3.8.98 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^{3/2}} \, dx\) [798]

Optimal. Leaf size=267 \[ -\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (1-a^2 x^2\right )^{3/2}}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x}+\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (2+2 n+n^2-a n (3+2 n) x\right ) \left (1-a^2 x^2\right )^{3/2}}{a^4 \left (1-n^2\right ) \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}-\frac {2^{\frac {1}{2} (-1+n)} n (1-a x)^{\frac {3-n}{2}} \left (1-a^2 x^2\right )^{3/2} \, _2F_1\left (\frac {3-n}{2},\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{a^4 (3-n) \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6295, 6285, 102, 150, 71} \begin {gather*} -\frac {\left (1-a^2 x^2\right )^{3/2} (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1}{2} (-n-1)}}{a^2 x \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}-\frac {2^{\frac {n-1}{2}} n \left (1-a^2 x^2\right )^{3/2} (1-a x)^{\frac {3-n}{2}} \, _2F_1\left (\frac {3-n}{2},\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{a^4 (3-n) x^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}+\frac {\left (1-a^2 x^2\right )^{3/2} (a x+1)^{\frac {n-1}{2}} \left (-a (2 n+3) n x+n^2+2 n+2\right ) (1-a x)^{\frac {1}{2} (-n-1)}}{a^4 \left (1-n^2\right ) x^3 \left (c-\frac {c}{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 - a*x)^((-1 - n)/2)*(1 + a*x)^((-1 + n)/2)*(1 - a^2*x^2)^(3/2))/(a^2*(c - c/(a^2*x^2))^(3/2)*x)) + ((1 -
a*x)^((-1 - n)/2)*(1 + a*x)^((-1 + n)/2)*(2 + 2*n + n^2 - a*n*(3 + 2*n)*x)*(1 - a^2*x^2)^(3/2))/(a^4*(1 - n^2
)*(c - c/(a^2*x^2))^(3/2)*x^3) - (2^((-1 + n)/2)*n*(1 - a*x)^((3 - n)/2)*(1 - a^2*x^2)^(3/2)*Hypergeometric2F1
[(3 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2])/(a^4*(3 - n)*(c - c/(a^2*x^2))^(3/2)*x^3)

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 102

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)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(
n + 1), x] + Dist[f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

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 \frac {e^{n \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2}} \, dx &=\frac {\left (1-a^2 x^2\right )^{3/2} \int \frac {e^{n \tanh ^{-1}(a x)} x^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}\\ &=\frac {\left (1-a^2 x^2\right )^{3/2} \int x^3 (1-a x)^{-\frac {3}{2}-\frac {n}{2}} (1+a x)^{-\frac {3}{2}+\frac {n}{2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (1-a^2 x^2\right )^{3/2}}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x}-\frac {\left (1-a^2 x^2\right )^{3/2} \int x (1-a x)^{-\frac {3}{2}-\frac {n}{2}} (1+a x)^{-\frac {3}{2}+\frac {n}{2}} (-2-a n x) \, dx}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (1-a^2 x^2\right )^{3/2}}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x}+\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (2+2 n+n^2-a n (3+2 n) x\right ) \left (1-a^2 x^2\right )^{3/2}}{a^4 \left (1-n^2\right ) \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}+\frac {\left (n \left (1-a^2 x^2\right )^{3/2}\right ) \int (1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{-\frac {3}{2}+\frac {n}{2}} \, dx}{a^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (1-a^2 x^2\right )^{3/2}}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x}+\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (2+2 n+n^2-a n (3+2 n) x\right ) \left (1-a^2 x^2\right )^{3/2}}{a^4 \left (1-n^2\right ) \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}-\frac {2^{\frac {1}{2} (-1+n)} n (1-a x)^{\frac {3-n}{2}} \left (1-a^2 x^2\right )^{3/2} \, _2F_1\left (\frac {3-n}{2},\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{a^4 (3-n) \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 186, normalized size = 0.70 \begin {gather*} \frac {(1-a x)^{\frac {1}{2} (-1-n)} \left (1-a^2 x^2\right )^{3/2} \left (-4 a^4 x^2 (1+a x)^{\frac {1}{2} (-1+n)}+\frac {4 a^2 (1+a x)^{\frac {1}{2} (-1+n)} \left (-2+n^2 (-1+2 a x)+n (-2+3 a x)\right )}{-1+n^2}+\frac {2^{\frac {3+n}{2}} a^2 n (-1+a x)^2 \, _2F_1\left (\frac {3}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2};\frac {5}{2}-\frac {n}{2};\frac {1}{2}-\frac {a x}{2}\right )}{-3+n}\right )}{4 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^((-1 - n)/2)*(1 - a^2*x^2)^(3/2)*(-4*a^4*x^2*(1 + a*x)^((-1 + n)/2) + (4*a^2*(1 + a*x)^((-1 + n)/2)
*(-2 + n^2*(-1 + 2*a*x) + n*(-2 + 3*a*x)))/(-1 + n^2) + (2^((3 + n)/2)*a^2*n*(-1 + a*x)^2*Hypergeometric2F1[3/
2 - n/2, 3/2 - n/2, 5/2 - n/2, 1/2 - (a*x)/2])/(-3 + n)))/(4*a^6*(c - c/(a^2*x^2))^(3/2)*x^3)

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

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {3}{2}}}\, dx \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]

Integral(exp(n*atanh(a*x))/(-c*(-1 + 1/(a*x))*(1 + 1/(a*x)))**(3/2), x)

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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 \frac {{\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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