3.5.31 \(\int e^{-3 \coth ^{-1}(a x)} (c-\frac {c}{a x})^2 \, dx\) [431]

Optimal. Leaf size=105 \[ \frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {5 c^2 \csc ^{-1}(a x)}{a}-\frac {5 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \]

[Out]

5*c^2*arccsc(a*x)/a-5*c^2*arctanh((1-1/a^2/x^2)^(1/2))/a+16*c^2*(a-1/x)/a^2/(1-1/a^2/x^2)^(1/2)+c^2*(1-1/a^2/x
^2)^(1/2)/a+c^2*x*(1-1/a^2/x^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.22, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {6312, 1819, 1821, 1823, 858, 222, 272, 65, 214} \begin {gather*} \frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 x \sqrt {1-\frac {1}{a^2 x^2}}+\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}-\frac {5 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}+\frac {5 c^2 \csc ^{-1}(a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[1 - 1/(a^2*x^2)])/a + (16*c^2*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + c^2*Sqrt[1 - 1/(a^2*x^2)]*
x + (5*c^2*ArcCsc[a*x])/a - (5*c^2*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/a

Rule 65

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 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 272

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 858

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 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
 a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
 x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1821

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 1823

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 6312

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

Rubi steps

\begin {align*} \int e^{-3 \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx &=-\frac {\text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^5}{x^2 \left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\text {Subst}\left (\int \frac {-c^5+\frac {5 c^5 x}{a}+\frac {5 c^5 x^2}{a^2}-\frac {c^5 x^3}{a^3}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {\text {Subst}\left (\int \frac {-\frac {5 c^5}{a}-\frac {5 c^5 x}{a^2}+\frac {c^5 x^2}{a^3}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {a^2 \text {Subst}\left (\int \frac {\frac {5 c^5}{a^3}+\frac {5 c^5 x}{a^4}}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^2}+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {5 c^2 \csc ^{-1}(a x)}{a}+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{2 a}\\ &=\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {5 c^2 \csc ^{-1}(a x)}{a}-\left (5 a c^2\right ) \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )\\ &=\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a}+\frac {16 c^2 \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+c^2 \sqrt {1-\frac {1}{a^2 x^2}} x+\frac {5 c^2 \csc ^{-1}(a x)}{a}-\frac {5 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.26, size = 424, normalized size = 4.04 \begin {gather*} \frac {c^2 \left (-35 a^2 \sqrt {1+\frac {1}{a x}} x^2+315 a^3 \sqrt {1+\frac {1}{a x}} x^3+280 a^4 \sqrt {1+\frac {1}{a x}} x^4-595 a^5 \sqrt {1+\frac {1}{a x}} x^5+35 a^6 \sqrt {1+\frac {1}{a x}} x^6+910 a^4 \sqrt {1-\frac {1}{a x}} x^4 \text {ArcSin}\left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )+910 a^5 \sqrt {1-\frac {1}{a x}} x^5 \text {ArcSin}\left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )-105 a^4 \sqrt {1-\frac {1}{a x}} x^4 \text {ArcSin}\left (\frac {1}{a x}\right )-105 a^5 \sqrt {1-\frac {1}{a x}} x^5 \text {ArcSin}\left (\frac {1}{a x}\right )-175 a^5 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {1+\frac {1}{a x}} x^5 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )+7 \sqrt {2} a x (-1+a x)^3 (1+a x) \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {1}{2} \left (1-\frac {1}{a x}\right )\right )+5 \sqrt {2} (-1+a x)^4 (1+a x) \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};\frac {1}{2} \left (1-\frac {1}{a x}\right )\right )\right )}{35 a^5 \sqrt {1-\frac {1}{a x}} x^4 (1+a x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(-35*a^2*Sqrt[1 + 1/(a*x)]*x^2 + 315*a^3*Sqrt[1 + 1/(a*x)]*x^3 + 280*a^4*Sqrt[1 + 1/(a*x)]*x^4 - 595*a^5*
Sqrt[1 + 1/(a*x)]*x^5 + 35*a^6*Sqrt[1 + 1/(a*x)]*x^6 + 910*a^4*Sqrt[1 - 1/(a*x)]*x^4*ArcSin[Sqrt[1 - 1/(a*x)]/
Sqrt[2]] + 910*a^5*Sqrt[1 - 1/(a*x)]*x^5*ArcSin[Sqrt[1 - 1/(a*x)]/Sqrt[2]] - 105*a^4*Sqrt[1 - 1/(a*x)]*x^4*Arc
Sin[1/(a*x)] - 105*a^5*Sqrt[1 - 1/(a*x)]*x^5*ArcSin[1/(a*x)] - 175*a^5*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[1 + 1/(a*x)]
*x^5*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]] + 7*Sqrt[2]*a*x*(-1 + a*x)^3*(1 + a*x)*Hypergeometric2F1[3/2, 5/2, 7/2, (1
 - 1/(a*x))/2] + 5*Sqrt[2]*(-1 + a*x)^4*(1 + a*x)*Hypergeometric2F1[3/2, 7/2, 9/2, (1 - 1/(a*x))/2]))/(35*a^5*
Sqrt[1 - 1/(a*x)]*x^4*(1 + a*x))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(599\) vs. \(2(97)=194\).
time = 0.12, size = 600, normalized size = 5.71

method result size
risch \(\frac {\left (a x +1\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}}{x \,a^{2}}+\frac {\left (a \sqrt {\left (a x +1\right ) \left (a x -1\right )}-\frac {5 a^{2} \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{\sqrt {a^{2}}}+\frac {16 \sqrt {a^{2} \left (x +\frac {1}{a}\right )^{2}-2 a \left (x +\frac {1}{a}\right )}}{x +\frac {1}{a}}+5 a \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{a^{2} \left (a x -1\right )}\) \(169\)
default \(-\frac {\left (-\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{4} x^{4}-4 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, a^{3} x^{3}+4 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}+\left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, a^{2} x^{2}-7 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{3} x^{3}-5 a^{3} x^{3} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{4} x^{3}+8 \sqrt {a^{2}}\, \left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} a x -8 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}+8 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+2 \sqrt {a^{2}}\, \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} a x -11 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{2} x^{2}-10 \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) a^{2} x^{2}+2 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}-4 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a x +4 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x +\left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}-5 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x -5 \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) a x +\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{2} x \right ) c^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a^{2} x \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \left (a x -1\right )}\) \(600\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*a^4*x^4-4*(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2)*a^3*x^3+4*ln((a^2*x+(a^2)^(1/2)
*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^4*x^3+(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*a^2*x^2-7*(a^2*x^2-1)^(1/2)*(a^2)
^(1/2)*a^3*x^3-5*a^3*x^3*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2
)^(1/2))*a^4*x^3+8*(a^2)^(1/2)*((a*x+1)*(a*x-1))^(3/2)*a*x-8*((a*x+1)*(a*x-1))^(1/2)*(a^2)^(1/2)*a^2*x^2+8*ln(
(a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^3*x^2+2*(a^2)^(1/2)*(a^2*x^2-1)^(3/2)*a*x-11*(a^2*x
^2-1)^(1/2)*(a^2)^(1/2)*a^2*x^2-10*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))*a^2*x^2+2*ln((a^2*x+(a^2*x^2-1)^(1/
2)*(a^2)^(1/2))/(a^2)^(1/2))*a^3*x^2-4*((a*x+1)*(a*x-1))^(1/2)*(a^2)^(1/2)*a*x+4*ln((a^2*x+(a^2)^(1/2)*((a*x+1
)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^2*x+(a^2*x^2-1)^(3/2)*(a^2)^(1/2)-5*(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*a*x-5*(a^2)
^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))*a*x+ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*a^2*x)/a^2*c^2*((
a*x-1)/(a*x+1))^(3/2)/x/(a^2)^(1/2)/((a*x+1)*(a*x-1))^(1/2)/(a*x-1)

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 149, normalized size = 1.42 \begin {gather*} -{\left (\frac {4 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} + \frac {10 \, c^{2} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} + \frac {5 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {5 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {16 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2}}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*c^2*sqrt((a*x - 1)/(a*x + 1))/((a*x - 1)^2*a^2/(a*x + 1)^2 - a^2) + 10*c^2*arctan(sqrt((a*x - 1)/(a*x + 1)
))/a^2 + 5*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 5*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 - 16*c^2*
sqrt((a*x - 1)/(a*x + 1))/a^2)*a

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 120, normalized size = 1.14 \begin {gather*} -\frac {10 \, a c^{2} x \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + 5 \, a c^{2} x \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 5 \, a c^{2} x \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (a^{2} c^{2} x^{2} + 18 \, a c^{2} x + c^{2}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {c^{2} \left (\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x^{3} + x^{2}}\right )\, dx + \int \frac {3 a \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x^{2} + x}\, dx + \int \left (- \frac {3 a^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\right )\, dx + \int \frac {a^{3} x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\, dx\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

________________________________________________________________________________________

Mupad [B]
time = 0.10, size = 117, normalized size = 1.11 \begin {gather*} \frac {16\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a}+\frac {4\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a-\frac {a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}}-\frac {10\,c^2\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {c^2\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\,1{}\mathrm {i}\right )\,10{}\mathrm {i}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16*c^2*((a*x - 1)/(a*x + 1))^(1/2))/a + (4*c^2*((a*x - 1)/(a*x + 1))^(1/2))/(a - (a*(a*x - 1)^2)/(a*x + 1)^2)
 - (10*c^2*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a + (c^2*atan(((a*x - 1)/(a*x + 1))^(1/2)*1i)*10i)/a

________________________________________________________________________________________