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

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {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*ArcTanh[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 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])

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]
)

Rubi steps

\begin {align*} \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 = 0.09, 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*ArcTanh[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

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 81, normalized size = 1.23 \[ -\frac {{\left ({\left (a^{2} p^{2} - a^{2} p\right )} x^{2} + p^{2} + 2 \, {\left (a p^{2} + a p - 2 \, a\right )} x + 3 \, p + 4\right )} {\left (-a c x + c\right )}^{p}}{a p^{3} - a p - {\left (a^{2} p^{3} - a^{2} p\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{4} {\left (-a c x + c\right )}^{p}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c)^p,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 74, normalized size = 1.12 \[ \frac {\left (a^{2} p^{2} x^{2}-a^{2} x^{2} p +2 a \,p^{2} x +2 a p x -4 a x +p^{2}+3 p +4\right ) \left (-a c x +c \right )^{p}}{\left (a x -1\right ) a p \left (p^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 77, normalized size = 1.17 \[ \frac {{\left ({\left (p^{2} - p\right )} a^{2} c^{p} x^{2} + 2 \, {\left (p^{2} + p - 2\right )} a c^{p} x + {\left (p^{2} + 3 \, p + 4\right )} c^{p}\right )} {\left (-a x + 1\right )}^{p}}{{\left (p^{3} - p\right )} a^{2} x - {\left (p^{3} - p\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(-a*c*x+c)^p,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.94, size = 57, normalized size = 0.86 \[ \frac {4\,{\left (c-a\,c\,x\right )}^p}{a\,\left (a\,x-1\right )\,\left (p-1\right )}+\frac {{\left (c-a\,c\,x\right )}^p\,\left (3\,p+a\,p\,x+4\right )}{a\,p\,\left (p+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 3.05, size = 541, normalized size = 8.20 \[ \begin {cases} c^{p} x & \text {for}\: a = 0 \\- \frac {a^{2} x^{2} \log {\left (x - \frac {1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {2 a x \log {\left (x - \frac {1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {4 a x}{a^{3} c x^{2} - 2 a^{2} c x + a c} - \frac {\log {\left (x - \frac {1}{a} \right )}}{a^{3} c x^{2} - 2 a^{2} c x + a c} - \frac {2}{a^{3} c x^{2} - 2 a^{2} c x + a c} & \text {for}\: p = -1 \\\frac {a^{2} x^{2}}{a^{2} x - a} + \frac {4 a x \log {\left (x - \frac {1}{a} \right )}}{a^{2} x - a} - \frac {a x}{a^{2} x - a} - \frac {4 \log {\left (x - \frac {1}{a} \right )}}{a^{2} x - a} - \frac {4}{a^{2} x - a} & \text {for}\: p = 0 \\- \frac {a c x^{2}}{2} - 3 c x - \frac {4 c \log {\left (x - \frac {1}{a} \right )}}{a} & \text {for}\: p = 1 \\\frac {a^{2} p^{2} x^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} - \frac {a^{2} p x^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {2 a p^{2} x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {2 a p x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} - \frac {4 a x \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {p^{2} \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {3 p \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} + \frac {4 \left (- a c x + c\right )^{p}}{a^{2} p^{3} x - a^{2} p x - a p^{3} + a p} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+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*x*log(x - 1/a)/(
a**3*c*x**2 - 2*a**2*c*x + a*c) + 4*a*x/(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) - 2/(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) - a*x/(a**2*x - a) - 4*log(x - 1/a)/(a**2*x - a) - 4/(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))

________________________________________________________________________________________