3.216 \(\int e^{-3 \coth ^{-1}(a x)} (c-a c x)^p \, dx\)

Optimal. Leaf size=94 \[ \frac {x \left (1-\frac {1}{a x}\right )^{3/2} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{-p-\frac {3}{2}} (c-a c x)^p \, _2F_1\left (-p-\frac {3}{2},-p-1;-p;\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(p+1) \sqrt {\frac {1}{a x}+1}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6176, 6181, 132} \[ \frac {x \left (1-\frac {1}{a x}\right )^{3/2} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{-p-\frac {3}{2}} (c-a c x)^p \, _2F_1\left (-p-\frac {3}{2},-p-1;-p;\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(p+1) \sqrt {\frac {1}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^p/E^(3*ArcCoth[a*x]),x]

[Out]

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

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

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

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Mathematica [A]  time = 0.07, size = 96, normalized size = 1.02 \[ \frac {\sqrt {1-\frac {1}{a x}} (a x+1) \left (\frac {a x-1}{a x+1}\right )^{-p-\frac {1}{2}} (c-a c x)^p \, _2F_1\left (-p-\frac {3}{2},-p-1;-p;\frac {2}{a x+1}\right )}{a (p+1) \sqrt {\frac {1}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a*c*x)^p/E^(3*ArcCoth[a*x]),x]

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a x - 1\right )} {\left (-a c x + c\right )}^{p} \sqrt {\frac {a x - 1}{a x + 1}}}{a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(a*x+1)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \left (-a c x +c \right )^{p} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-a c x + c\right )}^{p} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c-a\,c\,x\right )}^p\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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