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

Optimal. Leaf size=165 \[ \frac {\sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (a x+1)}{4 c \sqrt {c-a^2 c x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6153, 6150, 72} \[ \frac {\sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (a x+1)}{4 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 6150

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

Rule 6153

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{3/2}} \, dx &=\frac {\sqrt {1-a^2 x^2} \int \frac {e^{\tanh ^{-1}(a x)}}{x \left (1-a^2 x^2\right )^{3/2}} \, dx}{c \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {1}{x (1-a x)^2 (1+a x)} \, dx}{c \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{x}+\frac {a}{2 (-1+a x)^2}-\frac {3 a}{4 (-1+a x)}-\frac {a}{4 (1+a x)}\right ) \, dx}{c \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{2 c (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (x)}{c \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2} \log (1-a x)}{4 c \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \log (1+a x)}{4 c \sqrt {c-a^2 c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 65, normalized size = 0.39 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {1}{2-2 a x}-\frac {3}{4} \log (1-a x)-\frac {1}{4} \log (a x+1)+\log (x)\right )}{c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{a^{5} c^{2} x^{6} - a^{4} c^{2} x^{5} - 2 \, a^{3} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{3} + a c^{2} x^{2} - c^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 96, normalized size = 0.58 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (4 a \ln \relax (x ) x -3 \ln \left (a x -1\right ) x a -a x \ln \left (a x +1\right )-4 \ln \relax (x )+3 \ln \left (a x -1\right )+\ln \left (a x +1\right )-2\right )}{4 c^{2} \left (a x -1\right ) \left (a^{2} x^{2}-1\right )} \]

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)^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a\,x+1}{x\,{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)**(3/2),x)

[Out]

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

________________________________________________________________________________________