3.1087 \(\int e^{2 \tanh ^{-1}(a x)} x^2 (c-a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=136 \[ -\frac {2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}+\frac {3 c x \sqrt {c-a^2 c x^2}}{16 a^2}-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}+\frac {3 c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a^3}-\frac {(45 a x+32) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3} \]

[Out]

-2/5*x^2*(-a^2*c*x^2+c)^(3/2)/a-1/6*x^3*(-a^2*c*x^2+c)^(3/2)-1/120*(45*a*x+32)*(-a^2*c*x^2+c)^(3/2)/a^3+3/16*c
^(3/2)*arctan(a*x*c^(1/2)/(-a^2*c*x^2+c)^(1/2))/a^3+3/16*c*x*(-a^2*c*x^2+c)^(1/2)/a^2

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Rubi [A]  time = 0.31, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6151, 1809, 833, 780, 195, 217, 203} \[ \frac {3 c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a^3}-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}+\frac {3 c x \sqrt {c-a^2 c x^2}}{16 a^2}-\frac {(45 a x+32) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*c*x*Sqrt[c - a^2*c*x^2])/(16*a^2) - (2*x^2*(c - a^2*c*x^2)^(3/2))/(5*a) - (x^3*(c - a^2*c*x^2)^(3/2))/6 - (
(32 + 45*a*x)*(c - a^2*c*x^2)^(3/2))/(120*a^3) + (3*c^(3/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(16*a^3
)

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 780

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

Rule 833

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

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -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 e^{2 \tanh ^{-1}(a x)} x^2 \left (c-a^2 c x^2\right )^{3/2} \, dx &=c \int x^2 (1+a x)^2 \sqrt {c-a^2 c x^2} \, dx\\ &=-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {\int x^2 \left (-9 a^2 c-12 a^3 c x\right ) \sqrt {c-a^2 c x^2} \, dx}{6 a^2}\\ &=-\frac {2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}+\frac {\int x \left (24 a^3 c^2+45 a^4 c^2 x\right ) \sqrt {c-a^2 c x^2} \, dx}{30 a^4 c}\\ &=-\frac {2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac {(3 c) \int \sqrt {c-a^2 c x^2} \, dx}{8 a^2}\\ &=\frac {3 c x \sqrt {c-a^2 c x^2}}{16 a^2}-\frac {2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac {\left (3 c^2\right ) \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx}{16 a^2}\\ &=\frac {3 c x \sqrt {c-a^2 c x^2}}{16 a^2}-\frac {2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac {\left (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 )}{16 a^2}\\ &=\frac {3 c x \sqrt {c-a^2 c x^2}}{16 a^2}-\frac {2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac {1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac {(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac {3 c^{3/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{16 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 105, normalized size = 0.77 \[ \frac {c \left (40 a^5 x^5+96 a^4 x^4+50 a^3 x^3-32 a^2 x^2-45 a x-64\right ) \sqrt {c-a^2 c x^2}-45 c^{3/2} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )}{240 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c - a^2*c*x^2]*(-64 - 45*a*x - 32*a^2*x^2 + 50*a^3*x^3 + 96*a^4*x^4 + 40*a^5*x^5) - 45*c^(3/2)*ArcTan[
(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/(240*a^3)

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fricas [A]  time = 0.64, size = 216, normalized size = 1.59 \[ \left [\frac {45 \, \sqrt {-c} c \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) + 2 \, {\left (40 \, a^{5} c x^{5} + 96 \, a^{4} c x^{4} + 50 \, a^{3} c x^{3} - 32 \, a^{2} c x^{2} - 45 \, a c x - 64 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{480 \, a^{3}}, -\frac {45 \, c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - {\left (40 \, a^{5} c x^{5} + 96 \, a^{4} c x^{4} + 50 \, a^{3} c x^{3} - 32 \, a^{2} c x^{2} - 45 \, a c x - 64 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{240 \, a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/480*(45*sqrt(-c)*c*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*x^5 + 96*a^4*c*
x^4 + 50*a^3*c*x^3 - 32*a^2*c*x^2 - 45*a*c*x - 64*c)*sqrt(-a^2*c*x^2 + c))/a^3, -1/240*(45*c^(3/2)*arctan(sqrt
(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) - (40*a^5*c*x^5 + 96*a^4*c*x^4 + 50*a^3*c*x^3 - 32*a^2*c*x^2 - 4
5*a*c*x - 64*c)*sqrt(-a^2*c*x^2 + c))/a^3]

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giac [A]  time = 0.21, size = 107, normalized size = 0.79 \[ \frac {1}{240} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, a^{2} c x + 12 \, a c\right )} x + 25 \, c\right )} x - \frac {16 \, c}{a}\right )} x - \frac {45 \, c}{a^{2}}\right )} x - \frac {64 \, c}{a^{3}}\right )} - \frac {3 \, c^{2} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{16 \, a^{2} \sqrt {-c} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*sqrt(-a^2*c*x^2 + c)*((2*((4*(5*a^2*c*x + 12*a*c)*x + 25*c)*x - 16*c/a)*x - 45*c/a^2)*x - 64*c/a^3) - 3/
16*c^2*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a^2*sqrt(-c)*abs(a))

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maple [B]  time = 0.04, size = 244, normalized size = 1.79 \[ \frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{6 a^{2} c}-\frac {13 x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{24 a^{2}}-\frac {13 c x \sqrt {-a^{2} c \,x^{2}+c}}{16 a^{2}}-\frac {13 c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{16 a^{2} \sqrt {a^{2} c}}+\frac {2 \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{5 a^{3} c}-\frac {2 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3 a^{3}}+\frac {c \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}\, x}{a^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{a^{2} \sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*(-a^2*c*x^2+c)^(5/2)/a^2/c-13/24/a^2*x*(-a^2*c*x^2+c)^(3/2)-13/16*c*x*(-a^2*c*x^2+c)^(1/2)/a^2-13/16/a^2
*c^2/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))+2/5/a^3*(-a^2*c*x^2+c)^(5/2)/c-2/3/a^3*(-(x-1/
a)^2*a^2*c-2*a*c*(x-1/a))^(3/2)+1/a^2*c*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(1/2)*x+1/a^2*c^2/(a^2*c)^(1/2)*arcta
n((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 = 0.65, size = 189, normalized size = 1.39 \[ -\frac {1}{240} \, a {\left (\frac {130 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{a^{3}} - \frac {40 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x}{a^{3} c} - \frac {240 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c x}{a^{3}} + \frac {195 \, \sqrt {-a^{2} c x^{2} + c} c x}{a^{3}} + \frac {195 \, c^{\frac {3}{2}} \arcsin \left (a x\right )}{a^{4}} + \frac {160 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{a^{4}} - \frac {96 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}{a^{4} c} + \frac {480 \, \sqrt {a^{2} c x^{2} - 4 \, a c x + 3 \, c} c}{a^{4}} - \frac {240 \, c^{3} \arcsin \left (a x - 2\right )}{a^{7} \left (-\frac {c}{a^{2}}\right )^{\frac {3}{2}}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*a*(130*(-a^2*c*x^2 + c)^(3/2)*x/a^3 - 40*(-a^2*c*x^2 + c)^(5/2)*x/(a^3*c) - 240*sqrt(a^2*c*x^2 - 4*a*c*
x + 3*c)*c*x/a^3 + 195*sqrt(-a^2*c*x^2 + c)*c*x/a^3 + 195*c^(3/2)*arcsin(a*x)/a^4 + 160*(-a^2*c*x^2 + c)^(3/2)
/a^4 - 96*(-a^2*c*x^2 + c)^(5/2)/(a^4*c) + 480*sqrt(a^2*c*x^2 - 4*a*c*x + 3*c)*c/a^4 - 240*c^3*arcsin(a*x - 2)
/(a^7*(-c/a^2)^(3/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 29.50, size = 515, normalized size = 3.79 \[ a^{2} c \left (\begin {cases} \frac {i a^{2} \sqrt {c} x^{7}}{6 \sqrt {a^{2} x^{2} - 1}} - \frac {5 i \sqrt {c} x^{5}}{24 \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} x^{3}}{48 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {i \sqrt {c} x}{16 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} \operatorname {acosh}{\left (a x \right )}}{16 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {a^{2} \sqrt {c} x^{7}}{6 \sqrt {- a^{2} x^{2} + 1}} + \frac {5 \sqrt {c} x^{5}}{24 \sqrt {- a^{2} x^{2} + 1}} + \frac {\sqrt {c} x^{3}}{48 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {\sqrt {c} x}{16 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {\sqrt {c} \operatorname {asin}{\left (a x \right )}}{16 a^{5}} & \text {otherwise} \end {cases}\right ) + 2 a c \left (\begin {cases} \frac {x^{4} \sqrt {- a^{2} c x^{2} + c}}{5} - \frac {x^{2} \sqrt {- a^{2} c x^{2} + c}}{15 a^{2}} - \frac {2 \sqrt {- a^{2} c x^{2} + c}}{15 a^{4}} & \text {for}\: a \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \frac {i a^{2} \sqrt {c} x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \sqrt {c} x^{3}}{8 \sqrt {a^{2} x^{2} - 1}} + \frac {i \sqrt {c} x}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} \operatorname {acosh}{\left (a x \right )}}{8 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {a^{2} \sqrt {c} x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \sqrt {c} x^{3}}{8 \sqrt {- a^{2} x^{2} + 1}} - \frac {\sqrt {c} x}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\sqrt {c} \operatorname {asin}{\left (a x \right )}}{8 a^{3}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*c*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)*a
sin(a*x)/(16*a**5), True)) + 2*a*c*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)) + c*Piecewise((I*a**2*sqrt(c)*x
**5/(4*sqrt(a**2*x**2 - 1)) - 3*I*sqrt(c)*x**3/(8*sqrt(a**2*x**2 - 1)) + I*sqrt(c)*x/(8*a**2*sqrt(a**2*x**2 -
1)) - I*sqrt(c)*acosh(a*x)/(8*a**3), Abs(a**2*x**2) > 1), (-a**2*sqrt(c)*x**5/(4*sqrt(-a**2*x**2 + 1)) + 3*sqr
t(c)*x**3/(8*sqrt(-a**2*x**2 + 1)) - sqrt(c)*x/(8*a**2*sqrt(-a**2*x**2 + 1)) + sqrt(c)*asin(a*x)/(8*a**3), Tru
e))

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