3.474 \(\int e^{4 \tanh ^{-1}(a x)} (c-\frac{c}{a x})^p \, dx\)

Optimal. Leaf size=93 \[ \frac{(4-p) \left (c-\frac{c}{a x}\right )^p \text{Hypergeometric2F1}\left (1,p,p+1,1-\frac{1}{a x}\right )}{a p}-\frac{c (5-p) \left (c-\frac{c}{a x}\right )^{p-1}}{a (1-p)}+c x \left (c-\frac{c}{a x}\right )^{p-1} \]

[Out]

-((c*(5 - p)*(c - c/(a*x))^(-1 + p))/(a*(1 - p))) + c*(c - c/(a*x))^(-1 + p)*x + ((4 - p)*(c - c/(a*x))^p*Hype
rgeometric2F1[1, p, 1 + p, 1 - 1/(a*x)])/(a*p)

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Rubi [A]  time = 0.114792, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6133, 25, 514, 375, 89, 79, 65} \[ \frac{(4-p) \left (c-\frac{c}{a x}\right )^p \, _2F_1\left (1,p;p+1;1-\frac{1}{a x}\right )}{a p}-\frac{c (5-p) \left (c-\frac{c}{a x}\right )^{p-1}}{a (1-p)}+c x \left (c-\frac{c}{a x}\right )^{p-1} \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcTanh[a*x])*(c - c/(a*x))^p,x]

[Out]

-((c*(5 - p)*(c - c/(a*x))^(-1 + p))/(a*(1 - p))) + c*(c - c/(a*x))^(-1 + p)*x + ((4 - p)*(c - c/(a*x))^p*Hype
rgeometric2F1[1, p, 1 + p, 1 - 1/(a*x)])/(a*p)

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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])

Rubi steps

\begin{align*} \int e^{4 \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)^2}{(1-a x)^2} \, dx\\ &=\frac{c^2 \int \frac{\left (c-\frac{c}{a x}\right )^{-2+p} (1+a x)^2}{x^2} \, dx}{a^2}\\ &=\frac{c^2 \int \left (a+\frac{1}{x}\right )^2 \left (c-\frac{c}{a x}\right )^{-2+p} \, dx}{a^2}\\ &=-\frac{c^2 \operatorname{Subst}\left (\int \frac{(a+x)^2 \left (c-\frac{c x}{a}\right )^{-2+p}}{x^2} \, dx,x,\frac{1}{x}\right )}{a^2}\\ &=c \left (c-\frac{c}{a x}\right )^{-1+p} x-\frac{c \operatorname{Subst}\left (\int \frac{(a c (4-p)+c x) \left (c-\frac{c x}{a}\right )^{-2+p}}{x} \, dx,x,\frac{1}{x}\right )}{a^2}\\ &=-\frac{c (5-p) \left (c-\frac{c}{a x}\right )^{-1+p}}{a (1-p)}+c \left (c-\frac{c}{a x}\right )^{-1+p} x-\frac{(c (4-p)) \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{-1+p}}{x} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{c (5-p) \left (c-\frac{c}{a x}\right )^{-1+p}}{a (1-p)}+c \left (c-\frac{c}{a x}\right )^{-1+p} x+\frac{(4-p) \left (c-\frac{c}{a x}\right )^p \, _2F_1\left (1,p;1+p;1-\frac{1}{a x}\right )}{a p}\\ \end{align*}

Mathematica [A]  time = 0.0443018, size = 81, normalized size = 0.87 \[ \frac{\left (c-\frac{c}{a x}\right )^p \left (a p x (p (a x-1)-a x+5)-\left (p^2-5 p+4\right ) (a x-1) \text{Hypergeometric2F1}\left (1,p,p+1,1-\frac{1}{a x}\right )\right )}{a (p-1) p (a x-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcTanh[a*x])*(c - c/(a*x))^p,x]

[Out]

((c - c/(a*x))^p*(a*p*x*(5 - a*x + p*(-1 + a*x)) - (4 - 5*p + p^2)*(-1 + a*x)*Hypergeometric2F1[1, p, 1 + p, 1
- 1/(a*x)]))/(a*(-1 + p)*p*(-1 + a*x))

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Maple [F]  time = 0.353, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{4}}{ \left ( -{a}^{2}{x}^{2}+1 \right ) ^{2}} \left ( c-{\frac{c}{ax}} \right ) ^{p}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^p,x)

[Out]

int((a*x+1)^4/(-a^2*x^2+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 )}^{4}{\left (c - \frac{c}{a x}\right )}^{p}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^p,x, algorithm="maxima")

[Out]

integrate((a*x + 1)^4*(c - c/(a*x))^p/(a^2*x^2 - 1)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^p,x, algorithm="fricas")

[Out]

integral((a^2*x^2 + 2*a*x + 1)*((a*c*x - c)/(a*x))^p/(a^2*x^2 - 2*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 )^{2}}{\left (a x - 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**4/(-a**2*x**2+1)**2*(c-c/a/x)**p,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}^{4}{\left (c - \frac{c}{a x}\right )}^{p}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^4/(-a^2*x^2+1)^2*(c-c/a/x)^p,x, algorithm="giac")

[Out]

integrate((a*x + 1)^4*(c - c/(a*x))^p/(a^2*x^2 - 1)^2, x)