3.206 \(\int e^{-2 \coth ^{-1}(a x)} (c-a c x)^p \, dx\)

Optimal. Leaf size=44 \[ \frac {(c-a c x)^{p+2} \, _2F_1\left (1,p+2;p+3;\frac {1}{2} (1-a x)\right )}{2 a c^2 (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6167, 6130, 21, 68} \[ \frac {(c-a c x)^{p+2} \, _2F_1\left (1,p+2;p+3;\frac {1}{2} (1-a x)\right )}{2 a c^2 (p+2)} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^p/E^(2*ArcCoth[a*x]),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 6130

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

Rule 6167

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

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Mathematica [A]  time = 0.02, size = 44, normalized size = 1.00 \[ -\frac {(a x-1) (c-a c x)^p \left (\, _2F_1\left (1,p+1;p+2;\frac {1}{2} (1-a x)\right )-1\right )}{a (p+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a x - 1\right )} {\left (-a c x + c\right )}^{p}}{a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.51, size = 0, normalized size = 0.00 \[ \int \frac {\left (-a c x +c \right )^{p} \left (a x -1\right )}{a x +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (c-a\,c\,x\right )}^p\,\left (a\,x-1\right )}{a\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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