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

Optimal. Leaf size=84 \[ \frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {(5-3 a x) \sqrt {c-a^2 c x^2}}{3 a^2}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2} \]

[Out]

arctan(a*x*c^(1/2)/(-a^2*c*x^2+c)^(1/2))*c^(1/2)/a^2+1/3*x^2*(-a^2*c*x^2+c)^(1/2)+1/3*(-3*a*x+5)*(-a^2*c*x^2+c
)^(1/2)/a^2

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Rubi [A]  time = 0.25, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6167, 6152, 1809, 780, 217, 203} \[ \frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {(5-3 a x) \sqrt {c-a^2 c x^2}}{3 a^2}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[c - a^2*c*x^2])/3 + ((5 - 3*a*x)*Sqrt[c - a^2*c*x^2])/(3*a^2) + (Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[c
- a^2*c*x^2]])/a^2

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 6152

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/c^(n/2), Int[(x^m
*(c + d*x^2)^(p + n/2))/(1 - a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p
] || GtQ[c, 0]) && ILtQ[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)} x \sqrt {c-a^2 c x^2} \, dx &=-\int e^{-2 \tanh ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx\\ &=-\left (c \int \frac {x (1-a x)^2}{\sqrt {c-a^2 c x^2}} \, dx\right )\\ &=\frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {\int \frac {x \left (-5 a^2 c+6 a^3 c x\right )}{\sqrt {c-a^2 c x^2}} \, dx}{3 a^2}\\ &=\frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {(5-3 a x) \sqrt {c-a^2 c x^2}}{3 a^2}+\frac {c \int \frac {1}{\sqrt {c-a^2 c x^2}} \, dx}{a}\\ &=\frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {(5-3 a x) \sqrt {c-a^2 c x^2}}{3 a^2}+\frac {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 )}{a}\\ &=\frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {(5-3 a x) \sqrt {c-a^2 c x^2}}{3 a^2}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 79, normalized size = 0.94 \[ \frac {\left (a^2 x^2-3 a x+5\right ) \sqrt {c-a^2 c x^2}-3 \sqrt {c} \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5 - 3*a*x + a^2*x^2)*Sqrt[c - a^2*c*x^2] - 3*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2
))])/(3*a^2)

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fricas [A]  time = 0.60, size = 150, normalized size = 1.79 \[ \left [\frac {2 \, \sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 3 \, a x + 5\right )} + 3 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{6 \, a^{2}}, \frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 3 \, a x + 5\right )} - 3 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{3 \, a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 73, normalized size = 0.87 \[ \frac {1}{3} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (x - \frac {3}{a}\right )} x + \frac {5}{a^{2}}\right )} - \frac {c \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{a \sqrt {-c} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(-a^2*c*x^2 + c)*((x - 3/a)*x + 5/a^2) - c*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a*sqrt(-c
)*abs(a))

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maple [B]  time = 0.05, size = 156, normalized size = 1.86 \[ -\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3 a^{2} c}-\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{a}-\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{a \sqrt {a^{2} c}}+\frac {2 \sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}{a^{2}}+\frac {2 c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}\right )}{a \sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 70, normalized size = 0.83 \[ -\frac {\sqrt {-a^{2} c x^{2} + c} x}{a} + \frac {\sqrt {c} \arcsin \left (a x\right )}{a^{2}} + \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{2}} - \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{3 \, a^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x*(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 \[ \int \frac {x \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x - 1\right )}{a x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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