3.8.95 \(\int e^{n \tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{3/2} \, dx\) [795]
Optimal. Leaf size=430 \[ -\frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x (1-a x)^{\frac {5-n}{2}} (1+a x)^{\frac {1}{2} (-3+n)}}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac {a (4+n) \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^2 (1-a x)^{\frac {5-n}{2}} (1+a x)^{\frac {1}{2} (-3+n)}}{2 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3 (1-a x)^{\frac {5-n}{2}} (1+a x)^{\frac {1}{2} (-3+n)}}{(3-n) \left (1-a^2 x^2\right )^{3/2}}-\frac {a^2 \left (3-n^2\right ) \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3 (1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {1}{2} (-3+n)} \, _2F_1\left (1,\frac {1}{2} (-3+n);\frac {1}{2} (-1+n);\frac {1+a x}{1-a x}\right )}{(3-n) \left (1-a^2 x^2\right )^{3/2}}+\frac {2^{\frac {1}{2} (-1+n)} a^2 n \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3 (1-a x)^{\frac {5-n}{2}} \, _2F_1\left (\frac {3-n}{2},\frac {5-n}{2};\frac {7-n}{2};\frac {1}{2} (1-a x)\right )}{(3-n) (5-n) \left (1-a^2 x^2\right )^{3/2}} \]
[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 [A]
time = 0.35, antiderivative size = 430, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6295, 6285,
131, 80, 71, 1627, 156, 12, 133} \begin {gather*} -\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}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[E^(n*ArcTanh[a*x])*(c - c/(a^2*x^2))^(3/2),x]
[Out]
-1/2*((c - c/(a^2*x^2))^(3/2)*x*(1 - a*x)^((5 - n)/2)*(1 + a*x)^((-3 + n)/2))/(1 - a^2*x^2)^(3/2) - (a*(4 + n)
*(c - c/(a^2*x^2))^(3/2)*x^2*(1 - a*x)^((5 - n)/2)*(1 + a*x)^((-3 + n)/2))/(2*(1 - a^2*x^2)^(3/2)) - (3*a^2*(c
- c/(a^2*x^2))^(3/2)*x^3*(1 - a*x)^((5 - n)/2)*(1 + a*x)^((-3 + n)/2))/((3 - n)*(1 - a^2*x^2)^(3/2)) - (a^2*(
3 - n^2)*(c - c/(a^2*x^2))^(3/2)*x^3*(1 - a*x)^((3 - n)/2)*(1 + a*x)^((-3 + n)/2)*Hypergeometric2F1[1, (-3 + n
)/2, (-1 + n)/2, (1 + a*x)/(1 - a*x)])/((3 - n)*(1 - a^2*x^2)^(3/2)) + (2^((-1 + n)/2)*a^2*n*(c - c/(a^2*x^2))
^(3/2)*x^3*(1 - a*x)^((5 - n)/2)*Hypergeometric2F1[(3 - n)/2, (5 - n)/2, (7 - n)/2, (1 - a*x)/2])/((3 - n)*(5
- n)*(1 - a^2*x^2)^(3/2))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 80
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c
, d, e, f, n, p}, x] && !RationalQ[p] && SumSimplerQ[p, 1]
Rule 131
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[f^(p -
1)/d^p, Int[(a + b*x)^m*((d*e*p - c*f*(p - 1) + d*f*x)/(c + d*x)^(m + 1)), x], x] + Dist[f^(p - 1), Int[(a + b
*x)^m*((e + f*x)^p/(c + d*x)^(m + 1))*ExpandToSum[f^(-p + 1)*(c + d*x)^(-p + 1) - (d*e*p - c*f*(p - 1) + d*f*x
)/(d^p*(e + f*x)^p), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[m + n + p, 0] && ILtQ[p, 0] && (L
tQ[m, 0] || SumSimplerQ[m, 1] || !(LtQ[n, 0] || SumSimplerQ[n, 1]))
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 156
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, -1]
Rule 1627
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> With[{
Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[b*R*(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*ExpandToSum[(m + 1)*(b*c - a*d)*(b*e - a*f)*Qx + a*d*f
*R*(m + 1) - b*R*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*R*(m + n + p + 3)*x, x], x], x]] /; FreeQ[{a, b,
c, d, e, f, n, p}, x] && PolyQ[Px, x] && ILtQ[m, -1]
Rule 6285
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 6295
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/x^(2*p))*(1 + c*(x^2/d))^p*E^(n*ArcTanh[a*x]), 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 = 1.18, size = 190, normalized size = 0.44 \begin {gather*} \frac {c e^{n \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \text {csch}\left (\frac {1}{2} \tanh ^{-1}(a x)\right ) \text {sech}\left (\frac {1}{2} \tanh ^{-1}(a x)\right ) \left (8 a e^{\tanh ^{-1}(a x)} n x \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-e^{2 \tanh ^{-1}(a x)}\right )-4 a e^{\tanh ^{-1}(a x)} \left (-3+n^2\right ) x \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};e^{2 \tanh ^{-1}(a x)}\right )-(1+n) \left (a x (n+a x)+\left (1-a^2 x^2\right ) \cosh \left (2 \tanh ^{-1}(a x)\right )\right ) \text {csch}\left (\frac {1}{2} \tanh ^{-1}(a x)\right ) \text {sech}\left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right )}{8 (1+n) \left (-1+a^2 x^2\right )} \end {gather*}
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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Maple [F]
time = 0.01, 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 \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^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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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^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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}\,{\left (c-\frac {c}{a^2\,x^2}\right )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(exp(n*atanh(a*x))*(c - c/(a^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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