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

Optimal. Leaf size=127 \[ \frac{8 a (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{c^3 x}+\frac{a (79 a x+60)}{15 c^3 \sqrt{1-a^2 x^2}}+\frac{4 a (8 a x+5)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{4 a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^3} \]

[Out]

(8*a*(1 + a*x))/(5*c^3*(1 - a^2*x^2)^(5/2)) + (4*a*(5 + 8*a*x))/(15*c^3*(1 - a^2*x^2)^(3/2)) + (a*(60 + 79*a*x
))/(15*c^3*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(c^3*x) - (4*a*ArcTanh[Sqrt[1 - a^2*x^2]])/c^3

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Rubi [A]  time = 0.351343, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {6128, 852, 1805, 807, 266, 63, 208} \[ \frac{8 a (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{c^3 x}+\frac{a (79 a x+60)}{15 c^3 \sqrt{1-a^2 x^2}}+\frac{4 a (8 a x+5)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac{4 a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*(1 + a*x))/(5*c^3*(1 - a^2*x^2)^(5/2)) + (4*a*(5 + 8*a*x))/(15*c^3*(1 - a^2*x^2)^(3/2)) + (a*(60 + 79*a*x
))/(15*c^3*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(c^3*x) - (4*a*ArcTanh[Sqrt[1 - a^2*x^2]])/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 852

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 807

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)}}{x^2 (c-a c x)^3} \, dx &=c \int \frac{\sqrt{1-a^2 x^2}}{x^2 (c-a c x)^4} \, dx\\ &=\frac{\int \frac{(c+a c x)^4}{x^2 \left (1-a^2 x^2\right )^{7/2}} \, dx}{c^7}\\ &=\frac{8 a (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{\int \frac{-5 c^4-20 a c^4 x-27 a^2 c^4 x^2}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^7}\\ &=\frac{8 a (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a (5+8 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{\int \frac{15 c^4+60 a c^4 x+64 a^2 c^4 x^2}{x^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^7}\\ &=\frac{8 a (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a (5+8 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (60+79 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\int \frac{-15 c^4-60 a c^4 x}{x^2 \sqrt{1-a^2 x^2}} \, dx}{15 c^7}\\ &=\frac{8 a (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a (5+8 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (60+79 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^3 x}+\frac{(4 a) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{c^3}\\ &=\frac{8 a (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a (5+8 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (60+79 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^3 x}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{c^3}\\ &=\frac{8 a (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a (5+8 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (60+79 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^3 x}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{a c^3}\\ &=\frac{8 a (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a (5+8 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (60+79 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^3 x}-\frac{4 a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^3}\\ \end{align*}

Mathematica [A]  time = 0.0546973, size = 101, normalized size = 0.8 \[ \frac{94 a^4 x^4-128 a^3 x^3-73 a^2 x^2-60 a x (a x-1)^2 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+134 a x-15}{15 c^3 x (a x-1)^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15 + 134*a*x - 73*a^2*x^2 - 128*a^3*x^3 + 94*a^4*x^4 - 60*a*x*(-1 + a*x)^2*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1
- a^2*x^2]])/(15*c^3*x*(-1 + a*x)^2*Sqrt[1 - a^2*x^2])

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Maple [B]  time = 0.048, size = 248, normalized size = 2. \begin{align*} -{\frac{1}{{c}^{3}} \left ( 2\,{\frac{1}{a} \left ( 1/5\,{\frac{1}{a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-3}}-2/5\,a \left ( 1/3\,{\frac{1}{a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}-1/3\,{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \right ) \right ) }+{\frac{1}{x}\sqrt{-{a}^{2}{x}^{2}+1}}+4\,a{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) -{\frac{1}{a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+5\,{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \right ) } \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*(2/a*(1/5/a/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-2/5*a*(1/3/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-
1/a))^(1/2)-1/3/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)))+(-a^2*x^2+1)^(1/2)/x+4*a*arctanh(1/(-a^2*x^2+1)^(
1/2))-1/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+5/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))

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

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

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Fricas [A]  time = 1.48822, size = 338, normalized size = 2.66 \begin{align*} \frac{104 \, a^{4} x^{4} - 312 \, a^{3} x^{3} + 312 \, a^{2} x^{2} - 104 \, a x + 60 \,{\left (a^{4} x^{4} - 3 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - a x\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (94 \, a^{3} x^{3} - 222 \, a^{2} x^{2} + 149 \, a x - 15\right )} \sqrt{-a^{2} x^{2} + 1}}{15 \,{\left (a^{3} c^{3} x^{4} - 3 \, a^{2} c^{3} x^{3} + 3 \, a c^{3} x^{2} - c^{3} x\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]

1/15*(104*a^4*x^4 - 312*a^3*x^3 + 312*a^2*x^2 - 104*a*x + 60*(a^4*x^4 - 3*a^3*x^3 + 3*a^2*x^2 - a*x)*log((sqrt
(-a^2*x^2 + 1) - 1)/x) - (94*a^3*x^3 - 222*a^2*x^2 + 149*a*x - 15)*sqrt(-a^2*x^2 + 1))/(a^3*c^3*x^4 - 3*a^2*c^
3*x^3 + 3*a*c^3*x^2 - c^3*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a x}{a^{3} x^{5} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{4} \sqrt{- a^{2} x^{2} + 1} + 3 a x^{3} \sqrt{- a^{2} x^{2} + 1} - x^{2} \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{1}{a^{3} x^{5} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{4} \sqrt{- a^{2} x^{2} + 1} + 3 a x^{3} \sqrt{- a^{2} x^{2} + 1} - x^{2} \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(a*x/(a**3*x**5*sqrt(-a**2*x**2 + 1) - 3*a**2*x**4*sqrt(-a**2*x**2 + 1) + 3*a*x**3*sqrt(-a**2*x**2 +
1) - x**2*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a**3*x**5*sqrt(-a**2*x**2 + 1) - 3*a**2*x**4*sqrt(-a**2*x**
2 + 1) + 3*a*x**3*sqrt(-a**2*x**2 + 1) - x**2*sqrt(-a**2*x**2 + 1)), x))/c**3

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Giac [B]  time = 1.23041, size = 363, normalized size = 2.86 \begin{align*} -\frac{4 \, a^{2} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{c^{3}{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{2 \, c^{3} x{\left | a \right |}} - \frac{{\left (15 \, a^{2} - \frac{491 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{x} + \frac{1690 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{2} x^{2}} - \frac{2570 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{4} x^{3}} + \frac{1815 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{6} x^{4}} - \frac{555 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5}}{a^{8} x^{5}}\right )} a^{2} x}{30 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} 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]

-4*a^2*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c^3*abs(a)) - 1/2*(sqrt(-a^2*x^2 + 1)*ab
s(a) + a)/(c^3*x*abs(a)) - 1/30*(15*a^2 - 491*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/x + 1690*(sqrt(-a^2*x^2 + 1)*abs
(a) + a)^2/(a^2*x^2) - 2570*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^4*x^3) + 1815*(sqrt(-a^2*x^2 + 1)*abs(a) + a)
^4/(a^6*x^4) - 555*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^5/(a^8*x^5))*a^2*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^3*((s
qrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))