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

Optimal. Leaf size=131 \[ \frac {2 \sqrt {\frac {1}{a x}+1} x^{m+1} \sqrt {c-a c x}}{(2 m+3) \sqrt {1-\frac {1}{a x}}}-\frac {2 (4 m+5) x^m \sqrt {c-a c x} \, _2F_1\left (\frac {1}{2},-m-\frac {1}{2};\frac {1}{2}-m;-\frac {1}{a x}\right )}{a (2 m+1) (2 m+3) \sqrt {1-\frac {1}{a x}}} \]

[Out]

-2*(5+4*m)*x^m*hypergeom([1/2, -1/2-m],[1/2-m],-1/a/x)*(-a*c*x+c)^(1/2)/a/(4*m^2+8*m+3)/(1-1/a/x)^(1/2)+2*x^(1
+m)*(1+1/a/x)^(1/2)*(-a*c*x+c)^(1/2)/(3+2*m)/(1-1/a/x)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6176, 6181, 79, 64} \[ \frac {2 \sqrt {\frac {1}{a x}+1} x^{m+1} \sqrt {c-a c x}}{(2 m+3) \sqrt {1-\frac {1}{a x}}}-\frac {2 (4 m+5) x^m \sqrt {c-a c x} \, _2F_1\left (\frac {1}{2},-m-\frac {1}{2};\frac {1}{2}-m;-\frac {1}{a x}\right )}{a (2 m+1) (2 m+3) \sqrt {1-\frac {1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[1 + 1/(a*x)]*x^(1 + m)*Sqrt[c - a*c*x])/((3 + 2*m)*Sqrt[1 - 1/(a*x)]) - (2*(5 + 4*m)*x^m*Sqrt[c - a*c*
x]*Hypergeometric2F1[1/2, -1/2 - m, 1/2 - m, -(1/(a*x))])/(a*(1 + 2*m)*(3 + 2*m)*Sqrt[1 - 1/(a*x)])

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 6176

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

Rule 6181

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

Rubi steps

\begin {align*} \int e^{-\coth ^{-1}(a x)} x^m \sqrt {c-a c x} \, dx &=\frac {\sqrt {c-a c x} \int e^{-\coth ^{-1}(a x)} \sqrt {1-\frac {1}{a x}} x^{\frac {1}{2}+m} \, dx}{\sqrt {1-\frac {1}{a x}} \sqrt {x}}\\ &=-\frac {\left (\left (\frac {1}{x}\right )^{\frac {1}{2}+m} x^m \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {x^{-\frac {5}{2}-m} \left (1-\frac {x}{a}\right )}{\sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a x}}}\\ &=\frac {2 \sqrt {1+\frac {1}{a x}} x^{1+m} \sqrt {c-a c x}}{(3+2 m) \sqrt {1-\frac {1}{a x}}}+\frac {\left ((5+4 m) \left (\frac {1}{x}\right )^{\frac {1}{2}+m} x^m \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {x^{-\frac {3}{2}-m}}{\sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{a (3+2 m) \sqrt {1-\frac {1}{a x}}}\\ &=\frac {2 \sqrt {1+\frac {1}{a x}} x^{1+m} \sqrt {c-a c x}}{(3+2 m) \sqrt {1-\frac {1}{a x}}}-\frac {2 (5+4 m) x^m \sqrt {c-a c x} \, _2F_1\left (\frac {1}{2},-\frac {1}{2}-m;\frac {1}{2}-m;-\frac {1}{a x}\right )}{a (1+2 m) (3+2 m) \sqrt {1-\frac {1}{a x}}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 102, normalized size = 0.78 \[ \frac {2 x^m \sqrt {c-a c x} \left (a (2 m+1) x \sqrt {\frac {1}{a x}+1}-(4 m+5) \, _2F_1\left (\frac {1}{2},-m-\frac {1}{2};\frac {1}{2}-m;-\frac {1}{a x}\right )\right )}{a (2 m+1) (2 m+3) \sqrt {1-\frac {1}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^m*Sqrt[c - a*c*x]*(a*(1 + 2*m)*Sqrt[1 + 1/(a*x)]*x - (5 + 4*m)*Hypergeometric2F1[1/2, -1/2 - m, 1/2 - m,
-(1/(a*x))]))/(a*(1 + 2*m)*(3 + 2*m)*Sqrt[1 - 1/(a*x)])

________________________________________________________________________________________

fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a c x + c} x^{m} \sqrt {\frac {a x - 1}{a x + 1}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a c x + c} x^{m} \sqrt {\frac {a x - 1}{a x + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int x^{m} \sqrt {-a c x +c}\, \sqrt {\frac {a x -1}{a x +1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a c x + c} x^{m} \sqrt {\frac {a x - 1}{a x + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________