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

Optimal. Leaf size=101 \[ \frac {a x+1}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 a x+15}{15 c^3 \sqrt {1-a^2 x^2}}+\frac {4 a x+5}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6148, 823, 12, 266, 63, 208} \[ \frac {a x+1}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 a x+15}{15 c^3 \sqrt {1-a^2 x^2}}+\frac {4 a x+5}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 6148

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^3} \, dx &=\frac {\int \frac {1+a x}{x \left (1-a^2 x^2\right )^{7/2}} \, dx}{c^3}\\ &=\frac {1+a x}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {\int \frac {5 a^2+4 a^3 x}{x \left (1-a^2 x^2\right )^{5/2}} \, dx}{5 a^2 c^3}\\ &=\frac {1+a x}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {5+4 a x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {\int \frac {15 a^4+8 a^5 x}{x \left (1-a^2 x^2\right )^{3/2}} \, dx}{15 a^4 c^3}\\ &=\frac {1+a x}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {5+4 a x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {15+8 a x}{15 c^3 \sqrt {1-a^2 x^2}}+\frac {\int \frac {15 a^6}{x \sqrt {1-a^2 x^2}} \, dx}{15 a^6 c^3}\\ &=\frac {1+a x}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {5+4 a x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {15+8 a x}{15 c^3 \sqrt {1-a^2 x^2}}+\frac {\int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{c^3}\\ &=\frac {1+a x}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {5+4 a x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {15+8 a x}{15 c^3 \sqrt {1-a^2 x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{2 c^3}\\ &=\frac {1+a x}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {5+4 a x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {15+8 a x}{15 c^3 \sqrt {1-a^2 x^2}}-\frac {\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^2 c^3}\\ &=\frac {1+a x}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {5+4 a x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {15+8 a x}{15 c^3 \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 108, normalized size = 1.07 \[ \frac {8 a^4 x^4+7 a^3 x^3-27 a^2 x^2-15 (a x-1)^2 (a x+1) \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-8 a x+23}{15 c^3 (a x-1)^2 (a x+1) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23 - 8*a*x - 27*a^2*x^2 + 7*a^3*x^3 + 8*a^4*x^4 - 15*(-1 + a*x)^2*(1 + a*x)*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1
- a^2*x^2]])/(15*c^3*(-1 + a*x)^2*(1 + a*x)*Sqrt[1 - a^2*x^2])

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fricas [B]  time = 0.61, size = 198, normalized size = 1.96 \[ \frac {23 \, a^{5} x^{5} - 23 \, a^{4} x^{4} - 46 \, a^{3} x^{3} + 46 \, a^{2} x^{2} + 23 \, a x + 15 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (8 \, a^{4} x^{4} + 7 \, a^{3} x^{3} - 27 \, a^{2} x^{2} - 8 \, a x + 23\right )} \sqrt {-a^{2} x^{2} + 1} - 23}{15 \, {\left (a^{5} c^{3} x^{5} - a^{4} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{3} + 2 \, a^{2} c^{3} x^{2} + a c^{3} x - c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(23*a^5*x^5 - 23*a^4*x^4 - 46*a^3*x^3 + 46*a^2*x^2 + 23*a*x + 15*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x
^2 + a*x - 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (8*a^4*x^4 + 7*a^3*x^3 - 27*a^2*x^2 - 8*a*x + 23)*sqrt(-a^2*x^
2 + 1) - 23)/(a^5*c^3*x^5 - a^4*c^3*x^4 - 2*a^3*c^3*x^3 + 2*a^2*c^3*x^2 + a*c^3*x - c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 384, normalized size = 3.80 \[ -\frac {\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{2 a}+\frac {11 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{16 a \left (x -\frac {1}{a}\right )}+\frac {\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {2 a \left (\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{4 a^{2}}+\frac {-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 a \left (x +\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 \left (x +\frac {1}{a}\right )}}{8 a}-\frac {5 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{16 a \left (x +\frac {1}{a}\right )}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c^3*(arctanh(1/(-a^2*x^2+1)^(1/2))-1/2/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))+11/16/a/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+1/4/a^2*(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)))+1/8/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))-5/16/a/(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 \[ -\int \frac {a x + 1}{{\left (a^{2} c x^{2} - c\right )}^{3} \sqrt {-a^{2} x^{2} + 1} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 306, normalized size = 3.03 \[ \frac {a^2\,\sqrt {1-a^2\,x^2}}{5\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {a^2\,\sqrt {1-a^2\,x^2}}{24\,\left (a^4\,c^3\,x^2+2\,a^3\,c^3\,x+a^2\,c^3\right )}-\frac {17\,a\,\sqrt {1-a^2\,x^2}}{48\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}+\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {71\,a\,\sqrt {1-a^2\,x^2}}{80\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {a\,\sqrt {1-a^2\,x^2}}{20\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )}+\frac {\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((1 - a^2*x^2)^(1/2)*1i)*1i)/c^3 + (a^2*(1 - a^2*x^2)^(1/2))/(5*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) +
(a^2*(1 - a^2*x^2)^(1/2))/(24*(a^2*c^3 + 2*a^3*c^3*x + a^4*c^3*x^2)) - (17*a*(1 - a^2*x^2)^(1/2))/(48*(-a^2)^(
1/2)*(c^3*x*(-a^2)^(1/2) + (c^3*(-a^2)^(1/2))/a)) + (71*a*(1 - a^2*x^2)^(1/2))/(80*(-a^2)^(1/2)*(c^3*x*(-a^2)^
(1/2) - (c^3*(-a^2)^(1/2))/a)) + (a*(1 - a^2*x^2)^(1/2))/(20*(-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 \[ \frac {\int \frac {a}{- a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{- a^{6} x^{7} \sqrt {- a^{2} x^{2} + 1} + 3 a^{4} x^{5} \sqrt {- a^{2} x^{2} + 1} - 3 a^{2} x^{3} \sqrt {- a^{2} x^{2} + 1} + x \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(-a**6*x**6*sqrt(-a**2*x**2 + 1) + 3*a**4*x**4*sqrt(-a**2*x**2 + 1) - 3*a**2*x**2*sqrt(-a**2*x**2
+ 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(1/(-a**6*x**7*sqrt(-a**2*x**2 + 1) + 3*a**4*x**5*sqrt(-a**2*x**2 +
1) - 3*a**2*x**3*sqrt(-a**2*x**2 + 1) + x*sqrt(-a**2*x**2 + 1)), x))/c**3

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