3.420 \(\int \frac{e^{-\coth ^{-1}(a x)}}{(c-\frac{c}{a x})^4} \, dx\)

Optimal. Leaf size=138 \[ -\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{15 a+\frac{19}{x}}{5 a^2 c^4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c^4}+\frac{3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^4} \]

[Out]

(-4*(a + x^(-1)))/(5*a^2*c^4*(1 - 1/(a^2*x^2))^(5/2)) - (5*a + 7/x)/(5*a^2*c^4*(1 - 1/(a^2*x^2))^(3/2)) - (15*
a + 19/x)/(5*a^2*c^4*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[1 - 1/(a^2*x^2)]*x)/c^4 + (3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]
])/(a*c^4)

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Rubi [A]  time = 0.390016, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6177, 852, 1805, 807, 266, 63, 208} \[ -\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{15 a+\frac{19}{x}}{5 a^2 c^4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c^4}+\frac{3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcCoth[a*x]*(c - c/(a*x))^4),x]

[Out]

(-4*(a + x^(-1)))/(5*a^2*c^4*(1 - 1/(a^2*x^2))^(5/2)) - (5*a + 7/x)/(5*a^2*c^4*(1 - 1/(a^2*x^2))^(3/2)) - (15*
a + 19/x)/(5*a^2*c^4*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[1 - 1/(a^2*x^2)]*x)/c^4 + (3*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]
])/(a*c^4)

Rule 6177

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]

Rule 852

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

Rule 1805

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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 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 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 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]

Rubi steps

\begin{align*} \int \frac{e^{-\coth ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^4} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (c-\frac{c x}{a}\right )^3 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\left (c+\frac{c x}{a}\right )^3}{x^2 \left (1-\frac{x^2}{a^2}\right )^{7/2}} \, dx,x,\frac{1}{x}\right )}{c^7}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}+\frac{\operatorname{Subst}\left (\int \frac{-5 c^3-\frac{15 c^3 x}{a}-\frac{16 c^3 x^2}{a^2}}{x^2 \left (1-\frac{x^2}{a^2}\right )^{5/2}} \, dx,x,\frac{1}{x}\right )}{5 c^7}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{15 c^3+\frac{45 c^3 x}{a}+\frac{42 c^3 x^2}{a^2}}{x^2 \left (1-\frac{x^2}{a^2}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{15 c^7}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{15 a+\frac{19}{x}}{5 a^2 c^4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\operatorname{Subst}\left (\int \frac{-15 c^3-\frac{45 c^3 x}{a}}{x^2 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{15 c^7}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{15 a+\frac{19}{x}}{5 a^2 c^4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^4}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a c^4}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{15 a+\frac{19}{x}}{5 a^2 c^4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^4}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{2 a c^4}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{15 a+\frac{19}{x}}{5 a^2 c^4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^4}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{a^2-a^2 x^2} \, dx,x,\sqrt{1-\frac{1}{a^2 x^2}}\right )}{c^4}\\ &=-\frac{4 \left (a+\frac{1}{x}\right )}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{5 a+\frac{7}{x}}{5 a^2 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{15 a+\frac{19}{x}}{5 a^2 c^4 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c^4}+\frac{3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^4}\\ \end{align*}

Mathematica [A]  time = 0.0778986, size = 104, normalized size = 0.75 \[ \frac{5 a^4 x^4-34 a^3 x^3+18 a^2 x^2+15 a x \sqrt{1-\frac{1}{a^2 x^2}} (a x-1)^2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )+33 a x-24}{5 a^2 c^4 x \sqrt{1-\frac{1}{a^2 x^2}} (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - c/(a*x))^4),x]

[Out]

(-24 + 33*a*x + 18*a^2*x^2 - 34*a^3*x^3 + 5*a^4*x^4 + 15*a*Sqrt[1 - 1/(a^2*x^2)]*x*(-1 + a*x)^2*ArcTanh[Sqrt[1
- 1/(a^2*x^2)]])/(5*a^2*c^4*Sqrt[1 - 1/(a^2*x^2)]*x*(-1 + a*x)^2)

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Maple [B]  time = 0.148, size = 436, normalized size = 3.2 \begin{align*}{\frac{ax+1}{40\,a{c}^{4} \left ( ax-1 \right ) ^{4}}\sqrt{{\frac{ax-1}{ax+1}}} \left ( 120\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{4}{a}^{5}+125\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{4}{a}^{4}-480\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{3}{a}^{4}-85\,\sqrt{{a}^{2}} \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}{x}^{2}{a}^{2}-500\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{3}{a}^{3}+720\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{2}{a}^{3}+148\,\sqrt{{a}^{2}} \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}xa+750\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}-480\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}-67\, \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{3/2}\sqrt{{a}^{2}}-500\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa+120\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) +125\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x)^4,x)

[Out]

1/40*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)/a*(120*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^4*a^
5+125*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^4*a^4-480*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/
2))*x^3*a^4-85*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(3/2)*x^2*a^2-500*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^3*a^3+720
*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^2*a^3+148*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(3/2)*x
*a+750*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^2*a^2-480*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1
/2))*x*a^2-67*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)-500*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x*a+120*a*ln((a^2*x+
(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))+125*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/((a*x-1)*(a*x+1))^(
1/2)/c^4/(a*x-1)^4/(a^2)^(1/2)

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Maxima [A]  time = 1.09118, size = 207, normalized size = 1.5 \begin{align*} \frac{1}{20} \, a{\left (\frac{\frac{9 \,{\left (a x - 1\right )}}{a x + 1} + \frac{75 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - \frac{125 \,{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} + 1}{a^{2} c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{7}{2}} - a^{2} c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{2}}} + \frac{60 \, \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c^{4}} - \frac{60 \, \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2} c^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x)^4,x, algorithm="maxima")

[Out]

1/20*a*((9*(a*x - 1)/(a*x + 1) + 75*(a*x - 1)^2/(a*x + 1)^2 - 125*(a*x - 1)^3/(a*x + 1)^3 + 1)/(a^2*c^4*((a*x
- 1)/(a*x + 1))^(7/2) - a^2*c^4*((a*x - 1)/(a*x + 1))^(5/2)) + 60*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c^4)
- 60*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/(a^2*c^4))

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Fricas [A]  time = 1.93405, size = 385, normalized size = 2.79 \begin{align*} \frac{15 \,{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - 15 \,{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) +{\left (5 \, a^{4} x^{4} - 34 \, a^{3} x^{3} + 18 \, a^{2} x^{2} + 33 \, a x - 24\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{5 \,{\left (a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x)^4,x, algorithm="fricas")

[Out]

1/5*(15*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*(a^3*x^3 - 3*a^2*x^2 + 3*a*x
- 1)*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (5*a^4*x^4 - 34*a^3*x^3 + 18*a^2*x^2 + 33*a*x - 24)*sqrt((a*x - 1)/
(a*x + 1)))/(a^4*c^4*x^3 - 3*a^3*c^4*x^2 + 3*a^2*c^4*x - a*c^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{4} \int \frac{x^{4} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a^{4} x^{4} - 4 a^{3} x^{3} + 6 a^{2} x^{2} - 4 a x + 1}\, dx}{c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*Integral(x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**4*x**4 - 4*a**3*x**3 + 6*a**2*x**2 - 4*a*x + 1), x)/c
**4

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Giac [A]  time = 1.19832, size = 80, normalized size = 0.58 \begin{align*} -\frac{3 \, \log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} - 1} \right |}\right ) \mathrm{sgn}\left (a x + 1\right )}{c^{4}{\left | a \right |}} + \frac{\sqrt{a^{2} x^{2} - 1} \mathrm{sgn}\left (a x + 1\right )}{a c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a/x)^4,x, algorithm="giac")

[Out]

-3*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/(c^4*abs(a)) + sqrt(a^2*x^2 - 1)*sgn(a*x + 1)/(a*c^4)