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

Optimal. Leaf size=157 \[ -\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}+\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}-\frac{3 c^3 (a x+8) \sqrt{1-a^2 x^2}}{8 a^2 x}+\frac{3 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{8 a}-\frac{3 c^3 \sin ^{-1}(a x)}{a} \]

[Out]

(-3*c^3*(8 + a*x)*Sqrt[1 - a^2*x^2])/(8*a^2*x) + (c^3*(8 - a*x)*(1 - a^2*x^2)^(3/2))/(8*a^4*x^3) + (c^3*(1 - a
^2*x^2)^(5/2))/(5*a^6*x^5) - (3*c^3*(1 - a^2*x^2)^(5/2))/(4*a^5*x^4) - (3*c^3*ArcSin[a*x])/a + (3*c^3*ArcTanh[
Sqrt[1 - a^2*x^2]])/(8*a)

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Rubi [A]  time = 0.321341, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {6157, 6149, 1807, 811, 813, 844, 216, 266, 63, 208} \[ -\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}+\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}-\frac{3 c^3 (a x+8) \sqrt{1-a^2 x^2}}{8 a^2 x}+\frac{3 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{8 a}-\frac{3 c^3 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*c^3*(8 + a*x)*Sqrt[1 - a^2*x^2])/(8*a^2*x) + (c^3*(8 - a*x)*(1 - a^2*x^2)^(3/2))/(8*a^4*x^3) + (c^3*(1 - a
^2*x^2)^(5/2))/(5*a^6*x^5) - (3*c^3*(1 - a^2*x^2)^(5/2))/(4*a^5*x^4) - (3*c^3*ArcSin[a*x])/a + (3*c^3*ArcTanh[
Sqrt[1 - a^2*x^2]])/(8*a)

Rule 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rule 6149

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(x^m*(1 -
a^2*x^2)^(p + n/2))/(1 - a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || G
tQ[c, 0]) && ILtQ[(n - 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right )^3 \, dx &=-\frac{c^3 \int \frac{e^{-3 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^3}{x^6} \, dx}{a^6}\\ &=-\frac{c^3 \int \frac{(1-a x)^3 \left (1-a^2 x^2\right )^{3/2}}{x^6} \, dx}{a^6}\\ &=\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}+\frac{c^3 \int \frac{\left (1-a^2 x^2\right )^{3/2} \left (15 a-15 a^2 x+5 a^3 x^2\right )}{x^5} \, dx}{5 a^6}\\ &=\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}-\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}-\frac{c^3 \int \frac{\left (60 a^2-5 a^3 x\right ) \left (1-a^2 x^2\right )^{3/2}}{x^4} \, dx}{20 a^6}\\ &=\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}-\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}+\frac{c^3 \int \frac{\left (240 a^4-30 a^5 x\right ) \sqrt{1-a^2 x^2}}{x^2} \, dx}{80 a^6}\\ &=-\frac{3 c^3 (8+a x) \sqrt{1-a^2 x^2}}{8 a^2 x}+\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}-\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}-\frac{c^3 \int \frac{60 a^5+480 a^6 x}{x \sqrt{1-a^2 x^2}} \, dx}{160 a^6}\\ &=-\frac{3 c^3 (8+a x) \sqrt{1-a^2 x^2}}{8 a^2 x}+\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}-\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}-\left (3 c^3\right ) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx-\frac{\left (3 c^3\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{8 a}\\ &=-\frac{3 c^3 (8+a x) \sqrt{1-a^2 x^2}}{8 a^2 x}+\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}-\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}-\frac{3 c^3 \sin ^{-1}(a x)}{a}-\frac{\left (3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{16 a}\\ &=-\frac{3 c^3 (8+a x) \sqrt{1-a^2 x^2}}{8 a^2 x}+\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}-\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}-\frac{3 c^3 \sin ^{-1}(a x)}{a}+\frac{\left (3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{8 a^3}\\ &=-\frac{3 c^3 (8+a x) \sqrt{1-a^2 x^2}}{8 a^2 x}+\frac{c^3 (8-a x) \left (1-a^2 x^2\right )^{3/2}}{8 a^4 x^3}+\frac{c^3 \left (1-a^2 x^2\right )^{5/2}}{5 a^6 x^5}-\frac{3 c^3 \left (1-a^2 x^2\right )^{5/2}}{4 a^5 x^4}-\frac{3 c^3 \sin ^{-1}(a x)}{a}+\frac{3 c^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{8 a}\\ \end{align*}

Mathematica [C]  time = 0.0988295, size = 186, normalized size = 1.18 \[ \frac{c^3 \left (8 a^5 x^5 \left (a^2 x^2-1\right )^3 \text{Hypergeometric2F1}\left (2,\frac{5}{2},\frac{7}{2},1-a^2 x^2\right )+40 a^2 x^2 \sqrt{1-a^2 x^2} \text{Hypergeometric2F1}\left (-\frac{3}{2},-\frac{3}{2},-\frac{1}{2},a^2 x^2\right )-8 a^6 x^6-75 a^5 x^5+24 a^4 x^4+105 a^3 x^3-24 a^2 x^2-45 a^5 x^5 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-30 a x+8\right )}{40 a^6 x^5 \sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*(8 - 30*a*x - 24*a^2*x^2 + 105*a^3*x^3 + 24*a^4*x^4 - 75*a^5*x^5 - 8*a^6*x^6 - 45*a^5*x^5*Sqrt[1 - a^2*x^
2]*ArcTanh[Sqrt[1 - a^2*x^2]] + 40*a^2*x^2*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[-3/2, -3/2, -1/2, a^2*x^2] + 8*
a^5*x^5*(-1 + a^2*x^2)^3*Hypergeometric2F1[2, 5/2, 7/2, 1 - a^2*x^2]))/(40*a^6*x^5*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.066, size = 243, normalized size = 1.6 \begin{align*} -{\frac{3\,{c}^{3}}{4\,{a}^{5}{x}^{4}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{{c}^{3}}{8\,{x}^{2}{a}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{{c}^{3}}{8\,a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{c}^{3}}{8\,a}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{3\,{c}^{3}}{8\,a}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{{c}^{3}}{5\,{a}^{6}{x}^{5}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{{c}^{3}}{{a}^{4}{x}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-2\,{\frac{{c}^{3} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{5/2}}{{a}^{2}x}}-2\,{c}^{3}x \left ( -{a}^{2}{x}^{2}+1 \right ) ^{3/2}-3\,{c}^{3}x\sqrt{-{a}^{2}{x}^{2}+1}-3\,{\frac{{c}^{3}}{\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*c^3*(-a^2*x^2+1)^(5/2)/a^5/x^4-1/8*c^3/a^3/x^2*(-a^2*x^2+1)^(5/2)-1/8*c^3*(-a^2*x^2+1)^(3/2)/a-3/8*c^3*(-
a^2*x^2+1)^(1/2)/a+3/8*c^3/a*arctanh(1/(-a^2*x^2+1)^(1/2))+1/5*c^3*(-a^2*x^2+1)^(5/2)/a^6/x^5+c^3/a^4/x^3*(-a^
2*x^2+1)^(5/2)-2*c^3/a^2/x*(-a^2*x^2+1)^(5/2)-2*c^3*x*(-a^2*x^2+1)^(3/2)-3*c^3*x*(-a^2*x^2+1)^(1/2)-3*c^3/(a^2
)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{3}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.1836, size = 338, normalized size = 2.15 \begin{align*} \frac{240 \, a^{5} c^{3} x^{5} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) - 15 \, a^{5} c^{3} x^{5} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - 40 \, a^{5} c^{3} x^{5} -{\left (40 \, a^{5} c^{3} x^{5} + 152 \, a^{4} c^{3} x^{4} - 55 \, a^{3} c^{3} x^{3} - 24 \, a^{2} c^{3} x^{2} + 30 \, a c^{3} x - 8 \, c^{3}\right )} \sqrt{-a^{2} x^{2} + 1}}{40 \, a^{6} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(240*a^5*c^3*x^5*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - 15*a^5*c^3*x^5*log((sqrt(-a^2*x^2 + 1) - 1)/x)
- 40*a^5*c^3*x^5 - (40*a^5*c^3*x^5 + 152*a^4*c^3*x^4 - 55*a^3*c^3*x^3 - 24*a^2*c^3*x^2 + 30*a*c^3*x - 8*c^3)*s
qrt(-a^2*x^2 + 1))/(a^6*x^5)

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Sympy [C]  time = 45.8976, size = 695, normalized size = 4.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c**3*Piecewise((I*sqrt(a**2*x**2 - 1) - log(a*x) + log(a**2*x**2)/2 + I*asin(1/(a*x)), Abs(a**2*x**2) > 1), (
sqrt(-a**2*x**2 + 1) + log(a**2*x**2)/2 - log(sqrt(-a**2*x**2 + 1) + 1), True))/a + 3*c**3*Piecewise((-I*a**2*
x/sqrt(a**2*x**2 - 1) + I*a*acosh(a*x) + I/(x*sqrt(a**2*x**2 - 1)), Abs(a**2*x**2) > 1), (a**2*x/sqrt(-a**2*x*
*2 + 1) - a*asin(a*x) - 1/(x*sqrt(-a**2*x**2 + 1)), True))/a**2 - 2*c**3*Piecewise((a**2*acosh(1/(a*x))/2 + a/
(2*x*sqrt(-1 + 1/(a**2*x**2))) - 1/(2*a*x**3*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (-I*a**2*asin(1
/(a*x))/2 - I*a*sqrt(1 - 1/(a**2*x**2))/(2*x), True))/a**3 - 2*c**3*Piecewise((a**3*sqrt(-1 + 1/(a**2*x**2))/3
- a*sqrt(-1 + 1/(a**2*x**2))/(3*x**2), 1/Abs(a**2*x**2) > 1), (I*a**3*sqrt(1 - 1/(a**2*x**2))/3 - I*a*sqrt(1
- 1/(a**2*x**2))/(3*x**2), True))/a**4 + 3*c**3*Piecewise((a**4*acosh(1/(a*x))/8 - a**3/(8*x*sqrt(-1 + 1/(a**2
*x**2))) + 3*a/(8*x**3*sqrt(-1 + 1/(a**2*x**2))) - 1/(4*a*x**5*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1
), (-I*a**4*asin(1/(a*x))/8 + I*a**3/(8*x*sqrt(1 - 1/(a**2*x**2))) - 3*I*a/(8*x**3*sqrt(1 - 1/(a**2*x**2))) +
I/(4*a*x**5*sqrt(1 - 1/(a**2*x**2))), True))/a**5 - c**3*Piecewise((2*I*a**4*sqrt(a**2*x**2 - 1)/(15*x) + I*a*
*2*sqrt(a**2*x**2 - 1)/(15*x**3) - I*sqrt(a**2*x**2 - 1)/(5*x**5), Abs(a**2*x**2) > 1), (2*a**4*sqrt(-a**2*x**
2 + 1)/(15*x) + a**2*sqrt(-a**2*x**2 + 1)/(15*x**3) - sqrt(-a**2*x**2 + 1)/(5*x**5), True))/a**6

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Giac [B]  time = 1.27207, size = 521, normalized size = 3.32 \begin{align*} -\frac{{\left (2 \, c^{3} - \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{3}}{a^{2} x} + \frac{30 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{3}}{a^{4} x^{2}} + \frac{80 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c^{3}}{a^{6} x^{3}} - \frac{580 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} c^{3}}{a^{8} x^{4}}\right )} a^{10} x^{5}}{320 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5}{\left | a \right |}} - \frac{3 \, c^{3} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} + \frac{3 \, c^{3} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{8 \,{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{3}}{a} - \frac{\frac{580 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{2} c^{3}}{x} - \frac{80 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{3}}{x^{2}} - \frac{30 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c^{3}}{a^{2} x^{3}} + \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} c^{3}}{a^{4} x^{4}} - \frac{2 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5} c^{3}}{a^{6} x^{5}}}{320 \, a^{4}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/320*(2*c^3 - 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^3/(a^2*x) + 30*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^3/(a^4
*x^2) + 80*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c^3/(a^6*x^3) - 580*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^3/(a^8*x^
4))*a^10*x^5/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^5*abs(a)) - 3*c^3*arcsin(a*x)*sgn(a)/abs(a) + 3/8*c^3*log(1/2*ab
s(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - sqrt(-a^2*x^2 + 1)*c^3/a - 1/320*(580*(sqrt(-a^2*
x^2 + 1)*abs(a) + a)*a^2*c^3/x - 80*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^3/x^2 - 30*(sqrt(-a^2*x^2 + 1)*abs(a)
+ a)^3*c^3/(a^2*x^3) + 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^3/(a^4*x^4) - 2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^
5*c^3/(a^6*x^5))/(a^4*abs(a))