3.678 \(\int e^{-3 \tanh ^{-1}(a x)} (c-\frac{c}{a^2 x^2})^4 \, dx\)

Optimal. Leaf size=191 \[ \frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}+\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{3 c^4 (5 a x+16) \sqrt{1-a^2 x^2}}{16 a^2 x}+\frac{15 c^4 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{16 a}-\frac{3 c^4 \sin ^{-1}(a x)}{a} \]

[Out]

(-3*c^4*(16 + 5*a*x)*Sqrt[1 - a^2*x^2])/(16*a^2*x) + (c^4*(16 - 5*a*x)*(1 - a^2*x^2)^(3/2))/(16*a^4*x^3) - (c^
4*(24 - 5*a*x)*(1 - a^2*x^2)^(5/2))/(40*a^6*x^5) - (c^4*(1 - a^2*x^2)^(7/2))/(7*a^8*x^7) + (c^4*(1 - a^2*x^2)^
(7/2))/(2*a^7*x^6) - (3*c^4*ArcSin[a*x])/a + (15*c^4*ArcTanh[Sqrt[1 - a^2*x^2]])/(16*a)

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Rubi [A]  time = 0.361692, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ \frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}+\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{3 c^4 (5 a x+16) \sqrt{1-a^2 x^2}}{16 a^2 x}+\frac{15 c^4 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{16 a}-\frac{3 c^4 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a^2*x^2))^4/E^(3*ArcTanh[a*x]),x]

[Out]

(-3*c^4*(16 + 5*a*x)*Sqrt[1 - a^2*x^2])/(16*a^2*x) + (c^4*(16 - 5*a*x)*(1 - a^2*x^2)^(3/2))/(16*a^4*x^3) - (c^
4*(24 - 5*a*x)*(1 - a^2*x^2)^(5/2))/(40*a^6*x^5) - (c^4*(1 - a^2*x^2)^(7/2))/(7*a^8*x^7) + (c^4*(1 - a^2*x^2)^
(7/2))/(2*a^7*x^6) - (3*c^4*ArcSin[a*x])/a + (15*c^4*ArcTanh[Sqrt[1 - a^2*x^2]])/(16*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 )^4 \, dx &=\frac{c^4 \int \frac{e^{-3 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^4}{x^8} \, dx}{a^8}\\ &=\frac{c^4 \int \frac{(1-a x)^3 \left (1-a^2 x^2\right )^{5/2}}{x^8} \, dx}{a^8}\\ &=-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}-\frac{c^4 \int \frac{\left (1-a^2 x^2\right )^{5/2} \left (21 a-21 a^2 x+7 a^3 x^2\right )}{x^7} \, dx}{7 a^8}\\ &=-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}+\frac{c^4 \int \frac{\left (126 a^2-21 a^3 x\right ) \left (1-a^2 x^2\right )^{5/2}}{x^6} \, dx}{42 a^8}\\ &=-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\frac{c^4 \int \frac{\left (1008 a^4-210 a^5 x\right ) \left (1-a^2 x^2\right )^{3/2}}{x^4} \, dx}{336 a^8}\\ &=\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}+\frac{c^4 \int \frac{\left (4032 a^6-1260 a^7 x\right ) \sqrt{1-a^2 x^2}}{x^2} \, dx}{1344 a^8}\\ &=-\frac{3 c^4 (16+5 a x) \sqrt{1-a^2 x^2}}{16 a^2 x}+\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\frac{c^4 \int \frac{2520 a^7+8064 a^8 x}{x \sqrt{1-a^2 x^2}} \, dx}{2688 a^8}\\ &=-\frac{3 c^4 (16+5 a x) \sqrt{1-a^2 x^2}}{16 a^2 x}+\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\left (3 c^4\right ) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx-\frac{\left (15 c^4\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{16 a}\\ &=-\frac{3 c^4 (16+5 a x) \sqrt{1-a^2 x^2}}{16 a^2 x}+\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\frac{3 c^4 \sin ^{-1}(a x)}{a}-\frac{\left (15 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{32 a}\\ &=-\frac{3 c^4 (16+5 a x) \sqrt{1-a^2 x^2}}{16 a^2 x}+\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\frac{3 c^4 \sin ^{-1}(a x)}{a}+\frac{\left (15 c^4\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 )}{16 a^3}\\ &=-\frac{3 c^4 (16+5 a x) \sqrt{1-a^2 x^2}}{16 a^2 x}+\frac{c^4 (16-5 a x) \left (1-a^2 x^2\right )^{3/2}}{16 a^4 x^3}-\frac{c^4 (24-5 a x) \left (1-a^2 x^2\right )^{5/2}}{40 a^6 x^5}-\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{7 a^8 x^7}+\frac{c^4 \left (1-a^2 x^2\right )^{7/2}}{2 a^7 x^6}-\frac{3 c^4 \sin ^{-1}(a x)}{a}+\frac{15 c^4 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{16 a}\\ \end{align*}

Mathematica [C]  time = 0.18322, size = 191, normalized size = 1. \[ \frac{c^4 \left (\frac{5 \left (16 a^7 x^7 \left (a^2 x^2-1\right )^4 \text{Hypergeometric2F1}\left (3,\frac{7}{2},\frac{9}{2},1-a^2 x^2\right )-16 a^8 x^8-231 a^7 x^7+64 a^6 x^6+413 a^5 x^5-96 a^4 x^4-238 a^3 x^3+64 a^2 x^2-105 a^7 x^7 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+56 a x-16\right )}{\sqrt{1-a^2 x^2}}-336 a^2 x^2 \text{Hypergeometric2F1}\left (-\frac{5}{2},-\frac{5}{2},-\frac{3}{2},a^2 x^2\right )\right )}{560 a^8 x^7} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - c/(a^2*x^2))^4/E^(3*ArcTanh[a*x]),x]

[Out]

(c^4*(-336*a^2*x^2*Hypergeometric2F1[-5/2, -5/2, -3/2, a^2*x^2] + (5*(-16 + 56*a*x + 64*a^2*x^2 - 238*a^3*x^3
- 96*a^4*x^4 + 413*a^5*x^5 + 64*a^6*x^6 - 231*a^7*x^7 - 16*a^8*x^8 - 105*a^7*x^7*Sqrt[1 - a^2*x^2]*ArcTanh[Sqr
t[1 - a^2*x^2]] + 16*a^7*x^7*(-1 + a^2*x^2)^4*Hypergeometric2F1[3, 7/2, 9/2, 1 - a^2*x^2]))/Sqrt[1 - a^2*x^2])
)/(560*a^8*x^7)

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Maple [A]  time = 0.104, size = 289, normalized size = 1.5 \begin{align*} -{\frac{3\,{c}^{4}}{8\,{a}^{5}{x}^{4}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{c}^{4}}{16\,{x}^{2}{a}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{c}^{4}}{16\,a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{c}^{4}}{16\,a}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{15\,{c}^{4}}{16\,a}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }-{\frac{{c}^{4}}{7\,{a}^{8}{x}^{7}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{16\,{c}^{4}}{35\,{a}^{6}{x}^{5}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{{c}^{4}}{2\,{a}^{7}{x}^{6}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{{c}^{4}}{{a}^{4}{x}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-2\,{\frac{{c}^{4} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{5/2}}{{a}^{2}x}}-2\,{c}^{4}x \left ( -{a}^{2}{x}^{2}+1 \right ) ^{3/2}-3\,{c}^{4}x\sqrt{-{a}^{2}{x}^{2}+1}-3\,{\frac{{c}^{4}}{\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)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

-3/8*c^4/a^5/x^4*(-a^2*x^2+1)^(5/2)-5/16*c^4/a^3/x^2*(-a^2*x^2+1)^(5/2)-5/16*c^4*(-a^2*x^2+1)^(3/2)/a-15/16*c^
4*(-a^2*x^2+1)^(1/2)/a+15/16*c^4/a*arctanh(1/(-a^2*x^2+1)^(1/2))-1/7*c^4/a^8/x^7*(-a^2*x^2+1)^(5/2)-16/35*c^4/
a^6/x^5*(-a^2*x^2+1)^(5/2)+1/2*c^4/a^7/x^6*(-a^2*x^2+1)^(5/2)+c^4/a^4/x^3*(-a^2*x^2+1)^(5/2)-2*c^4/a^2/x*(-a^2
*x^2+1)^(5/2)-2*c^4*x*(-a^2*x^2+1)^(3/2)-3*c^4*x*(-a^2*x^2+1)^(1/2)-3*c^4/(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 )}^{4}}{{\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)^4/(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))^4/(a*x + 1)^3, x)

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Fricas [A]  time = 2.21505, size = 398, normalized size = 2.08 \begin{align*} \frac{3360 \, a^{7} c^{4} x^{7} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) - 525 \, a^{7} c^{4} x^{7} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) - 560 \, a^{7} c^{4} x^{7} -{\left (560 \, a^{7} c^{4} x^{7} + 2496 \, a^{6} c^{4} x^{6} - 525 \, a^{5} c^{4} x^{5} - 992 \, a^{4} c^{4} x^{4} + 770 \, a^{3} c^{4} x^{3} + 96 \, a^{2} c^{4} x^{2} - 280 \, a c^{4} x + 80 \, c^{4}\right )} \sqrt{-a^{2} x^{2} + 1}}{560 \, a^{8} x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*(3360*a^7*c^4*x^7*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - 525*a^7*c^4*x^7*log((sqrt(-a^2*x^2 + 1) - 1)/
x) - 560*a^7*c^4*x^7 - (560*a^7*c^4*x^7 + 2496*a^6*c^4*x^6 - 525*a^5*c^4*x^5 - 992*a^4*c^4*x^4 + 770*a^3*c^4*x
^3 + 96*a^2*c^4*x^2 - 280*a*c^4*x + 80*c^4)*sqrt(-a^2*x^2 + 1))/(a^8*x^7)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.25757, size = 683, normalized size = 3.58 \begin{align*} \frac{{\left (5 \, c^{4} - \frac{35 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} c^{4}}{a^{2} x} + \frac{49 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{4}}{a^{4} x^{2}} + \frac{245 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c^{4}}{a^{6} x^{3}} - \frac{875 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} c^{4}}{a^{8} x^{4}} - \frac{455 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5} c^{4}}{a^{10} x^{5}} + \frac{9065 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{6} c^{4}}{a^{12} x^{6}}\right )} a^{14} x^{7}}{4480 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{7}{\left | a \right |}} - \frac{3 \, c^{4} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{{\left | a \right |}} + \frac{15 \, c^{4} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{16 \,{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{4}}{a} - \frac{\frac{9065 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{4} c^{4}}{x} - \frac{455 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{2} c^{4}}{x^{2}} - \frac{875 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c^{4}}{x^{3}} + \frac{245 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4} c^{4}}{a^{2} x^{4}} + \frac{49 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5} c^{4}}{a^{4} x^{5}} - \frac{35 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{6} c^{4}}{a^{6} x^{6}} + \frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{7} c^{4}}{a^{8} x^{7}}}{4480 \, a^{6}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4480*(5*c^4 - 35*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^4/(a^2*x) + 49*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^4/(a^4
*x^2) + 245*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c^4/(a^6*x^3) - 875*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^4/(a^8*x
^4) - 455*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^5*c^4/(a^10*x^5) + 9065*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^6*c^4/(a^12*
x^6))*a^14*x^7/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^7*abs(a)) - 3*c^4*arcsin(a*x)*sgn(a)/abs(a) + 15/16*c^4*log(1/
2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - sqrt(-a^2*x^2 + 1)*c^4/a - 1/4480*(9065*(sqrt
(-a^2*x^2 + 1)*abs(a) + a)*a^4*c^4/x - 455*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2*c^4/x^2 - 875*(sqrt(-a^2*x^2
+ 1)*abs(a) + a)^3*c^4/x^3 + 245*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*c^4/(a^2*x^4) + 49*(sqrt(-a^2*x^2 + 1)*abs(
a) + a)^5*c^4/(a^4*x^5) - 35*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^6*c^4/(a^6*x^6) + 5*(sqrt(-a^2*x^2 + 1)*abs(a) +
a)^7*c^4/(a^8*x^7))/(a^6*abs(a))