3.682 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx\)

Optimal. Leaf size=97 \[ \frac {(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac {2 (1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{a c}-\frac {3 \sin ^{-1}(a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6157, 6149, 1635, 21, 669, 641, 216} \[ \frac {(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac {2 (1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{a c}-\frac {3 \sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 669

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

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 6149

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] || G
tQ[c, 0]) && ILtQ[(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^{-3 \tanh ^{-1}(a x)}}{c-\frac {c}{a^2 x^2}} \, dx &=-\frac {a^2 \int \frac {e^{-3 \tanh ^{-1}(a x)} x^2}{1-a^2 x^2} \, dx}{c}\\ &=-\frac {a^2 \int \frac {x^2 (1-a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c}\\ &=\frac {(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}+\frac {a^2 \int \frac {\left (\frac {3}{a^2}-\frac {3 x}{a}\right ) (1-a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c}\\ &=\frac {(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}+\frac {\int \frac {(1-a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=\frac {(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac {2 (1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {3 \int \frac {1-a x}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac {2 (1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{a c}-\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {(1-a x)^3}{3 a c \left (1-a^2 x^2\right )^{3/2}}-\frac {2 (1-a x)^2}{a c \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{a c}-\frac {3 \sin ^{-1}(a x)}{a c}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 78, normalized size = 0.80 \[ \frac {3 a^3 x^3+16 a^2 x^2-9 (a x+1) \sqrt {1-a^2 x^2} \sin ^{-1}(a x)-5 a x-14}{3 a c (a x+1) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a^2*x^2))),x]

[Out]

(-14 - 5*a*x + 16*a^2*x^2 + 3*a^3*x^3 - 9*(1 + a*x)*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(3*a*c*(1 + a*x)*Sqrt[1 - a
^2*x^2])

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fricas [A]  time = 0.63, size = 101, normalized size = 1.04 \[ -\frac {14 \, a^{2} x^{2} + 28 \, a x - 18 \, {\left (a^{2} x^{2} + 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{2} x^{2} + 19 \, a x + 14\right )} \sqrt {-a^{2} x^{2} + 1} + 14}{3 \, {\left (a^{3} c x^{2} + 2 \, a^{2} c x + a c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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)^3*(-a^2*x^2+1)^(3/2)/(c-c/a^2/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 [B]  time = 0.05, size = 330, normalized size = 3.40 \[ \frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{48 a c}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}\, x}{32 c}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{32 c \sqrt {a^{2}}}-\frac {47 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{24 c \,a^{3} \left (x +\frac {1}{a}\right )^{2}}-\frac {95 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{48 a c}-\frac {95 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{32 c}-\frac {95 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{32 c \sqrt {a^{2}}}+\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{6 a^{5} c \left (x +\frac {1}{a}\right )^{4}}-\frac {11 \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{12 a^{4} c \left (x +\frac {1}{a}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48/a/c*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)-1/32/c*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x-1/32/c/(a^2)^(1/2)*arc
tan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))-47/24/c/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)
-95/48/a/c*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-95/32/c*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-95/32/c/(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^5/c/(x+1/a)^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/
2)-11/12/a^4/c/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.85, size = 129, normalized size = 1.33 \[ \frac {2\,a\,\sqrt {1-a^2\,x^2}}{3\,\left (c\,a^4\,x^2+2\,c\,a^3\,x+c\,a^2\right )}+\frac {13\,\sqrt {1-a^2\,x^2}}{3\,\left (\frac {c\,\sqrt {-a^2}}{a}+c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{2} \left (\int \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} + 3 a^{4} x^{4} + 2 a^{3} x^{3} - 2 a^{2} x^{2} - 3 a x - 1}\, dx + \int \left (- \frac {a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} + 3 a^{4} x^{4} + 2 a^{3} x^{3} - 2 a^{2} x^{2} - 3 a x - 1}\right )\, dx\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(x**2*sqrt(-a**2*x**2 + 1)/(a**5*x**5 + 3*a**4*x**4 + 2*a**3*x**3 - 2*a**2*x**2 - 3*a*x - 1), x)
+ Integral(-a**2*x**4*sqrt(-a**2*x**2 + 1)/(a**5*x**5 + 3*a**4*x**4 + 2*a**3*x**3 - 2*a**2*x**2 - 3*a*x - 1),
x))/c

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