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

Optimal. Leaf size=128 \[ -\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {3 a \sqrt {1-a^2 x^2}}{2 c x^2}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}+\frac {a x+1}{c x^3 \sqrt {1-a^2 x^2}}-\frac {3 a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6148, 823, 835, 807, 266, 63, 208} \[ -\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {3 a \sqrt {1-a^2 x^2}}{2 c x^2}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}+\frac {a x+1}{c x^3 \sqrt {1-a^2 x^2}}-\frac {3 a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[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^4 \left (c-a^2 c x^2\right )} \, dx &=\frac {\int \frac {1+a x}{x^4 \left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=\frac {1+a x}{c x^3 \sqrt {1-a^2 x^2}}+\frac {\int \frac {4 a^2+3 a^3 x}{x^4 \sqrt {1-a^2 x^2}} \, dx}{a^2 c}\\ &=\frac {1+a x}{c x^3 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}-\frac {\int \frac {-9 a^3-8 a^4 x}{x^3 \sqrt {1-a^2 x^2}} \, dx}{3 a^2 c}\\ &=\frac {1+a x}{c x^3 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}-\frac {3 a \sqrt {1-a^2 x^2}}{2 c x^2}+\frac {\int \frac {16 a^4+9 a^5 x}{x^2 \sqrt {1-a^2 x^2}} \, dx}{6 a^2 c}\\ &=\frac {1+a x}{c x^3 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}-\frac {3 a \sqrt {1-a^2 x^2}}{2 c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}+\frac {\left (3 a^3\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{2 c}\\ &=\frac {1+a x}{c x^3 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}-\frac {3 a \sqrt {1-a^2 x^2}}{2 c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{4 c}\\ &=\frac {1+a x}{c x^3 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}-\frac {3 a \sqrt {1-a^2 x^2}}{2 c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {(3 a) \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}\\ &=\frac {1+a x}{c x^3 \sqrt {1-a^2 x^2}}-\frac {4 \sqrt {1-a^2 x^2}}{3 c x^3}-\frac {3 a \sqrt {1-a^2 x^2}}{2 c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {3 a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 c}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 91, normalized size = 0.71 \[ -\frac {-16 a^4 x^4-9 a^3 x^3+8 a^2 x^2+9 a^3 x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+3 a x+2}{6 c x^3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(2 + 3*a*x + 8*a^2*x^2 - 9*a^3*x^3 - 16*a^4*x^4 + 9*a^3*x^3*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])
/(c*x^3*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.64, size = 107, normalized size = 0.84 \[ \frac {6 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 9 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (16 \, a^{3} x^{3} - 7 \, a^{2} x^{2} - a x - 2\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a c x^{4} - c x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

1/6*(6*a^4*x^4 - 6*a^3*x^3 + 9*(a^4*x^4 - a^3*x^3)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (16*a^3*x^3 - 7*a^2*x^2 -
a*x - 2)*sqrt(-a^2*x^2 + 1))/(a*c*x^4 - c*x^3)

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giac [B]  time = 0.21, size = 283, normalized size = 2.21 \[ -\frac {{\left (a^{4} + \frac {2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{2}}{x} + \frac {18 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{x^{2}} - \frac {69 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{2} x^{3}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} - \frac {3 \, a^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, c {\left | a \right |}} - \frac {\frac {21 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4} c^{2}}{x} + \frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{2} c^{2}}{x^{2}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{2}}{x^{3}}}{24 \, a^{2} c^{3} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

-1/24*(a^4 + 2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^2/x + 18*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/x^2 - 69*(sqrt(-a^
2*x^2 + 1)*abs(a) + a)^3/(a^2*x^3))*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c*((sqrt(-a^2*x^2 + 1)*abs(a) +
a)/(a^2*x) - 1)*abs(a)) - 3/2*a^4*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a)) -
1/24*(21*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4*c^2/x + 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2*c^2/x^2 + (sqrt(-
a^2*x^2 + 1)*abs(a) + a)^3*c^2/x^3)/(a^2*c^3*abs(a))

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maple [A]  time = 0.05, size = 140, normalized size = 1.09 \[ -\frac {a^{3} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {5 a^{2} \sqrt {-a^{2} x^{2}+1}}{3 x}+\frac {\sqrt {-a^{2} x^{2}+1}}{3 x^{3}}+\frac {a^{2} \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}-a \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c),x)

[Out]

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

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maxima [A]  time = 0.33, size = 148, normalized size = 1.16 \[ -\frac {\frac {3 \, a^{4} \log \left (\sqrt {-a^{2} x^{2} + 1} + 1\right )}{c} - \frac {3 \, a^{4} \log \left (\sqrt {-a^{2} x^{2} + 1} - 1\right )}{c} + \frac {2 \, {\left (3 \, {\left (a^{2} x^{2} - 1\right )} a^{4} + 2 \, a^{4}\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c - \sqrt {-a^{2} x^{2} + 1} c}}{4 \, a} + \frac {8 \, a^{4} x^{4} - 4 \, a^{2} x^{2} - 1}{3 \, \sqrt {a x + 1} \sqrt {-a x + 1} c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

-1/4*(3*a^4*log(sqrt(-a^2*x^2 + 1) + 1)/c - 3*a^4*log(sqrt(-a^2*x^2 + 1) - 1)/c + 2*(3*(a^2*x^2 - 1)*a^4 + 2*a
^4)/((-a^2*x^2 + 1)^(3/2)*c - sqrt(-a^2*x^2 + 1)*c))/a + 1/3*(8*a^4*x^4 - 4*a^2*x^2 - 1)/(sqrt(a*x + 1)*sqrt(-
a*x + 1)*c*x^3)

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mupad [B]  time = 0.90, size = 140, normalized size = 1.09 \[ -\frac {\sqrt {1-a^2\,x^2}}{3\,c\,x^3}-\frac {a\,\sqrt {1-a^2\,x^2}}{2\,c\,x^2}-\frac {5\,a^2\,\sqrt {1-a^2\,x^2}}{3\,c\,x}-\frac {a^4\,\sqrt {1-a^2\,x^2}}{\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {a^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/(x^4*(c - a^2*c*x^2)*(1 - a^2*x^2)^(1/2)),x)

[Out]

(a^3*atan((1 - a^2*x^2)^(1/2)*1i)*3i)/(2*c) - (1 - a^2*x^2)^(1/2)/(3*c*x^3) - (a*(1 - a^2*x^2)^(1/2))/(2*c*x^2
) - (5*a^2*(1 - a^2*x^2)^(1/2))/(3*c*x) - (a^4*(1 - a^2*x^2)^(1/2))/(((c*(-a^2)^(1/2))/a - c*x*(-a^2)^(1/2))*(
-a^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{- a^{2} x^{5} \sqrt {- a^{2} x^{2} + 1} + x^{3} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{- a^{2} x^{6} \sqrt {- a^{2} x^{2} + 1} + x^{4} \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x**4/(-a**2*c*x**2+c),x)

[Out]

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

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