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

Optimal. Leaf size=114 \[ \frac{\left (c-\frac{c}{a x}\right )^{p+2} \text{Hypergeometric2F1}\left (1,p+2,p+3,\frac{a-\frac{1}{x}}{2 a}\right )}{2 a c^2 (p+2)}-\frac{\left (c-\frac{c}{a x}\right )^{p+2} \text{Hypergeometric2F1}\left (1,p+2,p+3,1-\frac{1}{a x}\right )}{a c^2}+\frac{x \left (c-\frac{c}{a x}\right )^{p+2}}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.152468, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {6167, 6133, 25, 514, 375, 103, 156, 65, 68} \[ \frac{\left (c-\frac{c}{a x}\right )^{p+2} \, _2F_1\left (1,p+2;p+3;\frac{a-\frac{1}{x}}{2 a}\right )}{2 a c^2 (p+2)}-\frac{\left (c-\frac{c}{a x}\right )^{p+2} \, _2F_1\left (1,p+2;p+3;1-\frac{1}{a x}\right )}{a c^2}+\frac{x \left (c-\frac{c}{a x}\right )^{p+2}}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6133

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, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[(u*(
a + b*x^n)^(m + p))/x^(n*p), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 65

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.0506705, size = 87, normalized size = 0.76 \[ \frac{(a x-1)^2 \left (c-\frac{c}{a x}\right )^p \left (\text{Hypergeometric2F1}\left (1,p+2,p+3,\frac{a-\frac{1}{x}}{2 a}\right )+2 (p+2) \left (a x-\text{Hypergeometric2F1}\left (1,p+2,p+3,1-\frac{1}{a x}\right )\right )\right )}{2 a^3 (p+2) x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - c/(a*x))^p*(-1 + a*x)^2*(Hypergeometric2F1[1, 2 + p, 3 + p, (a - x^(-1))/(2*a)] + 2*(2 + p)*(a*x - Hyper
geometric2F1[1, 2 + p, 3 + p, 1 - 1/(a*x)])))/(2*a^3*(2 + p)*x^2)

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Maple [F]  time = 0.333, size = 0, normalized size = 0. \begin{align*} \int{\frac{ax-1}{ax+1} \left ( c-{\frac{c}{ax}} \right ) ^{p}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c-c/a/x)^p/(a*x+1)*(a*x-1),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}}{a x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral((a*x - 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 )}{a x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-c*(-1 + 1/(a*x)))**p*(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}}{a x + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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