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]

-1/5*(-a^2*x^2+1)^(3/2)/a/c^3/(-a*x+1)^4+14/15*(-a^2*x^2+1)^(3/2)/a/c^3/(-a*x+1)^3+(-a^2*x^2+1)^(3/2)/a/c^3/(-
a*x+1)^2+4*arcsin(a*x)/a/c^3-8*(-a^2*x^2+1)^(1/2)/a/c^3/(-a*x+1)

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Rubi [A]  time = 0.28, antiderivative size = 136, 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 = {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 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]

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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.16, 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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fricas [A]  time = 0.47, size = 143, normalized size = 1.05 \[ -\frac {94 \, a^{3} x^{3} - 282 \, a^{2} x^{2} + 282 \, a x + 120 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{3} x^{3} - 149 \, a^{2} x^{2} + 222 \, a x - 94\right )} \sqrt {-a^{2} x^{2} + 1} - 94}{15 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \]

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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giac [A]  time = 0.20, size = 181, normalized size = 1.33 \[ \frac {4 \, \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{c^{3} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1}}{a c^{3}} + \frac {2 \, {\left (\frac {335 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {505 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {285 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {60 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 79\right )}}{15 \, c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \]

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))

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maple [A]  time = 0.05, size = 184, normalized size = 1.35 \[ -\frac {\sqrt {-a^{2} x^{2}+1}}{a \,c^{3}}+\frac {4 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c^{3} \sqrt {a^{2}}}+\frac {31 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 a^{3} c^{3} \left (x -\frac {1}{a}\right )^{2}}+\frac {104 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 a^{2} c^{3} \left (x -\frac {1}{a}\right )}+\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a^{4} c^{3} \left (x -\frac {1}{a}\right )^{3}} \]

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))+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)+2/5/a^4/c^3/(x-1/a
)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (c - \frac {c}{a x}\right )}^{3}}\,{d x} \]

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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mupad [B]  time = 0.82, size = 225, normalized size = 1.65 \[ \frac {31\,a\,\sqrt {1-a^2\,x^2}}{15\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^3\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^3}-\frac {104\,\sqrt {1-a^2\,x^2}}{15\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}-\frac {2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(31*a*(1 - a^2*x^2)^(1/2))/(15*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) + (4*asinh(x*(-a^2)^(1/2)))/(c^3*(-a^2)^
(1/2)) - (1 - a^2*x^2)^(1/2)/(a*c^3) - (104*(1 - a^2*x^2)^(1/2))/(15*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(
-a^2)^(1/2))/a)) - (2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(3*c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(1/2))/a + a^2*
c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x^2*(-a^2)^(1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{3} \left (\int \frac {x^{3}}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{4}}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx\right )}{c^{3}} \]

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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