3.453 \(\int \frac {e^{\tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^2} \, dx\)

Optimal. Leaf size=104 \[ \frac {\left (1-a^2 x^2\right )^{3/2}}{a c^2 (1-a x)^2}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c^2 (1-a x)^3}-\frac {6 \sqrt {1-a^2 x^2}}{a c^2 (1-a x)}+\frac {3 \sin ^{-1}(a x)}{a c^2} \]

[Out]

1/3*(-a^2*x^2+1)^(3/2)/a/c^2/(-a*x+1)^3+(-a^2*x^2+1)^(3/2)/a/c^2/(-a*x+1)^2+3*arcsin(a*x)/a/c^2-6*(-a^2*x^2+1)
^(1/2)/a/c^2/(-a*x+1)

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Rubi [A]  time = 0.19, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6131, 6128, 1639, 793, 663, 216} \[ \frac {\left (1-a^2 x^2\right )^{3/2}}{a c^2 (1-a x)^2}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c^2 (1-a x)^3}-\frac {6 \sqrt {1-a^2 x^2}}{a c^2 (1-a x)}+\frac {3 \sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(c - c/(a*x))^2,x]

[Out]

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

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 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 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 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 )^2} \, dx &=\frac {a^2 \int \frac {e^{\tanh ^{-1}(a x)} x^2}{(1-a x)^2} \, dx}{c^2}\\ &=\frac {a^2 \int \frac {x^2 \sqrt {1-a^2 x^2}}{(1-a x)^3} \, dx}{c^2}\\ &=\frac {\left (1-a^2 x^2\right )^{3/2}}{a c^2 (1-a x)^2}-\frac {\int \frac {\left (2 a^2-3 a^3 x\right ) \sqrt {1-a^2 x^2}}{(1-a x)^3} \, dx}{a^2 c^2}\\ &=\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c^2 (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{a c^2 (1-a x)^2}-\frac {3 \int \frac {\sqrt {1-a^2 x^2}}{(1-a x)^2} \, dx}{c^2}\\ &=-\frac {6 \sqrt {1-a^2 x^2}}{a c^2 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c^2 (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{a c^2 (1-a x)^2}+\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{c^2}\\ &=-\frac {6 \sqrt {1-a^2 x^2}}{a c^2 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a c^2 (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{a c^2 (1-a x)^2}+\frac {3 \sin ^{-1}(a x)}{a c^2}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 53, normalized size = 0.51 \[ \frac {\frac {\sqrt {1-a^2 x^2} \left (-3 a^2 x^2+19 a x-14\right )}{(a x-1)^2}+9 \sin ^{-1}(a x)}{3 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-14 + 19*a*x - 3*a^2*x^2)*Sqrt[1 - a^2*x^2])/(-1 + a*x)^2 + 9*ArcSin[a*x])/(3*a*c^2)

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fricas [A]  time = 0.49, size = 107, normalized size = 1.03 \[ -\frac {14 \, a^{2} x^{2} - 28 \, a x + 18 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{2} x^{2} - 19 \, a x + 14\right )} \sqrt {-a^{2} x^{2} + 1} + 14}{3 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

-1/3*(14*a^2*x^2 - 28*a*x + 18*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^2*x^2 - 19*
a*x + 14)*sqrt(-a^2*x^2 + 1) + 14)/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 140, normalized size = 1.35 \[ -\frac {\sqrt {-a^{2} x^{2}+1}}{a \,c^{2}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c^{2} \sqrt {a^{2}}}+\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{3} c^{2} \left (x -\frac {1}{a}\right )^{2}}+\frac {13 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{2} c^{2} \left (x -\frac {1}{a}\right )} \]

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)^2,x)

[Out]

-(-a^2*x^2+1)^(1/2)/a/c^2+3/c^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2/3/a^3/c^2/(x-1/a)^2*(-a
^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+13/3/a^2/c^2/(x-1/a)*(-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 \[ \int \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (c - \frac {c}{a x}\right )}^{2}}\,{d x} \]

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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.82, size = 140, normalized size = 1.35 \[ \frac {2\,a\,\sqrt {1-a^2\,x^2}}{3\,\left (a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2\right )}+\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^2}-\frac {13\,\sqrt {1-a^2\,x^2}}{3\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*(1 - a^2*x^2)^(1/2))/(3*(a^2*c^2 - 2*a^3*c^2*x + a^4*c^2*x^2)) + (3*asinh(x*(-a^2)^(1/2)))/(c^2*(-a^2)^(1
/2)) - (1 - a^2*x^2)^(1/2)/(a*c^2) - (13*(1 - a^2*x^2)^(1/2))/(3*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) - (c^2*(-a^2
)^(1/2))/a))

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

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)**2,x)

[Out]

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

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