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

Optimal. Leaf size=137 \[ \frac {x \left (1-a^2 x^2\right )^{-p} \left (c-\frac {c}{a^2 x^2}\right )^p \, _2F_1\left (\frac {1}{2} (1-2 p),\frac {1}{2}-p;\frac {1}{2} (3-2 p);a^2 x^2\right )}{1-2 p}+\frac {a x^2 \left (1-a^2 x^2\right )^{-p} \left (c-\frac {c}{a^2 x^2}\right )^p \, _2F_1\left (\frac {1}{2}-p,1-p;2-p;a^2 x^2\right )}{2 (1-p)} \]

[Out]

(c-c/a^2/x^2)^p*x*hypergeom([1/2-p, 1/2-p],[3/2-p],a^2*x^2)/(1-2*p)/((-a^2*x^2+1)^p)+1/2*a*(c-c/a^2/x^2)^p*x^2
*hypergeom([1-p, 1/2-p],[2-p],a^2*x^2)/(1-p)/((-a^2*x^2+1)^p)

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Rubi [A]  time = 0.13, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6160, 6148, 808, 364} \[ \frac {x \left (1-a^2 x^2\right )^{-p} \left (c-\frac {c}{a^2 x^2}\right )^p \, _2F_1\left (\frac {1}{2} (1-2 p),\frac {1}{2}-p;\frac {1}{2} (3-2 p);a^2 x^2\right )}{1-2 p}+\frac {a x^2 \left (1-a^2 x^2\right )^{-p} \left (c-\frac {c}{a^2 x^2}\right )^p \, _2F_1\left (\frac {1}{2}-p,1-p;2-p;a^2 x^2\right )}{2 (1-p)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c/(a^2*x^2))^p*x*Hypergeometric2F1[(1 - 2*p)/2, 1/2 - p, (3 - 2*p)/2, a^2*x^2])/((1 - 2*p)*(1 - a^2*x^2)
^p) + (a*(c - c/(a^2*x^2))^p*x^2*Hypergeometric2F1[1/2 - p, 1 - p, 2 - p, a^2*x^2])/(2*(1 - p)*(1 - a^2*x^2)^p
)

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 6148

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

Rule 6160

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^p \, dx &=\left (\left (c-\frac {c}{a^2 x^2}\right )^p x^{2 p} \left (1-a^2 x^2\right )^{-p}\right ) \int e^{\tanh ^{-1}(a x)} x^{-2 p} \left (1-a^2 x^2\right )^p \, dx\\ &=\left (\left (c-\frac {c}{a^2 x^2}\right )^p x^{2 p} \left (1-a^2 x^2\right )^{-p}\right ) \int x^{-2 p} (1+a x) \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=\left (\left (c-\frac {c}{a^2 x^2}\right )^p x^{2 p} \left (1-a^2 x^2\right )^{-p}\right ) \int x^{-2 p} \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx+\left (a \left (c-\frac {c}{a^2 x^2}\right )^p x^{2 p} \left (1-a^2 x^2\right )^{-p}\right ) \int x^{1-2 p} \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=\frac {\left (c-\frac {c}{a^2 x^2}\right )^p x \left (1-a^2 x^2\right )^{-p} \, _2F_1\left (\frac {1}{2} (1-2 p),\frac {1}{2}-p;\frac {1}{2} (3-2 p);a^2 x^2\right )}{1-2 p}+\frac {a \left (c-\frac {c}{a^2 x^2}\right )^p x^2 \left (1-a^2 x^2\right )^{-p} \, _2F_1\left (\frac {1}{2}-p,1-p;2-p;a^2 x^2\right )}{2 (1-p)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 112, normalized size = 0.82 \[ -\frac {x \left (1-a^2 x^2\right )^{-p} \left (c-\frac {c}{a^2 x^2}\right )^p \left (2 (p-1) \, _2F_1\left (\frac {1}{2}-p,\frac {1}{2}-p;\frac {3}{2}-p;a^2 x^2\right )+a (2 p-1) x \, _2F_1\left (\frac {1}{2}-p,1-p;2-p;a^2 x^2\right )\right )}{2 (p-1) (2 p-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((c - c/(a^2*x^2))^p*x*(2*(-1 + p)*Hypergeometric2F1[1/2 - p, 1/2 - p, 3/2 - p, a^2*x^2] + a*(-1 + 2*p)*x
*Hypergeometric2F1[1/2 - p, 1 - p, 2 - p, a^2*x^2]))/((-1 + p)*(-1 + 2*p)*(1 - a^2*x^2)^p)

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \left (\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}\right )^{p}}{a x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right ) \left (c -\frac {c}{a^{2} x^{2}}\right )^{p}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-\frac {c}{a^2\,x^2}\right )}^p\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 14.93, size = 178, normalized size = 1.30 \[ \frac {a c^{p} x^{2} \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, 1 \\ 2, p + 1 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (p + 1\right )} + \frac {c^{p} x \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} - \frac {1}{2}, 1, - p \\ \frac {1}{2}, \frac {1}{2} \end {matrix}\middle | {\frac {e^{2 i \pi }}{a^{2} x^{2}}} \right )}}{\sqrt {\pi } \Gamma \left (p + 1\right )} + \frac {c^{p} x \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, \frac {1}{2}, 1 \\ \frac {3}{2}, p + 1 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{\sqrt {\pi } \Gamma \left (p + 1\right )} - \frac {c^{p} {G_{3, 3}^{2, 2}\left (\begin {matrix} -1, p & 1 \\-1, 0 & - \frac {1}{2} \end {matrix} \middle | {\frac {e^{i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 a \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**p*x**2*gamma(p + 1/2)*hyper((1/2, 1, 1), (2, p + 1), a**2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)*gamma(p + 1
)) + c**p*x*gamma(p + 1/2)*hyper((-1/2, 1, -p), (1/2, 1/2), exp_polar(2*I*pi)/(a**2*x**2))/(sqrt(pi)*gamma(p +
1)) + c**p*x*gamma(p + 1/2)*hyper((1/2, 1/2, 1), (3/2, p + 1), a**2*x**2*exp_polar(2*I*pi))/(sqrt(pi)*gamma(p
+ 1)) - c**p*meijerg(((-1, p), (1,)), ((-1, 0), (-1/2,)), exp_polar(I*pi)/(a**2*x**2))*gamma(p + 1/2)/(2*a*ga
mma(-p)*gamma(p + 1))

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