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

Optimal. Leaf size=137 \[ \frac {4 c (1-a x)^{1-\frac {n}{2}} (a x+1)^{\frac {n-2}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {1-a x}{a x+1}\right )}{a (2-n)}-\frac {c 2^{\frac {n}{2}+1} (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a (2-n)} \]

[Out]

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

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Rubi [C]  time = 0.10, antiderivative size = 70, normalized size of antiderivative = 0.51, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6157, 6150, 136} \[ -\frac {c 2^{2-\frac {n}{2}} (a x+1)^{\frac {n+4}{2}} F_1\left (\frac {n+4}{2};\frac {n-2}{2},2;\frac {n+6}{2};\frac {1}{2} (a x+1),a x+1\right )}{a (n+4)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(2 - n/2)*c*(1 + a*x)^((4 + n)/2)*AppellF1[(4 + n)/2, (-2 + n)/2, 2, (6 + n)/2, (1 + a*x)/2, 1 + a*x])/(a
*(4 + 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 ) \, dx &=-\frac {c \int \frac {e^{n \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2}\\ &=-\frac {c \int \frac {(1-a x)^{1-\frac {n}{2}} (1+a x)^{1+\frac {n}{2}}}{x^2} \, dx}{a^2}\\ &=-\frac {2^{2-\frac {n}{2}} c (1+a x)^{\frac {4+n}{2}} F_1\left (\frac {4+n}{2};\frac {1}{2} (-2+n),2;\frac {6+n}{2};\frac {1}{2} (1+a x),1+a x\right )}{a (4+n)}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 126, normalized size = 0.92 \[ \frac {c e^{n \tanh ^{-1}(a x)} \left (a n x e^{2 \tanh ^{-1}(a x)} \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;e^{2 \tanh ^{-1}(a x)}\right )+a (n+2) x \, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;e^{2 \tanh ^{-1}(a x)}\right )+4 a x e^{2 \tanh ^{-1}(a x)} \, _2F_1\left (2,\frac {n}{2}+1;\frac {n}{2}+2;-e^{2 \tanh ^{-1}(a x)}\right )+n+2\right )}{a^2 (n+2) x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*E^(n*ArcTanh[a*x])*(2 + n + a*E^(2*ArcTanh[a*x])*n*x*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, E^(2*ArcTanh[a*
x])] + a*(2 + n)*x*Hypergeometric2F1[1, n/2, 1 + n/2, E^(2*ArcTanh[a*x])] + 4*a*E^(2*ArcTanh[a*x])*x*Hypergeom
etric2F1[2, 1 + n/2, 2 + n/2, -E^(2*ArcTanh[a*x])]))/(a^2*(2 + n)*x)

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fricas [F]  time = 1.37, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} - c\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \arctanh \left (a x \right )} \left (c -\frac {c}{a^{2} x^{2}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c - c/(a^2*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 \[ \int {\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}\,\left (c-\frac {c}{a^2\,x^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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