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

Optimal. Leaf size=162 \[ \frac{a^2 (164 a x+135)}{15 c^3 \sqrt{1-a^2 x^2}}+\frac{4 a^2 (13 a x+10)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{8 a^2 (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{4 a \sqrt{1-a^2 x^2}}{c^3 x}-\frac{\sqrt{1-a^2 x^2}}{2 c^3 x^2}-\frac{19 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^3} \]

[Out]

(8*a^2*(1 + a*x))/(5*c^3*(1 - a^2*x^2)^(5/2)) + (4*a^2*(10 + 13*a*x))/(15*c^3*(1 - a^2*x^2)^(3/2)) + (a^2*(135
+ 164*a*x))/(15*c^3*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(2*c^3*x^2) - (4*a*Sqrt[1 - a^2*x^2])/(c^3*x) - (1
9*a^2*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.430019, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {6128, 852, 1805, 1807, 807, 266, 63, 208} \[ \frac{a^2 (164 a x+135)}{15 c^3 \sqrt{1-a^2 x^2}}+\frac{4 a^2 (13 a x+10)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{8 a^2 (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac{4 a \sqrt{1-a^2 x^2}}{c^3 x}-\frac{\sqrt{1-a^2 x^2}}{2 c^3 x^2}-\frac{19 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*(1 + a*x))/(5*c^3*(1 - a^2*x^2)^(5/2)) + (4*a^2*(10 + 13*a*x))/(15*c^3*(1 - a^2*x^2)^(3/2)) + (a^2*(135
+ 164*a*x))/(15*c^3*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(2*c^3*x^2) - (4*a*Sqrt[1 - a^2*x^2])/(c^3*x) - (1
9*a^2*ArcTanh[Sqrt[1 - a^2*x^2]])/(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 1807

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

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^3 (c-a c x)^3} \, dx &=c \int \frac{\sqrt{1-a^2 x^2}}{x^3 (c-a c x)^4} \, dx\\ &=\frac{\int \frac{(c+a c x)^4}{x^3 \left (1-a^2 x^2\right )^{7/2}} \, dx}{c^7}\\ &=\frac{8 a^2 (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-35 a^2 c^4 x^2-32 a^3 c^4 x^3}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^7}\\ &=\frac{8 a^2 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a^2 (10+13 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+120 a^2 c^4 x^2+104 a^3 c^4 x^3}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^7}\\ &=\frac{8 a^2 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a^2 (10+13 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (135+164 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\int \frac{-15 c^4-60 a c^4 x-135 a^2 c^4 x^2}{x^3 \sqrt{1-a^2 x^2}} \, dx}{15 c^7}\\ &=\frac{8 a^2 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a^2 (10+13 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (135+164 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^3 x^2}+\frac{\int \frac{120 a c^4+285 a^2 c^4 x}{x^2 \sqrt{1-a^2 x^2}} \, dx}{30 c^7}\\ &=\frac{8 a^2 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a^2 (10+13 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (135+164 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^3 x^2}-\frac{4 a \sqrt{1-a^2 x^2}}{c^3 x}+\frac{\left (19 a^2\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{2 c^3}\\ &=\frac{8 a^2 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a^2 (10+13 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (135+164 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^3 x^2}-\frac{4 a \sqrt{1-a^2 x^2}}{c^3 x}+\frac{\left (19 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{4 c^3}\\ &=\frac{8 a^2 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a^2 (10+13 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (135+164 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^3 x^2}-\frac{4 a \sqrt{1-a^2 x^2}}{c^3 x}-\frac{19 \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{2 c^3}\\ &=\frac{8 a^2 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac{4 a^2 (10+13 a x)}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (135+164 a x)}{15 c^3 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^3 x^2}-\frac{4 a \sqrt{1-a^2 x^2}}{c^3 x}-\frac{19 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0593456, size = 113, normalized size = 0.7 \[ \frac{448 a^5 x^5-611 a^4 x^4-346 a^3 x^3+638 a^2 x^2-285 a^2 x^2 (a x-1)^2 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-90 a x-15}{30 c^3 x^2 (a x-1)^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15 - 90*a*x + 638*a^2*x^2 - 346*a^3*x^3 - 611*a^4*x^4 + 448*a^5*x^5 - 285*a^2*x^2*(-1 + a*x)^2*Sqrt[1 - a^2*
x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(30*c^3*x^2*(-1 + a*x)^2*Sqrt[1 - a^2*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.055, size = 223, normalized size = 1.4 \begin{align*} -{\frac{1}{{c}^{3}} \left ({\frac{2}{5\,a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-3}}-{\frac{29\,a}{5} \left ({\frac{1}{3\,a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}-{\frac{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 ) }+4\,{\frac{a\sqrt{-{a}^{2}{x}^{2}+1}}{x}}+{\frac{19\,{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+9\,{a\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}}+{\frac{1}{2\,{x}^{2}}\sqrt{-{a}^{2}{x}^{2}+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^3/(-a*c*x+c)^3,x)

[Out]

-1/c^3*(2/5/a/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-29/5*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))+4*a*(-a^2*x^2+1)^(1/2)/x+19/2*a^2*arctanh(1/(-a^2*x^2+
1)^(1/2))+9*a/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+1/2*(-a^2*x^2+1)^(1/2)/x^2)

________________________________________________________________________________________

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^{3}}\,{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^3/(-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^3), x)

________________________________________________________________________________________

Fricas [A]  time = 1.63086, size = 375, normalized size = 2.31 \begin{align*} \frac{398 \, a^{5} x^{5} - 1194 \, a^{4} x^{4} + 1194 \, a^{3} x^{3} - 398 \, a^{2} x^{2} + 285 \,{\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 3 \, a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (448 \, a^{4} x^{4} - 1059 \, a^{3} x^{3} + 713 \, a^{2} x^{2} - 75 \, a x - 15\right )} \sqrt{-a^{2} x^{2} + 1}}{30 \,{\left (a^{3} c^{3} x^{5} - 3 \, a^{2} c^{3} x^{4} + 3 \, a c^{3} x^{3} - c^{3} x^{2}\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^3/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

1/30*(398*a^5*x^5 - 1194*a^4*x^4 + 1194*a^3*x^3 - 398*a^2*x^2 + 285*(a^5*x^5 - 3*a^4*x^4 + 3*a^3*x^3 - a^2*x^2
)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (448*a^4*x^4 - 1059*a^3*x^3 + 713*a^2*x^2 - 75*a*x - 15)*sqrt(-a^2*x^2 + 1
))/(a^3*c^3*x^5 - 3*a^2*c^3*x^4 + 3*a*c^3*x^3 - c^3*x^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a x}{a^{3} x^{6} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{5} \sqrt{- a^{2} x^{2} + 1} + 3 a x^{4} \sqrt{- a^{2} x^{2} + 1} - x^{3} \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{1}{a^{3} x^{6} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{5} \sqrt{- a^{2} x^{2} + 1} + 3 a x^{4} \sqrt{- a^{2} x^{2} + 1} - x^{3} \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**3/(-a*c*x+c)**3,x)

[Out]

-(Integral(a*x/(a**3*x**6*sqrt(-a**2*x**2 + 1) - 3*a**2*x**5*sqrt(-a**2*x**2 + 1) + 3*a*x**4*sqrt(-a**2*x**2 +
1) - x**3*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a**3*x**6*sqrt(-a**2*x**2 + 1) - 3*a**2*x**5*sqrt(-a**2*x**
2 + 1) + 3*a*x**4*sqrt(-a**2*x**2 + 1) - x**3*sqrt(-a**2*x**2 + 1)), x))/c**3

________________________________________________________________________________________

Giac [B]  time = 1.26039, size = 456, normalized size = 2.81 \begin{align*} -\frac{19 \, a^{3} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{2 \, c^{3}{\left | a \right |}} - \frac{{\left (15 \, a^{3} + \frac{165 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a}{x} - \frac{4234 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a x^{2}} + \frac{14330 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{3} x^{3}} - \frac{20965 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{4}}{a^{5} x^{4}} + \frac{14385 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{5}}{a^{7} x^{5}} - \frac{4080 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{6}}{a^{9} x^{6}}\right )} a^{4} x^{2}}{120 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{3}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}^{5}{\left | a \right |}} - \frac{\frac{16 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a c^{3}{\left | a \right |}}{x} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c^{3}{\left | a \right |}}{a x^{2}}}{8 \, a^{2} c^{6}} \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^3/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

-19/2*a^3*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c^3*abs(a)) - 1/120*(15*a^3 + 165*(sq
rt(-a^2*x^2 + 1)*abs(a) + a)*a/x - 4234*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a*x^2) + 14330*(sqrt(-a^2*x^2 + 1)*
abs(a) + a)^3/(a^3*x^3) - 20965*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^5*x^4) + 14385*(sqrt(-a^2*x^2 + 1)*abs(a)
+ a)^5/(a^7*x^5) - 4080*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^6/(a^9*x^6))*a^4*x^2/((sqrt(-a^2*x^2 + 1)*abs(a) + a)
^2*c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a)) - 1/8*(16*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a*c^3
*abs(a)/x + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^3*abs(a)/(a*x^2))/(a^2*c^6)