### 3.167 $$\int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx$$

Optimal. Leaf size=42 $\frac{2 (c-a c x)^p}{a p}-\frac{(c-a c x)^{p+1}}{a c (p+1)}$

[Out]

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

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Rubi [A]  time = 0.0666535, antiderivative size = 42, normalized size of antiderivative = 1., 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 = {6167, 6130, 21, 43} $\frac{2 (c-a c x)^p}{a p}-\frac{(c-a c x)^{p+1}}{a c (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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]

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 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A]  time = 0.022598, size = 28, normalized size = 0.67 $\frac{(a p x+p+2) (c-a c x)^p}{a p (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 29, normalized size = 0.7 \begin{align*}{\frac{ \left ( apx+p+2 \right ) \left ( -acx+c \right ) ^{p}}{ap \left ( 1+p \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06866, size = 66, normalized size = 1.57 \begin{align*} \frac{{\left (a c^{p} p x + c^{p}\right )}{\left (-a x + 1\right )}^{p}}{{\left (p^{2} + p\right )} a} + \frac{{\left (-a x + 1\right )}^{p} c^{p}}{a p} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.59576, size = 62, normalized size = 1.48 \begin{align*} \frac{{\left (a p x + p + 2\right )}{\left (-a c x + c\right )}^{p}}{a p^{2} + a p} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*p*x + p + 2)*(-a*c*x + c)^p/(a*p^2 + a*p)

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Sympy [A]  time = 0.83473, size = 124, normalized size = 2.95 \begin{align*} \begin{cases} - c^{p} x & \text{for}\: a = 0 \\- \frac{a x \log{\left (x - \frac{1}{a} \right )}}{a^{2} c x - a c} + \frac{\log{\left (x - \frac{1}{a} \right )}}{a^{2} c x - a c} + \frac{2}{a^{2} c x - a c} & \text{for}\: p = -1 \\x + \frac{2 \log{\left (x - \frac{1}{a} \right )}}{a} & \text{for}\: p = 0 \\\frac{a p x \left (- a c x + c\right )^{p}}{a p^{2} + a p} + \frac{p \left (- a c x + c\right )^{p}}{a p^{2} + a p} + \frac{2 \left (- a c x + c\right )^{p}}{a p^{2} + a p} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-c**p*x, Eq(a, 0)), (-a*x*log(x - 1/a)/(a**2*c*x - a*c) + log(x - 1/a)/(a**2*c*x - a*c) + 2/(a**2*c
*x - a*c), Eq(p, -1)), (x + 2*log(x - 1/a)/a, Eq(p, 0)), (a*p*x*(-a*c*x + c)**p/(a*p**2 + a*p) + p*(-a*c*x + c
)**p/(a*p**2 + a*p) + 2*(-a*c*x + c)**p/(a*p**2 + a*p), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}{\left (-a c x + c\right )}^{p}}{a x - 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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