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

Optimal. Leaf size=84 \[ \frac {(a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a c^3 \left (n^2+6 n+8\right )}+\frac {(a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-2}}{a c^3 (n+4)} \]

[Out]

(-a*x+1)^(-1-1/2*n)*(a*x+1)^(1+1/2*n)/a/c^3/(n^2+6*n+8)+(-a*x+1)^(-2-1/2*n)*(a*x+1)^(1+1/2*n)/a/c^3/(4+n)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6129, 45, 37} \[ \frac {(a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a c^3 \left (n^2+6 n+8\right )}+\frac {(a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-2}}{a c^3 (n+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(-2 - n/2)*(1 + a*x)^((2 + n)/2))/(a*c^3*(4 + n)) + ((1 - a*x)^(-1 - n/2)*(1 + a*x)^((2 + n)/2))/(a
*c^3*(8 + 6*n + n^2))

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 6129

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[(u*(1 + (d*x)/c)
^p*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 51, normalized size = 0.61 \[ \frac {(1-a x)^{-\frac {n}{2}-2} (-a x+n+3) (a x+1)^{\frac {n}{2}+1}}{a c^3 (n+2) (n+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(-2 - n/2)*(3 + n - a*x)*(1 + a*x)^(1 + n/2))/(a*c^3*(2 + n)*(4 + n))

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 128, normalized size = 1.52 \[ -\frac {{\left (a^{2} x^{2} - {\left (a n + 2 \, a\right )} x - n - 3\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{3} n^{2} + 6 \, a c^{3} n + 8 \, a c^{3} + {\left (a^{3} c^{3} n^{2} + 6 \, a^{3} c^{3} n + 8 \, a^{3} c^{3}\right )} x^{2} - 2 \, {\left (a^{2} c^{3} n^{2} + 6 \, a^{2} c^{3} n + 8 \, a^{2} c^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*x^2 - (a*n + 2*a)*x - n - 3)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a*c^3*n^2 + 6*a*c^3*n + 8*a*c^3 + (a^3*c^3*n
^2 + 6*a^3*c^3*n + 8*a^3*c^3)*x^2 - 2*(a^2*c^3*n^2 + 6*a^2*c^3*n + 8*a^2*c^3)*x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 46, normalized size = 0.55 \[ -\frac {{\mathrm e}^{n \arctanh \left (a x \right )} \left (a x -n -3\right ) \left (a x +1\right )}{\left (a x -1\right )^{2} c^{3} \left (n^{2}+6 n +8\right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(n*arctanh(a*x))*(a*x-n-3)*(a*x+1)/(a*x-1)^2/c^3/(n^2+6*n+8)/a

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.27, size = 100, normalized size = 1.19 \[ \frac {{\mathrm {e}}^{\frac {n\,\ln \left (a\,x+1\right )}{2}-\frac {n\,\ln \left (1-a\,x\right )}{2}}\,\left (\frac {n+3}{a^3\,c^3\,\left (n^2+6\,n+8\right )}-\frac {x^2}{a\,c^3\,\left (n^2+6\,n+8\right )}+\frac {x\,\left (n+2\right )}{a^2\,c^3\,\left (n^2+6\,n+8\right )}\right )}{\frac {1}{a^2}-\frac {2\,x}{a}+x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp((n*log(a*x + 1))/2 - (n*log(1 - a*x))/2)*((n + 3)/(a^3*c^3*(6*n + n^2 + 8)) - x^2/(a*c^3*(6*n + n^2 + 8))
+ (x*(n + 2))/(a^2*c^3*(6*n + n^2 + 8))))/(1/a^2 - (2*x)/a + x^2)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*Integral(exp(n*atanh(a*x)), x), Eq(c, 0)), (a**2*x**2*atanh(a*x)/(2*a**3*c**3*x**2*exp(4*atanh(
a*x)) - 4*a**2*c**3*x*exp(4*atanh(a*x)) + 2*a*c**3*exp(4*atanh(a*x))) + 2*a*x*atanh(a*x)/(2*a**3*c**3*x**2*exp
(4*atanh(a*x)) - 4*a**2*c**3*x*exp(4*atanh(a*x)) + 2*a*c**3*exp(4*atanh(a*x))) - a*x/(2*a**3*c**3*x**2*exp(4*a
tanh(a*x)) - 4*a**2*c**3*x*exp(4*atanh(a*x)) + 2*a*c**3*exp(4*atanh(a*x))) + atanh(a*x)/(2*a**3*c**3*x**2*exp(
4*atanh(a*x)) - 4*a**2*c**3*x*exp(4*atanh(a*x)) + 2*a*c**3*exp(4*atanh(a*x))) - 1/(2*a**3*c**3*x**2*exp(4*atan
h(a*x)) - 4*a**2*c**3*x*exp(4*atanh(a*x)) + 2*a*c**3*exp(4*atanh(a*x))), Eq(n, -4)), (-a**2*x**2*atanh(a*x)/(2
*a**3*c**3*x**2*exp(2*atanh(a*x)) - 4*a**2*c**3*x*exp(2*atanh(a*x)) + 2*a*c**3*exp(2*atanh(a*x))) + a*x/(2*a**
3*c**3*x**2*exp(2*atanh(a*x)) - 4*a**2*c**3*x*exp(2*atanh(a*x)) + 2*a*c**3*exp(2*atanh(a*x))) + atanh(a*x)/(2*
a**3*c**3*x**2*exp(2*atanh(a*x)) - 4*a**2*c**3*x*exp(2*atanh(a*x)) + 2*a*c**3*exp(2*atanh(a*x))) + 1/(2*a**3*c
**3*x**2*exp(2*atanh(a*x)) - 4*a**2*c**3*x*exp(2*atanh(a*x)) + 2*a*c**3*exp(2*atanh(a*x))), Eq(n, -2)), (-a**2
*x**2*exp(n*atanh(a*x))/(a**3*c**3*n**2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12
*a**2*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2 + 6*a*c**3*n + 8*a*c**3) + a*n*x*exp(n*atanh(a*x))/(a**3*c**3*n*
*2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2*c**3*n*x - 16*a**2*c**3*x + a*c
**3*n**2 + 6*a*c**3*n + 8*a*c**3) + 2*a*x*exp(n*atanh(a*x))/(a**3*c**3*n**2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3
*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2 + 6*a*c**3*n + 8*a*c**3) + n
*exp(n*atanh(a*x))/(a**3*c**3*n**2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2
*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2 + 6*a*c**3*n + 8*a*c**3) + 3*exp(n*atanh(a*x))/(a**3*c**3*n**2*x**2 +
6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2
+ 6*a*c**3*n + 8*a*c**3), True))

________________________________________________________________________________________