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

Optimal. Leaf size=107 \[ -\frac{3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^4}+\frac{2 \sqrt{1-a^2 x^2}}{a^3 c^3 (1-a x)}-\frac{\sin ^{-1}(a x)}{a^3 c^3} \]

[Out]

(2*Sqrt[1 - a^2*x^2])/(a^3*c^3*(1 - a*x)) + (1 - a^2*x^2)^(3/2)/(5*a^3*c^3*(1 - a*x)^4) - (3*(1 - a^2*x^2)^(3/
2))/(5*a^3*c^3*(1 - a*x)^3) - ArcSin[a*x]/(a^3*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.220088, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {6128, 1637, 659, 651, 663, 216} \[ -\frac{3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^3}+\frac{\left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^4}+\frac{2 \sqrt{1-a^2 x^2}}{a^3 c^3 (1-a x)}-\frac{\sin ^{-1}(a x)}{a^3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 - a^2*x^2])/(a^3*c^3*(1 - a*x)) + (1 - a^2*x^2)^(3/2)/(5*a^3*c^3*(1 - a*x)^4) - (3*(1 - a^2*x^2)^(3/
2))/(5*a^3*c^3*(1 - a*x)^3) - ArcSin[a*x]/(a^3*c^3)

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 1637

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[m, 0]

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]

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 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]

Rubi steps

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

Mathematica [C]  time = 0.0583067, size = 77, normalized size = 0.72 \[ \frac{20 \sqrt{2} (a x-1) \text{Hypergeometric2F1}\left (-\frac{3}{2},-\frac{3}{2},-\frac{1}{2},\frac{1}{2} (1-a x)\right )+\sqrt{a x+1} \left (-a^2 x^2+3 a x+4\right )}{15 a^3 c^3 (1-a x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[1 + a*x]*(4 + 3*a*x - a^2*x^2) + 20*Sqrt[2]*(-1 + a*x)*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 - a*x)/2])
/(15*a^3*c^3*(1 - a*x)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.058, size = 167, normalized size = 1.6 \begin{align*} -{\frac{1}{{c}^{3}{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{2}{5\,{a}^{6}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-3}}-{\frac{7}{5\,{c}^{3}{a}^{5}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}-{\frac{13}{5\,{a}^{4}{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c^3/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-2/5/c^3/a^6/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1
/a))^(1/2)-7/5/c^3/a^5/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-13/5/c^3/a^4/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(
x-1/a))^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.56742, size = 300, normalized size = 2.8 \begin{align*} \frac{8 \, a^{3} x^{3} - 24 \, a^{2} x^{2} + 24 \, a x + 10 \,{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) -{\left (13 \, a^{2} x^{2} - 19 \, a x + 8\right )} \sqrt{-a^{2} x^{2} + 1} - 8}{5 \,{\left (a^{6} c^{3} x^{3} - 3 \, a^{5} c^{3} x^{2} + 3 \, a^{4} c^{3} x - a^{3} c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(8*a^3*x^3 - 24*a^2*x^2 + 24*a*x + 10*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a
*x)) - (13*a^2*x^2 - 19*a*x + 8)*sqrt(-a^2*x^2 + 1) - 8)/(a^6*c^3*x^3 - 3*a^5*c^3*x^2 + 3*a^4*c^3*x - a^3*c^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{x^{2}}{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 + \int \frac{a x^{3}}{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^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.18566, size = 225, normalized size = 2.1 \begin{align*} -\frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{a^{2} c^{3}{\left | a \right |}} - \frac{2 \,{\left (\frac{35 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{55 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac{25 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac{5 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 8\right )}}{5 \, a^{2} c^{3}{\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((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

-arcsin(a*x)*sgn(a)/(a^2*c^3*abs(a)) - 2/5*(35*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 55*(sqrt(-a^2*x^2 + 1
)*abs(a) + a)^2/(a^4*x^2) + 25*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)
^4/(a^8*x^4) - 8)/(a^2*c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))