### 3.187 $$\int e^{4 \coth ^{-1}(a x)} (c-a c x)^p \, dx$$

Optimal. Leaf size=66 $\frac{4 c (c-a c x)^{p-1}}{a (1-p)}+\frac{4 (c-a c x)^p}{a p}-\frac{(c-a c x)^{p+1}}{a c (p+1)}$

[Out]

(4*c*(c - a*c*x)^(-1 + p))/(a*(1 - p)) + (4*(c - a*c*x)^p)/(a*p) - (c - a*c*x)^(1 + p)/(a*c*(1 + p))

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Rubi [A]  time = 0.0810893, antiderivative size = 66, 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} $\frac{4 c (c-a c x)^{p-1}}{a (1-p)}+\frac{4 (c-a c x)^p}{a p}-\frac{(c-a c x)^{p+1}}{a c (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(4*ArcCoth[a*x])*(c - a*c*x)^p,x]

[Out]

(4*c*(c - a*c*x)^(-1 + p))/(a*(1 - p)) + (4*(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^{4 \coth ^{-1}(a x)} (c-a c x)^p \, dx &=\int e^{4 \tanh ^{-1}(a x)} (c-a c x)^p \, dx\\ &=\int \frac{(1+a x)^2 (c-a c x)^p}{(1-a x)^2} \, dx\\ &=c^2 \int (1+a x)^2 (c-a c x)^{-2+p} \, dx\\ &=c^2 \int \left (4 (c-a c x)^{-2+p}-\frac{4 (c-a c x)^{-1+p}}{c}+\frac{(c-a c x)^p}{c^2}\right ) \, dx\\ &=\frac{4 c (c-a c x)^{-1+p}}{a (1-p)}+\frac{4 (c-a c x)^p}{a p}-\frac{(c-a c x)^{1+p}}{a c (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0763295, size = 50, normalized size = 0.76 $\frac{\left (\frac{a x}{p+1}+\frac{4}{(p-1) (a x-1)}+\frac{3 p+4}{p (p+1)}\right ) (c-a c x)^p}{a}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - a*c*x)^p,x]

[Out]

((c - a*c*x)^p*((4 + 3*p)/(p*(1 + p)) + (a*x)/(1 + p) + 4/((-1 + p)*(-1 + a*x))))/a

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Maple [A]  time = 0.046, size = 74, normalized size = 1.1 \begin{align*}{\frac{ \left ( -acx+c \right ) ^{p} \left ({a}^{2}{p}^{2}{x}^{2}-{a}^{2}{x}^{2}p+2\,a{p}^{2}x+2\,apx-4\,ax+{p}^{2}+3\,p+4 \right ) }{ \left ({p}^{2}-1 \right ) ap \left ( ax-1 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a*c*x+c)^p*(a^2*p^2*x^2-a^2*p*x^2+2*a*p^2*x+2*a*p*x-4*a*x+p^2+3*p+4)/(p^2-1)/a/p/(a*x-1)

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Maxima [B]  time = 1.07685, size = 207, normalized size = 3.14 \begin{align*} \frac{{\left ({\left (p^{2} - p\right )} a^{2} c^{p} x^{2} + 2 \, a c^{p}{\left (p - 1\right )} x + 2 \, c^{p}\right )}{\left (-a x + 1\right )}^{p} a^{2}}{{\left (p^{3} - p\right )} a^{4} x -{\left (p^{3} - p\right )} a^{3}} + \frac{2 \,{\left (a c^{p}{\left (p - 1\right )} x + c^{p}\right )}{\left (-a x + 1\right )}^{p} a}{{\left (p^{2} - p\right )} a^{3} x -{\left (p^{2} - p\right )} a^{2}} + \frac{{\left (-a x + 1\right )}^{p} c^{p}}{a^{2}{\left (p - 1\right )} x - a{\left (p - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((p^2 - p)*a^2*c^p*x^2 + 2*a*c^p*(p - 1)*x + 2*c^p)*(-a*x + 1)^p*a^2/((p^3 - p)*a^4*x - (p^3 - p)*a^3) + 2*(a*
c^p*(p - 1)*x + c^p)*(-a*x + 1)^p*a/((p^2 - p)*a^3*x - (p^2 - p)*a^2) + (-a*x + 1)^p*c^p/(a^2*(p - 1)*x - a*(p
- 1))

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Fricas [A]  time = 1.52886, size = 161, normalized size = 2.44 \begin{align*} -\frac{{\left ({\left (a^{2} p^{2} - a^{2} p\right )} x^{2} + p^{2} + 2 \,{\left (a p^{2} + a p - 2 \, a\right )} x + 3 \, p + 4\right )}{\left (-a c x + c\right )}^{p}}{a p^{3} - a p -{\left (a^{2} p^{3} - a^{2} p\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2*p^2 - a^2*p)*x^2 + p^2 + 2*(a*p^2 + a*p - 2*a)*x + 3*p + 4)*(-a*c*x + c)^p/(a*p^3 - a*p - (a^2*p^3 - a^
2*p)*x)

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Sympy [A]  time = 1.96831, size = 512, normalized size = 7.76 \begin{align*} \begin{cases} c^{p} x & \text{for}\: a = 0 \\- \frac{a^{2} x^{2} \log{\left (x - \frac{1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac{2 a^{2} x^{2}}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac{2 a x \log{\left (x - \frac{1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} - \frac{\log{\left (x - \frac{1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} & \text{for}\: p = -1 \\\frac{a^{2} x^{2}}{a^{2} x - a} + \frac{4 a x \log{\left (x - \frac{1}{a} \right )}}{a^{2} x - a} - \frac{4 \log{\left (x - \frac{1}{a} \right )}}{a^{2} x - a} - \frac{5}{a^{2} x - a} & \text{for}\: p = 0 \\- \frac{a c x^{2}}{2} - 3 c x - \frac{4 c \log{\left (x - \frac{1}{a} \right )}}{a} & \text{for}\: p = 1 \\\frac{a^{2} p^{2} x^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} - \frac{a^{2} p x^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac{2 a p^{2} x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac{2 a p x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} - \frac{4 a x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac{p^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac{3 p \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac{4 \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c**p*x, Eq(a, 0)), (-a**2*x**2*log(x - 1/a)/(a**3*c*x**2 - 2*a**2*c*x + a*c) + 2*a**2*x**2/(a**3*c*
x**2 - 2*a**2*c*x + a*c) + 2*a*x*log(x - 1/a)/(a**3*c*x**2 - 2*a**2*c*x + a*c) - log(x - 1/a)/(a**3*c*x**2 - 2
*a**2*c*x + a*c), Eq(p, -1)), (a**2*x**2/(a**2*x - a) + 4*a*x*log(x - 1/a)/(a**2*x - a) - 4*log(x - 1/a)/(a**2
*x - a) - 5/(a**2*x - a), Eq(p, 0)), (-a*c*x**2/2 - 3*c*x - 4*c*log(x - 1/a)/a, Eq(p, 1)), (a**2*p**2*x**2*(-a
*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) - a**2*p*x**2*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a
*p**3 + a*p) + 2*a*p**2*x*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) + 2*a*p*x*(-a*c*x + c)**p/(a
**2*p**3*x - a**2*p*x - a*p**3 + a*p) - 4*a*x*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) + p**2*(
-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + a*p) + 3*p*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3
+ a*p) + 4*(-a*c*x + c)**p/(a**2*p**3*x - a**2*p*x - a*p**3 + 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 )}^{2}{\left (-a c x + c\right )}^{p}}{{\left (a x - 1\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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