3.14.40 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{x^3 \sqrt {c-a^2 c x^2}} \, dx\) [1340]

Optimal. Leaf size=242 \[ -\frac {(1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1+n}{2}} \sqrt {1-a^2 x^2}}{2 x^2 \sqrt {c-a^2 c x^2}}-\frac {a n (1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1+n}{2}} \sqrt {1-a^2 x^2}}{2 x \sqrt {c-a^2 c x^2}}-\frac {a^2 \left (1+n^2\right ) (1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {1-a x}{1+a x}\right )}{(1-n) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6288, 6285, 105, 156, 12, 133} \begin {gather*} -\frac {a^2 \left (n^2+1\right ) \sqrt {1-a^2 x^2} (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {1-a x}{a x+1}\right )}{(1-n) \sqrt {c-a^2 c x^2}}-\frac {a n \sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{2 x \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{2 x^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((1 - a*x)^((1 - n)/2)*(1 + a*x)^((1 + n)/2)*Sqrt[1 - a^2*x^2])/(x^2*Sqrt[c - a^2*c*x^2]) - (a*n*(1 - a*x
)^((1 - n)/2)*(1 + a*x)^((1 + n)/2)*Sqrt[1 - a^2*x^2])/(2*x*Sqrt[c - a^2*c*x^2]) - (a^2*(1 + n^2)*(1 - a*x)^((
1 - n)/2)*(1 + a*x)^((-1 + n)/2)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1, (1 - n)/2, (3 - n)/2, (1 - a*x)/(1 + a
*x)])/((1 - n)*Sqrt[c - a^2*c*x^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 105

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, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

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 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 6288

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

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 134, normalized size = 0.55 \begin {gather*} \frac {(1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \left (-((-1+n) (1+a x) (1+a n x))+2 a^2 \left (1+n^2\right ) x^2 \, _2F_1\left (1,\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};\frac {1-a x}{1+a x}\right )\right )}{2 (-1+n) x^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{n \arctanh \left (a x \right )}}{x^{3} \sqrt {-a^{2} c \,x^{2}+c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))/x^3/(-a^2*c*x^2+c)^(1/2),x)

[Out]

int(exp(n*arctanh(a*x))/x^3/(-a^2*c*x^2+c)^(1/2),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))/x^3/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((-(a*x + 1)/(a*x - 1))^(1/2*n)/(sqrt(-a^2*c*x^2 + c)*x^3), 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))/x^3/(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*c*x^2 + c)*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a^2*c*x^5 - c*x^3), 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 )}}}{x^{3} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [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))/x^3/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

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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 )}}{x^3\,\sqrt {c-a^2\,c\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))/(x^3*(c - a^2*c*x^2)^(1/2)),x)

[Out]

int(exp(n*atanh(a*x))/(x^3*(c - a^2*c*x^2)^(1/2)), x)

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