\(\int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx\) [222]

Optimal result

Integrand size = 17, antiderivative size = 148 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {32 b^3 \text {arctanh}(\tanh (a+b x))^{3/2}}{315 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^4}+\frac {16 b^2 \text {arctanh}(\tanh (a+b x))^{3/2}}{105 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{3/2}}{21 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))} \] Output:

32/315*b^3*arctanh(tanh(b*x+a))^(3/2)/x^(3/2)/(b*x-arctanh(tanh(b*x+a)))^4 
+16/105*b^2*arctanh(tanh(b*x+a))^(3/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^ 
3+4/21*b*arctanh(tanh(b*x+a))^(3/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^2+2 
/9*arctanh(tanh(b*x+a))^(3/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2} \left (105 b^3 x^3-189 b^2 x^2 \text {arctanh}(\tanh (a+b x))+135 b x \text {arctanh}(\tanh (a+b x))^2-35 \text {arctanh}(\tanh (a+b x))^3\right )}{315 x^{9/2} (-b x+\text {arctanh}(\tanh (a+b x)))^4} \] Input:

Integrate[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(11/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(3/2)*(105*b^3*x^3 - 189*b^2*x^2*ArcTanh[Tanh[a 
+ b*x]] + 135*b*x*ArcTanh[Tanh[a + b*x]]^2 - 35*ArcTanh[Tanh[a + b*x]]^3)) 
/(315*x^(9/2)*(-(b*x) + ArcTanh[Tanh[a + b*x]])^4)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2602, 2602, 2602, 2598}

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.

\(\displaystyle \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx\)

\(\Big \downarrow \) 2602

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

\(\Big \downarrow \) 2602

\(\displaystyle \frac {2 b \left (\frac {4 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{7/2}}dx}{7 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{3 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\)

\(\Big \downarrow \) 2602

\(\displaystyle \frac {2 b \left (\frac {4 b \left (\frac {2 b \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{5/2}}dx}{5 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{7 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{3 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\)

\(\Big \downarrow \) 2598

\(\displaystyle \frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{3/2}}{5 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{3/2}}{15 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}\right )}{7 (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{3 (b x-\text {arctanh}(\tanh (a+b x)))}\)

Input:

Int[Sqrt[ArcTanh[Tanh[a + b*x]]]/x^(11/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(3/2))/(9*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]]) 
) + (2*b*((2*ArcTanh[Tanh[a + b*x]]^(3/2))/(7*x^(7/2)*(b*x - ArcTanh[Tanh[ 
a + b*x]])) + (4*b*((4*b*ArcTanh[Tanh[a + b*x]]^(3/2))/(15*x^(3/2)*(b*x - 
ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(3/2))/(5*x^(5/2)*( 
b*x - ArcTanh[Tanh[a + b*x]]))))/(7*(b*x - ArcTanh[Tanh[a + b*x]]))))/(3*( 
b*x - ArcTanh[Tanh[a + b*x]]))
 

Defintions of rubi rules used

rule 2598

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp 
lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] /; 
NeQ[b*u - a*v, 0]] /; FreeQ[{m, n}, x] && PiecewiseLinearQ[u, v, x] && EqQ[ 
m + n + 2, 0] && NeQ[m, -1]
 

rule 2602

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simp 
lify[D[v, x]]}, Simp[(-u^(m + 1))*(v^(n + 1)/((m + 1)*(b*u - a*v))), x] + S 
imp[b*((m + n + 2)/((m + 1)*(b*u - a*v)))   Int[u^(m + 1)*v^n, x], x] /; Ne 
Q[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m 
, -1]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(151\)
default	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{5 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {5}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {3}{2}}}{15 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {3}{2}}}\right )}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{3 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(151\)

Input:

int(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x,method=_RETURNVERBOSE)
 

Output:

-2/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(3/2)-4/3*b/( 
arctanh(tanh(b*x+a))-b*x)*(-1/7/(arctanh(tanh(b*x+a))-b*x)/x^(7/2)*arctanh 
(tanh(b*x+a))^(3/2)-4/7*b/(arctanh(tanh(b*x+a))-b*x)*(-1/5/(arctanh(tanh(b 
*x+a))-b*x)/x^(5/2)*arctanh(tanh(b*x+a))^(3/2)+2/15*b/(arctanh(tanh(b*x+a) 
)-b*x)^2/x^(3/2)*arctanh(tanh(b*x+a))^(3/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {2 \, {\left (16 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{4} x^{\frac {9}{2}}} \] Input:

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x, algorithm="fricas")
 

Output:

2/315*(16*b^4*x^4 - 8*a*b^3*x^3 + 6*a^2*b^2*x^2 - 5*a^3*b*x - 35*a^4)*sqrt 
(b*x + a)/(a^4*x^(9/2))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\text {Timed out} \] Input:

integrate(atanh(tanh(b*x+a))**(1/2)/x**(11/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {2 \, {\left (16 \, b^{4} x^{4} - 8 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{4} x^{\frac {9}{2}}} \] Input:

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x, algorithm="maxima")
 

Output:

2/315*(16*b^4*x^4 - 8*a*b^3*x^3 + 6*a^2*b^2*x^2 - 5*a^3*b*x - 35*a^4)*sqrt 
(b*x + a)/(a^4*x^(9/2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {64 \, {\left (315 \, b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{10} + 189 \, a b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{8} + 84 \, a^{2} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{6} - 36 \, a^{3} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{4} + 9 \, a^{4} b^{\frac {9}{2}} {\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a^{5} b^{\frac {9}{2}}\right )}}{315 \, {\left ({\left (\sqrt {b} \sqrt {x} - \sqrt {b x + a}\right )}^{2} - a\right )}^{9}} \] Input:

integrate(arctanh(tanh(b*x+a))^(1/2)/x^(11/2),x, algorithm="giac")
 

Output:

64/315*(315*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^10 + 189*a*b^(9/2)*( 
sqrt(b)*sqrt(x) - sqrt(b*x + a))^8 + 84*a^2*b^(9/2)*(sqrt(b)*sqrt(x) - sqr 
t(b*x + a))^6 - 36*a^3*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^4 + 9*a^4 
*b^(9/2)*(sqrt(b)*sqrt(x) - sqrt(b*x + a))^2 - a^5*b^(9/2))/((sqrt(b)*sqrt 
(x) - sqrt(b*x + a))^2 - a)^9
 

Mupad [B] (verification not implemented)

Time = 3.74 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {16\,b^2\,x^2}{105\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^3\,x^3}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^4\,x^4}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}+\frac {4\,b\,x}{63\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}-\frac {2}{9}\right )}{x^{9/2}} \] Input:

int(atanh(tanh(a + b*x))^(1/2)/x^(11/2),x)
 

Output:

((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 - log(2/(exp(2* 
a)*exp(2*b*x) + 1))/2)^(1/2)*((16*b^2*x^2)/(105*(log(2/(exp(2*a)*exp(2*b*x 
) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^ 
2) + (128*b^3*x^3)/(315*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a 
)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3) + (512*b^4*x^4)/(315* 
(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)* 
exp(2*b*x) + 1)) + 2*b*x)^4) + (4*b*x)/(63*(log(2/(exp(2*a)*exp(2*b*x) + 1 
)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)) - 2/ 
9))/x^(9/2)
 

Reduce [F]

\[ \int \frac {\sqrt {\text {arctanh}(\tanh (a+b x))}}{x^{11/2}} \, dx=\frac {-2 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}+\sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{5}}d x \right ) b \,x^{4}}{9 \sqrt {x}\, x^{4}} \] Input:

int(atanh(tanh(b*x+a))^(1/2)/x^(11/2),x)
 

Output:

( - 2*sqrt(atanh(tanh(a + b*x))) + sqrt(x)*int((sqrt(x)*sqrt(atanh(tanh(a 
+ b*x))))/(atanh(tanh(a + b*x))*x**5),x)*b*x**4)/(9*sqrt(x)*x**4)