3.5.84 \(\int e^{-\tanh ^{-1}(a x)} (c-\frac {c}{a x})^p \, dx\) [484]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 = {6269, 6264, 138} \begin {gather*} \frac {x (1-a x)^{-p} F_1\left (1-p;-p-\frac {1}{2},\frac {1}{2};2-p;a x,-a x\right ) \left (c-\frac {c}{a x}\right )^p}{1-p} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))^p/E^ArcTanh[a*x],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 138

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

Rule 6264

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 6269

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]

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 \frac {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*}

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Mathematica [F]
time = 1.51, size = 0, normalized size = 0.00 \begin {gather*} \int e^{-\tanh ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.70, size = 0, normalized size = 0.00 \[\int \frac {\left (c -\frac {c}{a x}\right )^{p} \sqrt {-a^{2} x^{2}+1}}{a x +1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^p/(a*x+1)*(-a^2*x^2+1)^(1/2),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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- c \left (-1 + \frac {1}{a x}\right )\right )^{p} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{a x + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c-\frac {c}{a\,x}\right )}^p\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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