3.1319 \(\int \frac {e^{n \tanh ^{-1}(a x)} x^3}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=311 \[ \frac {2^{\frac {n}{2}+2} (1-a x)^{-\frac {n}{2}-1} \, _2F_1\left (-\frac {n}{2}-1,-\frac {n}{2}-1;-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a^4 c^2 (n+2)}+\frac {2 (a x+1)^{\frac {n-2}{2}} (1-a x)^{1-\frac {n}{2}}}{a^4 c^2 n \left (4-n^2\right )}-\frac {(a x+1)^{\frac {n-2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a^4 c^2 (n+2)}+\frac {3 (a x+1)^{n/2} (1-a x)^{-\frac {n}{2}-1}}{a^4 c^2 (n+2)}-\frac {3 (a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a^4 c^2 (n+2)}-\frac {2 (a x+1)^{\frac {n-2}{2}} (1-a x)^{-n/2}}{a^4 c^2 n (n+2)}+\frac {3 (a x+1)^{n/2} (1-a x)^{-n/2}}{a^4 c^2 n (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6150, 128, 45, 37, 69} \[ \frac {2^{\frac {n}{2}+2} (1-a x)^{-\frac {n}{2}-1} \, _2F_1\left (-\frac {n}{2}-1,-\frac {n}{2}-1;-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{a^4 c^2 (n+2)}+\frac {2 (a x+1)^{\frac {n-2}{2}} (1-a x)^{1-\frac {n}{2}}}{a^4 c^2 n \left (4-n^2\right )}-\frac {(a x+1)^{\frac {n-2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a^4 c^2 (n+2)}+\frac {3 (a x+1)^{n/2} (1-a x)^{-\frac {n}{2}-1}}{a^4 c^2 (n+2)}-\frac {3 (a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a^4 c^2 (n+2)}-\frac {2 (a x+1)^{\frac {n-2}{2}} (1-a x)^{-n/2}}{a^4 c^2 n (n+2)}+\frac {3 (a x+1)^{n/2} (1-a x)^{-n/2}}{a^4 c^2 n (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((1 - a*x)^(-1 - n/2)*(1 + a*x)^((-2 + n)/2))/(a^4*c^2*(2 + n))) + (2*(1 - a*x)^(1 - n/2)*(1 + a*x)^((-2 + n
)/2))/(a^4*c^2*n*(4 - n^2)) - (2*(1 + a*x)^((-2 + n)/2))/(a^4*c^2*n*(2 + n)*(1 - a*x)^(n/2)) + (3*(1 - a*x)^(-
1 - n/2)*(1 + a*x)^(n/2))/(a^4*c^2*(2 + n)) + (3*(1 + a*x)^(n/2))/(a^4*c^2*n*(2 + n)*(1 - a*x)^(n/2)) - (3*(1
- a*x)^(-1 - n/2)*(1 + a*x)^((2 + n)/2))/(a^4*c^2*(2 + n)) + (2^(2 + n/2)*(1 - a*x)^(-1 - n/2)*Hypergeometric2
F1[-1 - n/2, -1 - n/2, -n/2, (1 - a*x)/2])/(a^4*c^2*(2 + n))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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 128

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (IGtQ[m, 0] || (
ILtQ[m, 0] && ILtQ[n, 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])

Rubi steps

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

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Mathematica [A]  time = 5.40, size = 145, normalized size = 0.47 \[ -\frac {e^{n \tanh ^{-1}(a x)} \left (-\left (n^2-4\right ) \left (a^2 x^2-1\right ) \, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;-e^{2 \tanh ^{-1}(a x)}\right )+(n-2) n \left (a^2 x^2-1\right ) e^{2 \tanh ^{-1}(a x)} \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;-e^{2 \tanh ^{-1}(a x)}\right )+n \left (-a^2 x^2+a n x-1\right )\right )}{a^4 c^2 (n-2) n (n+2) \left (a^2 x^2-1\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((E^(n*ArcTanh[a*x])*(n*(-1 + a*n*x - a^2*x^2) + E^(2*ArcTanh[a*x])*(-2 + n)*n*(-1 + a^2*x^2)*Hypergeometric2
F1[1, 1 + n/2, 2 + n/2, -E^(2*ArcTanh[a*x])] - (-4 + n^2)*(-1 + a^2*x^2)*Hypergeometric2F1[1, n/2, 1 + n/2, -E
^(2*ArcTanh[a*x])]))/(a^4*c^2*(-2 + n)*n*(2 + n)*(-1 + a^2*x^2)))

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

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)^2,x, algorithm="fricas")

[Out]

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

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)^2,x, algorithm="giac")

[Out]

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

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

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)^2,x)

[Out]

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

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

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)^2,x, algorithm="maxima")

[Out]

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

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

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)^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{4} x^{4} - 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

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)**2,x)

[Out]

Integral(x**3*exp(n*atanh(a*x))/(a**4*x**4 - 2*a**2*x**2 + 1), x)/c**2

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