3.1246 \(\int e^{-2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{5/2} \, dx\)

Optimal. Leaf size=131 \[ \frac {7 c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a}+\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}+\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}+\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a} \]

[Out]

7/24*c*x*(-a^2*c*x^2+c)^(3/2)+7/30*(-a^2*c*x^2+c)^(5/2)/a+1/6*(-a*x+1)*(-a^2*c*x^2+c)^(5/2)/a+7/16*c^(5/2)*arc
tan(a*x*c^(1/2)/(-a^2*c*x^2+c)^(1/2))/a+7/16*c^2*x*(-a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6142, 671, 641, 195, 217, 203} \[ \frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}+\frac {7 c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a}+\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}+\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*c^2*x*Sqrt[c - a^2*c*x^2])/16 + (7*c*x*(c - a^2*c*x^2)^(3/2))/24 + (7*(c - a^2*c*x^2)^(5/2))/(30*a) + ((1 -
a*x)*(c - a^2*c*x^2)^(5/2))/(6*a) + (7*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(16*a)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 671

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

Rule 6142

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

Rubi steps

\begin {align*} \int e^{-2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx &=c \int (1-a x)^2 \left (c-a^2 c x^2\right )^{3/2} \, dx\\ &=\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {1}{6} (7 c) \int (1-a x) \left (c-a^2 c x^2\right )^{3/2} \, dx\\ &=\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}+\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {1}{6} (7 c) \int \left (c-a^2 c x^2\right )^{3/2} \, dx\\ &=\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}+\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}+\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {1}{8} \left (7 c^2\right ) \int \sqrt {c-a^2 c x^2} \, dx\\ &=\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}+\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}+\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}+\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {1}{16} \left (7 c^3\right ) \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {7}{16} c^2 x \sqrt {c-a^2 c x^2}+\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}+\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}+\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {1}{16} \left (7 c^3\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 {7}{16} c^2 x \sqrt {c-a^2 c x^2}+\frac {7}{24} c x \left (c-a^2 c x^2\right )^{3/2}+\frac {7 \left (c-a^2 c x^2\right )^{5/2}}{30 a}+\frac {(1-a x) \left (c-a^2 c x^2\right )^{5/2}}{6 a}+\frac {7 c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 135, normalized size = 1.03 \[ \frac {c^2 \sqrt {c-a^2 c x^2} \left (\sqrt {a x+1} \left (40 a^6 x^6-136 a^5 x^5+86 a^4 x^4+202 a^3 x^3-327 a^2 x^2+39 a x+96\right )-210 \sqrt {1-a x} \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{240 a \sqrt {1-a x} \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*Sqrt[c - a^2*c*x^2]*(Sqrt[1 + a*x]*(96 + 39*a*x - 327*a^2*x^2 + 202*a^3*x^3 + 86*a^4*x^4 - 136*a^5*x^5 +
40*a^6*x^6) - 210*Sqrt[1 - a*x]*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(240*a*Sqrt[1 - a*x]*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.60, size = 241, normalized size = 1.84 \[ \left [\frac {105 \, \sqrt {-c} c^{2} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 2 \, {\left (40 \, a^{5} c^{2} x^{5} - 96 \, a^{4} c^{2} x^{4} - 10 \, a^{3} c^{2} x^{3} + 192 \, a^{2} c^{2} x^{2} - 135 \, a c^{2} x - 96 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{480 \, a}, -\frac {105 \, c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + {\left (40 \, a^{5} c^{2} x^{5} - 96 \, a^{4} c^{2} x^{4} - 10 \, a^{3} c^{2} x^{3} + 192 \, a^{2} c^{2} x^{2} - 135 \, a c^{2} x - 96 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}}{240 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/480*(105*sqrt(-c)*c^2*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 2*(40*a^5*c^2*x^5 - 96*a
^4*c^2*x^4 - 10*a^3*c^2*x^3 + 192*a^2*c^2*x^2 - 135*a*c^2*x - 96*c^2)*sqrt(-a^2*c*x^2 + c))/a, -1/240*(105*c^(
5/2)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + (40*a^5*c^2*x^5 - 96*a^4*c^2*x^4 - 10*a^3*c^2*
x^3 + 192*a^2*c^2*x^2 - 135*a*c^2*x - 96*c^2)*sqrt(-a^2*c*x^2 + c))/a]

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giac [B]  time = 0.28, size = 320, normalized size = 2.44 \[ -\frac {{\left (6720 \, a^{7} c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) - \frac {{\left (105 \, a^{7} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{5} c^{3} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) - 595 \, a^{7} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{4} c^{4} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) - 1686 \, a^{7} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{3} c^{5} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) + 1386 \, a^{7} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{2} c^{6} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) + 105 \, a^{7} c^{8} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) + 595 \, a^{7} c^{7} {\left (-c + \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a)\right )} {\left (a x + 1\right )}^{6}}{c^{6}}\right )} {\left | a \right |}}{7680 \, a^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7680*(6720*a^7*c^(5/2)*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(c))*sgn(1/(a*x + 1))*sgn(a) - (105*a^7*(c - 2*c
/(a*x + 1))^5*c^3*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) - 595*a^7*(c - 2*c/(a*x + 1))^4*c^4*sqrt(-c
+ 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) - 1686*a^7*(c - 2*c/(a*x + 1))^3*c^5*sqrt(-c + 2*c/(a*x + 1))*sgn(1/
(a*x + 1))*sgn(a) + 1386*a^7*(c - 2*c/(a*x + 1))^2*c^6*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) + 105*
a^7*c^8*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) + 595*a^7*c^7*(-c + 2*c/(a*x + 1))^(3/2)*sgn(1/(a*x +
1))*sgn(a))*(a*x + 1)^6/c^6)*abs(a)/a^9

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maple [B]  time = 0.04, size = 226, normalized size = 1.73 \[ -\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{6}-\frac {5 c x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{24}-\frac {5 c^{2} x \sqrt {-a^{2} c \,x^{2}+c}}{16}-\frac {5 c^{3} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{16 \sqrt {a^{2} c}}+\frac {2 \left (-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{5 a}+\frac {c \left (-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}} x}{2}+\frac {3 c^{2} \sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )}\, x}{4}+\frac {3 c^{3} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )}}\right )}{4 \sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*x*(-a^2*c*x^2+c)^(5/2)-5/24*c*x*(-a^2*c*x^2+c)^(3/2)-5/16*c^2*x*(-a^2*c*x^2+c)^(1/2)-5/16*c^3/(a^2*c)^(1/
2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))+2/5/a*(-(x+1/a)^2*a^2*c+2*a*c*(x+1/a))^(5/2)+1/2*c*(-(x+1/a)^2
*a^2*c+2*a*c*(x+1/a))^(3/2)*x+3/4*c^2*(-(x+1/a)^2*a^2*c+2*a*c*(x+1/a))^(1/2)*x+3/4*c^3/(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))

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maxima [A]  time = 0.42, size = 154, normalized size = 1.18 \[ -\frac {1}{6} \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x + \frac {7}{24} \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c x + \frac {3}{4} \, \sqrt {a^{2} c x^{2} + 4 \, a c x + 3 \, c} c^{2} x - \frac {5}{16} \, \sqrt {-a^{2} c x^{2} + c} c^{2} x - \frac {3 \, c^{4} \arcsin \left (a x + 2\right )}{4 \, a \left (-c\right )^{\frac {3}{2}}} - \frac {5 \, c^{\frac {5}{2}} \arcsin \left (a x\right )}{16 \, a} + \frac {2 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}{5 \, a} + \frac {3 \, \sqrt {a^{2} c x^{2} + 4 \, a c x + 3 \, c} c^{2}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(-a^2*c*x^2 + c)^(5/2)*x + 7/24*(-a^2*c*x^2 + c)^(3/2)*c*x + 3/4*sqrt(a^2*c*x^2 + 4*a*c*x + 3*c)*c^2*x -
5/16*sqrt(-a^2*c*x^2 + c)*c^2*x - 3/4*c^4*arcsin(a*x + 2)/(a*(-c)^(3/2)) - 5/16*c^(5/2)*arcsin(a*x)/a + 2/5*(-
a^2*c*x^2 + c)^(5/2)/a + 3/2*sqrt(a^2*c*x^2 + 4*a*c*x + 3*c)*c^2/a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\left (a^2\,x^2-1\right )}{{\left (a\,x+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 10.27, size = 478, normalized size = 3.65 \[ - a^{4} c^{2} \left (\begin {cases} \frac {i a^{2} \sqrt {c} x^{7}}{6 \sqrt {a^{2} x^{2} - 1}} - \frac {5 i \sqrt {c} x^{5}}{24 \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} x^{3}}{48 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {i \sqrt {c} x}{16 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} \operatorname {acosh}{\left (a x \right )}}{16 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {a^{2} \sqrt {c} x^{7}}{6 \sqrt {- a^{2} x^{2} + 1}} + \frac {5 \sqrt {c} x^{5}}{24 \sqrt {- a^{2} x^{2} + 1}} + \frac {\sqrt {c} x^{3}}{48 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {\sqrt {c} x}{16 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {\sqrt {c} \operatorname {asin}{\left (a x \right )}}{16 a^{5}} & \text {otherwise} \end {cases}\right ) + 2 a^{3} c^{2} \left (\begin {cases} \frac {x^{4} \sqrt {- a^{2} c x^{2} + c}}{5} - \frac {x^{2} \sqrt {- a^{2} c x^{2} + c}}{15 a^{2}} - \frac {2 \sqrt {- a^{2} c x^{2} + c}}{15 a^{4}} & \text {for}\: a \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) - 2 a c^{2} \left (\begin {cases} 0 & \text {for}\: c = 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\left (- a^{2} c x^{2} + c\right )^{\frac {3}{2}}}{3 a^{2} c} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} \frac {i a^{2} \sqrt {c} x^{3}}{2 \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} x}{2 \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} \operatorname {acosh}{\left (a x \right )}}{2 a} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {\sqrt {c} x \sqrt {- a^{2} x^{2} + 1}}{2} + \frac {\sqrt {c} \operatorname {asin}{\left (a x \right )}}{2 a} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**4*c**2*Piecewise((I*a**2*sqrt(c)*x**7/(6*sqrt(a**2*x**2 - 1)) - 5*I*sqrt(c)*x**5/(24*sqrt(a**2*x**2 - 1))
- I*sqrt(c)*x**3/(48*a**2*sqrt(a**2*x**2 - 1)) + I*sqrt(c)*x/(16*a**4*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*acosh(a
*x)/(16*a**5), Abs(a**2*x**2) > 1), (-a**2*sqrt(c)*x**7/(6*sqrt(-a**2*x**2 + 1)) + 5*sqrt(c)*x**5/(24*sqrt(-a*
*2*x**2 + 1)) + sqrt(c)*x**3/(48*a**2*sqrt(-a**2*x**2 + 1)) - sqrt(c)*x/(16*a**4*sqrt(-a**2*x**2 + 1)) + sqrt(
c)*asin(a*x)/(16*a**5), True)) + 2*a**3*c**2*Piecewise((x**4*sqrt(-a**2*c*x**2 + c)/5 - x**2*sqrt(-a**2*c*x**2
+ c)/(15*a**2) - 2*sqrt(-a**2*c*x**2 + c)/(15*a**4), Ne(a, 0)), (sqrt(c)*x**4/4, True)) - 2*a*c**2*Piecewise(
(0, Eq(c, 0)), (sqrt(c)*x**2/2, Eq(a**2, 0)), (-(-a**2*c*x**2 + c)**(3/2)/(3*a**2*c), True)) + c**2*Piecewise(
(I*a**2*sqrt(c)*x**3/(2*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*x/(2*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*acosh(a*x)/(2*a
), Abs(a**2*x**2) > 1), (sqrt(c)*x*sqrt(-a**2*x**2 + 1)/2 + sqrt(c)*asin(a*x)/(2*a), True))

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