\(\int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx\) [167]

Optimal result

Integrand size = 18, antiderivative size = 42 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\frac {2 (c-a c x)^p}{a p}-\frac {(c-a c x)^{1+p}}{a c (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6302, 6265, 21, 45} \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\frac {2 (c-a c x)^p}{a p}-\frac {(c-a c x)^{p+1}}{a c (p+1)} \]

[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 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 45

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])

Rule 6265

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 6302

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*} \text {integral}& = -\int e^{2 \text {arctanh}(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] (verified)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\frac {(c-a c x)^p (2+p+a p x)}{a p (1+p)} \]

[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))

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {\left (p a x +p +2\right ) \left (-a c x +c \right )^{p}}{a p \left (p +1\right )}\)	\(29\)
risch	\(\frac {\left (p a x +p +2\right ) \left (-a c x +c \right )^{p}}{a p \left (p +1\right )}\)	\(29\)
norman	\(\frac {x \,{\mathrm e}^{p \ln \left (-a c x +c \right )}}{p +1}+\frac {\left (2+p \right ) {\mathrm e}^{p \ln \left (-a c x +c \right )}}{a p \left (p +1\right )}\)	\(46\)
parallelrisch	\(\frac {x \left (-a c x +c \right )^{p} a p +\left (-a c x +c \right )^{p} p +2 \left (-a c x +c \right )^{p}}{a p \left (p +1\right )}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\frac {{\left (a p x + p + 2\right )} {\left (-a c x + c\right )}^{p}}{a p^{2} + a p} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (29) = 58\).

Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.95 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\begin {cases} - c^{p} x & \text {for}\: a = 0 \\- \frac {a x \log {\left (x - \frac {1}{a} \right )}}{a^{2} c x - a c} + \frac {\log {\left (x - \frac {1}{a} \right )}}{a^{2} c x - a c} + \frac {2}{a^{2} c x - a c} & \text {for}\: p = -1 \\x + \frac {2 \log {\left (x - \frac {1}{a} \right )}}{a} & \text {for}\: p = 0 \\\frac {a p x \left (- a c x + c\right )^{p}}{a p^{2} + a p} + \frac {p \left (- a c x + c\right )^{p}}{a p^{2} + a p} + \frac {2 \left (- a c x + c\right )^{p}}{a p^{2} + a p} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.17 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\frac {{\left (a c^{p} p x + c^{p}\right )} {\left (-a x + 1\right )}^{p}}{{\left (p^{2} + p\right )} a} + \frac {{\left (-a x + 1\right )}^{p} c^{p}}{a p} \]

[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)

Giac [F]

\[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\int { \frac {{\left (a x + 1\right )} {\left (-a c x + c\right )}^{p}}{a x - 1} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 4.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^p \, dx=\frac {{\left (c-a\,c\,x\right )}^p\,\left (p+a\,p\,x+2\right )}{a\,p\,\left (p+1\right )} \]

[In]

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

[Out]

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