\(\int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx\) [730]

Optimal result

Integrand size = 27, antiderivative size = 136 \[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\frac {3 x^m \sqrt {c-a^2 c x^2}}{a (1+m) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^{1+m} \sqrt {c-a^2 c x^2}}{(2+m) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 x^m \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x)}{a (1+m) \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

3*x^m*(-a^2*c*x^2+c)^(1/2)/a/(1+m)/(1-1/a^2/x^2)^(1/2)+x^(1+m)*(-a^2*c*x^2+c)^(1/2)/(2+m)/(1-1/a^2/x^2)^(1/2)-
4*x^m*hypergeom([1, 1+m],[2+m],a*x)*(-a^2*c*x^2+c)^(1/2)/a/(1+m)/(1-1/a^2/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6327, 6328, 90, 66} \[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=-\frac {4 x^m \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}(1,m+1,m+2,a x)}{a (m+1) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^{m+1} \sqrt {c-a^2 c x^2}}{(m+2) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x^m \sqrt {c-a^2 c x^2}}{a (m+1) \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Int[E^(3*ArcCoth[a*x])*x^m*Sqrt[c - a^2*c*x^2],x]

[Out]

(3*x^m*Sqrt[c - a^2*c*x^2])/(a*(1 + m)*Sqrt[1 - 1/(a^2*x^2)]) + (x^(1 + m)*Sqrt[c - a^2*c*x^2])/((2 + m)*Sqrt[
1 - 1/(a^2*x^2)]) - (4*x^m*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[1, 1 + m, 2 + m, a*x])/(a*(1 + m)*Sqrt[1 - 1/
(a^2*x^2)])

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6327

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

Rule 6328

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {c-a^2 c x^2} \int e^{3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^{1+m} \, dx}{\sqrt {1-\frac {1}{a^2 x^2}} x} \\ & = \frac {\sqrt {c-a^2 c x^2} \int \frac {x^m (1+a x)^2}{-1+a x} \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}} x} \\ & = \frac {\sqrt {c-a^2 c x^2} \int \left (3 x^m+a x^{1+m}+\frac {4 x^m}{-1+a x}\right ) \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}} x} \\ & = \frac {3 x^m \sqrt {c-a^2 c x^2}}{a (1+m) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^{1+m} \sqrt {c-a^2 c x^2}}{(2+m) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\left (4 \sqrt {c-a^2 c x^2}\right ) \int \frac {x^m}{-1+a x} \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}} x} \\ & = \frac {3 x^m \sqrt {c-a^2 c x^2}}{a (1+m) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^{1+m} \sqrt {c-a^2 c x^2}}{(2+m) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {4 x^m \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x)}{a (1+m) \sqrt {1-\frac {1}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.54 \[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\frac {x^m \sqrt {c-a^2 c x^2} (6+a x+m (3+a x)-4 (2+m) \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x))}{a (1+m) (2+m) \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Integrate[E^(3*ArcCoth[a*x])*x^m*Sqrt[c - a^2*c*x^2],x]

[Out]

(x^m*Sqrt[c - a^2*c*x^2]*(6 + a*x + m*(3 + a*x) - 4*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, a*x]))/(a*(1 +
m)*(2 + m)*Sqrt[1 - 1/(a^2*x^2)])

Maple [F]

\[\int \frac {x^{m} \sqrt {-a^{2} c \,x^{2}+c}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}d x\]

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)*x^m*(-a^2*c*x^2+c)^(1/2),x)

[Out]

int(1/((a*x-1)/(a*x+1))^(3/2)*x^m*(-a^2*c*x^2+c)^(1/2),x)

Fricas [F]

\[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} x^{m}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*x^m*(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 2*a*x + 1)*x^m*sqrt((a*x - 1)/(a*x + 1))/(a^2*x^2 - 2*a*x + 1), x)

Sympy [F(-1)]

Timed out. \[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\text {Timed out} \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(3/2)*x**m*(-a**2*c*x**2+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} x^{m}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*x^m*(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*x^m/((a*x - 1)/(a*x + 1))^(3/2), x)

Giac [F(-2)]

Exception generated. \[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*x^m*(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int e^{3 \coth ^{-1}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\int \frac {x^m\,\sqrt {c-a^2\,c\,x^2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]

[In]

int((x^m*(c - a^2*c*x^2)^(1/2))/((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

int((x^m*(c - a^2*c*x^2)^(1/2))/((a*x - 1)/(a*x + 1))^(3/2), x)