3.7.22 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx\) [622]

Optimal. Leaf size=111 \[ -\frac {(1-a x)^{-n/2} (1+a x)^{\frac {2+n}{2}}}{a c n}-\frac {2^{1+\frac {n}{2}} (1+n) (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a c (2-n) n} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 = {6266, 6264, 80, 71} \begin {gather*} -\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} \end {gather*}

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, -1/2*n, 2 - n/2, (1 - a*x)/2])/(a*c*(2 - n)*n)

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 80

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 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 6266

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]

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*}

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Mathematica [A]
time = 0.03, size = 95, normalized size = 0.86 \begin {gather*} \frac {(1-a x)^{-n/2} \left (-\left ((-2+n) (1+a x)^{1+\frac {n}{2}}\right )-2^{1+\frac {n}{2}} (1+n) (-1+a x) \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )\right )}{a c (-2+n) n} \end {gather*}

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{n \arctanh \left (a x \right )}}{c -\frac {c}{a x}}\, dx\]

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)

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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(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)

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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(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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {a \int \frac {x e^{n \operatorname {atanh}{\left (a x \right )}}}{a x - 1}\, dx}{c} \end {gather*}

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

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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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{c-\frac {c}{a\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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