3.7.20 \(\int e^{n \tanh ^{-1}(a x)} (c-\frac {c}{a x})^2 \, dx\) [620]

Optimal. Leaf size=130 \[ \frac {4 c^2 (1-a x)^{-n/2} (1+a x)^{n/2} \, _2F_1\left (2,\frac {n}{2};\frac {2+n}{2};\frac {1+a x}{1-a x}\right )}{a n}+\frac {2^{n/2} c^2 (1-a x)^{2-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},2-\frac {n}{2};3-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a (4-n)} \]

[Out]

4*c^2*(a*x+1)^(1/2*n)*hypergeom([2, 1/2*n],[1+1/2*n],(a*x+1)/(-a*x+1))/a/n/((-a*x+1)^(1/2*n))+2^(1/2*n)*c^2*(-
a*x+1)^(2-1/2*n)*hypergeom([1-1/2*n, 2-1/2*n],[3-1/2*n],-1/2*a*x+1/2)/a/(4-n)

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Rubi [A]
time = 0.11, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6266, 6264, 130, 71, 133} \begin {gather*} \frac {c^2 2^{n/2} (1-a x)^{2-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},2-\frac {n}{2};3-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a (4-n)}+\frac {4 c^2 (a x+1)^{n/2} (1-a x)^{-n/2} \, _2F_1\left (2,\frac {n}{2};\frac {n+2}{2};\frac {a x+1}{1-a x}\right )}{a n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a*x])*(c - c/(a*x))^2,x]

[Out]

(4*c^2*(1 + a*x)^(n/2)*Hypergeometric2F1[2, n/2, (2 + n)/2, (1 + a*x)/(1 - a*x)])/(a*n*(1 - a*x)^(n/2)) + (2^(
n/2)*c^2*(1 - a*x)^(2 - n/2)*Hypergeometric2F1[1 - n/2, 2 - n/2, 3 - n/2, (1 - a*x)/2])/(a*(4 - 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 130

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_))^2, x_Symbol] :> Dist[b*(d/f^2),
Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x] + Dist[(b*e - a*f)*((d*e - c*f)/f^2), Int[(a + b*x)^(m - 1)*(
(c + d*x)^(n - 1)/(e + f*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[m + n, 0] && EqQ[2*b*d*e
- f*(b*c + a*d), 0]

Rule 133

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*c - a
*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 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 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 e^{n \tanh ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^2 \, dx &=\frac {c^2 \int \frac {e^{n \tanh ^{-1}(a x)} (1-a x)^2}{x^2} \, dx}{a^2}\\ &=\frac {c^2 \int \frac {(1-a x)^{2-\frac {n}{2}} (1+a x)^{n/2}}{x^2} \, dx}{a^2}\\ &=\frac {2^{3-\frac {n}{2}} c^2 (1+a x)^{\frac {2+n}{2}} F_1\left (\frac {2+n}{2};\frac {1}{2} (-4+n),2;\frac {4+n}{2};\frac {1}{2} (1+a x),1+a x\right )}{a (2+n)}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(262\) vs. \(2(130)=260\).
time = 0.31, size = 262, normalized size = 2.02 \begin {gather*} -\frac {c^2 e^{n \tanh ^{-1}(a x)} \left (2 n+n^2-2 a e^{2 \tanh ^{-1}(a x)} n x \, _2F_1\left (1,1+\frac {n}{2};2+\frac {n}{2};-e^{2 \tanh ^{-1}(a x)}\right )+a e^{2 \tanh ^{-1}(a x)} (-2+n) n x \, _2F_1\left (1,1+\frac {n}{2};2+\frac {n}{2};e^{2 \tanh ^{-1}(a x)}\right )+4 a x \, _2F_1\left (1,\frac {n}{2};1+\frac {n}{2};-e^{2 \tanh ^{-1}(a x)}\right )+2 a n x \, _2F_1\left (1,\frac {n}{2};1+\frac {n}{2};-e^{2 \tanh ^{-1}(a x)}\right )-4 a x \, _2F_1\left (1,\frac {n}{2};1+\frac {n}{2};e^{2 \tanh ^{-1}(a x)}\right )+a n^2 x \, _2F_1\left (1,\frac {n}{2};1+\frac {n}{2};e^{2 \tanh ^{-1}(a x)}\right )-4 a e^{2 \tanh ^{-1}(a x)} n x \, _2F_1\left (2,1+\frac {n}{2};2+\frac {n}{2};-e^{2 \tanh ^{-1}(a x)}\right )\right )}{a^2 n (2+n) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a*x])*(c - c/(a*x))^2,x]

[Out]

-((c^2*E^(n*ArcTanh[a*x])*(2*n + n^2 - 2*a*E^(2*ArcTanh[a*x])*n*x*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, -E^(2
*ArcTanh[a*x])] + a*E^(2*ArcTanh[a*x])*(-2 + n)*n*x*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, E^(2*ArcTanh[a*x])]
+ 4*a*x*Hypergeometric2F1[1, n/2, 1 + n/2, -E^(2*ArcTanh[a*x])] + 2*a*n*x*Hypergeometric2F1[1, n/2, 1 + n/2,
-E^(2*ArcTanh[a*x])] - 4*a*x*Hypergeometric2F1[1, n/2, 1 + n/2, E^(2*ArcTanh[a*x])] + a*n^2*x*Hypergeometric2F
1[1, n/2, 1 + n/2, E^(2*ArcTanh[a*x])] - 4*a*E^(2*ArcTanh[a*x])*n*x*Hypergeometric2F1[2, 1 + n/2, 2 + n/2, -E^
(2*ArcTanh[a*x])]))/(a^2*n*(2 + n)*x))

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{n \arctanh \left (a x \right )} \left (c -\frac {c}{a x}\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))*(c-c/a/x)^2,x)

[Out]

int(exp(n*arctanh(a*x))*(c-c/a/x)^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(exp(n*arctanh(a*x))*(c-c/a/x)^2,x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {c^{2} \left (\int a^{2} e^{n \operatorname {atanh}{\left (a x \right )}}\, dx + \int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{x^{2}}\, dx + \int \left (- \frac {2 a e^{n \operatorname {atanh}{\left (a x \right )}}}{x}\right )\, dx\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))*(c-c/a/x)**2,x)

[Out]

c**2*(Integral(a**2*exp(n*atanh(a*x)), x) + Integral(exp(n*atanh(a*x))/x**2, x) + Integral(-2*a*exp(n*atanh(a*
x))/x, x))/a**2

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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)^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}\,{\left (c-\frac {c}{a\,x}\right )}^2 \,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))^2,x)

[Out]

int(exp(n*atanh(a*x))*(c - c/(a*x))^2, x)

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