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

Optimal. Leaf size=100 \[ \frac {2 a^2 (a x+1)}{c \sqrt {1-a^2 x^2}}-\frac {2 a \sqrt {1-a^2 x^2}}{c x}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}-\frac {5 a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {2 a^2 (a x+1)}{c \sqrt {1-a^2 x^2}}-\frac {2 a \sqrt {1-a^2 x^2}}{c x}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}-\frac {5 a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(1 + a*x))/(c*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(2*c*x^2) - (2*a*Sqrt[1 - a^2*x^2])/(c*x) - (5*a^2
*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 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 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^3 (c-a c x)} \, dx &=c \int \frac {\sqrt {1-a^2 x^2}}{x^3 (c-a c x)^2} \, dx\\ &=\frac {\int \frac {(c+a c x)^2}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {2 a^2 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\int \frac {-c^2-2 a c^2 x-2 a^2 c^2 x^2}{x^3 \sqrt {1-a^2 x^2}} \, dx}{c^3}\\ &=\frac {2 a^2 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}+\frac {\int \frac {4 a c^2+5 a^2 c^2 x}{x^2 \sqrt {1-a^2 x^2}} \, dx}{2 c^3}\\ &=\frac {2 a^2 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}-\frac {2 a \sqrt {1-a^2 x^2}}{c x}+\frac {\left (5 a^2\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{2 c}\\ &=\frac {2 a^2 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}-\frac {2 a \sqrt {1-a^2 x^2}}{c x}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{4 c}\\ &=\frac {2 a^2 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}-\frac {2 a \sqrt {1-a^2 x^2}}{c x}-\frac {5 \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 {2 a^2 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c x^2}-\frac {2 a \sqrt {1-a^2 x^2}}{c x}-\frac {5 a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{2 c}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 83, normalized size = 0.83 \[ -\frac {-8 a^3 x^3-5 a^2 x^2+5 a^2 x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+4 a x+1}{2 c x^2 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 99, normalized size = 0.99 \[ \frac {4 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + 5 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (8 \, a^{2} x^{2} - 3 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, {\left (a c x^{3} - c x^{2}\right )}} \]

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),x, algorithm="fricas")

[Out]

1/2*(4*a^3*x^3 - 4*a^2*x^2 + 5*(a^3*x^3 - a^2*x^2)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (8*a^2*x^2 - 3*a*x - 1)*s
qrt(-a^2*x^2 + 1))/(a*c*x^3 - c*x^2)

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giac [B]  time = 0.20, size = 224, normalized size = 2.24 \[ -\frac {{\left (a^{3} + \frac {7 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a}{x} - \frac {40 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a x^{2}}\right )} a^{4} x^{2}}{8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} - \frac {5 \, a^{3} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, c {\left | a \right |}} - \frac {\frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a c {\left | a \right |}}{x} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c {\left | a \right |}}{a x^{2}}}{8 \, a^{2} c^{2}} \]

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),x, algorithm="giac")

[Out]

-1/8*(a^3 + 7*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a/x - 40*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a*x^2))*a^4*x^2/((sq
rt(-a^2*x^2 + 1)*abs(a) + a)^2*c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a)) - 5/2*a^3*log(1/2*abs(-
2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a)) - 1/8*(8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a*c*abs(a
)/x + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c*abs(a)/(a*x^2))/(a^2*c^2)

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maple [A]  time = 0.04, size = 99, normalized size = 0.99 \[ -\frac {\frac {5 a^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {2 a \sqrt {-a^{2} x^{2}+1}}{x}+\frac {2 a \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}+\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}}{c} \]

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),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.83, size = 117, normalized size = 1.17 \[ -\frac {\sqrt {1-a^2\,x^2}}{2\,c\,x^2}-\frac {2\,a\,\sqrt {1-a^2\,x^2}}{c\,x}-\frac {2\,a^3\,\sqrt {1-a^2\,x^2}}{\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {a^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*atan((1 - a^2*x^2)^(1/2)*1i)*5i)/(2*c) - (1 - a^2*x^2)^(1/2)/(2*c*x^2) - (2*a*(1 - a^2*x^2)^(1/2))/(c*x)
- (2*a^3*(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 \[ - \frac {\int \frac {a x}{a x^{4} \sqrt {- a^{2} x^{2} + 1} - x^{3} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a x^{4} \sqrt {- a^{2} x^{2} + 1} - x^{3} \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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