3.791 \(\int e^{n \tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^2 \, dx\)

Optimal. Leaf size=331 \[ -\frac {c^2 (a x+1)^{\frac {n-4}{2}} (1-a x)^{3-\frac {n}{2}}}{3 a^4 x^3}-\frac {c^2 (n+10) (a x+1)^{\frac {n-4}{2}} (1-a x)^{3-\frac {n}{2}}}{6 a^3 x^2}-\frac {c^2 \left (n^2+5 n+14\right ) (a x+1)^{\frac {n-4}{2}} (1-a x)^{3-\frac {n}{2}}}{6 a^2 x}-\frac {c^2 n \left (10-n^2\right ) (a x+1)^{\frac {n-4}{2}} (1-a x)^{2-\frac {n}{2}} \, _2F_1\left (1,\frac {n-4}{2};\frac {n-2}{2};\frac {a x+1}{1-a x}\right )}{3 a (4-n)}+\frac {c^2 2^{\frac {n}{2}-1} n (1-a x)^{3-\frac {n}{2}} \, _2F_1\left (\frac {4-n}{2},3-\frac {n}{2};4-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a \left (n^2-10 n+24\right )}-\frac {4 c^2 (a x+1)^{\frac {n-4}{2}} (1-a x)^{3-\frac {n}{2}}}{a (4-n)} \]

[Out]

-4*c^2*(-a*x+1)^(3-1/2*n)*(a*x+1)^(-2+1/2*n)/a/(4-n)-1/3*c^2*(-a*x+1)^(3-1/2*n)*(a*x+1)^(-2+1/2*n)/a^4/x^3-1/6
*c^2*(10+n)*(-a*x+1)^(3-1/2*n)*(a*x+1)^(-2+1/2*n)/a^3/x^2-1/6*c^2*(n^2+5*n+14)*(-a*x+1)^(3-1/2*n)*(a*x+1)^(-2+
1/2*n)/a^2/x-1/3*c^2*n*(-n^2+10)*(-a*x+1)^(2-1/2*n)*(a*x+1)^(-2+1/2*n)*hypergeom([1, -2+1/2*n],[-1+1/2*n],(a*x
+1)/(-a*x+1))/a/(4-n)+2^(-1+1/2*n)*c^2*n*(-a*x+1)^(3-1/2*n)*hypergeom([3-1/2*n, 2-1/2*n],[4-1/2*n],-1/2*a*x+1/
2)/a/(n^2-10*n+24)

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Rubi [C]  time = 0.13, antiderivative size = 71, normalized size of antiderivative = 0.21, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6157, 6150, 136} \[ \frac {c^2 2^{3-\frac {n}{2}} (a x+1)^{\frac {n+6}{2}} F_1\left (\frac {n+6}{2};\frac {n-4}{2},4;\frac {n+8}{2};\frac {1}{2} (a x+1),a x+1\right )}{a (n+6)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(3 - n/2)*c^2*(1 + a*x)^((6 + n)/2)*AppellF1[(6 + n)/2, (-4 + n)/2, 4, (8 + n)/2, (1 + a*x)/2, 1 + a*x])/(a
*(6 + n))

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int e^{n \tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^2 \, dx &=\frac {c^2 \int \frac {e^{n \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^2}{x^4} \, dx}{a^4}\\ &=\frac {c^2 \int \frac {(1-a x)^{2-\frac {n}{2}} (1+a x)^{2+\frac {n}{2}}}{x^4} \, dx}{a^4}\\ &=\frac {2^{3-\frac {n}{2}} c^2 (1+a x)^{\frac {6+n}{2}} F_1\left (\frac {6+n}{2};\frac {1}{2} (-4+n),4;\frac {8+n}{2};\frac {1}{2} (1+a x),1+a x\right )}{a (6+n)}\\ \end {align*}

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Mathematica [A]  time = 0.85, size = 229, normalized size = 0.69 \[ -\frac {c^2 e^{n \tanh ^{-1}(a x)} \left (a^3 \left (n^2-10\right ) n x^3 e^{2 \tanh ^{-1}(a x)} \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;e^{2 \tanh ^{-1}(a x)}\right )+a^3 \left (n^3+2 n^2-10 n-20\right ) x^3 \, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;e^{2 \tanh ^{-1}(a x)}\right )-24 a^3 x^3 e^{2 \tanh ^{-1}(a x)} \, _2F_1\left (2,\frac {n}{2}+1;\frac {n}{2}+2;-e^{2 \tanh ^{-1}(a x)}\right )-a^3 n^2 x^3-2 a^3 n x^3+a^2 n^3 x^2+2 a^2 n^2 x^2-12 a^2 n x^2-24 a^2 x^2+a n^2 x+2 a n x+2 n+4\right )}{6 a^4 (n+2) x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*(c^2*E^(n*ArcTanh[a*x])*(4 + 2*n + 2*a*n*x + a*n^2*x - 24*a^2*x^2 - 12*a^2*n*x^2 + 2*a^2*n^2*x^2 + a^2*n^
3*x^2 - 2*a^3*n*x^3 - a^3*n^2*x^3 + a^3*E^(2*ArcTanh[a*x])*n*(-10 + n^2)*x^3*Hypergeometric2F1[1, 1 + n/2, 2 +
n/2, E^(2*ArcTanh[a*x])] + a^3*(-20 - 10*n + 2*n^2 + n^3)*x^3*Hypergeometric2F1[1, n/2, 1 + n/2, E^(2*ArcTanh
[a*x])] - 24*a^3*E^(2*ArcTanh[a*x])*x^3*Hypergeometric2F1[2, 1 + n/2, 2 + n/2, -E^(2*ArcTanh[a*x])]))/(a^4*(2
+ n)*x^3)

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \arctanh \left (a x \right )} \left (c -\frac {c}{a^{2} x^{2}}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {c^{2} \left (\int a^{4} e^{n \operatorname {atanh}{\left (a x \right )}}\, dx + \int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{x^{4}}\, dx + \int \left (- \frac {2 a^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{x^{2}}\right )\, dx\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*(Integral(a**4*exp(n*atanh(a*x)), x) + Integral(exp(n*atanh(a*x))/x**4, x) + Integral(-2*a**2*exp(n*atanh
(a*x))/x**2, x))/a**4

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