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

Optimal. Leaf size=111 \[ -\frac{2^{\frac{n}{2}+1} (n+1) (1-a x)^{1-\frac{n}{2}} \text{Hypergeometric2F1}\left (1-\frac{n}{2},-\frac{n}{2},2-\frac{n}{2},\frac{1}{2} (1-a x)\right )}{a c (2-n) n}-\frac{(a x+1)^{\frac{n+2}{2}} (1-a x)^{-n/2}}{a c n} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.131442, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {6131, 6129, 79, 69} \[ -\frac{2^{\frac{n}{2}+1} (n+1) (1-a x)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-a x)\right )}{a c (2-n) n}-\frac{(a x+1)^{\frac{n+2}{2}} (1-a x)^{-n/2}}{a c n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6131

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 + (c*x)/d)
^p*E^(n*ArcTanh[a*x]))/x^p, x], x] /; FreeQ[{a, c, d, n}, 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 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 69

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

Rubi steps

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

Mathematica [A]  time = 0.0399761, size = 95, normalized size = 0.86 \[ \frac{(1-a x)^{-n/2} \left (-2^{\frac{n}{2}+1} (n+1) (a x-1) \text{Hypergeometric2F1}\left (1-\frac{n}{2},-\frac{n}{2},2-\frac{n}{2},\frac{1}{2} (1-a x)\right )-(n-2) (a x+1)^{\frac{n}{2}+1}\right )}{a c (n-2) n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n{\it Artanh} \left ( ax \right ) }} \left ( c-{\frac{c}{ax}} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))/(c-c/a/x),x)

[Out]

int(exp(n*arctanh(a*x))/(c-c/a/x),x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(c-c/a/x),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(c-c/a/x),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))/(c-c/a/x),x)

[Out]

a*Integral(x*exp(n*atanh(a*x))/(a*x - 1), x)/c

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/(c-c/a/x),x, algorithm="giac")

[Out]

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