3.1092 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{3/2}}{x^3} \, dx\)

Optimal. Leaf size=121 \[ -2 a^2 c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-\frac {1}{2} a^2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )-\frac {a c (4-a x) \sqrt {c-a^2 c x^2}}{2 x}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6151, 1807, 813, 844, 217, 203, 266, 63, 208} \[ -2 a^2 c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-\frac {1}{2} a^2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )-\frac {a c (4-a x) \sqrt {c-a^2 c x^2}}{2 x}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c*(4 - a*x)*Sqrt[c - a^2*c*x^2])/(2*x) - (c - a^2*c*x^2)^(3/2)/(2*x^2) - 2*a^2*c^(3/2)*ArcTan[(a*Sqrt[c]*x
)/Sqrt[c - a^2*c*x^2]] - (a^2*c^(3/2)*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/2

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 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 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 217

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

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 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 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 6151

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^(n/2), Int[x^m*(c
+ d*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] ||
GtQ[c, 0]) && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2}}{x^3} \, dx &=c \int \frac {(1+a x)^2 \sqrt {c-a^2 c x^2}}{x^3} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2}-\frac {1}{2} \int \frac {\left (-4 a c-a^2 c x\right ) \sqrt {c-a^2 c x^2}}{x^2} \, dx\\ &=-\frac {a c (4-a x) \sqrt {c-a^2 c x^2}}{2 x}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2}+\frac {1}{4} \int \frac {2 a^2 c^2-8 a^3 c^2 x}{x \sqrt {c-a^2 c x^2}} \, dx\\ &=-\frac {a c (4-a x) \sqrt {c-a^2 c x^2}}{2 x}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2}+\frac {1}{2} \left (a^2 c^2\right ) \int \frac {1}{x \sqrt {c-a^2 c x^2}} \, dx-\left (2 a^3 c^2\right ) \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx\\ &=-\frac {a c (4-a x) \sqrt {c-a^2 c x^2}}{2 x}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2}+\frac {1}{4} \left (a^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-a^2 c x}} \, dx,x,x^2\right )-\left (2 a^3 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c-a^2 c x^2}}\right )\\ &=-\frac {a c (4-a x) \sqrt {c-a^2 c x^2}}{2 x}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2}-2 a^2 c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-\frac {1}{2} c \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c-a^2 c x^2}\right )\\ &=-\frac {a c (4-a x) \sqrt {c-a^2 c x^2}}{2 x}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{2 x^2}-2 a^2 c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-\frac {1}{2} a^2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 129, normalized size = 1.07 \[ \frac {1}{2} c \left (\frac {\left (2 a^2 x^2-4 a x-1\right ) \sqrt {c-a^2 c x^2}}{x^2}-a^2 \sqrt {c} \log \left (\sqrt {c} \sqrt {c-a^2 c x^2}+c\right )+4 a^2 \sqrt {c} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )+a^2 \sqrt {c} \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*(((-1 - 4*a*x + 2*a^2*x^2)*Sqrt[c - a^2*c*x^2])/x^2 + 4*a^2*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[
c]*(-1 + a^2*x^2))] + a^2*Sqrt[c]*Log[x] - a^2*Sqrt[c]*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]]))/2

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fricas [A]  time = 0.66, size = 267, normalized size = 2.21 \[ \left [\frac {8 \, a^{2} c^{\frac {3}{2}} x^{2} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + a^{2} c^{\frac {3}{2}} x^{2} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + 2 \, {\left (2 \, a^{2} c x^{2} - 4 \, a c x - c\right )} \sqrt {-a^{2} c x^{2} + c}}{4 \, x^{2}}, -\frac {a^{2} \sqrt {-c} c x^{2} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) - 2 \, a^{2} \sqrt {-c} c x^{2} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - {\left (2 \, a^{2} c x^{2} - 4 \, a c x - c\right )} \sqrt {-a^{2} c x^{2} + c}}{2 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(8*a^2*c^(3/2)*x^2*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + a^2*c^(3/2)*x^2*log(-(a^2*c
*x^2 + 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) + 2*(2*a^2*c*x^2 - 4*a*c*x - c)*sqrt(-a^2*c*x^2 + c))/x^2, -
1/2*(a^2*sqrt(-c)*c*x^2*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c*x^2 - c)) - 2*a^2*sqrt(-c)*c*x^2*log(2*a^2
*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - (2*a^2*c*x^2 - 4*a*c*x - c)*sqrt(-a^2*c*x^2 + c))/x^2]

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giac [B]  time = 0.23, size = 274, normalized size = 2.26 \[ \frac {a^{2} c^{2} \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {2 \, a^{3} \sqrt {-c} c \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}} + \sqrt {-a^{2} c x^{2} + c} a^{2} c - \frac {{\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{3} a^{2} c^{2} {\left | a \right |} - 4 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} a^{3} \sqrt {-c} c^{2} + {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a^{2} c^{3} {\left | a \right |} + 4 \, a^{3} \sqrt {-c} c^{3}}{{\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{2} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*c^2*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) - 2*a^3*sqrt(-c)*c*log(abs(-sqrt(-a
^2*c)*x + sqrt(-a^2*c*x^2 + c)))/abs(a) + sqrt(-a^2*c*x^2 + c)*a^2*c - ((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c)
)^3*a^2*c^2*abs(a) - 4*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2*a^3*sqrt(-c)*c^2 + (sqrt(-a^2*c)*x - sqrt(-a^
2*c*x^2 + c))*a^2*c^3*abs(a) + 4*a^3*sqrt(-c)*c^3)/(((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)^2*abs(a))

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maple [B]  time = 0.05, size = 316, normalized size = 2.61 \[ \frac {a^{2} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{6}-\frac {a^{2} c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{2}+\frac {a^{2} \sqrt {-a^{2} c \,x^{2}+c}\, c}{2}-\frac {2 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{c x}-2 a^{3} x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}-3 a^{3} c x \sqrt {-a^{2} c \,x^{2}+c}-\frac {3 a^{3} c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}-\frac {2 a^{2} \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a^{3} c \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}\, x +\frac {a^{3} c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2} c}}-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{2 c \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (c-a^2\,c\,x^2\right )}^{3/2}\,{\left (a\,x+1\right )}^2}{x^3\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 8.22, size = 366, normalized size = 3.02 \[ a^{2} c \left (\begin {cases} i \sqrt {c} \sqrt {a^{2} x^{2} - 1} - \sqrt {c} \log {\left (a x \right )} + \frac {\sqrt {c} \log {\left (a^{2} x^{2} \right )}}{2} + i \sqrt {c} \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\sqrt {c} \sqrt {- a^{2} x^{2} + 1} + \frac {\sqrt {c} \log {\left (a^{2} x^{2} \right )}}{2} - \sqrt {c} \log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )} & \text {otherwise} \end {cases}\right ) + 2 a c \left (\begin {cases} - \frac {i a^{2} \sqrt {c} x}{\sqrt {a^{2} x^{2} - 1}} + i a \sqrt {c} \operatorname {acosh}{\left (a x \right )} + \frac {i \sqrt {c}}{x \sqrt {a^{2} x^{2} - 1}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {a^{2} \sqrt {c} x}{\sqrt {- a^{2} x^{2} + 1}} - a \sqrt {c} \operatorname {asin}{\left (a x \right )} - \frac {\sqrt {c}}{x \sqrt {- a^{2} x^{2} + 1}} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \frac {a^{2} \sqrt {c} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} + \frac {a \sqrt {c}}{2 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {\sqrt {c}}{2 a x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac {i a^{2} \sqrt {c} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a \sqrt {c} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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