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

Optimal. Leaf size=96 \[ \frac {16 x}{35 c^4 \sqrt {1-a^2 x^2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {a x+1}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}} \]

[Out]

1/7*(a*x+1)/a/c^4/(-a^2*x^2+1)^(7/2)+6/35*x/c^4/(-a^2*x^2+1)^(5/2)+8/35*x/c^4/(-a^2*x^2+1)^(3/2)+16/35*x/c^4/(
-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6138, 639, 192, 191} \[ \frac {16 x}{35 c^4 \sqrt {1-a^2 x^2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {a x+1}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)/(7*a*c^4*(1 - a^2*x^2)^(7/2)) + (6*x)/(35*c^4*(1 - a^2*x^2)^(5/2)) + (8*x)/(35*c^4*(1 - a^2*x^2)^(3/
2)) + (16*x)/(35*c^4*Sqrt[1 - a^2*x^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 6138

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx &=\frac {\int \frac {1+a x}{\left (1-a^2 x^2\right )^{9/2}} \, dx}{c^4}\\ &=\frac {1+a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 \int \frac {1}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{7 c^4}\\ &=\frac {1+a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {24 \int \frac {1}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{35 c^4}\\ &=\frac {1+a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {16 \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{35 c^4}\\ &=\frac {1+a x}{7 a c^4 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x}{35 c^4 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x}{35 c^4 \left (1-a^2 x^2\right )^{3/2}}+\frac {16 x}{35 c^4 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 75, normalized size = 0.78 \[ \frac {-16 a^6 x^6+16 a^5 x^5+40 a^4 x^4-40 a^3 x^3-30 a^2 x^2+30 a x+5}{35 a c^4 (1-a x)^{7/2} (a x+1)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(5 + 30*a*x - 30*a^2*x^2 - 40*a^3*x^3 + 40*a^4*x^4 + 16*a^5*x^5 - 16*a^6*x^6)/(35*a*c^4*(1 - a*x)^(7/2)*(1 + a
*x)^(5/2))

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fricas [B]  time = 0.74, size = 198, normalized size = 2.06 \[ \frac {5 \, a^{7} x^{7} - 5 \, a^{6} x^{6} - 15 \, a^{5} x^{5} + 15 \, a^{4} x^{4} + 15 \, a^{3} x^{3} - 15 \, a^{2} x^{2} - 5 \, a x - {\left (16 \, a^{6} x^{6} - 16 \, a^{5} x^{5} - 40 \, a^{4} x^{4} + 40 \, a^{3} x^{3} + 30 \, a^{2} x^{2} - 30 \, a x - 5\right )} \sqrt {-a^{2} x^{2} + 1} + 5}{35 \, {\left (a^{8} c^{4} x^{7} - a^{7} c^{4} x^{6} - 3 \, a^{6} c^{4} x^{5} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^4,x, algorithm="fricas")

[Out]

1/35*(5*a^7*x^7 - 5*a^6*x^6 - 15*a^5*x^5 + 15*a^4*x^4 + 15*a^3*x^3 - 15*a^2*x^2 - 5*a*x - (16*a^6*x^6 - 16*a^5
*x^5 - 40*a^4*x^4 + 40*a^3*x^3 + 30*a^2*x^2 - 30*a*x - 5)*sqrt(-a^2*x^2 + 1) + 5)/(a^8*c^4*x^7 - a^7*c^4*x^6 -
3*a^6*c^4*x^5 + 3*a^5*c^4*x^4 + 3*a^4*c^4*x^3 - 3*a^3*c^4*x^2 - a^2*c^4*x + a*c^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 74, normalized size = 0.77 \[ \frac {16 x^{6} a^{6}-16 x^{5} a^{5}-40 x^{4} a^{4}+40 x^{3} a^{3}+30 a^{2} x^{2}-30 a x -5}{35 \left (a x -1\right ) c^{4} \left (-a^{2} x^{2}+1\right )^{\frac {5}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(16*a^6*x^6-16*a^5*x^5-40*a^4*x^4+40*a^3*x^3+30*a^2*x^2-30*a*x-5)/(a*x-1)/c^4/(-a^2*x^2+1)^(5/2)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.93, size = 477, normalized size = 4.97 \[ \frac {7\,a\,\sqrt {1-a^2\,x^2}}{80\,\left (a^4\,c^4\,x^2-2\,a^3\,c^4\,x+a^2\,c^4\right )}-\frac {a\,\sqrt {1-a^2\,x^2}}{20\,\left (a^4\,c^4\,x^2+2\,a^3\,c^4\,x+a^2\,c^4\right )}+\frac {a^3\,\sqrt {1-a^2\,x^2}}{140\,\left (a^6\,c^4\,x^2-2\,a^5\,c^4\,x+a^4\,c^4\right )}+\frac {a\,\sqrt {1-a^2\,x^2}}{56\,\left (a^6\,c^4\,x^4-4\,a^5\,c^4\,x^3+6\,a^4\,c^4\,x^2-4\,a^3\,c^4\,x+a^2\,c^4\right )}+\frac {33\,\sqrt {1-a^2\,x^2}}{160\,\sqrt {-a^2}\,\left (c^4\,x\,\sqrt {-a^2}+\frac {c^4\,\sqrt {-a^2}}{a}\right )}+\frac {281\,\sqrt {1-a^2\,x^2}}{1120\,\sqrt {-a^2}\,\left (c^4\,x\,\sqrt {-a^2}-\frac {c^4\,\sqrt {-a^2}}{a}\right )}+\frac {\sqrt {1-a^2\,x^2}}{80\,\sqrt {-a^2}\,\left (3\,c^4\,x\,\sqrt {-a^2}+\frac {c^4\,\sqrt {-a^2}}{a}+a^2\,c^4\,x^3\,\sqrt {-a^2}+3\,a\,c^4\,x^2\,\sqrt {-a^2}\right )}+\frac {27\,\sqrt {1-a^2\,x^2}}{560\,\sqrt {-a^2}\,\left (3\,c^4\,x\,\sqrt {-a^2}-\frac {c^4\,\sqrt {-a^2}}{a}+a^2\,c^4\,x^3\,\sqrt {-a^2}-3\,a\,c^4\,x^2\,\sqrt {-a^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*a*(1 - a^2*x^2)^(1/2))/(80*(a^2*c^4 - 2*a^3*c^4*x + a^4*c^4*x^2)) - (a*(1 - a^2*x^2)^(1/2))/(20*(a^2*c^4 +
2*a^3*c^4*x + a^4*c^4*x^2)) + (a^3*(1 - a^2*x^2)^(1/2))/(140*(a^4*c^4 - 2*a^5*c^4*x + a^6*c^4*x^2)) + (a*(1 -
a^2*x^2)^(1/2))/(56*(a^2*c^4 - 4*a^3*c^4*x + 6*a^4*c^4*x^2 - 4*a^5*c^4*x^3 + a^6*c^4*x^4)) + (33*(1 - a^2*x^2)
^(1/2))/(160*(-a^2)^(1/2)*(c^4*x*(-a^2)^(1/2) + (c^4*(-a^2)^(1/2))/a)) + (281*(1 - a^2*x^2)^(1/2))/(1120*(-a^2
)^(1/2)*(c^4*x*(-a^2)^(1/2) - (c^4*(-a^2)^(1/2))/a)) + (1 - a^2*x^2)^(1/2)/(80*(-a^2)^(1/2)*(3*c^4*x*(-a^2)^(1
/2) + (c^4*(-a^2)^(1/2))/a + a^2*c^4*x^3*(-a^2)^(1/2) + 3*a*c^4*x^2*(-a^2)^(1/2))) + (27*(1 - a^2*x^2)^(1/2))/
(560*(-a^2)^(1/2)*(3*c^4*x*(-a^2)^(1/2) - (c^4*(-a^2)^(1/2))/a + a^2*c^4*x^3*(-a^2)^(1/2) - 3*a*c^4*x^2*(-a^2)
^(1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x}{a^{8} x^{8} \sqrt {- a^{2} x^{2} + 1} - 4 a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 6 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 4 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a^{8} x^{8} \sqrt {- a^{2} x^{2} + 1} - 4 a^{6} x^{6} \sqrt {- a^{2} x^{2} + 1} + 6 a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 4 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x/(a**8*x**8*sqrt(-a**2*x**2 + 1) - 4*a**6*x**6*sqrt(-a**2*x**2 + 1) + 6*a**4*x**4*sqrt(-a**2*x**2
+ 1) - 4*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(1/(a**8*x**8*sqrt(-a**2*x**2 +
1) - 4*a**6*x**6*sqrt(-a**2*x**2 + 1) + 6*a**4*x**4*sqrt(-a**2*x**2 + 1) - 4*a**2*x**2*sqrt(-a**2*x**2 + 1) +
sqrt(-a**2*x**2 + 1)), x))/c**4

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