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

Optimal. Leaf size=135 \[ \frac {2 (a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a c^4 (n+6) \left (n^2+6 n+8\right )}+\frac {(a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-3}}{a c^4 (n+6)}+\frac {2 (a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-2}}{a c^4 (n+4) (n+6)} \]

[Out]

2*(-a*x+1)^(-2-1/2*n)*(a*x+1)^(1+1/2*n)/a/c^4/(n^2+10*n+24)+2*(-a*x+1)^(-1-1/2*n)*(a*x+1)^(1+1/2*n)/a/c^4/(n^3
+12*n^2+44*n+48)+(-a*x+1)^(-3-1/2*n)*(a*x+1)^(1+1/2*n)/a/c^4/(6+n)

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Rubi [A]  time = 0.08, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6129, 45, 37} \[ \frac {2 (a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a c^4 (n+6) \left (n^2+6 n+8\right )}+\frac {(a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-3}}{a c^4 (n+6)}+\frac {2 (a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-2}}{a c^4 (n+4) (n+6)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(-3 - n/2)*(1 + a*x)^((2 + n)/2))/(a*c^4*(6 + n)) + (2*(1 - a*x)^(-2 - n/2)*(1 + a*x)^((2 + n)/2))/
(a*c^4*(4 + n)*(6 + n)) + (2*(1 - a*x)^(-1 - n/2)*(1 + a*x)^((2 + n)/2))/(a*c^4*(6 + n)*(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)^4} \, dx &=\frac {\int (1-a x)^{-4-\frac {n}{2}} (1+a x)^{n/2} \, dx}{c^4}\\ &=\frac {(1-a x)^{-3-\frac {n}{2}} (1+a x)^{\frac {2+n}{2}}}{a c^4 (6+n)}+\frac {2 \int (1-a x)^{-3-\frac {n}{2}} (1+a x)^{n/2} \, dx}{c^4 (6+n)}\\ &=\frac {(1-a x)^{-3-\frac {n}{2}} (1+a x)^{\frac {2+n}{2}}}{a c^4 (6+n)}+\frac {2 (1-a x)^{-2-\frac {n}{2}} (1+a x)^{\frac {2+n}{2}}}{a c^4 (4+n) (6+n)}+\frac {2 \int (1-a x)^{-2-\frac {n}{2}} (1+a x)^{n/2} \, dx}{c^4 (4+n) (6+n)}\\ &=\frac {(1-a x)^{-3-\frac {n}{2}} (1+a x)^{\frac {2+n}{2}}}{a c^4 (6+n)}+\frac {2 (1-a x)^{-2-\frac {n}{2}} (1+a x)^{\frac {2+n}{2}}}{a c^4 (4+n) (6+n)}+\frac {2 (1-a x)^{-1-\frac {n}{2}} (1+a x)^{\frac {2+n}{2}}}{a c^4 (2+n) (4+n) (6+n)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 74, normalized size = 0.55 \[ \frac {(1-a x)^{-\frac {n}{2}-3} (a x+1)^{\frac {n}{2}+1} \left (2 a^2 x^2-2 a n x-8 a x+n^2+8 n+14\right )}{a c^4 (n+2) (n+4) (n+6)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(-3 - n/2)*(1 + a*x)^(1 + n/2)*(14 + 8*n + n^2 - 8*a*x - 2*a*n*x + 2*a^2*x^2))/(a*c^4*(2 + n)*(4 +
n)*(6 + n))

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fricas [A]  time = 0.73, size = 228, normalized size = 1.69 \[ \frac {{\left (2 \, a^{3} x^{3} - 2 \, {\left (a^{2} n + 3 \, a^{2}\right )} x^{2} + n^{2} + {\left (a n^{2} + 6 \, a n + 6 \, a\right )} x + 8 \, n + 14\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{4} n^{3} + 12 \, a c^{4} n^{2} + 44 \, a c^{4} n + 48 \, a c^{4} - {\left (a^{4} c^{4} n^{3} + 12 \, a^{4} c^{4} n^{2} + 44 \, a^{4} c^{4} n + 48 \, a^{4} c^{4}\right )} x^{3} + 3 \, {\left (a^{3} c^{4} n^{3} + 12 \, a^{3} c^{4} n^{2} + 44 \, a^{3} c^{4} n + 48 \, a^{3} c^{4}\right )} x^{2} - 3 \, {\left (a^{2} c^{4} n^{3} + 12 \, a^{2} c^{4} n^{2} + 44 \, a^{2} c^{4} n + 48 \, a^{2} c^{4}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^3*x^3 - 2*(a^2*n + 3*a^2)*x^2 + n^2 + (a*n^2 + 6*a*n + 6*a)*x + 8*n + 14)*((a*x + 1)/(a*x - 1))^(1/2*n)/(
a*c^4*n^3 + 12*a*c^4*n^2 + 44*a*c^4*n + 48*a*c^4 - (a^4*c^4*n^3 + 12*a^4*c^4*n^2 + 44*a^4*c^4*n + 48*a^4*c^4)*
x^3 + 3*(a^3*c^4*n^3 + 12*a^3*c^4*n^2 + 44*a^3*c^4*n + 48*a^3*c^4)*x^2 - 3*(a^2*c^4*n^3 + 12*a^2*c^4*n^2 + 44*
a^2*c^4*n + 48*a^2*c^4)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 68, normalized size = 0.50 \[ -\frac {\left (a x +1\right ) \left (2 a^{2} x^{2}-2 a n x -8 a x +n^{2}+8 n +14\right ) {\mathrm e}^{n \arctanh \left (a x \right )}}{\left (a x -1\right )^{3} c^{4} a \left (n^{2}+8 n +12\right ) \left (4+n \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x+1)*(2*a^2*x^2-2*a*n*x-8*a*x+n^2+8*n+14)*exp(n*arctanh(a*x))/(a*x-1)^3/c^4/a/(n^2+8*n+12)/(4+n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.34, size = 167, normalized size = 1.24 \[ -\frac {{\left (a\,x+1\right )}^{n/2}\,\left (\frac {2\,x^3}{a\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}+\frac {n^2+8\,n+14}{a^4\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}-\frac {x^2\,\left (2\,n+6\right )}{a^2\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}+\frac {x\,\left (n^2+6\,n+6\right )}{a^3\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}\right )}{{\left (1-a\,x\right )}^{n/2}\,\left (\frac {3\,x}{a^2}-\frac {1}{a^3}+x^3-\frac {3\,x^2}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a*x + 1)^(n/2)*((2*x^3)/(a*c^4*(44*n + 12*n^2 + n^3 + 48)) + (8*n + n^2 + 14)/(a^4*c^4*(44*n + 12*n^2 + n^3
+ 48)) - (x^2*(2*n + 6))/(a^2*c^4*(44*n + 12*n^2 + n^3 + 48)) + (x*(6*n + n^2 + 6))/(a^3*c^4*(44*n + 12*n^2 +
n^3 + 48))))/((1 - a*x)^(n/2)*((3*x)/a^2 - 1/a^3 + x^3 - (3*x^2)/a))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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