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

Optimal. Leaf size=125 \[ \frac{(a x+1)^3}{5 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (a x+1)^2}{5 a c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (a x+1)}{5 a c^4 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{3 \sin ^{-1}(a x)}{a c^4} \]

[Out]

(1 + a*x)^3/(5*a*c^4*(1 - a^2*x^2)^(5/2)) - (6*(1 + a*x)^2)/(5*a*c^4*(1 - a^2*x^2)^(3/2)) + (24*(1 + a*x))/(5*
a*c^4*Sqrt[1 - a^2*x^2]) + Sqrt[1 - a^2*x^2]/(a*c^4) - (3*ArcSin[a*x])/(a*c^4)

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Rubi [A]  time = 0.353745, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6131, 6128, 852, 1635, 641, 216} \[ \frac{(a x+1)^3}{5 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (a x+1)^2}{5 a c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (a x+1)}{5 a c^4 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{3 \sin ^{-1}(a x)}{a c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^3/(5*a*c^4*(1 - a^2*x^2)^(5/2)) - (6*(1 + a*x)^2)/(5*a*c^4*(1 - a^2*x^2)^(3/2)) + (24*(1 + a*x))/(5*
a*c^4*Sqrt[1 - a^2*x^2]) + Sqrt[1 - a^2*x^2]/(a*c^4) - (3*ArcSin[a*x])/(a*c^4)

Rule 6131

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

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && 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 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 641

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

Rule 216

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]

Rubi steps

\begin{align*} \int \frac{e^{-\tanh ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^4} \, dx &=\frac{a^4 \int \frac{e^{-\tanh ^{-1}(a x)} x^4}{(1-a x)^4} \, dx}{c^4}\\ &=\frac{a^4 \int \frac{x^4}{(1-a x)^3 \sqrt{1-a^2 x^2}} \, dx}{c^4}\\ &=\frac{a^4 \int \frac{x^4 (1+a x)^3}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^4}\\ &=\frac{(1+a x)^3}{5 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{a^4 \int \frac{(1+a x)^2 \left (\frac{3}{a^4}+\frac{5 x}{a^3}+\frac{5 x^2}{a^2}+\frac{5 x^3}{a}\right )}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^4}\\ &=\frac{(1+a x)^3}{5 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^4 \int \frac{(1+a x) \left (\frac{27}{a^4}+\frac{30 x}{a^3}+\frac{15 x^2}{a^2}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^4}\\ &=\frac{(1+a x)^3}{5 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (1+a x)}{5 a c^4 \sqrt{1-a^2 x^2}}-\frac{a^4 \int \frac{\frac{45}{a^4}+\frac{15 x}{a^3}}{\sqrt{1-a^2 x^2}} \, dx}{15 c^4}\\ &=\frac{(1+a x)^3}{5 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (1+a x)}{5 a c^4 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^4}\\ &=\frac{(1+a x)^3}{5 a c^4 \left (1-a^2 x^2\right )^{5/2}}-\frac{6 (1+a x)^2}{5 a c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac{24 (1+a x)}{5 a c^4 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^4}-\frac{3 \sin ^{-1}(a x)}{a c^4}\\ \end{align*}

Mathematica [A]  time = 0.158374, size = 61, normalized size = 0.49 \[ \frac{\frac{\sqrt{1-a^2 x^2} \left (5 a^3 x^3-39 a^2 x^2+57 a x-24\right )}{(a x-1)^3}-15 \sin ^{-1}(a x)}{5 a c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[1 - a^2*x^2]*(-24 + 57*a*x - 39*a^2*x^2 + 5*a^3*x^3))/(-1 + a*x)^3 - 15*ArcSin[a*x])/(5*a*c^4)

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Maple [B]  time = 0.053, size = 286, normalized size = 2.3 \begin{align*}{\frac{1}{10\,{a}^{5}{c}^{4}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-4}}+{\frac{11}{20\,{a}^{4}{c}^{4}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-3}}+{\frac{17}{8\,{a}^{3}{c}^{4}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{49}{16\,a{c}^{4}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}-{\frac{49}{16\,{c}^{4}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{16\,a{c}^{4}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{1}{16\,{c}^{4}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10/a^5/c^4/(x-1/a)^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+11/20/a^4/c^4/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))
^(3/2)+17/8/a^3/c^4/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+49/16/a/c^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2
)-49/16/c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))+1/16/a/c^4*(-a^2*(x+1/a)^2+2*
a*(x+1/a))^(1/2)+1/16/c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} x^{2} + 1}}{{\left (a x + 1\right )}{\left (c - \frac{c}{a x}\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.10625, size = 317, normalized size = 2.54 \begin{align*} \frac{24 \, a^{3} x^{3} - 72 \, a^{2} x^{2} + 72 \, a x + 30 \,{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (5 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 57 \, a x - 24\right )} \sqrt{-a^{2} x^{2} + 1} - 24}{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(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^4,x, algorithm="fricas")

[Out]

1/5*(24*a^3*x^3 - 72*a^2*x^2 + 72*a*x + 30*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(
a*x)) + (5*a^3*x^3 - 39*a^2*x^2 + 57*a*x - 24)*sqrt(-a^2*x^2 + 1) - 24)/(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{- a^{2} x^{2} + 1}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\, dx}{c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*Integral(x**4*sqrt(-a**2*x**2 + 1)/(a**5*x**5 - 3*a**4*x**4 + 2*a**3*x**3 + 2*a**2*x**2 - 3*a*x + 1), x)/
c**4

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Giac [A]  time = 1.18799, size = 243, normalized size = 1.94 \begin{align*} -\frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c^{4}{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a c^{4}} - \frac{2 \,{\left (\frac{80 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{120 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{70 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 19\right )}}{5 \, c^{4}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}^{5}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*arcsin(a*x)*sgn(a)/(c^4*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c^4) - 2/5*(80*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2
*x) - 120*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 70*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 15*(s
qrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 19)/(c^4*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))