3.795 \(\int e^{n \tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{3/2} \, dx\)

Optimal. Leaf size=430 \[ -\frac {a^2 \left (3-n^2\right ) x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac {n-3}{2}} (1-a x)^{\frac {3-n}{2}} \, _2F_1\left (1,\frac {n-3}{2};\frac {n-1}{2};\frac {a x+1}{1-a x}\right )}{(3-n) \left (1-a^2 x^2\right )^{3/2}}+\frac {a^2 2^{\frac {n-1}{2}} n x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (1-a x)^{\frac {5-n}{2}} \, _2F_1\left (\frac {3-n}{2},\frac {5-n}{2};\frac {7-n}{2};\frac {1}{2} (1-a x)\right )}{(3-n) (5-n) \left (1-a^2 x^2\right )^{3/2}}-\frac {a (n+4) x^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac {n-3}{2}} (1-a x)^{\frac {5-n}{2}}}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac {x \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac {n-3}{2}} (1-a x)^{\frac {5-n}{2}}}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 a^2 x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac {n-3}{2}} (1-a x)^{\frac {5-n}{2}}}{(3-n) \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

-1/2*(c-c/a^2/x^2)^(3/2)*x*(-a*x+1)^(5/2-1/2*n)*(a*x+1)^(-3/2+1/2*n)/(-a^2*x^2+1)^(3/2)-1/2*a*(4+n)*(c-c/a^2/x
^2)^(3/2)*x^2*(-a*x+1)^(5/2-1/2*n)*(a*x+1)^(-3/2+1/2*n)/(-a^2*x^2+1)^(3/2)-3*a^2*(c-c/a^2/x^2)^(3/2)*x^3*(-a*x
+1)^(5/2-1/2*n)*(a*x+1)^(-3/2+1/2*n)/(3-n)/(-a^2*x^2+1)^(3/2)-a^2*(-n^2+3)*(c-c/a^2/x^2)^(3/2)*x^3*(-a*x+1)^(3
/2-1/2*n)*(a*x+1)^(-3/2+1/2*n)*hypergeom([1, -3/2+1/2*n],[-1/2+1/2*n],(a*x+1)/(-a*x+1))/(3-n)/(-a^2*x^2+1)^(3/
2)+2^(-1/2+1/2*n)*a^2*n*(c-c/a^2/x^2)^(3/2)*x^3*(-a*x+1)^(5/2-1/2*n)*hypergeom([5/2-1/2*n, 3/2-1/2*n],[7/2-1/2
*n],-1/2*a*x+1/2)/(3-n)/(5-n)/(-a^2*x^2+1)^(3/2)

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Rubi [C]  time = 0.21, antiderivative size = 103, normalized size of antiderivative = 0.24, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6160, 6150, 136} \[ -\frac {a^2 2^{\frac {5}{2}-\frac {n}{2}} x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (a x+1)^{\frac {n+5}{2}} F_1\left (\frac {n+5}{2};\frac {n-3}{2},3;\frac {n+7}{2};\frac {1}{2} (a x+1),a x+1\right )}{(n+5) \left (1-a^2 x^2\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Int[E^(n*ArcTanh[a*x])*(c - c/(a^2*x^2))^(3/2),x]

[Out]

-((2^(5/2 - n/2)*a^2*(c - c/(a^2*x^2))^(3/2)*x^3*(1 + a*x)^((5 + n)/2)*AppellF1[(5 + n)/2, (-3 + n)/2, 3, (7 +
n)/2, (1 + a*x)/2, 1 + a*x])/((5 + n)*(1 - a^2*x^2)^(3/2)))

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 6160

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

Rubi steps

\begin {align*} \int e^{n \tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} \, dx &=\frac {\left (\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3\right ) \int \frac {e^{n \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^{3/2}}{x^3} \, dx}{\left (1-a^2 x^2\right )^{3/2}}\\ &=\frac {\left (\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3\right ) \int \frac {(1-a x)^{\frac {3}{2}-\frac {n}{2}} (1+a x)^{\frac {3}{2}+\frac {n}{2}}}{x^3} \, dx}{\left (1-a^2 x^2\right )^{3/2}}\\ &=-\frac {2^{\frac {5}{2}-\frac {n}{2}} a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3 (1+a x)^{\frac {5+n}{2}} F_1\left (\frac {5+n}{2};\frac {1}{2} (-3+n),3;\frac {7+n}{2};\frac {1}{2} (1+a x),1+a x\right )}{(5+n) \left (1-a^2 x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.69, size = 190, normalized size = 0.44 \[ \frac {c x \sqrt {c-\frac {c}{a^2 x^2}} e^{n \tanh ^{-1}(a x)} \text {csch}\left (\frac {1}{2} \tanh ^{-1}(a x)\right ) \text {sech}\left (\frac {1}{2} \tanh ^{-1}(a x)\right ) \left (-(n+1) \text {csch}\left (\frac {1}{2} \tanh ^{-1}(a x)\right ) \text {sech}\left (\frac {1}{2} \tanh ^{-1}(a x)\right ) \left (\left (1-a^2 x^2\right ) \cosh \left (2 \tanh ^{-1}(a x)\right )+a x (a x+n)\right )-4 a \left (n^2-3\right ) x e^{\tanh ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};e^{2 \tanh ^{-1}(a x)}\right )+8 a n x e^{\tanh ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-e^{2 \tanh ^{-1}(a x)}\right )\right )}{8 (n+1) \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a*x])*(c - c/(a^2*x^2))^(3/2),x]

[Out]

(c*E^(n*ArcTanh[a*x])*Sqrt[c - c/(a^2*x^2)]*x*Csch[ArcTanh[a*x]/2]*Sech[ArcTanh[a*x]/2]*(8*a*E^ArcTanh[a*x]*n*
x*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -E^(2*ArcTanh[a*x])] - 4*a*E^ArcTanh[a*x]*(-3 + n^2)*x*Hypergeome
tric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*ArcTanh[a*x])] - (1 + n)*(a*x*(n + a*x) + (1 - a^2*x^2)*Cosh[2*ArcTanh[a
*x]])*Csch[ArcTanh[a*x]/2]*Sech[ArcTanh[a*x]/2]))/(8*(1 + n)*(-1 + a^2*x^2))

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} - c\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*(c-c/a^2/x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*(c-c/a^2/x^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \arctanh \left (a x \right )} \left (c -\frac {c}{a^{2} x^{2}}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))*(c-c/a^2/x^2)^(3/2),x)

[Out]

int(exp(n*arctanh(a*x))*(c-c/a^2/x^2)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {3}{2}} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*(c-c/a^2/x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((c - c/(a^2*x^2))^(3/2)*((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 \[ \int {\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}\,{\left (c-\frac {c}{a^2\,x^2}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))*(c - c/(a^2*x^2))^(3/2),x)

[Out]

int(exp(n*atanh(a*x))*(c - c/(a^2*x^2))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))*(c-c/a**2/x**2)**(3/2),x)

[Out]

Timed out

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