3.1327 \(\int e^{n \tanh ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=256 \[ -\frac {x^2 \sqrt {c-a^2 c x^2} (a x+1)^{\frac {n+3}{2}} (1-a x)^{\frac {3-n}{2}}}{5 a^2 \sqrt {1-a^2 x^2}}-\frac {2^{\frac {n-1}{2}} n \left (n^2+11\right ) \sqrt {c-a^2 c x^2} (1-a x)^{\frac {3-n}{2}} \, _2F_1\left (\frac {1}{2} (-n-1),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{15 a^4 (3-n) \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-a^2 c x^2} (a x+1)^{\frac {n+3}{2}} \left (3 a n x+n^2+8\right ) (1-a x)^{\frac {3-n}{2}}}{60 a^4 \sqrt {1-a^2 x^2}} \]

[Out]

-1/5*x^2*(-a*x+1)^(3/2-1/2*n)*(a*x+1)^(3/2+1/2*n)*(-a^2*c*x^2+c)^(1/2)/a^2/(-a^2*x^2+1)^(1/2)-1/60*(-a*x+1)^(3
/2-1/2*n)*(a*x+1)^(3/2+1/2*n)*(3*a*n*x+n^2+8)*(-a^2*c*x^2+c)^(1/2)/a^4/(-a^2*x^2+1)^(1/2)-1/15*2^(-1/2+1/2*n)*
n*(n^2+11)*(-a*x+1)^(3/2-1/2*n)*hypergeom([3/2-1/2*n, -1/2-1/2*n],[5/2-1/2*n],-1/2*a*x+1/2)*(-a^2*c*x^2+c)^(1/
2)/a^4/(3-n)/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 256, 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, 147, 69} \[ -\frac {2^{\frac {n-1}{2}} n \left (n^2+11\right ) \sqrt {c-a^2 c x^2} (1-a x)^{\frac {3-n}{2}} \, _2F_1\left (\frac {1}{2} (-n-1),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{15 a^4 (3-n) \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-a^2 c x^2} (a x+1)^{\frac {n+3}{2}} \left (3 a n x+n^2+8\right ) (1-a x)^{\frac {3-n}{2}}}{60 a^4 \sqrt {1-a^2 x^2}}-\frac {x^2 \sqrt {c-a^2 c x^2} (a x+1)^{\frac {n+3}{2}} (1-a x)^{\frac {3-n}{2}}}{5 a^2 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*(1 - a*x)^((3 - n)/2)*(1 + a*x)^((3 + n)/2)*Sqrt[c - a^2*c*x^2])/(5*a^2*Sqrt[1 - a^2*x^2]) - ((1 - a*x)^
((3 - n)/2)*(1 + a*x)^((3 + n)/2)*(8 + n^2 + 3*a*n*x)*Sqrt[c - a^2*c*x^2])/(60*a^4*Sqrt[1 - a^2*x^2]) - (2^((-
1 + n)/2)*n*(11 + n^2)*(1 - a*x)^((3 - n)/2)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[(-1 - n)/2, (3 - n)/2, (5 -
n)/2, (1 - a*x)/2])/(15*a^4*(3 - n)*Sqrt[1 - a^2*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 147

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

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 e^{n \tanh ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int e^{n \tanh ^{-1}(a x)} x^3 \sqrt {1-a^2 x^2} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int x^3 (1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{\frac {1}{2}+\frac {n}{2}} \, dx}{\sqrt {1-a^2 x^2}}\\ &=-\frac {x^2 (1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \sqrt {c-a^2 c x^2}}{5 a^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {c-a^2 c x^2} \int x (1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{\frac {1}{2}+\frac {n}{2}} (-2-a n x) \, dx}{5 a^2 \sqrt {1-a^2 x^2}}\\ &=-\frac {x^2 (1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \sqrt {c-a^2 c x^2}}{5 a^2 \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \left (8+n^2+3 a n x\right ) \sqrt {c-a^2 c x^2}}{60 a^4 \sqrt {1-a^2 x^2}}+\frac {\left (n \left (11+n^2\right ) \sqrt {c-a^2 c x^2}\right ) \int (1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{\frac {1}{2}+\frac {n}{2}} \, dx}{60 a^3 \sqrt {1-a^2 x^2}}\\ &=-\frac {x^2 (1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \sqrt {c-a^2 c x^2}}{5 a^2 \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{\frac {3-n}{2}} (1+a x)^{\frac {3+n}{2}} \left (8+n^2+3 a n x\right ) \sqrt {c-a^2 c x^2}}{60 a^4 \sqrt {1-a^2 x^2}}-\frac {2^{\frac {1}{2} (-1+n)} n \left (11+n^2\right ) (1-a x)^{\frac {3-n}{2}} \sqrt {c-a^2 c x^2} \, _2F_1\left (\frac {1}{2} (-1-n),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )}{15 a^4 (3-n) \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 236, normalized size = 0.92 \[ \frac {\sqrt {c-a^2 c x^2} (1-a x)^{\frac {3}{2}-\frac {n}{2}} \left (a^2 (n-3) x^2 (a x+1)^{\frac {n+3}{2}}-2^{\frac {n+7}{2}} n \, _2F_1\left (\frac {1}{2} (-n-5),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )+2^{\frac {n+7}{2}} (n-1) \, _2F_1\left (\frac {1}{2} (-n-3),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )-2^{\frac {n+3}{2}} (n-2) \, _2F_1\left (\frac {1}{2} (-n-1),\frac {3-n}{2};\frac {5-n}{2};\frac {1}{2} (1-a x)\right )\right )}{5 a^4 (3-n) \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - a*x)^(3/2 - n/2)*Sqrt[c - a^2*c*x^2]*(a^2*(-3 + n)*x^2*(1 + a*x)^((3 + n)/2) - 2^((7 + n)/2)*n*Hypergeom
etric2F1[(-5 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2] + 2^((7 + n)/2)*(-1 + n)*Hypergeometric2F1[(-3 - n)/2,
(3 - n)/2, (5 - n)/2, (1 - a*x)/2] - 2^((3 + n)/2)*(-2 + n)*Hypergeometric2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/
2, (1 - a*x)/2]))/(5*a^4*(3 - n)*Sqrt[1 - a^2*x^2])

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

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)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*x^3*((a*x + 1)/(a*x - 1))^(1/2*n), 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)^(1/2),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 = 0.26, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \arctanh \left (a x \right )} x^{3} \sqrt {-a^{2} c \,x^{2}+c}\, dx \]

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 \[ \int \sqrt {-a^{2} c x^{2} + c} x^{3} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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)^(1/2),x)

[Out]

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

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

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)**(1/2),x)

[Out]

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

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