3.1259 \(\int \frac{e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=94 \[ -\frac{2 \sqrt{1-a^2 x^2}}{15 a c^2 (a x+1)}-\frac{2 \sqrt{1-a^2 x^2}}{15 a c^2 (a x+1)^2}-\frac{\sqrt{1-a^2 x^2}}{5 a c^2 (a x+1)^3} \]

[Out]

-Sqrt[1 - a^2*x^2]/(5*a*c^2*(1 + a*x)^3) - (2*Sqrt[1 - a^2*x^2])/(15*a*c^2*(1 + a*x)^2) - (2*Sqrt[1 - a^2*x^2]
)/(15*a*c^2*(1 + a*x))

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Rubi [A]  time = 0.0797878, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6139, 655, 659, 651} \[ -\frac{2 \sqrt{1-a^2 x^2}}{15 a c^2 (a x+1)}-\frac{2 \sqrt{1-a^2 x^2}}{15 a c^2 (a x+1)^2}-\frac{\sqrt{1-a^2 x^2}}{5 a c^2 (a x+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - a^2*x^2]/(5*a*c^2*(1 + a*x)^3) - (2*Sqrt[1 - a^2*x^2])/(15*a*c^2*(1 + a*x)^2) - (2*Sqrt[1 - a^2*x^2]
)/(15*a*c^2*(1 + a*x))

Rule 6139

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

Rule 655

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

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 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]

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx &=\frac{\int \frac{(1-a x)^3}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^2}\\ &=\frac{\int \frac{1}{(1+a x)^3 \sqrt{1-a^2 x^2}} \, dx}{c^2}\\ &=-\frac{\sqrt{1-a^2 x^2}}{5 a c^2 (1+a x)^3}+\frac{2 \int \frac{1}{(1+a x)^2 \sqrt{1-a^2 x^2}} \, dx}{5 c^2}\\ &=-\frac{\sqrt{1-a^2 x^2}}{5 a c^2 (1+a x)^3}-\frac{2 \sqrt{1-a^2 x^2}}{15 a c^2 (1+a x)^2}+\frac{2 \int \frac{1}{(1+a x) \sqrt{1-a^2 x^2}} \, dx}{15 c^2}\\ &=-\frac{\sqrt{1-a^2 x^2}}{5 a c^2 (1+a x)^3}-\frac{2 \sqrt{1-a^2 x^2}}{15 a c^2 (1+a x)^2}-\frac{2 \sqrt{1-a^2 x^2}}{15 a c^2 (1+a x)}\\ \end{align*}

Mathematica [A]  time = 0.0185846, size = 43, normalized size = 0.46 \[ -\frac{\sqrt{1-a x} \left (2 a^2 x^2+6 a x+7\right )}{15 a c^2 (a x+1)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sqrt[1 - a*x]*(7 + 6*a*x + 2*a^2*x^2))/(15*a*c^2*(1 + a*x)^(5/2))

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Maple [A]  time = 0.029, size = 42, normalized size = 0.5 \begin{align*} -{\frac{2\,{a}^{2}{x}^{2}+6\,ax+7}{15\, \left ( ax+1 \right ) ^{3}{c}^{2}a}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(-a^2*x^2+1)^(1/2)*(2*a^2*x^2+6*a*x+7)/(a*x+1)^3/c^2/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a^{2} c x^{2} - c\right )}^{2}{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.11633, size = 192, normalized size = 2.04 \begin{align*} -\frac{7 \, a^{3} x^{3} + 21 \, a^{2} x^{2} + 21 \, a x +{\left (2 \, a^{2} x^{2} + 6 \, a x + 7\right )} \sqrt{-a^{2} x^{2} + 1} + 7}{15 \,{\left (a^{4} c^{2} x^{3} + 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x + a c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(7*a^3*x^3 + 21*a^2*x^2 + 21*a*x + (2*a^2*x^2 + 6*a*x + 7)*sqrt(-a^2*x^2 + 1) + 7)/(a^4*c^2*x^3 + 3*a^3*
c^2*x^2 + 3*a^2*c^2*x + a*c^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{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}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(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) + s
qrt(-a**2*x**2 + 1)), x)/c**2

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Giac [A]  time = 1.1756, size = 196, normalized size = 2.09 \begin{align*} \frac{2 \,{\left (\frac{20 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} + \frac{40 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{30 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} + \frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} + 7\right )}}{15 \, c^{2}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}^{5}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(20*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 40*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 30*(sqrt(-
a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) + 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) + 7)/(c^2*((sqrt(-a^2*x^
2 + 1)*abs(a) + a)/(a^2*x) + 1)^5*abs(a))