3.5.76 \(\int \frac {\text {arctanh}(a x)^3}{(1-a^2 x^2)^{5/2}} \, dx\) [476]

3.5.76.1 Optimal result

Integrand size = 21, antiderivative size = 191 \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx=-\frac {2}{27 a \left (1-a^2 x^2\right )^{3/2}}-\frac {40}{9 a \sqrt {1-a^2 x^2}}+\frac {2 x \text {arctanh}(a x)}{9 \left (1-a^2 x^2\right )^{3/2}}+\frac {40 x \text {arctanh}(a x)}{9 \sqrt {1-a^2 x^2}}-\frac {\text {arctanh}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}-\frac {2 \text {arctanh}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {x \text {arctanh}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \text {arctanh}(a x)^3}{3 \sqrt {1-a^2 x^2}} \]

output

-2/27/a/(-a^2*x^2+1)^(3/2)+2/9*x*arctanh(a*x)/(-a^2*x^2+1)^(3/2)-1/3*arcta 
nh(a*x)^2/a/(-a^2*x^2+1)^(3/2)+1/3*x*arctanh(a*x)^3/(-a^2*x^2+1)^(3/2)-40/ 
9/a/(-a^2*x^2+1)^(1/2)+40/9*x*arctanh(a*x)/(-a^2*x^2+1)^(1/2)-2*arctanh(a* 
x)^2/a/(-a^2*x^2+1)^(1/2)+2/3*x*arctanh(a*x)^3/(-a^2*x^2+1)^(1/2)
 

3.5.76.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.46 \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {-122+120 a^2 x^2-6 a x \left (-21+20 a^2 x^2\right ) \text {arctanh}(a x)+9 \left (-7+6 a^2 x^2\right ) \text {arctanh}(a x)^2-9 a x \left (-3+2 a^2 x^2\right ) \text {arctanh}(a x)^3}{27 a \left (1-a^2 x^2\right )^{3/2}} \]

input

Integrate[ArcTanh[a*x]^3/(1 - a^2*x^2)^(5/2),x]
 

output

(-122 + 120*a^2*x^2 - 6*a*x*(-21 + 20*a^2*x^2)*ArcTanh[a*x] + 9*(-7 + 6*a^ 
2*x^2)*ArcTanh[a*x]^2 - 9*a*x*(-3 + 2*a^2*x^2)*ArcTanh[a*x]^3)/(27*a*(1 - 
a^2*x^2)^(3/2))
 

3.5.76.3 Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6526, 6522, 6520, 6524, 6520}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6526

\(\displaystyle \frac {2}{3} \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{5/2}}dx+\frac {2}{3} \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x \text {arctanh}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\text {arctanh}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 6522

\(\displaystyle \frac {2}{3} \left (\frac {2}{3} \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x \text {arctanh}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {2}{3} \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x \text {arctanh}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\text {arctanh}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 6520

\(\displaystyle \frac {2}{3} \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x \text {arctanh}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\text {arctanh}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{3} \left (\frac {x \text {arctanh}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{3} \left (\frac {x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}-\frac {1}{a \sqrt {1-a^2 x^2}}\right )-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 6524

\(\displaystyle \frac {2}{3} \left (6 \int \frac {\text {arctanh}(a x)}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 \text {arctanh}(a x)^2}{a \sqrt {1-a^2 x^2}}\right )+\frac {x \text {arctanh}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\text {arctanh}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{3} \left (\frac {x \text {arctanh}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{3} \left (\frac {x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}-\frac {1}{a \sqrt {1-a^2 x^2}}\right )-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}\right )\)

\(\Big \downarrow \) 6520

\(\displaystyle \frac {x \text {arctanh}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {\text {arctanh}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{3} \left (\frac {x \text {arctanh}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{3} \left (\frac {x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}-\frac {1}{a \sqrt {1-a^2 x^2}}\right )-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {2}{3} \left (\frac {x \text {arctanh}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 \text {arctanh}(a x)^2}{a \sqrt {1-a^2 x^2}}+6 \left (\frac {x \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}-\frac {1}{a \sqrt {1-a^2 x^2}}\right )\right )\)

input

Int[ArcTanh[a*x]^3/(1 - a^2*x^2)^(5/2),x]
 

output

-1/3*ArcTanh[a*x]^2/(a*(1 - a^2*x^2)^(3/2)) + (x*ArcTanh[a*x]^3)/(3*(1 - a 
^2*x^2)^(3/2)) + (2*(-1/9*1/(a*(1 - a^2*x^2)^(3/2)) + (x*ArcTanh[a*x])/(3* 
(1 - a^2*x^2)^(3/2)) + (2*(-(1/(a*Sqrt[1 - a^2*x^2])) + (x*ArcTanh[a*x])/S 
qrt[1 - a^2*x^2]))/3))/3 + (2*((-3*ArcTanh[a*x]^2)/(a*Sqrt[1 - a^2*x^2]) + 
 (x*ArcTanh[a*x]^3)/Sqrt[1 - a^2*x^2] + 6*(-(1/(a*Sqrt[1 - a^2*x^2])) + (x 
*ArcTanh[a*x])/Sqrt[1 - a^2*x^2])))/3
 

3.5.76.3.1 Defintions of rubi rules used

rule 6520

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symb 
ol] :> Simp[-b/(c*d*Sqrt[d + e*x^2]), x] + Simp[x*((a + b*ArcTanh[c*x])/(d* 
Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0]
 

rule 6522

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbo 
l] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c*d*(q + 1)^2)), x] + (-Simp[x*(d + 
 e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])/(2*d*(q + 1))), x] + Simp[(2*q + 3)/( 
2*d*(q + 1))   Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x]) /; Fre 
eQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]
 

rule 6524

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[(-b)*p*((a + b*ArcTanh[c*x])^(p - 1)/(c*d*Sqrt[d + e*x^2]) 
), x] + (Simp[x*((a + b*ArcTanh[c*x])^p/(d*Sqrt[d + e*x^2])), x] + Simp[b^2 
*p*(p - 1)   Int[(a + b*ArcTanh[c*x])^(p - 2)/(d + e*x^2)^(3/2), x], x]) /; 
 FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 1]
 

rule 6526

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_ 
Symbol] :> Simp[(-b)*p*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^(p - 1)/(4 
*c*d*(q + 1)^2)), x] + (-Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^p 
/(2*d*(q + 1))), x] + Simp[(2*q + 3)/(2*d*(q + 1))   Int[(d + e*x^2)^(q + 1 
)*(a + b*ArcTanh[c*x])^p, x], x] + Simp[b^2*p*((p - 1)/(4*(q + 1)^2))   Int 
[(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 2), x], x]) /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && GtQ[p, 1] && NeQ[q, -3/2]
 

3.5.76.4 Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.55

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (18 \operatorname {arctanh}\left (a x \right )^{3} a^{3} x^{3}+120 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )-54 a^{2} x^{2} \operatorname {arctanh}\left (a x \right )^{2}-27 \operatorname {arctanh}\left (a x \right )^{3} a x -120 a^{2} x^{2}-126 a x \,\operatorname {arctanh}\left (a x \right )+63 \operatorname {arctanh}\left (a x \right )^{2}+122\right )}{27 a \left (a^{2} x^{2}-1\right )^{2}}\)	\(105\)

input

int(arctanh(a*x)^3/(-a^2*x^2+1)^(5/2),x,method=_RETURNVERBOSE)
 

output

-1/27/a*(-a^2*x^2+1)^(1/2)*(18*arctanh(a*x)^3*a^3*x^3+120*a^3*x^3*arctanh( 
a*x)-54*a^2*x^2*arctanh(a*x)^2-27*arctanh(a*x)^3*a*x-120*a^2*x^2-126*a*x*a 
rctanh(a*x)+63*arctanh(a*x)^2+122)/(a^2*x^2-1)^2
 

3.5.76.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.70 \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\frac {{\left (960 \, a^{2} x^{2} - 9 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 18 \, {\left (6 \, a^{2} x^{2} - 7\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 24 \, {\left (20 \, a^{3} x^{3} - 21 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 976\right )} \sqrt {-a^{2} x^{2} + 1}}{216 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \]

input

integrate(arctanh(a*x)^3/(-a^2*x^2+1)^(5/2),x, algorithm="fricas")
 

output

1/216*(960*a^2*x^2 - 9*(2*a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1))^3 + 1 
8*(6*a^2*x^2 - 7)*log(-(a*x + 1)/(a*x - 1))^2 - 24*(20*a^3*x^3 - 21*a*x)*l 
og(-(a*x + 1)/(a*x - 1)) - 976)*sqrt(-a^2*x^2 + 1)/(a^5*x^4 - 2*a^3*x^2 + 
a)
 

3.5.76.6 Sympy [F]

\[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

input

integrate(atanh(a*x)**3/(-a**2*x**2+1)**(5/2),x)
 

output

Integral(atanh(a*x)**3/(-(a*x - 1)*(a*x + 1))**(5/2), x)
 

3.5.76.7 Maxima [F]

\[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { \frac {\operatorname {artanh}\left (a x\right )^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(arctanh(a*x)^3/(-a^2*x^2+1)^(5/2),x, algorithm="maxima")
 

output

integrate(arctanh(a*x)^3/(-a^2*x^2 + 1)^(5/2), x)
 

3.5.76.8 Giac [F]

\[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int { \frac {\operatorname {artanh}\left (a x\right )^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(arctanh(a*x)^3/(-a^2*x^2+1)^(5/2),x, algorithm="giac")
 

output

integrate(arctanh(a*x)^3/(-a^2*x^2 + 1)^(5/2), x)
 

3.5.76.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx=\int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \]

input

int(atanh(a*x)^3/(1 - a^2*x^2)^(5/2),x)
 

output

int(atanh(a*x)^3/(1 - a^2*x^2)^(5/2), x)