3.1136 \(\int e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=55 \[ -\frac{2^{p+1} (a x+1)^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-p-1,p,p+1,\frac{1}{2} (1-a x)\right )}{a p} \]

[Out]

-((2^(1 + p)*(c - a^2*c*x^2)^p*Hypergeometric2F1[-1 - p, p, 1 + p, (1 - a*x)/2])/(a*p*(1 + a*x)^p))

________________________________________________________________________________________

Rubi [A]  time = 0.0629494, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6141, 678, 69} \[ -\frac{2^{p+1} (a x+1)^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-p-1,p;p+1;\frac{1}{2} (1-a x)\right )}{a p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(1 + p)*(c - a^2*c*x^2)^p*Hypergeometric2F1[-1 - p, p, 1 + p, (1 - a*x)/2])/(a*p*(1 + a*x)^p))

Rule 6141

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^(n/2), Int[(c + d*x^2)^(p -
n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) && IGt
Q[n/2, 0]

Rule 678

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int e^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx &=c \int (1+a x)^2 \left (c-a^2 c x^2\right )^{-1+p} \, dx\\ &=\left (c (1+a x)^{-p} (c-a c x)^{-p} \left (c-a^2 c x^2\right )^p\right ) \int (1+a x)^{1+p} (c-a c x)^{-1+p} \, dx\\ &=-\frac{2^{1+p} (1+a x)^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-1-p,p;1+p;\frac{1}{2} (1-a x)\right )}{a p}\\ \end{align*}

Mathematica [A]  time = 0.0178815, size = 68, normalized size = 1.24 \[ -\frac{2^{p+1} (1-a x)^p \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-p-1,p,p+1,\frac{1}{2} (1-a x)\right )}{a p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(1 + p)*(1 - a*x)^p*(c - a^2*c*x^2)^p*Hypergeometric2F1[-1 - p, p, 1 + p, (1 - a*x)/2])/(a*p*(1 - a^2*x^2
)^p))

________________________________________________________________________________________

Maple [F]  time = 0.423, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{2} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p}}{-{a}^{2}{x}^{2}+1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (a x + 1\right )}^{2}{\left (-a^{2} c x^{2} + c\right )}^{p}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a x + 1\right )}{\left (-a^{2} c x^{2} + c\right )}^{p}}{a x - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 10.1833, size = 653, normalized size = 11.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*Piecewise((0**p*x/a - 0**p*log(1/(a**2*x**2))/(2*a**2) + 0**p*log(-1 + 1/(a**2*x**2))/(2*a**2) - 0**p*acoth
(1/(a*x))/a**2 + c**p*x**2*gamma(p)*gamma(1 - p)*hyper((2, 1, 1 - p), (2, 2), a**2*x**2*exp_polar(2*I*pi))/(2*
gamma(-p)*gamma(p + 1)) - a**(2*p)*c**p*p*x*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(-p - 1/2)*hyper((1 - p, -p - 1
/2), (1/2 - p,), 1/(a**2*x**2))/(2*a*gamma(1/2 - p)*gamma(p + 1)), 1/Abs(a**2*x**2) > 1), (0**p*x/a - 0**p*log
(1/(a**2*x**2))/(2*a**2) + 0**p*log(1 - 1/(a**2*x**2))/(2*a**2) - 0**p*atanh(1/(a*x))/a**2 + c**p*x**2*gamma(p
)*gamma(1 - p)*hyper((2, 1, 1 - p), (2, 2), a**2*x**2*exp_polar(2*I*pi))/(2*gamma(-p)*gamma(p + 1)) - a**(2*p)
*c**p*p*x*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(-p - 1/2)*hyper((1 - p, -p - 1/2), (1/2 - p,), 1/(a**2*x**2))/(2
*a*gamma(1/2 - p)*gamma(p + 1)), True)) - Piecewise((0**p*log(a**2*x**2 - 1)/(2*a) - 0**p*acoth(a*x)/a + a*c**
p*x**2*gamma(p)*gamma(1 - p)*hyper((2, 1, 1 - p), (2, 2), a**2*x**2*exp_polar(2*I*pi))/(2*gamma(-p)*gamma(p +
1)) - a**(2*p)*c**p*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(1/2 - p)*hyper((1 - p, 1/2 - p), (3/2 - p,), 1/(a**2
*x**2))/(2*a**2*x*gamma(3/2 - p)*gamma(p + 1)), Abs(a**2*x**2) > 1), (0**p*log(-a**2*x**2 + 1)/(2*a) - 0**p*at
anh(a*x)/a + a*c**p*x**2*gamma(p)*gamma(1 - p)*hyper((2, 1, 1 - p), (2, 2), a**2*x**2*exp_polar(2*I*pi))/(2*ga
mma(-p)*gamma(p + 1)) - a**(2*p)*c**p*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(1/2 - p)*hyper((1 - p, 1/2 - p), (
3/2 - p,), 1/(a**2*x**2))/(2*a**2*x*gamma(3/2 - p)*gamma(p + 1)), True))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (a x + 1\right )}^{2}{\left (-a^{2} c x^{2} + c\right )}^{p}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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