3.357 \(\int \frac{e^{\tanh ^{-1}(a x)} x^2}{(c-a c x)^4} \, dx\)

Optimal. Leaf size=97 \[ \frac{23 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 c^4 (1-a x)^3}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 c^4 (1-a x)^4}+\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 c^4 (1-a x)^5} \]

[Out]

(1 - a^2*x^2)^(3/2)/(7*a^3*c^4*(1 - a*x)^5) - (12*(1 - a^2*x^2)^(3/2))/(35*a^3*c^4*(1 - a*x)^4) + (23*(1 - a^2
*x^2)^(3/2))/(105*a^3*c^4*(1 - a*x)^3)

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Rubi [A]  time = 0.208041, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {6128, 1639, 793, 659, 651} \[ \frac{23 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 c^4 (1-a x)^3}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 c^4 (1-a x)^4}+\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 c^4 (1-a x)^5} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^2)/(c - a*c*x)^4,x]

[Out]

(1 - a^2*x^2)^(3/2)/(7*a^3*c^4*(1 - a*x)^5) - (12*(1 - a^2*x^2)^(3/2))/(35*a^3*c^4*(1 - a*x)^4) + (23*(1 - a^2
*x^2)^(3/2))/(105*a^3*c^4*(1 - a*x)^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 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 793

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

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)} x^2}{(c-a c x)^4} \, dx &=c \int \frac{x^2 \sqrt{1-a^2 x^2}}{(c-a c x)^5} \, dx\\ &=-\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 c^4 (1-a x)^4}+\frac{\int \frac{\left (4 a^2 c^2-3 a^3 c^2 x\right ) \sqrt{1-a^2 x^2}}{(c-a c x)^5} \, dx}{a^4 c}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 c^4 (1-a x)^5}-\frac{\left (1-a^2 x^2\right )^{3/2}}{a^3 c^4 (1-a x)^4}+\frac{23 \int \frac{\sqrt{1-a^2 x^2}}{(c-a c x)^4} \, dx}{7 a^2}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 c^4 (1-a x)^5}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 c^4 (1-a x)^4}+\frac{23 \int \frac{\sqrt{1-a^2 x^2}}{(c-a c x)^3} \, dx}{35 a^2 c}\\ &=\frac{\left (1-a^2 x^2\right )^{3/2}}{7 a^3 c^4 (1-a x)^5}-\frac{12 \left (1-a^2 x^2\right )^{3/2}}{35 a^3 c^4 (1-a x)^4}+\frac{23 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 c^4 (1-a x)^3}\\ \end{align*}

Mathematica [A]  time = 0.0263635, size = 43, normalized size = 0.44 \[ -\frac{(a x+1)^{3/2} \left (-23 a^2 x^2+10 a x-2\right )}{105 a^3 c^4 (1-a x)^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcTanh[a*x]*x^2)/(c - a*c*x)^4,x]

[Out]

-((1 + a*x)^(3/2)*(-2 + 10*a*x - 23*a^2*x^2))/(105*a^3*c^4*(1 - a*x)^(7/2))

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Maple [A]  time = 0.034, size = 49, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 23\,{a}^{2}{x}^{2}-10\,ax+2 \right ) \left ( ax+1 \right ) ^{2}}{105\,{c}^{4} \left ( ax-1 \right ) ^{3}{a}^{3}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+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)^4,x)

[Out]

-1/105*(23*a^2*x^2-10*a*x+2)*(a*x+1)^2/(a*x-1)^3/c^4/(-a^2*x^2+1)^(1/2)/a^3

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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)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79434, size = 250, normalized size = 2.58 \begin{align*} \frac{2 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 8 \, a x +{\left (23 \, a^{3} x^{3} + 13 \, a^{2} x^{2} - 8 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} + 2}{105 \,{\left (a^{7} c^{4} x^{4} - 4 \, a^{6} c^{4} x^{3} + 6 \, a^{5} c^{4} x^{2} - 4 \, a^{4} c^{4} x + a^{3} c^{4}\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)^4,x, algorithm="fricas")

[Out]

1/105*(2*a^4*x^4 - 8*a^3*x^3 + 12*a^2*x^2 - 8*a*x + (23*a^3*x^3 + 13*a^2*x^2 - 8*a*x + 2)*sqrt(-a^2*x^2 + 1) +
2)/(a^7*c^4*x^4 - 4*a^6*c^4*x^3 + 6*a^5*c^4*x^2 - 4*a^4*c^4*x + a^3*c^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{2}}{a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 4 a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} + 6 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} - 4 a x \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{a x^{3}}{a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 4 a^{3} x^{3} \sqrt{- a^{2} x^{2} + 1} + 6 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} - 4 a x \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx}{c^{4}} \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)**4,x)

[Out]

(Integral(x**2/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 4*a**3*x**3*sqrt(-a**2*x**2 + 1) + 6*a**2*x**2*sqrt(-a**2*x**
2 + 1) - 4*a*x*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**3/(a**4*x**4*sqrt(-a**2*x**2 +
1) - 4*a**3*x**3*sqrt(-a**2*x**2 + 1) + 6*a**2*x**2*sqrt(-a**2*x**2 + 1) - 4*a*x*sqrt(-a**2*x**2 + 1) + sqrt(
-a**2*x**2 + 1)), x))/c**4

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Giac [A]  time = 1.29527, size = 200, normalized size = 2.06 \begin{align*} -\frac{4 \,{\left (\frac{7 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{21 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} - \frac{35 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac{70 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 1\right )}}{105 \, a^{2} c^{4}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}^{7}{\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)^4,x, algorithm="giac")

[Out]

-4/105*(7*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 21*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) - 35*(sqrt(
-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 70*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 1)/(a^2*c^4*((sqrt(-a
^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^7*abs(a))