3.8.74 \(\int e^{\coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^2 \, dx\) [774]

Optimal. Leaf size=194 \[ -\frac {5 c^2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}{2 a}-\frac {7 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/2}}{6 a}+\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{5/2}}{3 a}+c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x+\frac {3 c^2 \csc ^{-1}(a x)}{2 a}+\frac {c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}\right )}{a} \]

[Out]

c^2*(1-1/a/x)^(3/2)*(1+1/a/x)^(5/2)*x+3/2*c^2*arccsc(a*x)/a+c^2*arctanh((1-1/a/x)^(1/2)*(1+1/a/x)^(1/2))/a-7/6
*c^2*(1+1/a/x)^(3/2)*(1-1/a/x)^(1/2)/a+4/3*c^2*(1+1/a/x)^(5/2)*(1-1/a/x)^(1/2)/a-5/2*c^2*(1-1/a/x)^(1/2)*(1+1/
a/x)^(1/2)/a

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Rubi [A]
time = 0.09, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6329, 99, 159, 163, 41, 222, 94, 214} \begin {gather*} c^2 x \left (1-\frac {1}{a x}\right )^{3/2} \left (\frac {1}{a x}+1\right )^{5/2}+\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{5/2}}{3 a}-\frac {7 c^2 \sqrt {1-\frac {1}{a x}} \left (\frac {1}{a x}+1\right )^{3/2}}{6 a}-\frac {5 c^2 \sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}}{2 a}+\frac {3 c^2 \csc ^{-1}(a x)}{2 a}+\frac {c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*c^2*Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)])/(2*a) - (7*c^2*Sqrt[1 - 1/(a*x)]*(1 + 1/(a*x))^(3/2))/(6*a) + (4*
c^2*Sqrt[1 - 1/(a*x)]*(1 + 1/(a*x))^(5/2))/(3*a) + c^2*(1 - 1/(a*x))^(3/2)*(1 + 1/(a*x))^(5/2)*x + (3*c^2*ArcC
sc[a*x])/(2*a) + (c^2*ArcTanh[Sqrt[1 - 1/(a*x)]*Sqrt[1 + 1/(a*x)]])/a

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 222

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 6329

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

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx &=-\left (c^2 \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{3/2} \left (1+\frac {x}{a}\right )^{5/2}}{x^2} \, dx,x,\frac {1}{x}\right )\right )\\ &=c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x-c^2 \text {Subst}\left (\int \frac {\left (\frac {1}{a}-\frac {4 x}{a^2}\right ) \sqrt {1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/2}}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{5/2}}{3 a}+c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x-\frac {1}{3} \left (a c^2\right ) \text {Subst}\left (\int \frac {\left (\frac {3}{a^2}-\frac {7 x}{a^3}\right ) \left (1+\frac {x}{a}\right )^{3/2}}{x \sqrt {1-\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {7 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/2}}{6 a}+\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{5/2}}{3 a}+c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x+\frac {1}{6} \left (a^2 c^2\right ) \text {Subst}\left (\int \frac {\left (-\frac {6}{a^3}+\frac {15 x}{a^4}\right ) \sqrt {1+\frac {x}{a}}}{x \sqrt {1-\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {5 c^2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}{2 a}-\frac {7 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/2}}{6 a}+\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{5/2}}{3 a}+c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x-\frac {1}{6} \left (a^3 c^2\right ) \text {Subst}\left (\int \frac {\frac {6}{a^4}-\frac {9 x}{a^5}}{x \sqrt {1-\frac {x}{a}} \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {5 c^2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}{2 a}-\frac {7 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/2}}{6 a}+\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{5/2}}{3 a}+c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x+\frac {\left (3 c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x}{a}} \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a^2}-\frac {c^2 \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a}} \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {5 c^2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}{2 a}-\frac {7 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/2}}{6 a}+\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{5/2}}{3 a}+c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x+\frac {c^2 \text {Subst}\left (\int \frac {1}{\frac {1}{a}-\frac {x^2}{a}} \, dx,x,\sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}\right )}{a^2}+\frac {\left (3 c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a^2}\\ &=-\frac {5 c^2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}}{2 a}-\frac {7 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{3/2}}{6 a}+\frac {4 c^2 \sqrt {1-\frac {1}{a x}} \left (1+\frac {1}{a x}\right )^{5/2}}{3 a}+c^2 \left (1-\frac {1}{a x}\right )^{3/2} \left (1+\frac {1}{a x}\right )^{5/2} x+\frac {3 c^2 \csc ^{-1}(a x)}{2 a}+\frac {c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}}\right )}{a}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 94, normalized size = 0.48 \begin {gather*} \frac {c^2 \left (\sqrt {1-\frac {1}{a^2 x^2}} \left (2+3 a x-8 a^2 x^2+6 a^3 x^3\right )+9 a^2 x^2 \text {ArcSin}\left (\frac {1}{a x}\right )+6 a^2 x^2 \log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{6 a^3 x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(Sqrt[1 - 1/(a^2*x^2)]*(2 + 3*a*x - 8*a^2*x^2 + 6*a^3*x^3) + 9*a^2*x^2*ArcSin[1/(a*x)] + 6*a^2*x^2*Log[(1
+ Sqrt[1 - 1/(a^2*x^2)])*x]))/(6*a^3*x^2)

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Maple [A]
time = 0.07, size = 224, normalized size = 1.15

method	result	size
risch	\(-\frac {\left (a x -1\right ) \left (8 a^{2} x^{2}-3 a x -2\right ) c^{2}}{6 x^{3} a^{4} \sqrt {\frac {a x -1}{a x +1}}}+\frac {\left (a^{3} \sqrt {\left (a x +1\right ) \left (a x -1\right )}+\frac {a^{4} \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{\sqrt {a^{2}}}+\frac {3 a^{3} \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )}{2}\right ) c^{2} \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{a^{4} \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(156\)
default	\(\frac {\left (a x -1\right ) c^{2} \left (-6 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{4} x^{4}+6 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, a^{2} x^{2}+9 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{3} x^{3}+6 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}+9 a^{3} x^{3} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )-3 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x -2 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\right )}{6 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, a^{4} x^{3} \sqrt {a^{2}}}\)	\(224\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*(c-c/a^2/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/6*(a*x-1)*c^2*(-6*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*a^4*x^4+6*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*a^2*x^2+9*(a^2*x^2-1
)^(1/2)*(a^2)^(1/2)*a^3*x^3+6*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*a^4*x^3+9*a^3*x^3*(a^2)^(1
/2)*arctan(1/(a^2*x^2-1)^(1/2))-3*(a^2)^(1/2)*(a^2*x^2-1)^(3/2)*a*x-2*(a^2*x^2-1)^(3/2)*(a^2)^(1/2))/((a*x-1)/
(a*x+1))^(1/2)/((a*x+1)*(a*x-1))^(1/2)/a^4/x^3/(a^2)^(1/2)

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Maxima [A]
time = 0.46, size = 223, normalized size = 1.15 \begin {gather*} -\frac {1}{3} \, a {\left (\frac {9 \, c^{2} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} - \frac {3 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac {3 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {15 \, c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} + 29 \, c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 3 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {2 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {2 \, {\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} + a^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(9*c^2*arctan(sqrt((a*x - 1)/(a*x + 1)))/a^2 - 3*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 + 3*c^2*log
(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 - (15*c^2*((a*x - 1)/(a*x + 1))^(7/2) + 29*c^2*((a*x - 1)/(a*x + 1))^(5/2)
+ c^2*((a*x - 1)/(a*x + 1))^(3/2) + 3*c^2*sqrt((a*x - 1)/(a*x + 1)))/(2*(a*x - 1)*a^2/(a*x + 1) - 2*(a*x - 1)
^3*a^2/(a*x + 1)^3 - (a*x - 1)^4*a^2/(a*x + 1)^4 + a^2))

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Fricas [A]
time = 0.38, size = 157, normalized size = 0.81 \begin {gather*} -\frac {18 \, a^{3} c^{2} x^{3} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - 6 \, a^{3} c^{2} x^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + 6 \, a^{3} c^{2} x^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (6 \, a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} - 5 \, a^{2} c^{2} x^{2} + 5 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{6 \, a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(18*a^3*c^2*x^3*arctan(sqrt((a*x - 1)/(a*x + 1))) - 6*a^3*c^2*x^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) + 6*
a^3*c^2*x^3*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (6*a^4*c^2*x^4 - 2*a^3*c^2*x^3 - 5*a^2*c^2*x^2 + 5*a*c^2*x +
2*c^2)*sqrt((a*x - 1)/(a*x + 1)))/(a^4*x^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {c^{2} \left (\int \frac {a^{4}}{\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx + \int \frac {1}{x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx + \int \left (- \frac {2 a^{2}}{x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\right )\, dx\right )}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*(Integral(a**4/sqrt(a*x/(a*x + 1) - 1/(a*x + 1)), x) + Integral(1/(x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))
), x) + Integral(-2*a**2/(x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))), x))/a**4

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Giac [A]
time = 0.44, size = 249, normalized size = 1.28 \begin {gather*} -\frac {3 \, c^{2} \arctan \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1}\right )}{a \mathrm {sgn}\left (a x + 1\right )} - \frac {c^{2} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{{\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} + \frac {\sqrt {a^{2} x^{2} - 1} c^{2}}{a \mathrm {sgn}\left (a x + 1\right )} - \frac {3 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{5} c^{2} {\left | a \right |} + 12 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{4} a c^{2} + 12 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} a c^{2} - 3 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )} c^{2} {\left | a \right |} + 8 \, a c^{2}}{3 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} + 1\right )}^{3} a {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*c^2*arctan(-x*abs(a) + sqrt(a^2*x^2 - 1))/(a*sgn(a*x + 1)) - c^2*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))/(a
bs(a)*sgn(a*x + 1)) + sqrt(a^2*x^2 - 1)*c^2/(a*sgn(a*x + 1)) - 1/3*(3*(x*abs(a) - sqrt(a^2*x^2 - 1))^5*c^2*abs
(a) + 12*(x*abs(a) - sqrt(a^2*x^2 - 1))^4*a*c^2 + 12*(x*abs(a) - sqrt(a^2*x^2 - 1))^2*a*c^2 - 3*(x*abs(a) - sq
rt(a^2*x^2 - 1))*c^2*abs(a) + 8*a*c^2)/(((x*abs(a) - sqrt(a^2*x^2 - 1))^2 + 1)^3*a*abs(a)*sgn(a*x + 1))

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Mupad [B]
time = 1.33, size = 183, normalized size = 0.94 \begin {gather*} \frac {c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}+\frac {c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}+\frac {29\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{3}+5\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{a+\frac {2\,a\,\left (a\,x-1\right )}{a\,x+1}-\frac {2\,a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {a\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}}-\frac {3\,c^2\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {2\,c^2\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^2*((a*x - 1)/(a*x + 1))^(1/2) + (c^2*((a*x - 1)/(a*x + 1))^(3/2))/3 + (29*c^2*((a*x - 1)/(a*x + 1))^(5/2))/
3 + 5*c^2*((a*x - 1)/(a*x + 1))^(7/2))/(a + (2*a*(a*x - 1))/(a*x + 1) - (2*a*(a*x - 1)^3)/(a*x + 1)^3 - (a*(a*
x - 1)^4)/(a*x + 1)^4) - (3*c^2*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a + (2*c^2*atanh(((a*x - 1)/(a*x + 1))^(1/2
)))/a

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