3.2.77 \(\int e^{3 \coth ^{-1}(a x)} (c-a c x)^p \, dx\) [177]

Optimal. Leaf size=202 \[ \frac {3 \sqrt {1+\frac {1}{a x}} (c-a c x)^p}{a p (1+p) \sqrt {1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{3/2} x (c-a c x)^p}{(1+p) \sqrt {1-\frac {1}{a x}}}-\frac {3 \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3}{2}-p} \sqrt {1+\frac {1}{a x}} (c-a c x)^p \, _2F_1\left (1-p,\frac {3}{2}-p;2-p;\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{a^2 p \left (1-p^2\right ) \left (1-\frac {1}{a x}\right )^{3/2} x} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6311, 6316, 96, 134} \begin {gather*} -\frac {3 \sqrt {\frac {1}{a x}+1} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3}{2}-p} (c-a c x)^p \, _2F_1\left (1-p,\frac {3}{2}-p;2-p;\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{a^2 p \left (1-p^2\right ) x \left (1-\frac {1}{a x}\right )^{3/2}}+\frac {x \left (\frac {1}{a x}+1\right )^{3/2} (c-a c x)^p}{(p+1) \sqrt {1-\frac {1}{a x}}}+\frac {3 \sqrt {\frac {1}{a x}+1} (c-a c x)^p}{a p (p+1) \sqrt {1-\frac {1}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[1 + 1/(a*x)]*(c - a*c*x)^p)/(a*p*(1 + p)*Sqrt[1 - 1/(a*x)]) + ((1 + 1/(a*x))^(3/2)*x*(c - a*c*x)^p)/((
1 + p)*Sqrt[1 - 1/(a*x)]) - (3*((a - x^(-1))/(a + x^(-1)))^(3/2 - p)*Sqrt[1 + 1/(a*x)]*(c - a*c*x)^p*Hypergeom
etric2F1[1 - p, 3/2 - p, 2 - p, 2/((a + x^(-1))*x)])/(a^2*p*(1 - p^2)*(1 - 1/(a*x))^(3/2)*x)

Rule 96

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)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 134

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)/((b*e - a*f)*(m + 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)*((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 6311

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 6316

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m, S
ubst[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 ) \text {Subst}\left (\int 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 (1+\frac {1}{a x}\right )^{3/2} x (c-a c x)^p}{(1+p) \sqrt {1-\frac {1}{a x}}}-\frac {\left (3 \left (1-\frac {1}{a x}\right )^{-p} \left (\frac {1}{x}\right )^p (c-a c x)^p\right ) \text {Subst}\left (\int x^{-1-p} \left (1-\frac {x}{a}\right )^{-\frac {3}{2}+p} \sqrt {1+\frac {x}{a}} \, dx,x,\frac {1}{x}\right )}{a (1+p)}\\ &=\frac {3 \sqrt {1+\frac {1}{a x}} (c-a c x)^p}{a p (1+p) \sqrt {1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{3/2} x (c-a c x)^p}{(1+p) \sqrt {1-\frac {1}{a x}}}-\frac {\left (3 \left (1-\frac {1}{a x}\right )^{-p} \left (\frac {1}{x}\right )^p (c-a c x)^p\right ) \text {Subst}\left (\int \frac {x^{-p} \left (1-\frac {x}{a}\right )^{-\frac {3}{2}+p}}{\sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{a^2 p (1+p)}\\ &=\frac {3 \sqrt {1+\frac {1}{a x}} (c-a c x)^p}{a p (1+p) \sqrt {1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{3/2} x (c-a c x)^p}{(1+p) \sqrt {1-\frac {1}{a x}}}-\frac {3 \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3}{2}-p} \sqrt {1+\frac {1}{a x}} (c-a c x)^p \, _2F_1\left (1-p,\frac {3}{2}-p;2-p;\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{a^2 p \left (1-p^2\right ) \left (1-\frac {1}{a x}\right )^{3/2} x}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 155, normalized size = 0.77 \begin {gather*} \frac {\sqrt {1+\frac {1}{a x}} \left (\frac {-1+a x}{1+a x}\right )^{-p} (c-a c x)^p \left ((-1+p) \left (\frac {-1+a x}{1+a x}\right )^p (1+a x) (3+p+a p x)+3 \sqrt {\frac {-1+a x}{1+a x}} \, _2F_1\left (1-p,\frac {3}{2}-p;2-p;\frac {2}{1+a x}\right )\right )}{a (-1+p) p (1+p) \sqrt {1-\frac {1}{a x}} (1+a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-a c x +c \right )^{p}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

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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