3.1102 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{5/2}}{x} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6151, 1809, 815, 844, 217, 203, 266, 63, 208} \[ \frac {1}{4} c^2 (3 a x+4) \sqrt {c-a^2 c x^2}+\frac {3}{4} c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )+\frac {1}{6} c (3 a x+2) \left (c-a^2 c x^2\right )^{3/2}-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(4 + 3*a*x)*Sqrt[c - a^2*c*x^2])/4 + (c*(2 + 3*a*x)*(c - a^2*c*x^2)^(3/2))/6 - (c - a^2*c*x^2)^(5/2)/5 +
(3*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/4 - c^(5/2)*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]]

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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 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 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 \frac {e^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2}}{x} \, dx &=c \int \frac {(1+a x)^2 \left (c-a^2 c x^2\right )^{3/2}}{x} \, dx\\ &=-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2}-\frac {\int \frac {\left (-5 a^2 c-10 a^3 c x\right ) \left (c-a^2 c x^2\right )^{3/2}}{x} \, dx}{5 a^2}\\ &=\frac {1}{6} c (2+3 a x) \left (c-a^2 c x^2\right )^{3/2}-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2}+\frac {\int \frac {\left (20 a^4 c^3+30 a^5 c^3 x\right ) \sqrt {c-a^2 c x^2}}{x} \, dx}{20 a^4 c}\\ &=\frac {1}{4} c^2 (4+3 a x) \sqrt {c-a^2 c x^2}+\frac {1}{6} c (2+3 a x) \left (c-a^2 c x^2\right )^{3/2}-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2}-\frac {\int \frac {-40 a^6 c^5-30 a^7 c^5 x}{x \sqrt {c-a^2 c x^2}} \, dx}{40 a^6 c^2}\\ &=\frac {1}{4} c^2 (4+3 a x) \sqrt {c-a^2 c x^2}+\frac {1}{6} c (2+3 a x) \left (c-a^2 c x^2\right )^{3/2}-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2}+c^3 \int \frac {1}{x \sqrt {c-a^2 c x^2}} \, dx+\frac {1}{4} \left (3 a c^3\right ) \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {1}{4} c^2 (4+3 a x) \sqrt {c-a^2 c x^2}+\frac {1}{6} c (2+3 a x) \left (c-a^2 c x^2\right )^{3/2}-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2}+\frac {1}{2} c^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-a^2 c x}} \, dx,x,x^2\right )+\frac {1}{4} \left (3 a 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 {1}{4} c^2 (4+3 a x) \sqrt {c-a^2 c x^2}+\frac {1}{6} c (2+3 a x) \left (c-a^2 c x^2\right )^{3/2}-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2}+\frac {3}{4} c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-\frac {c^2 \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 )}{a^2}\\ &=\frac {1}{4} c^2 (4+3 a x) \sqrt {c-a^2 c x^2}+\frac {1}{6} c (2+3 a x) \left (c-a^2 c x^2\right )^{3/2}-\frac {1}{5} \left (c-a^2 c x^2\right )^{5/2}+\frac {3}{4} c^{5/2} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 136, normalized size = 1.00 \[ -c^{5/2} \log \left (\sqrt {c} \sqrt {c-a^2 c x^2}+c\right )-\frac {3}{4} c^{5/2} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )-\frac {1}{60} c^2 \left (12 a^4 x^4+30 a^3 x^3-4 a^2 x^2-75 a x-68\right ) \sqrt {c-a^2 c x^2}+c^{5/2} \log (x) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/60*(c^2*Sqrt[c - a^2*c*x^2]*(-68 - 75*a*x - 4*a^2*x^2 + 30*a^3*x^3 + 12*a^4*x^4)) - (3*c^(5/2)*ArcTan[(a*x*
Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/4 + c^(5/2)*Log[x] - c^(5/2)*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^
2]]

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fricas [A]  time = 0.54, size = 295, normalized size = 2.17 \[ \left [-\frac {3}{4} \, c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + \frac {1}{2} \, c^{\frac {5}{2}} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - \frac {1}{60} \, {\left (12 \, a^{4} c^{2} x^{4} + 30 \, a^{3} c^{2} x^{3} - 4 \, a^{2} c^{2} x^{2} - 75 \, a c^{2} x - 68 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}, -\sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + \frac {3}{8} \, \sqrt {-c} c^{2} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - \frac {1}{60} \, {\left (12 \, a^{4} c^{2} x^{4} + 30 \, a^{3} c^{2} x^{3} - 4 \, a^{2} c^{2} x^{2} - 75 \, a c^{2} x - 68 \, c^{2}\right )} \sqrt {-a^{2} c x^{2} + c}\right ] \]

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)^(5/2)/x,x, algorithm="fricas")

[Out]

[-3/4*c^(5/2)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + 1/2*c^(5/2)*log(-(a^2*c*x^2 + 2*sqrt(
-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - 1/60*(12*a^4*c^2*x^4 + 30*a^3*c^2*x^3 - 4*a^2*c^2*x^2 - 75*a*c^2*x - 68*
c^2)*sqrt(-a^2*c*x^2 + c), -sqrt(-c)*c^2*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c*x^2 - c)) + 3/8*sqrt(-c)*
c^2*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 1/60*(12*a^4*c^2*x^4 + 30*a^3*c^2*x^3 - 4*a^2
*c^2*x^2 - 75*a*c^2*x - 68*c^2)*sqrt(-a^2*c*x^2 + c)]

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giac [A]  time = 0.70, size = 150, normalized size = 1.10 \[ \frac {2 \, c^{3} \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {3 \, a \sqrt {-c} c^{2} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{4 \, {\left | a \right |}} + \frac {1}{60} \, \sqrt {-a^{2} c x^{2} + c} {\left (68 \, c^{2} + {\left (75 \, a c^{2} + 2 \, {\left (2 \, a^{2} c^{2} - 3 \, {\left (2 \, a^{4} c^{2} x + 5 \, a^{3} c^{2}\right )} x\right )} x\right )} x\right )} \]

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)^(5/2)/x,x, algorithm="giac")

[Out]

2*c^3*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) + 3/4*a*sqrt(-c)*c^2*log(abs(-sqrt(-a
^2*c)*x + sqrt(-a^2*c*x^2 + c)))/abs(a) + 1/60*sqrt(-a^2*c*x^2 + c)*(68*c^2 + (75*a*c^2 + 2*(2*a^2*c^2 - 3*(2*
a^4*c^2*x + 5*a^3*c^2)*x)*x)*x)

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maple [B]  time = 0.04, size = 235, normalized size = 1.73 \[ \frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{5}+\frac {c \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3}-c^{\frac {5}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )+\sqrt {-a^{2} c \,x^{2}+c}\, c^{2}-\frac {2 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {5}{2}}}{5}+\frac {a c \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}} x}{2}+\frac {3 a \,c^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}\, x}{4}+\frac {3 a \,c^{3} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{4 \sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(-a^2*c*x^2+c)^(5/2)+1/3*c*(-a^2*c*x^2+c)^(3/2)-c^(5/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)+(-a^2*c
*x^2+c)^(1/2)*c^2-2/5*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(5/2)+1/2*a*c*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(3/2)*x+
3/4*a*c^2*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(1/2)*x+3/4*a*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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}^{2}}{{\left (a^{2} x^{2} - 1\right )} x}\,{d x} \]

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)^(5/2)/x,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 33.54, size = 508, normalized size = 3.74 \[ - a^{4} c^{2} \left (\begin {cases} \frac {x^{4} \sqrt {- a^{2} c x^{2} + c}}{5} - \frac {x^{2} \sqrt {- a^{2} c x^{2} + c}}{15 a^{2}} - \frac {2 \sqrt {- a^{2} c x^{2} + c}}{15 a^{4}} & \text {for}\: a \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) - 2 a^{3} c^{2} \left (\begin {cases} \frac {i a^{2} \sqrt {c} x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \sqrt {c} x^{3}}{8 \sqrt {a^{2} x^{2} - 1}} + \frac {i \sqrt {c} x}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} \operatorname {acosh}{\left (a x \right )}}{8 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {a^{2} \sqrt {c} x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \sqrt {c} x^{3}}{8 \sqrt {- a^{2} x^{2} + 1}} - \frac {\sqrt {c} x}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\sqrt {c} \operatorname {asin}{\left (a x \right )}}{8 a^{3}} & \text {otherwise} \end {cases}\right ) + 2 a c^{2} \left (\begin {cases} \frac {i a^{2} \sqrt {c} x^{3}}{2 \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} x}{2 \sqrt {a^{2} x^{2} - 1}} - \frac {i \sqrt {c} \operatorname {acosh}{\left (a x \right )}}{2 a} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {\sqrt {c} x \sqrt {- a^{2} x^{2} + 1}}{2} + \frac {\sqrt {c} \operatorname {asin}{\left (a x \right )}}{2 a} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} i \sqrt {c} \sqrt {a^{2} x^{2} - 1} - \sqrt {c} \log {\left (a x \right )} + \frac {\sqrt {c} \log {\left (a^{2} x^{2} \right )}}{2} + i \sqrt {c} \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\sqrt {c} \sqrt {- a^{2} x^{2} + 1} + \frac {\sqrt {c} \log {\left (a^{2} x^{2} \right )}}{2} - \sqrt {c} \log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )} & \text {otherwise} \end {cases}\right ) \]

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)**(5/2)/x,x)

[Out]

-a**4*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**3*c**2*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*sqrt(c)*x**3/(8*s
qrt(-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), True)) + 2*a*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)*acos
h(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)) + c**2*
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))

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