\(\int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx\) [675]

Optimal result

Integrand size = 25, antiderivative size = 85 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {(5+3 a x) \sqrt {c-a^2 c x^2}}{3 a^2}-\frac {\sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2} \]

[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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6302, 6286, 1823, 794, 223, 209} \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=-\frac {\sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2}+\frac {1}{3} x^2 \sqrt {c-a^2 c x^2}+\frac {(3 a x+5) \sqrt {c-a^2 c x^2}}{3 a^2} \]

[In]

Int[E^(2*ArcCoth[a*x])*x*Sqrt[c - a^2*c*x^2],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 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 794

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 1823

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 6286

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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]) && IGtQ[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*} \text {integral}& = -\int e^{2 \text {arctanh}(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 \text {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} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {\left (5+3 a x+a^2 x^2\right ) \sqrt {c-a^2 c x^2}+3 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{3 a^2} \]

[In]

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

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {\left (a^{2} x^{2}+3 a x +5\right ) \left (a^{2} x^{2}-1\right ) c}{3 a^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {\arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c}{a \sqrt {a^{2} c}}\)	\(80\)
default	\(-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3 a^{2} c}+\frac {x \sqrt {-a^{2} c \,x^{2}+c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}}{a}+\frac {2 \sqrt {-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c}-\frac {2 a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}}{a^{2}}\)	\(163\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.76 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\left [\frac {2 \, \sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 3 \, a x + 5\right )} + 3 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{6 \, a^{2}}, \frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 3 \, a x + 5\right )} + 3 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{3 \, a^{2}}\right ] \]

[In]

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

Sympy [F]

\[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int \frac {x \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )}{a x - 1}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {-a^{2} c x^{2} + c} x}{a} - \frac {\sqrt {c} \arcsin \left (a x\right )}{a^{2}} + \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{2}} - \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{3 \, a^{2} c} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {1}{3} \, \sqrt {-a^{2} c x^{2} + c} {\left ({\left (x + \frac {3}{a}\right )} x + \frac {5}{a^{2}}\right )} + \frac {c \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{a \sqrt {-c} {\left | a \right |}} \]

[In]

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

Mupad [F(-1)]

Timed out. \[ \int e^{2 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int \frac {x\,\sqrt {c-a^2\,c\,x^2}\,\left (a\,x+1\right )}{a\,x-1} \,d x \]

[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)