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]

8/5*a*(a*x+1)/c^3/(-a^2*x^2+1)^(5/2)+4/15*a*(8*a*x+5)/c^3/(-a^2*x^2+1)^(3/2)-4*a*arctanh((-a^2*x^2+1)^(1/2))/c
^3+1/15*a*(79*a*x+60)/c^3/(-a^2*x^2+1)^(1/2)-(-a^2*x^2+1)^(1/2)/c^3/x

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Rubi [A]  time = 0.35, antiderivative size = 127, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 0.06, size = 101, normalized size = 0.80 \[ \frac {94 a^4 x^4-128 a^3 x^3-73 a^2 x^2-60 a x (a x-1)^2 \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+134 a x-15}{15 c^3 x (a x-1)^2 \sqrt {1-a^2 x^2}} \]

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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fricas [A]  time = 0.61, size = 155, normalized size = 1.22 \[ \frac {104 \, a^{4} x^{4} - 312 \, a^{3} x^{3} + 312 \, a^{2} x^{2} - 104 \, a x + 60 \, {\left (a^{4} x^{4} - 3 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - a x\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (94 \, a^{3} x^{3} - 222 \, a^{2} x^{2} + 149 \, a x - 15\right )} \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{3} c^{3} x^{4} - 3 \, a^{2} c^{3} x^{3} + 3 \, a c^{3} x^{2} - c^{3} x\right )}} \]

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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giac [B]  time = 0.25, size = 269, normalized size = 2.12 \[ -\frac {4 \, a^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c^{3} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{2 \, c^{3} x {\left | a \right |}} - \frac {{\left (15 \, a^{2} - \frac {491 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{x} + \frac {1690 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{2} x^{2}} - \frac {2570 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{4} x^{3}} + \frac {1815 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{6} x^{4}} - \frac {555 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{5}}{a^{8} x^{5}}\right )} a^{2} x}{30 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \]

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))

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maple [B]  time = 0.05, size = 248, normalized size = 1.95 \[ -\frac {4 a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\sqrt {-a^{2} x^{2}+1}}{x}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )^{2}}+\frac {5 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}+\frac {\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a}}{c^{3}} \]

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*(4*a*arctanh(1/(-a^2*x^2+1)^(1/2))+(-a^2*x^2+1)^(1/2)/x-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)+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))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )}^{3} x^{2}}\,{d x} \]

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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mupad [B]  time = 0.83, size = 234, normalized size = 1.84 \[ \frac {19\,a^3\,\sqrt {1-a^2\,x^2}}{15\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}-\frac {\sqrt {1-a^2\,x^2}}{c^3\,x}+\frac {79\,a^2\,\sqrt {1-a^2\,x^2}}{15\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {2\,a^2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )}+\frac {a\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(19*a^3*(1 - a^2*x^2)^(1/2))/(15*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) - (1 - a^2*x^2)^(1/2)/(c^3*x) + (a*ata
n((1 - a^2*x^2)^(1/2)*1i)*4i)/c^3 + (79*a^2*(1 - a^2*x^2)^(1/2))/(15*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(
-a^2)^(1/2))/a)) + (2*a^2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(3*c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(1/2))/a +
a^2*c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x^2*(-a^2)^(1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x}{a^{3} x^{5} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1} + 3 a x^{3} \sqrt {- a^{2} x^{2} + 1} - x^{2} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a^{3} x^{5} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1} + 3 a x^{3} \sqrt {- a^{2} x^{2} + 1} - x^{2} \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \]

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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