3.627 \(\int e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=86 \[ \frac {(a x+1) \sqrt {c-a^2 c x^2}}{2 a}+\frac {3 \sqrt {c-a^2 c x^2}}{2 a}-\frac {3 \sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6167, 6141, 671, 641, 217, 203} \[ \frac {(a x+1) \sqrt {c-a^2 c x^2}}{2 a}+\frac {3 \sqrt {c-a^2 c x^2}}{2 a}-\frac {3 \sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 641

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

Rule 671

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

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]

Rule 6167

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

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Mathematica [A]  time = 0.06, size = 76, normalized size = 0.88 \[ \frac {\sqrt {c-a^2 c x^2} \left (\sqrt {1-a^2 x^2} (a x+4)+6 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{2 a \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*((4 + a*x)*Sqrt[1 - a^2*x^2] + 6*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(2*a*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.42, size = 134, normalized size = 1.56 \[ \left [\frac {2 \, \sqrt {-a^{2} c x^{2} + c} {\left (a x + 4\right )} + 3 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{4 \, a}, \frac {\sqrt {-a^{2} c x^{2} + c} {\left (a x + 4\right )} + 3 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{2 \, a}\right ] \]

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]

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

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giac [A]  time = 0.15, size = 62, normalized size = 0.72 \[ \frac {1}{2} \, \sqrt {-a^{2} c x^{2} + c} {\left (x + \frac {4}{a}\right )} + \frac {3 \, c \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{2 \, \sqrt {-c} {\left | a \right |}} \]

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]

1/2*sqrt(-a^2*c*x^2 + c)*(x + 4/a) + 3/2*c*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(sqrt(-c)*abs(a))

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maple [A]  time = 0.04, size = 134, normalized size = 1.56 \[ \frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}{a}-\frac {2 c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 47, normalized size = 0.55 \[ \frac {1}{2} \, \sqrt {-a^{2} c x^{2} + c} x - \frac {3 \, \sqrt {c} \arcsin \left (a x\right )}{2 \, a} + \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a} \]

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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