3.7.28 \(\int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\) [628]

Optimal. Leaf size=59 \[ -\frac {2 (1+a x)}{a \sqrt {c-a^2 c x^2}}+\frac {\text {ArcTan}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}} \]

[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 = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6302, 6276, 667, 223, 209} \begin {gather*} \frac {\text {ArcTan}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}}-\frac {2 (a x+1)}{a \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[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 209

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

Rule 223

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 667

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 6276

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]

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{2 \coth ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx &=-\int \frac {e^{2 \tanh ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\\ &=-\left (c \int \frac {(1+a x)^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\right )\\ &=-\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}}+\text {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 = 0.03, size = 82, normalized size = 1.39 \begin {gather*} -\frac {2 \sqrt {1-a^2 x^2} \left (\sqrt {1+a x}+\sqrt {1-a x} \text {ArcSin}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{a \sqrt {1-a x} \sqrt {c-a^2 c x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcCoth[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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Maple [A]
time = 0.14, size = 79, normalized size = 1.34

method result size
default \(\frac {\arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}+\frac {2 \sqrt {-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c}}{a^{2} c \left (x -\frac {1}{a}\right )}\) \(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*(-a^2*c*(x-1/a)^2-2*(x-1/a)*a*c)^
(1/2)

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Maxima [A]
time = 0.47, size = 40, normalized size = 0.68 \begin {gather*} \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{2} c x - a c} + \frac {\arcsin \left (a x\right )}{a \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 153, normalized size = 2.59 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x + 1}{\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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