2 + 1))**(5/2), x)

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