3.454 \(\int \frac{e^{\tanh ^{-1}(a x)}}{(c-\frac{c}{a x})^3} \, dx\)

Optimal. Leaf size=136 \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^2}+\frac{14 \left (1-a^2 x^2\right )^{3/2}}{15 a c^3 (1-a x)^3}-\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a c^3 (1-a x)^4}-\frac{8 \sqrt{1-a^2 x^2}}{a c^3 (1-a x)}+\frac{4 \sin ^{-1}(a x)}{a c^3} \]

[Out]

(-8*Sqrt[1 - a^2*x^2])/(a*c^3*(1 - a*x)) - (1 - a^2*x^2)^(3/2)/(5*a*c^3*(1 - a*x)^4) + (14*(1 - a^2*x^2)^(3/2)
)/(15*a*c^3*(1 - a*x)^3) + (1 - a^2*x^2)^(3/2)/(a*c^3*(1 - a*x)^2) + (4*ArcSin[a*x])/(a*c^3)

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Rubi [A]  time = 0.281984, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6131, 6128, 1639, 1637, 659, 651, 663, 216} \[ \frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^2}+\frac{14 \left (1-a^2 x^2\right )^{3/2}}{15 a c^3 (1-a x)^3}-\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a c^3 (1-a x)^4}-\frac{8 \sqrt{1-a^2 x^2}}{a c^3 (1-a x)}+\frac{4 \sin ^{-1}(a x)}{a c^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(c - c/(a*x))^3,x]

[Out]

(-8*Sqrt[1 - a^2*x^2])/(a*c^3*(1 - a*x)) - (1 - a^2*x^2)^(3/2)/(5*a*c^3*(1 - a*x)^4) + (14*(1 - a^2*x^2)^(3/2)
)/(15*a*c^3*(1 - a*x)^3) + (1 - a^2*x^2)^(3/2)/(a*c^3*(1 - a*x)^2) + (4*ArcSin[a*x])/(a*c^3)

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 1639

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

Rule 1637

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[m, 0]

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 663

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

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 )^3} \, dx &=-\frac{a^3 \int \frac{e^{\tanh ^{-1}(a x)} x^3}{(1-a x)^3} \, dx}{c^3}\\ &=-\frac{a^3 \int \frac{x^3 \sqrt{1-a^2 x^2}}{(1-a x)^4} \, dx}{c^3}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^2}-\frac{\int \frac{\sqrt{1-a^2 x^2} \left (2 a^2-5 a^3 x+4 a^4 x^2\right )}{(1-a x)^4} \, dx}{a^2 c^3}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^2}-\frac{\int \left (\frac{a^2 \sqrt{1-a^2 x^2}}{(-1+a x)^4}+\frac{3 a^2 \sqrt{1-a^2 x^2}}{(-1+a x)^3}+\frac{4 a^2 \sqrt{1-a^2 x^2}}{(-1+a x)^2}\right ) \, dx}{a^2 c^3}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^2}-\frac{\int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^4} \, dx}{c^3}-\frac{3 \int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^3} \, dx}{c^3}-\frac{4 \int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^2} \, dx}{c^3}\\ &=-\frac{8 \sqrt{1-a^2 x^2}}{a c^3 (1-a x)}-\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a c^3 (1-a x)^4}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^2}+\frac{\int \frac{\sqrt{1-a^2 x^2}}{(-1+a x)^3} \, dx}{5 c^3}+\frac{4 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^3}\\ &=-\frac{8 \sqrt{1-a^2 x^2}}{a c^3 (1-a x)}-\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a c^3 (1-a x)^4}+\frac{14 \left (1-a^2 x^2\right )^{3/2}}{15 a c^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{a c^3 (1-a x)^2}+\frac{4 \sin ^{-1}(a x)}{a c^3}\\ \end{align*}

Mathematica [A]  time = 0.146847, size = 61, normalized size = 0.45 \[ \frac{\frac{\sqrt{1-a^2 x^2} \left (-15 a^3 x^3+149 a^2 x^2-222 a x+94\right )}{(a x-1)^3}+60 \sin ^{-1}(a x)}{15 a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[1 - a^2*x^2]*(94 - 222*a*x + 149*a^2*x^2 - 15*a^3*x^3))/(-1 + a*x)^3 + 60*ArcSin[a*x])/(15*a*c^3)

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Maple [A]  time = 0.046, size = 184, normalized size = 1.4 \begin{align*} -{\frac{1}{a{c}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}+4\,{\frac{1}{{c}^{3}\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{2}{5\,{a}^{4}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-3}}+{\frac{31}{15\,{a}^{3}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{104}{15\,{a}^{2}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-a^2*x^2+1)^(1/2)/a/c^3+4/c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2/5/a^4/c^3/(x-1/a)^3*(-a
^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+31/15/a^3/c^3/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+104/15/a^2/c^3/(x-1
/a)*(-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{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (c - \frac{c}{a x}\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.2657, size = 328, normalized size = 2.41 \begin{align*} -\frac{94 \, a^{3} x^{3} - 282 \, a^{2} x^{2} + 282 \, a x + 120 \,{\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 (15 \, a^{3} x^{3} - 149 \, a^{2} x^{2} + 222 \, a x - 94\right )} \sqrt{-a^{2} x^{2} + 1} - 94}{15 \,{\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(94*a^3*x^3 - 282*a^2*x^2 + 282*a*x + 120*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) -
1)/(a*x)) + (15*a^3*x^3 - 149*a^2*x^2 + 222*a*x - 94)*sqrt(-a^2*x^2 + 1) - 94)/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 +
3*a^2*c^3*x - a*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17342, size = 244, normalized size = 1.79 \begin{align*} \frac{4 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c^{3}{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a c^{3}} + \frac{2 \,{\left (\frac{335 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{505 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{285 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac{60 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 79\right )}}{15 \, c^{3}{\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((a*x+1)/(-a^2*x^2+1)^(1/2)/(c-c/a/x)^3,x, algorithm="giac")

[Out]

4*arcsin(a*x)*sgn(a)/(c^3*abs(a)) - sqrt(-a^2*x^2 + 1)/(a*c^3) + 2/15*(335*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^
2*x) - 505*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 285*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 60*
(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 79)/(c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))