3.1169 \(\int \frac {e^{3 \tanh ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=83 \[ \frac {2 \sqrt {1-a^2 x^2}}{a (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{a \sqrt {c-a^2 c x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6143, 6140, 43} \[ \frac {2 \sqrt {1-a^2 x^2}}{a (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \log (1-a x)}{a \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6140

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

Rule 6143

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 51, normalized size = 0.61 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {2}{1-a x}+\log (1-a x)\right )}{a \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.61, size = 382, normalized size = 4.60 \[ \left [\frac {4 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x + {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {c} \log \left (\frac {a^{6} c x^{6} - 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} + 4 \, a c x - {\left (a^{4} x^{4} - 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 4 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - 2 \, c}{a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1}\right )}{2 \, {\left (a^{4} c x^{3} - a^{3} c x^{2} - a^{2} c x + a c\right )}}, \frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a x + {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 2 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c}}{a^{4} c x^{4} - 2 \, a^{3} c x^{3} - a^{2} c x^{2} + 2 \, a c x}\right )}{a^{4} c x^{3} - a^{3} c x^{2} - a^{2} c x + a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(4*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*x + (a^3*x^3 - a^2*x^2 - a*x + 1)*sqrt(c)*log((a^6*c*x^6 - 4
*a^5*c*x^5 + 5*a^4*c*x^4 - 4*a^2*c*x^2 + 4*a*c*x - (a^4*x^4 - 4*a^3*x^3 + 6*a^2*x^2 - 4*a*x)*sqrt(-a^2*c*x^2 +
c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - 2*c)/(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)))/(a^4*c*x^3 - a^3*c*x^2 - a^2*c*x + a
*c), (2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*x + (a^3*x^3 - a^2*x^2 - a*x + 1)*sqrt(-c)*arctan(sqrt(-a^2*
c*x^2 + c)*(a^2*x^2 - 2*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt(-c)/(a^4*c*x^4 - 2*a^3*c*x^3 - a^2*c*x^2 + 2*a*c*x)))
/(a^4*c*x^3 - a^3*c*x^2 - a^2*c*x + a*c)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 70, normalized size = 0.84 \[ \frac {\left (-\ln \left (a x -1\right ) x a +\ln \left (a x -1\right )+2\right ) \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}}{\left (a^{2} x^{2}-1\right ) c a \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________