3.1323 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=190 \[ -\frac {2 (a x+1)^{n/2} (1-a x)^{-n/2} \, _2F_1\left (1,\frac {n}{2};\frac {n+2}{2};\frac {a x+1}{1-a x}\right )}{c^2 n}-\frac {\left (-n^2-n+4\right ) (a x+1)^{\frac {n-2}{2}} (1-a x)^{1-\frac {n}{2}}}{c^2 n \left (4-n^2\right )}+\frac {(a x+1)^{\frac {n-2}{2}} (1-a x)^{-\frac {n}{2}-1}}{c^2 (n+2)}+\frac {(n+4) (a x+1)^{\frac {n-2}{2}} (1-a x)^{-n/2}}{c^2 n (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 200, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6150, 129, 155, 12, 131} \[ -\frac {2 (a x+1)^{\frac {n-2}{2}} (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1,1-\frac {n}{2};2-\frac {n}{2};\frac {1-a x}{a x+1}\right )}{c^2 (2-n)}-\frac {\left (-n^2-n+4\right ) (a x+1)^{\frac {n-2}{2}} (1-a x)^{1-\frac {n}{2}}}{c^2 n \left (4-n^2\right )}+\frac {(a x+1)^{\frac {n-2}{2}} (1-a x)^{-\frac {n}{2}-1}}{c^2 (n+2)}+\frac {(n+4) (a x+1)^{\frac {n-2}{2}} (1-a x)^{-n/2}}{c^2 n (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 129

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

Rule 131

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rule 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

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])

Rubi steps

\begin {align*} \int \frac {e^{n \tanh ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^2} \, dx &=\frac {\int \frac {(1-a x)^{-2-\frac {n}{2}} (1+a x)^{-2+\frac {n}{2}}}{x} \, dx}{c^2}\\ &=\frac {(1-a x)^{-1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 (2+n)}-\frac {\int \frac {(1-a x)^{-1-\frac {n}{2}} (1+a x)^{-2+\frac {n}{2}} \left (-a (2+n)-2 a^2 x\right )}{x} \, dx}{a c^2 (2+n)}\\ &=\frac {(1-a x)^{-1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 (2+n)}+\frac {(4+n) (1-a x)^{-n/2} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 n (2+n)}+\frac {\int \frac {(1-a x)^{-n/2} (1+a x)^{-2+\frac {n}{2}} \left (a^2 n (2+n)+a^3 (4+n) x\right )}{x} \, dx}{a^2 c^2 n (2+n)}\\ &=\frac {(1-a x)^{-1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 (2+n)}-\frac {\left (4-n-n^2\right ) (1-a x)^{1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 n \left (4-n^2\right )}+\frac {(4+n) (1-a x)^{-n/2} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 n (2+n)}+\frac {\int \frac {a^3 (2-n) n (2+n) (1-a x)^{-n/2} (1+a x)^{-1+\frac {n}{2}}}{x} \, dx}{a^3 c^2 n \left (4-n^2\right )}\\ &=\frac {(1-a x)^{-1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 (2+n)}-\frac {\left (4-n-n^2\right ) (1-a x)^{1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 n \left (4-n^2\right )}+\frac {(4+n) (1-a x)^{-n/2} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 n (2+n)}+\frac {\int \frac {(1-a x)^{-n/2} (1+a x)^{-1+\frac {n}{2}}}{x} \, dx}{c^2}\\ &=\frac {(1-a x)^{-1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 (2+n)}-\frac {\left (4-n-n^2\right ) (1-a x)^{1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 n \left (4-n^2\right )}+\frac {(4+n) (1-a x)^{-n/2} (1+a x)^{\frac {1}{2} (-2+n)}}{c^2 n (2+n)}-\frac {2 (1-a x)^{1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)} \, _2F_1\left (1,1-\frac {n}{2};2-\frac {n}{2};\frac {1-a x}{1+a x}\right )}{c^2 (2-n)}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 121, normalized size = 0.64 \[ -\frac {(1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n}{2}-1} \left (n^2 \left (a^2 x^2-a x-1\right )+a^2 n x^2-4 a^2 x^2-2 (n+2) n (a x-1)^2 \, _2F_1\left (1,1-\frac {n}{2};2-\frac {n}{2};\frac {1-a x}{a x+1}\right )+n+4\right )}{c^2 n \left (n^2-4\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{4} c^{2} x^{5} - 2 \, a^{2} c^{2} x^{3} + c^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*c^2*x^5 - 2*a^2*c^2*x^3 + c^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )}}{x \left (-a^{2} c \,x^{2}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{4} x^{5} - 2 a^{2} x^{3} + x}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(n*atanh(a*x))/(a**4*x**5 - 2*a**2*x**3 + x), x)/c**2

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