3.1311 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^4}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=209 \[ \frac {2^{\frac {n}{2}-1} n \left (n^2+8\right ) (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},1-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{3 a^5 c (2-n)}+\frac {(a x+1)^{n/2} \left (-a \left (n^2+6\right ) n x+n^3+n^2+8 n+6\right ) (1-a x)^{-n/2}}{6 a^5 c n}-\frac {n x^2 (a x+1)^{n/2} (1-a x)^{-n/2}}{6 a^3 c}-\frac {x^3 (a x+1)^{n/2} (1-a x)^{-n/2}}{3 a^2 c} \]

[Out]

-1/6*n*x^2*(a*x+1)^(1/2*n)/a^3/c/((-a*x+1)^(1/2*n))-1/3*x^3*(a*x+1)^(1/2*n)/a^2/c/((-a*x+1)^(1/2*n))+1/6*(a*x+
1)^(1/2*n)*(6+8*n+n^2+n^3-a*n*(n^2+6)*x)/a^5/c/n/((-a*x+1)^(1/2*n))+1/3*2^(-1+1/2*n)*n*(n^2+8)*(-a*x+1)^(1-1/2
*n)*hypergeom([1-1/2*n, 1-1/2*n],[2-1/2*n],-1/2*a*x+1/2)/a^5/c/(2-n)

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Rubi [A]  time = 0.26, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6150, 100, 153, 143, 69} \[ \frac {2^{\frac {n}{2}-1} n \left (n^2+8\right ) (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},1-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{3 a^5 c (2-n)}+\frac {(a x+1)^{n/2} \left (-a \left (n^2+6\right ) n x+n^3+n^2+8 n+6\right ) (1-a x)^{-n/2}}{6 a^5 c n}-\frac {x^3 (a x+1)^{n/2} (1-a x)^{-n/2}}{3 a^2 c}-\frac {n x^2 (a x+1)^{n/2} (1-a x)^{-n/2}}{6 a^3 c} \]

Antiderivative was successfully verified.

[In]

Int[(E^(n*ArcTanh[a*x])*x^4)/(c - a^2*c*x^2),x]

[Out]

-(n*x^2*(1 + a*x)^(n/2))/(6*a^3*c*(1 - a*x)^(n/2)) - (x^3*(1 + a*x)^(n/2))/(3*a^2*c*(1 - a*x)^(n/2)) + ((1 + a
*x)^(n/2)*(6 + 8*n + n^2 + n^3 - a*n*(6 + n^2)*x))/(6*a^5*c*n*(1 - a*x)^(n/2)) + (2^(-1 + n/2)*n*(8 + n^2)*(1
- a*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2, 1 - n/2, 2 - n/2, (1 - a*x)/2])/(3*a^5*c*(2 - n))

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 143

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

Rule 153

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

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])

Rubi steps

\begin {align*} \int \frac {e^{n \tanh ^{-1}(a x)} x^4}{c-a^2 c x^2} \, dx &=\frac {\int x^4 (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} \, dx}{c}\\ &=-\frac {x^3 (1-a x)^{-n/2} (1+a x)^{n/2}}{3 a^2 c}-\frac {\int x^2 (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} (-3-a n x) \, dx}{3 a^2 c}\\ &=-\frac {n x^2 (1-a x)^{-n/2} (1+a x)^{n/2}}{6 a^3 c}-\frac {x^3 (1-a x)^{-n/2} (1+a x)^{n/2}}{3 a^2 c}+\frac {\int x (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} \left (2 a n+a^2 \left (6+n^2\right ) x\right ) \, dx}{6 a^4 c}\\ &=-\frac {n x^2 (1-a x)^{-n/2} (1+a x)^{n/2}}{6 a^3 c}-\frac {x^3 (1-a x)^{-n/2} (1+a x)^{n/2}}{3 a^2 c}+\frac {(1-a x)^{-n/2} (1+a x)^{n/2} \left (6+8 n+n^2+n^3-a n \left (6+n^2\right ) x\right )}{6 a^5 c n}-\frac {\left (n \left (8+n^2\right )\right ) \int (1-a x)^{-n/2} (1+a x)^{-1+\frac {n}{2}} \, dx}{6 a^4 c}\\ &=-\frac {n x^2 (1-a x)^{-n/2} (1+a x)^{n/2}}{6 a^3 c}-\frac {x^3 (1-a x)^{-n/2} (1+a x)^{n/2}}{3 a^2 c}+\frac {(1-a x)^{-n/2} (1+a x)^{n/2} \left (6+8 n+n^2+n^3-a n \left (6+n^2\right ) x\right )}{6 a^5 c n}+\frac {2^{-1+\frac {n}{2}} n \left (8+n^2\right ) (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},1-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{3 a^5 c (2-n)}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 172, normalized size = 0.82 \[ \frac {(1-a x)^{-n/2} \left (2^{\frac {n}{2}+1} n \left (n^2-2 n+2\right ) (a x-1) \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )-(n-2) \left ((a x+1)^{n/2} \left (2 n \left (a^3 x^3+2 a x+3\right )+(a n x+n)^2-6\right )-2^{\frac {n}{2}+2} n (n+3) \, _2F_1\left (-\frac {n}{2},-\frac {n}{2};1-\frac {n}{2};\frac {1}{2} (1-a x)\right )\right )\right )}{6 a^5 c (n-2) n} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(n*ArcTanh[a*x])*x^4)/(c - a^2*c*x^2),x]

[Out]

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

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x^{4} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*x^4/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{4} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*x^4/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )} x^{4}}{-a^{2} c \,x^{2}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))*x^4/(-a^2*c*x^2+c),x)

[Out]

int(exp(n*arctanh(a*x))*x^4/(-a^2*c*x^2+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {x^{4} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))*x^4/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{c-a^2\,c\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*exp(n*atanh(a*x)))/(c - a^2*c*x^2),x)

[Out]

int((x^4*exp(n*atanh(a*x)))/(c - a^2*c*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {x^{4} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{2} x^{2} - 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))*x**4/(-a**2*c*x**2+c),x)

[Out]

-Integral(x**4*exp(n*atanh(a*x))/(a**2*x**2 - 1), x)/c

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