3.1312 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^3}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=160 \[ \frac {2^{\frac {n}{2}-1} \left (n^2+2\right ) (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},1-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a^4 c (2-n)}+\frac {(a x+1)^{n/2} \left (-a n^2 x+n^2+n+2\right ) (1-a x)^{-n/2}}{2 a^4 c n}-\frac {x^2 (a x+1)^{n/2} (1-a x)^{-n/2}}{2 a^2 c} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6150, 100, 143, 69} \[ \frac {2^{\frac {n}{2}-1} \left (n^2+2\right ) (1-a x)^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},1-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a^4 c (2-n)}+\frac {(a x+1)^{n/2} \left (-a n^2 x+n^2+n+2\right ) (1-a x)^{-n/2}}{2 a^4 c n}-\frac {x^2 (a x+1)^{n/2} (1-a x)^{-n/2}}{2 a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*(1 + a*x)^(n/2))/(2*a^2*c*(1 - a*x)^(n/2)) + ((1 + a*x)^(n/2)*(2 + n + n^2 - a*n^2*x))/(2*a^4*c*n*(1 - a
*x)^(n/2)) + (2^(-1 + n/2)*(2 + n^2)*(1 - a*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2, 1 - n/2, 2 - n/2, (1 - a*x
)/2])/(a^4*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 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^3}{c-a^2 c x^2} \, dx &=\frac {\int x^3 (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} \, dx}{c}\\ &=-\frac {x^2 (1-a x)^{-n/2} (1+a x)^{n/2}}{2 a^2 c}-\frac {\int x (1-a x)^{-1-\frac {n}{2}} (1+a x)^{-1+\frac {n}{2}} (-2-a n x) \, dx}{2 a^2 c}\\ &=-\frac {x^2 (1-a x)^{-n/2} (1+a x)^{n/2}}{2 a^2 c}+\frac {(1-a x)^{-n/2} (1+a x)^{n/2} \left (2+n+n^2-a n^2 x\right )}{2 a^4 c n}-\frac {\left (2+n^2\right ) \int (1-a x)^{-n/2} (1+a x)^{-1+\frac {n}{2}} \, dx}{2 a^3 c}\\ &=-\frac {x^2 (1-a x)^{-n/2} (1+a x)^{n/2}}{2 a^2 c}+\frac {(1-a x)^{-n/2} (1+a x)^{n/2} \left (2+n+n^2-a n^2 x\right )}{2 a^4 c n}+\frac {2^{-1+\frac {n}{2}} \left (2+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 )}{a^4 c (2-n)}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 120, normalized size = 0.75 \[ \frac {(1-a x)^{-n/2} \left (2^{n/2} n \left (n^2+2\right ) (a x-1) \, _2F_1\left (1-\frac {n}{2},1-\frac {n}{2};2-\frac {n}{2};\frac {1}{2} (1-a x)\right )-(n-2) (a x+1)^{n/2} \left (n \left (a^2 x^2-1\right )+n^2 (a x-1)-2\right )\right )}{2 a^4 c (n-2) n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((-2 + n)*(1 + a*x)^(n/2)*(-2 + n^2*(-1 + a*x) + n*(-1 + a^2*x^2))) + 2^(n/2)*n*(2 + n^2)*(-1 + a*x)*Hyperge
ometric2F1[1 - n/2, 1 - n/2, 2 - n/2, (1 - a*x)/2])/(2*a^4*c*(-2 + n)*n*(1 - a*x)^(n/2))

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x^{3} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )} x^{3}}{-a^{2} c \,x^{2}+c}\, dx \]

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),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(x^3*((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 = -0.01 \[ \int \frac {x^3\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{c-a^2\,c\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^3*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 \[ - \frac {\int \frac {x^{3} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{2} x^{2} - 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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