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

Optimal. Leaf size=94 \[ -\frac {(a x+1)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 (a x+1)}{3 a c^3 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {2 \sin ^{-1}(a x)}{a c^3} \]

[Out]

-1/3*(a*x+1)^2/a/c^3/(-a^2*x^2+1)^(3/2)-2*arcsin(a*x)/a/c^3+8/3*(a*x+1)/a/c^3/(-a^2*x^2+1)^(1/2)+(-a^2*x^2+1)^
(1/2)/a/c^3

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Rubi [A]  time = 0.26, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {(a x+1)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 (a x+1)}{3 a c^3 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {2 \sin ^{-1}(a x)}{a c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - c/(a*x))^3),x]

[Out]

-(1 + a*x)^2/(3*a*c^3*(1 - a^2*x^2)^(3/2)) + (8*(1 + a*x))/(3*a*c^3*Sqrt[1 - a^2*x^2]) + Sqrt[1 - a^2*x^2]/(a*
c^3) - (2*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 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 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 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}{(1-a x)^2 \sqrt {1-a^2 x^2}} \, dx}{c^3}\\ &=-\frac {a^3 \int \frac {x^3 (1+a x)^2}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^3}\\ &=-\frac {(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {a^3 \int \frac {(1+a x) \left (\frac {2}{a^3}+\frac {3 x}{a^2}+\frac {3 x^2}{a}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^3}\\ &=-\frac {(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 (1+a x)}{3 a c^3 \sqrt {1-a^2 x^2}}-\frac {a^3 \int \frac {\frac {6}{a^3}+\frac {3 x}{a^2}}{\sqrt {1-a^2 x^2}} \, dx}{3 c^3}\\ &=-\frac {(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 (1+a x)}{3 a c^3 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{c^3}\\ &=-\frac {(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 (1+a x)}{3 a c^3 \sqrt {1-a^2 x^2}}+\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {2 \sin ^{-1}(a x)}{a c^3}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 53, normalized size = 0.56 \[ \frac {\frac {\sqrt {1-a^2 x^2} \left (3 a^2 x^2-14 a x+10\right )}{(a x-1)^2}-6 \sin ^{-1}(a x)}{3 a c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a*x))^3),x]

[Out]

((Sqrt[1 - a^2*x^2]*(10 - 14*a*x + 3*a^2*x^2))/(-1 + a*x)^2 - 6*ArcSin[a*x])/(3*a*c^3)

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fricas [A]  time = 0.64, size = 107, normalized size = 1.14 \[ \frac {10 \, a^{2} x^{2} - 20 \, a x + 12 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{2} x^{2} - 14 \, a x + 10\right )} \sqrt {-a^{2} x^{2} + 1} + 10}{3 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

1/3*(10*a^2*x^2 - 20*a*x + 12*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^2*x^2 - 14*a
*x + 10)*sqrt(-a^2*x^2 + 1) + 10)/(a^3*c^3*x^2 - 2*a^2*c^3*x + a*c^3)

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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(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^3,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 [B]  time = 0.05, size = 242, normalized size = 2.57 \[ \frac {5 \left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{4 a^{3} c^{3} \left (x -\frac {1}{a}\right )^{2}}+\frac {17 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{8 a \,c^{3}}-\frac {17 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{8 c^{3} \sqrt {a^{2}}}+\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{6 a^{4} c^{3} \left (x -\frac {1}{a}\right )^{3}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{8 a \,c^{3}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{8 c^{3} \sqrt {a^{2}}} \]

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)^3,x)

[Out]

5/4/a^3/c^3/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+17/8/a/c^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-17/8/c^
3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))+1/6/a^4/c^3/(x-1/a)^3*(-a^2*(x-1/a)^2-2
*a*(x-1/a))^(3/2)+1/8/a/c^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/8/c^3/(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a x}\right )}^{3}}\,{d x} \]

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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 139, normalized size = 1.48 \[ \frac {\sqrt {1-a^2\,x^2}}{a\,c^3}-\frac {2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^3\,\sqrt {-a^2}}-\frac {a\,\sqrt {1-a^2\,x^2}}{3\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {8\,\sqrt {1-a^2\,x^2}}{3\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - a^2*x^2)^(1/2)/(a*c^3) - (2*asinh(x*(-a^2)^(1/2)))/(c^3*(-a^2)^(1/2)) - (a*(1 - a^2*x^2)^(1/2))/(3*(a^2*c
^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) + (8*(1 - a^2*x^2)^(1/2))/(3*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(
1/2))/a))

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

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)**3,x)

[Out]

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

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