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

Optimal. Leaf size=41 \[ \frac {(c-a c x)^{p+1}}{a c (p+1)}-\frac {2 (c-a c x)^p}{a p} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 41, 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} \[ \frac {(c-a c x)^{p+1}}{a c (p+1)}-\frac {2 (c-a c x)^p}{a p} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [A]  time = 0.03, size = 29, normalized size = 0.71 \[ -\frac {(a p x+p+2) (c-a c x)^p}{a p (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[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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fricas [A]  time = 0.52, size = 29, normalized size = 0.71 \[ -\frac {{\left (a p x + p + 2\right )} {\left (-a c x + c\right )}^{p}}{a p^{2} + a p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 30, normalized size = 0.73 \[ -\frac {\left (-a c x +c \right )^{p} \left (a p x +p +2\right )}{a p \left (1+p \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 35, normalized size = 0.85 \[ -\frac {{\left (a c^{p} p x + c^{p} {\left (p + 2\right )}\right )} {\left (-a x + 1\right )}^{p}}{{\left (p^{2} + p\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.84, size = 29, normalized size = 0.71 \[ -\frac {{\left (c-a\,c\,x\right )}^p\,\left (p+a\,p\,x+2\right )}{a\,p\,\left (p+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.96, size = 126, normalized size = 3.07 \[ \begin {cases} c^{p} x & \text {for}\: a = 0 \\\frac {a x \log {\left (x - \frac {1}{a} \right )}}{a^{2} c x - a c} - \frac {\log {\left (x - \frac {1}{a} \right )}}{a^{2} c x - a c} - \frac {2}{a^{2} c x - a c} & \text {for}\: p = -1 \\- x - \frac {2 \log {\left (x - \frac {1}{a} \right )}}{a} & \text {for}\: p = 0 \\- \frac {a p x \left (- a c x + c\right )^{p}}{a p^{2} + a p} - \frac {p \left (- a c x + c\right )^{p}}{a p^{2} + a p} - \frac {2 \left (- a c x + c\right )^{p}}{a p^{2} + a p} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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