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

Optimal. Leaf size=60 \[ \frac{x (1-a x)^{-p} F_1\left (1-p;\frac{1}{2}-p,-\frac{1}{2};2-p;a x,-a x\right ) \left (c-\frac{c}{a x}\right )^p}{1-p} \]

[Out]

((c - c/(a*x))^p*x*AppellF1[1 - p, 1/2 - p, -1/2, 2 - p, a*x, -(a*x)])/((1 - p)*(1 - a*x)^p)

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Rubi [A]  time = 0.101507, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6134, 6129, 133} \[ \frac{x (1-a x)^{-p} F_1\left (1-p;\frac{1}{2}-p,-\frac{1}{2};2-p;a x,-a x\right ) \left (c-\frac{c}{a x}\right )^p}{1-p} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*(c - c/(a*x))^p,x]

[Out]

((c - c/(a*x))^p*x*AppellF1[1 - p, 1/2 - p, -1/2, 2 - p, a*x, -(a*x)])/((1 - p)*(1 - a*x)^p)

Rule 6134

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

Rule 6129

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

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^p \, dx &=\left (\left (c-\frac{c}{a x}\right )^p x^p (1-a x)^{-p}\right ) \int e^{\tanh ^{-1}(a x)} x^{-p} (1-a x)^p \, dx\\ &=\left (\left (c-\frac{c}{a x}\right )^p x^p (1-a x)^{-p}\right ) \int x^{-p} (1-a x)^{-\frac{1}{2}+p} \sqrt{1+a x} \, dx\\ &=\frac{\left (c-\frac{c}{a x}\right )^p x (1-a x)^{-p} F_1\left (1-p;\frac{1}{2}-p,-\frac{1}{2};2-p;a x,-a x\right )}{1-p}\\ \end{align*}

Mathematica [F]  time = 0.802792, size = 0, normalized size = 0. \[ \int e^{\tanh ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.369, size = 0, normalized size = 0. \begin{align*} \int{(ax+1) \left ( c-{\frac{c}{ax}} \right ) ^{p}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}\, dx \end{align*}

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/x)^p,x)

[Out]

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

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

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/x)^p,x, algorithm="maxima")

[Out]

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

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

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/x)^p,x, algorithm="fricas")

[Out]

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

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

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/x)**p,x)

[Out]

Integral((-c*(-1 + 1/(a*x)))**p*(a*x + 1)/sqrt(-(a*x - 1)*(a*x + 1)), x)

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

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/x)^p,x, algorithm="giac")

[Out]

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