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

Optimal. Leaf size=136 \[ \frac {c^3 (8-15 a x) \sqrt {1-a^2 x^2}}{8 a^2 x}+\frac {15 c^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{8 a}+\frac {c^3 (5 a x+4) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}-\frac {c^3 (15 a x+8) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 \sin ^{-1}(a x)}{a} \]

[Out]

-1/24*c^3*(15*a*x+8)*(-a^2*x^2+1)^(3/2)/a^4/x^3+1/20*c^3*(5*a*x+4)*(-a^2*x^2+1)^(5/2)/a^6/x^5+c^3*arcsin(a*x)/
a+15/8*c^3*arctanh((-a^2*x^2+1)^(1/2))/a+1/8*c^3*(-15*a*x+8)*(-a^2*x^2+1)^(1/2)/a^2/x

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Rubi [A]  time = 0.19, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6157, 6148, 811, 813, 844, 216, 266, 63, 208} \[ \frac {c^3 (5 a x+4) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}-\frac {c^3 (15 a x+8) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 (8-15 a x) \sqrt {1-a^2 x^2}}{8 a^2 x}+\frac {15 c^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{8 a}+\frac {c^3 \sin ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(8 - 15*a*x)*Sqrt[1 - a^2*x^2])/(8*a^2*x) - (c^3*(8 + 15*a*x)*(1 - a^2*x^2)^(3/2))/(24*a^4*x^3) + (c^3*(4
+ 5*a*x)*(1 - a^2*x^2)^(5/2))/(20*a^6*x^5) + (c^3*ArcSin[a*x])/a + (15*c^3*ArcTanh[Sqrt[1 - a^2*x^2]])/(8*a)

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]

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 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 6148

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] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]

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]

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx &=-\frac {c^3 \int \frac {e^{\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) \left (1-a^2 x^2\right )^{5/2}}{x^6} \, dx}{a^6}\\ &=\frac {c^3 (4+5 a x) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}+\frac {c^3 \int \frac {\left (8 a^2+10 a^3 x\right ) \left (1-a^2 x^2\right )^{3/2}}{x^4} \, dx}{8 a^6}\\ &=-\frac {c^3 (8+15 a x) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 (4+5 a x) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}-\frac {c^3 \int \frac {\left (32 a^4+60 a^5 x\right ) \sqrt {1-a^2 x^2}}{x^2} \, dx}{32 a^6}\\ &=\frac {c^3 (8-15 a x) \sqrt {1-a^2 x^2}}{8 a^2 x}-\frac {c^3 (8+15 a x) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 (4+5 a x) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}+\frac {c^3 \int \frac {-120 a^5+64 a^6 x}{x \sqrt {1-a^2 x^2}} \, dx}{64 a^6}\\ &=\frac {c^3 (8-15 a x) \sqrt {1-a^2 x^2}}{8 a^2 x}-\frac {c^3 (8+15 a x) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 (4+5 a x) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}+c^3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx-\frac {\left (15 c^3\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{8 a}\\ &=\frac {c^3 (8-15 a x) \sqrt {1-a^2 x^2}}{8 a^2 x}-\frac {c^3 (8+15 a x) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 (4+5 a x) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}+\frac {c^3 \sin ^{-1}(a x)}{a}-\frac {\left (15 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{16 a}\\ &=\frac {c^3 (8-15 a x) \sqrt {1-a^2 x^2}}{8 a^2 x}-\frac {c^3 (8+15 a x) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 (4+5 a x) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}+\frac {c^3 \sin ^{-1}(a x)}{a}+\frac {\left (15 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 {c^3 (8-15 a x) \sqrt {1-a^2 x^2}}{8 a^2 x}-\frac {c^3 (8+15 a x) \left (1-a^2 x^2\right )^{3/2}}{24 a^4 x^3}+\frac {c^3 (4+5 a x) \left (1-a^2 x^2\right )^{5/2}}{20 a^6 x^5}+\frac {c^3 \sin ^{-1}(a x)}{a}+\frac {15 c^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{8 a}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 70, normalized size = 0.51 \[ \frac {c^3 \left (\frac {7 \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};a^2 x^2\right )}{x^5}+5 a^5 \left (1-a^2 x^2\right )^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};1-a^2 x^2\right )\right )}{35 a^6} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*((7*Hypergeometric2F1[-5/2, -5/2, -3/2, a^2*x^2])/x^5 + 5*a^5*(1 - a^2*x^2)^(7/2)*Hypergeometric2F1[3, 7/
2, 9/2, 1 - a^2*x^2]))/(35*a^6)

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fricas [A]  time = 0.62, size = 153, normalized size = 1.12 \[ -\frac {240 \, a^{5} c^{3} x^{5} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + 225 \, a^{5} c^{3} x^{5} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + 120 \, a^{5} c^{3} x^{5} + {\left (120 \, a^{5} c^{3} x^{5} - 184 \, a^{4} c^{3} x^{4} + 135 \, a^{3} c^{3} x^{3} + 88 \, a^{2} c^{3} x^{2} - 30 \, a c^{3} x - 24 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{120 \, a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(240*a^5*c^3*x^5*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 225*a^5*c^3*x^5*log((sqrt(-a^2*x^2 + 1) - 1)/
x) + 120*a^5*c^3*x^5 + (120*a^5*c^3*x^5 - 184*a^4*c^3*x^4 + 135*a^3*c^3*x^3 + 88*a^2*c^3*x^2 - 30*a*c^3*x - 24
*c^3)*sqrt(-a^2*x^2 + 1))/(a^6*x^5)

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giac [B]  time = 0.27, size = 385, normalized size = 2.83 \[ -\frac {{\left (6 \, c^{3} + \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{3}}{a^{2} x} - \frac {70 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{3}}{a^{4} x^{2}} - \frac {240 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{3}}{a^{6} x^{3}} + \frac {660 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} c^{3}}{a^{8} x^{4}}\right )} a^{10} x^{5}}{960 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{5} {\left | a \right |}} + \frac {c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {15 \, c^{3} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{8 \, {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{3}}{a} + \frac {\frac {660 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{2} c^{3}}{x} - \frac {240 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{3}}{x^{2}} - \frac {70 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{3}}{a^{2} x^{3}} + \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} c^{3}}{a^{4} x^{4}} + \frac {6 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{5} c^{3}}{a^{6} x^{5}}}{960 \, a^{4} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(6*c^3 + 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^3/(a^2*x) - 70*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^3/(a^4
*x^2) - 240*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c^3/(a^6*x^3) + 660*(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)) + c^3*arcsin(a*x)*sgn(a)/abs(a) + 15/8*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 + 1/960*(660*(sqrt(-a^2*
x^2 + 1)*abs(a) + a)*a^2*c^3/x - 240*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^3/x^2 - 70*(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) + 6*(sqrt(-a^2*x^2 + 1)*abs(a) + a)
^5*c^3/(a^6*x^5))/(a^4*abs(a))

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maple [A]  time = 0.05, size = 187, normalized size = 1.38 \[ -\frac {c^{3} \sqrt {-a^{2} x^{2}+1}}{a}+\frac {c^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+\frac {15 c^{3} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a}+\frac {23 c^{3} \sqrt {-a^{2} x^{2}+1}}{15 a^{2} x}-\frac {11 c^{3} \sqrt {-a^{2} x^{2}+1}}{15 a^{4} x^{3}}-\frac {9 c^{3} \sqrt {-a^{2} x^{2}+1}}{8 x^{2} a^{3}}+\frac {c^{3} \sqrt {-a^{2} x^{2}+1}}{4 a^{5} x^{4}}+\frac {c^{3} \sqrt {-a^{2} x^{2}+1}}{5 a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^3*(-a^2*x^2+1)^(1/2)/a+c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+15/8*c^3/a*arctanh(1/(-a^2*
x^2+1)^(1/2))+23/15*c^3*(-a^2*x^2+1)^(1/2)/a^2/x-11/15*c^3/a^4/x^3*(-a^2*x^2+1)^(1/2)-9/8*c^3*(-a^2*x^2+1)^(1/
2)/x^2/a^3+1/4*c^3/a^5/x^4*(-a^2*x^2+1)^(1/2)+1/5*c^3/a^6/x^5*(-a^2*x^2+1)^(1/2)

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maxima [B]  time = 0.42, size = 332, normalized size = 2.44 \[ \frac {c^{3} \arcsin \left (a x\right )}{a} + \frac {3 \, c^{3} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right )}{a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{3}}{a} - \frac {3 \, {\left (a^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-a^{2} x^{2} + 1}}{x^{2}}\right )} c^{3}}{2 \, a^{3}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{a^{2} x} - \frac {{\left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x} + \frac {\sqrt {-a^{2} x^{2} + 1}}{x^{3}}\right )} c^{3}}{a^{4}} + \frac {{\left (3 \, a^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {3 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{x^{4}}\right )} c^{3}}{8 \, a^{5}} + \frac {{\left (\frac {8 \, \sqrt {-a^{2} x^{2} + 1} a^{4}}{x} + \frac {4 \, \sqrt {-a^{2} x^{2} + 1} a^{2}}{x^{3}} + \frac {3 \, \sqrt {-a^{2} x^{2} + 1}}{x^{5}}\right )} c^{3}}{15 \, a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*arcsin(a*x)/a + 3*c^3*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x))/a - sqrt(-a^2*x^2 + 1)*c^3/a - 3/2*(a^2*
log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-a^2*x^2 + 1)/x^2)*c^3/a^3 + 3*sqrt(-a^2*x^2 + 1)*c^3/(a^2*
x) - (2*sqrt(-a^2*x^2 + 1)*a^2/x + sqrt(-a^2*x^2 + 1)/x^3)*c^3/a^4 + 1/8*(3*a^4*log(2*sqrt(-a^2*x^2 + 1)/abs(x
) + 2/abs(x)) + 3*sqrt(-a^2*x^2 + 1)*a^2/x^2 + 2*sqrt(-a^2*x^2 + 1)/x^4)*c^3/a^5 + 1/15*(8*sqrt(-a^2*x^2 + 1)*
a^4/x + 4*sqrt(-a^2*x^2 + 1)*a^2/x^3 + 3*sqrt(-a^2*x^2 + 1)/x^5)*c^3/a^6

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mupad [B]  time = 0.05, size = 182, normalized size = 1.34 \[ \frac {c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}-\frac {c^3\,\sqrt {1-a^2\,x^2}}{a}+\frac {23\,c^3\,\sqrt {1-a^2\,x^2}}{15\,a^2\,x}-\frac {9\,c^3\,\sqrt {1-a^2\,x^2}}{8\,a^3\,x^2}-\frac {11\,c^3\,\sqrt {1-a^2\,x^2}}{15\,a^4\,x^3}+\frac {c^3\,\sqrt {1-a^2\,x^2}}{4\,a^5\,x^4}+\frac {c^3\,\sqrt {1-a^2\,x^2}}{5\,a^6\,x^5}-\frac {c^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,15{}\mathrm {i}}{8\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^3*asinh(x*(-a^2)^(1/2)))/(-a^2)^(1/2) - (c^3*atan((1 - a^2*x^2)^(1/2)*1i)*15i)/(8*a) - (c^3*(1 - a^2*x^2)^(
1/2))/a + (23*c^3*(1 - a^2*x^2)^(1/2))/(15*a^2*x) - (9*c^3*(1 - a^2*x^2)^(1/2))/(8*a^3*x^2) - (11*c^3*(1 - a^2
*x^2)^(1/2))/(15*a^4*x^3) + (c^3*(1 - a^2*x^2)^(1/2))/(4*a^5*x^4) + (c^3*(1 - a^2*x^2)^(1/2))/(5*a^6*x^5)

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sympy [A]  time = 16.35, size = 687, normalized size = 5.05 \[ a c^{3} \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) + c^{3} \left (\begin {cases} \sqrt {\frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \\\sqrt {- \frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \end {cases}\right ) - \frac {3 c^{3} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right )}{a} - \frac {3 c^{3} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right )}{a^{2}} + \frac {3 c^{3} \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a}{2 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{2 a x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{a^{3}} + \frac {3 c^{3} \left (\begin {cases} - \frac {2 i a^{2} \sqrt {a^{2} x^{2} - 1}}{3 x} - \frac {i \sqrt {a^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {2 a^{2} \sqrt {- a^{2} x^{2} + 1}}{3 x} - \frac {\sqrt {- a^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right )}{a^{4}} - \frac {c^{3} \left (\begin {cases} - \frac {3 a^{4} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} + \frac {3 a^{3}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {a}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {3 i a^{4} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} - \frac {3 i a^{3}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i a}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{a^{5}} - \frac {c^{3} \left (\begin {cases} - \frac {8 a^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{15} - \frac {4 a^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{15 x^{2}} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{5 x^{4}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac {8 i a^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{15} - \frac {4 i a^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{15 x^{2}} - \frac {i a \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{5 x^{4}} & \text {otherwise} \end {cases}\right )}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**3*Piecewise((x**2/2, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True)) + c**3*Piecewise((sqrt(a**(-2))*as
in(x*sqrt(a**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) - 3*c**3*Piecewise((-acosh(1/(a*x
)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True))/a - 3*c**3*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x
**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True))/a**2 + 3*c**3*Piecewise((-a**2*acosh(1/(a*x))/2 - a*sqrt(-1 + 1/(a
**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*x))/2 - I*a/(2*x*sqrt(1 - 1/(a**2*x**2))) + I/(2*a*
x**3*sqrt(1 - 1/(a**2*x**2))), True))/a**3 + 3*c**3*Piecewise((-2*I*a**2*sqrt(a**2*x**2 - 1)/(3*x) - I*sqrt(a*
*2*x**2 - 1)/(3*x**3), Abs(a**2*x**2) > 1), (-2*a**2*sqrt(-a**2*x**2 + 1)/(3*x) - sqrt(-a**2*x**2 + 1)/(3*x**3
), True))/a**4 - c**3*Piecewise((-3*a**4*acosh(1/(a*x))/8 + 3*a**3/(8*x*sqrt(-1 + 1/(a**2*x**2))) - 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), (3*I*a**4*asin(1/(a*
x))/8 - 3*I*a**3/(8*x*sqrt(1 - 1/(a**2*x**2))) + 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((-8*a**5*sqrt(-1 + 1/(a**2*x**2))/15 - 4*a**3*sqrt(-1 + 1/(a**2*x
**2))/(15*x**2) - a*sqrt(-1 + 1/(a**2*x**2))/(5*x**4), 1/Abs(a**2*x**2) > 1), (-8*I*a**5*sqrt(1 - 1/(a**2*x**2
))/15 - 4*I*a**3*sqrt(1 - 1/(a**2*x**2))/(15*x**2) - I*a*sqrt(1 - 1/(a**2*x**2))/(5*x**4), True))/a**6

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