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

Optimal. Leaf size=96 \[ -\frac {x (4 a x+3)}{3 c^2 \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {1-a^2 x^2}}{3 a c^2}+\frac {a^2 x^3 (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {\sin ^{-1}(a x)}{a c^2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6157, 6148, 819, 641, 216} \[ \frac {a^2 x^3 (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {x (4 a x+3)}{3 c^2 \sqrt {1-a^2 x^2}}-\frac {8 \sqrt {1-a^2 x^2}}{3 a c^2}+\frac {\sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x^3*(1 + a*x))/(3*c^2*(1 - a^2*x^2)^(3/2)) - (x*(3 + 4*a*x))/(3*c^2*Sqrt[1 - a^2*x^2]) - (8*Sqrt[1 - a^2*
x^2])/(3*a*c^2) + 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 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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 6148

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

Rule 6157

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 78, normalized size = 0.81 \[ \frac {3 a^3 x^3-7 a^2 x^2+3 (a x-1) \sqrt {1-a^2 x^2} \sin ^{-1}(a x)-5 a x+8}{3 a c^2 (a x-1) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8 - 5*a*x - 7*a^2*x^2 + 3*a^3*x^3 + 3*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(3*a*c^2*(-1 + a*x)*Sqrt[1 -
a^2*x^2])

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fricas [A]  time = 0.42, size = 142, normalized size = 1.48 \[ -\frac {8 \, a^{3} x^{3} - 8 \, a^{2} x^{2} - 8 \, a x + 6 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{3} x^{3} - 7 \, a^{2} x^{2} - 5 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} + 8}{3 \, {\left (a^{4} c^{2} x^{3} - a^{3} c^{2} x^{2} - a^{2} c^{2} x + a c^{2}\right )}} \]

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^2/x^2)^2,x, algorithm="fricas")

[Out]

-1/3*(8*a^3*x^3 - 8*a^2*x^2 - 8*a*x + 6*(a^3*x^3 - a^2*x^2 - a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) +
(3*a^3*x^3 - 7*a^2*x^2 - 5*a*x + 8)*sqrt(-a^2*x^2 + 1) + 8)/(a^4*c^2*x^3 - a^3*c^2*x^2 - a^2*c^2*x + a*c^2)

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

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^2/x^2)^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 177, normalized size = 1.84 \[ -\frac {\sqrt {-a^{2} x^{2}+1}}{a \,c^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c^{2} \sqrt {a^{2}}}+\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{6 a^{3} c^{2} \left (x -\frac {1}{a}\right )^{2}}+\frac {19 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{12 a^{2} c^{2} \left (x -\frac {1}{a}\right )}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2} c^{2} \left (x +\frac {1}{a}\right )} \]

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^2/x^2)^2,x)

[Out]

-(-a^2*x^2+1)^(1/2)/a/c^2+1/c^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+1/6/a^3/c^2/(x-1/a)^2*(-a
^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+19/12/a^2/c^2/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-1/4/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 \[ \int \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{2}}\,{d x} \]

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^2/x^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 188, normalized size = 1.96 \[ \frac {a\,\sqrt {1-a^2\,x^2}}{6\,\left (a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2\right )}+\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^2}+\frac {\sqrt {1-a^2\,x^2}}{4\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}+\frac {c^2\,\sqrt {-a^2}}{a}\right )}-\frac {19\,\sqrt {1-a^2\,x^2}}{12\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*(1 - a^2*x^2)^(1/2))/(6*(a^2*c^2 - 2*a^3*c^2*x + a^4*c^2*x^2)) + asinh(x*(-a^2)^(1/2))/(c^2*(-a^2)^(1/2)) -
(1 - a^2*x^2)^(1/2)/(a*c^2) + (1 - a^2*x^2)^(1/2)/(4*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) + (c^2*(-a^2)^(1/2))/a)
) - (19*(1 - a^2*x^2)^(1/2))/(12*(-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 \[ \frac {a^{4} \int \frac {x^{4}}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} - a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \]

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**2/x**2)**2,x)

[Out]

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

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