3.3.36 \(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx\) [236]

3.3.36.1 Optimal result

Integrand size = 17, antiderivative size = 93 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx=2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {2 b^2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}}-\frac {2 b \text {arctanh}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}-\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^{5/2}} \]

output

2*b^(5/2)*arctanh(b^(1/2)*x^(1/2)/arctanh(tanh(b*x+a))^(1/2))-2/3*b*arctan 
h(tanh(b*x+a))^(3/2)/x^(3/2)-2/5*arctanh(tanh(b*x+a))^(5/2)/x^(5/2)-2*b^2* 
arctanh(tanh(b*x+a))^(1/2)/x^(1/2)
 

3.3.36.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx=-\frac {2 \left (15 b^2 x^2 \sqrt {\text {arctanh}(\tanh (a+b x))}+5 b x \text {arctanh}(\tanh (a+b x))^{3/2}+3 \text {arctanh}(\tanh (a+b x))^{5/2}-15 b^{5/2} x^{5/2} \log \left (b \sqrt {x}+\sqrt {b} \sqrt {\text {arctanh}(\tanh (a+b x))}\right )\right )}{15 x^{5/2}} \]

input

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(7/2),x]
 

output

(-2*(15*b^2*x^2*Sqrt[ArcTanh[Tanh[a + b*x]]] + 5*b*x*ArcTanh[Tanh[a + b*x] 
]^(3/2) + 3*ArcTanh[Tanh[a + b*x]]^(5/2) - 15*b^(5/2)*x^(5/2)*Log[b*Sqrt[x 
] + Sqrt[b]*Sqrt[ArcTanh[Tanh[a + b*x]]]]))/(15*x^(5/2))
 

3.3.36.3 Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2599, 2599, 2599, 2596}

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.

\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx\)

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2599

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

\(\Big \downarrow \) 2596

\(\displaystyle b \left (b \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {\text {arctanh}(\tanh (a+b x))}}\right )-\frac {2 \sqrt {\text {arctanh}(\tanh (a+b x))}}{\sqrt {x}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{3 x^{3/2}}\right )-\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2}}{5 x^{5/2}}\)

input

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(7/2),x]
 

output

(-2*ArcTanh[Tanh[a + b*x]]^(5/2))/(5*x^(5/2)) + b*(b*(2*Sqrt[b]*ArcTanh[(S 
qrt[b]*Sqrt[x])/Sqrt[ArcTanh[Tanh[a + b*x]]]] - (2*Sqrt[ArcTanh[Tanh[a + b 
*x]]])/Sqrt[x]) - (2*ArcTanh[Tanh[a + b*x]]^(3/2))/(3*x^(3/2)))
 

3.3.36.3.1 Defintions of rubi rules used

rule 2596

Int[1/(Sqrt[u_]*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Si 
mplify[D[v, x]]}, Simp[(2/Rt[a*b, 2])*ArcTanh[Rt[a*b, 2]*(Sqrt[u]/(a*Sqrt[v 
]))], x] /; NeQ[b*u - a*v, 0] && PosQ[a*b]] /; PiecewiseLinearQ[u, v, x]
 

rule 2599

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Sim 
plify[D[v, x]]}, Simp[u^(m + 1)*(v^n/(a*(m + 1))), x] - Simp[b*(n/(a*(m + 1 
)))   Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m, n} 
, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0 
] &&  !(ILtQ[m + n, -2] && (FractionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ 
[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]) || (ILt 
Q[m, 0] &&  !IntegerQ[n]))
 

3.3.36.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(246\) vs. \(2(69)=138\).

Time = 0.18 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.66

method	result	size
derivativedivides	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {6 b \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{6}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(247\)
default	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {3}{2}}}+\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \sqrt {x}}+\frac {6 b \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{6}+\frac {5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{4}+\frac {3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \left (\frac {\sqrt {x}\, \sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}}{2}+\frac {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) \ln \left (\sqrt {b}\, \sqrt {x}+\sqrt {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(247\)

input

int(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x,method=_RETURNVERBOSE)
 

output

-2/5/(arctanh(tanh(b*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(7/2)+4/5*b/( 
arctanh(tanh(b*x+a))-b*x)*(-1/3/(arctanh(tanh(b*x+a))-b*x)/x^(3/2)*arctanh 
(tanh(b*x+a))^(7/2)+4/3*b/(arctanh(tanh(b*x+a))-b*x)*(-1/(arctanh(tanh(b*x 
+a))-b*x)/x^(1/2)*arctanh(tanh(b*x+a))^(7/2)+6*b/(arctanh(tanh(b*x+a))-b*x 
)*(1/6*x^(1/2)*arctanh(tanh(b*x+a))^(5/2)+5/6*(arctanh(tanh(b*x+a))-b*x)*( 
1/4*x^(1/2)*arctanh(tanh(b*x+a))^(3/2)+3/4*(arctanh(tanh(b*x+a))-b*x)*(1/2 
*x^(1/2)*arctanh(tanh(b*x+a))^(1/2)+1/2/b^(1/2)*(arctanh(tanh(b*x+a))-b*x) 
*ln(b^(1/2)*x^(1/2)+arctanh(tanh(b*x+a))^(1/2)))))))
 

3.3.36.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.47 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx=\left [\frac {15 \, b^{\frac {5}{2}} x^{3} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) - 2 \, {\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}}{15 \, x^{3}}, -\frac {2 \, {\left (15 \, \sqrt {-b} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt {b x + a} \sqrt {x}\right )}}{15 \, x^{3}}\right ] \]

input

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x, algorithm="fricas")
 

output

[1/15*(15*b^(5/2)*x^3*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2 
*(23*b^2*x^2 + 11*a*b*x + 3*a^2)*sqrt(b*x + a)*sqrt(x))/x^3, -2/15*(15*sqr 
t(-b)*b^2*x^3*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (23*b^2*x^2 + 1 
1*a*b*x + 3*a^2)*sqrt(b*x + a)*sqrt(x))/x^3]
 

3.3.36.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx=\text {Timed out} \]

input

integrate(atanh(tanh(b*x+a))**(5/2)/x**(7/2),x)
 

output

Timed out
 

3.3.36.7 Maxima [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{\frac {5}{2}}}{x^{\frac {7}{2}}} \,d x } \]

input

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x, algorithm="maxima")
 

output

integrate(arctanh(tanh(b*x + a))^(5/2)/x^(7/2), x)
 

3.3.36.8 Giac [A] (verification not implemented)

Time = 75.95 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.19 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx=-\frac {\sqrt {2} {\left (15 \, \sqrt {2} b^{\frac {5}{2}} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} + \sqrt {{\left (b x + a\right )} b - a b} \right |}\right ) + \frac {{\left (15 \, \sqrt {2} a^{2} b^{5} + {\left (23 \, \sqrt {2} {\left (b x + a\right )} b^{5} - 35 \, \sqrt {2} a b^{5}\right )} {\left (b x + a\right )}\right )} \sqrt {b x + a}}{{\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}}\right )} b}{15 \, {\left | b \right |}} \]

input

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(7/2),x, algorithm="giac")
 

output

-1/15*sqrt(2)*(15*sqrt(2)*b^(5/2)*log(abs(-sqrt(b*x + a)*sqrt(b) + sqrt((b 
*x + a)*b - a*b))) + (15*sqrt(2)*a^2*b^5 + (23*sqrt(2)*(b*x + a)*b^5 - 35* 
sqrt(2)*a*b^5)*(b*x + a))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(5/2))*b/abs(b 
)
 

3.3.36.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{7/2}} \, dx=\int \frac {{\mathrm {atanh}\left (\mathrm {tanh}\left (a+b\,x\right )\right )}^{5/2}}{x^{7/2}} \,d x \]

input

int(atanh(tanh(a + b*x))^(5/2)/x^(7/2),x)
 

output

int(atanh(tanh(a + b*x))^(5/2)/x^(7/2), x)