3.1341 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^3}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=270 \[ -\frac {x^2 \sqrt {1-a^2 x^2} (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1}{2} (-n-1)}}{a^2 c \sqrt {c-a^2 c x^2}}-\frac {2^{\frac {n-1}{2}} n \sqrt {1-a^2 x^2} (1-a x)^{\frac {3-n}{2}} \, _2F_1\left (\frac {3-n}{2},\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{a^4 c (3-n) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n-1}{2}} \left (-a (2 n+3) n x+n^2+2 n+2\right ) (1-a x)^{\frac {1}{2} (-n-1)}}{a^4 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6153, 6150, 100, 145, 69} \[ -\frac {2^{\frac {n-1}{2}} n \sqrt {1-a^2 x^2} (1-a x)^{\frac {3-n}{2}} \, _2F_1\left (\frac {3-n}{2},\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{a^4 c (3-n) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n-1}{2}} \left (-a (2 n+3) n x+n^2+2 n+2\right ) (1-a x)^{\frac {1}{2} (-n-1)}}{a^4 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {x^2 \sqrt {1-a^2 x^2} (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1}{2} (-n-1)}}{a^2 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 100

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 145

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)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), 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 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 6153

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

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Mathematica [A]  time = 0.29, size = 186, normalized size = 0.69 \[ \frac {\sqrt {1-a^2 x^2} (1-a x)^{-\frac {n}{2}-\frac {1}{2}} \left (-4 a^4 x^2 (a x+1)^{\frac {n-1}{2}}+\frac {a^2 2^{\frac {n+3}{2}} n (a x-1)^2 \, _2F_1\left (\frac {3}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2};\frac {5}{2}-\frac {n}{2};\frac {1}{2}-\frac {a x}{2}\right )}{n-3}+\frac {4 a^2 \left (n^2 (2 a x-1)+n (3 a x-2)-2\right ) (a x+1)^{\frac {n-1}{2}}}{n^2-1}\right )}{4 a^6 c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(-1/2 - n/2)*Sqrt[1 - a^2*x^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*Sqrt[c - a^2*c*x^2])

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} x^{3} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*x^3*((a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^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))*x^3/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )} x^{3}}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

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)^(3/2),x)

[Out]

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

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)**(3/2),x)

[Out]

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

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