3.1111 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=60 \[ \frac {2 (a x+1)}{a \sqrt {c-a^2 c x^2}}-\frac {\tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6141, 653, 217, 203} \[ \frac {2 (a x+1)}{a \sqrt {c-a^2 c x^2}}-\frac {\tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 6141

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 82, normalized size = 1.37 \[ \frac {2 \sqrt {1-a^2 x^2} \left (\sqrt {a x+1}+\sqrt {1-a x} \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{a \sqrt {1-a x} \sqrt {c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 152, normalized size = 2.53 \[ \left [-\frac {{\left (a x - 1\right )} \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) + 4 \, \sqrt {-a^{2} c x^{2} + c}}{2 \, {\left (a^{2} c x - a c\right )}}, \frac {{\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{2} c x - a c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.04, size = 80, normalized size = 1.33 \[ -\frac {\arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}{a^{2} c \left (x -\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 45, normalized size = 0.75 \[ -a {\left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{3} c x - a^{2} c} + \frac {\arcsin \left (a x\right )}{a^{2} \sqrt {c}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(2*sqrt(-a^2*c*x^2 + c)/(a^3*c*x - a^2*c) + arcsin(a*x)/(a^2*sqrt(c)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int -\frac {{\left (a\,x+1\right )}^2}{\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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