\(\int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx\) [240]

Optimal result

Integrand size = 17, antiderivative size = 148 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=\frac {32 b^3 \text {arctanh}(\tanh (a+b x))^{7/2}}{3003 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^4}+\frac {16 b^2 \text {arctanh}(\tanh (a+b x))^{7/2}}{429 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {12 b \text {arctanh}(\tanh (a+b x))^{7/2}}{143 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{13 x^{13/2} (b x-\text {arctanh}(\tanh (a+b x)))} \] Output:

32/3003*b^3*arctanh(tanh(b*x+a))^(7/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^ 
4+16/429*b^2*arctanh(tanh(b*x+a))^(7/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a))) 
^3+12/143*b*arctanh(tanh(b*x+a))^(7/2)/x^(11/2)/(b*x-arctanh(tanh(b*x+a))) 
^2+2/13*arctanh(tanh(b*x+a))^(7/2)/x^(13/2)/(b*x-arctanh(tanh(b*x+a)))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.55 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2} \left (429 b^3 x^3-1001 b^2 x^2 \text {arctanh}(\tanh (a+b x))+819 b x \text {arctanh}(\tanh (a+b x))^2-231 \text {arctanh}(\tanh (a+b x))^3\right )}{3003 x^{13/2} (-b x+\text {arctanh}(\tanh (a+b x)))^4} \] Input:

Integrate[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(15/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(429*b^3*x^3 - 1001*b^2*x^2*ArcTanh[Tanh[a 
 + b*x]] + 819*b*x*ArcTanh[Tanh[a + b*x]]^2 - 231*ArcTanh[Tanh[a + b*x]]^3 
))/(3003*x^(13/2)*(-(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.

\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx\)

\(\Big \downarrow \) 2602

\(\displaystyle \frac {6 b \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{13/2}}dx}{13 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{13 x^{13/2} (b x-\text {arctanh}(\tanh (a+b x)))}\)

\(\Big \downarrow \) 2602

\(\displaystyle \frac {6 b \left (\frac {4 b \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{11/2}}dx}{11 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{11 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{13 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{13 x^{13/2} (b x-\text {arctanh}(\tanh (a+b x)))}\)

\(\Big \downarrow \) 2602

\(\displaystyle \frac {6 b \left (\frac {4 b \left (\frac {2 b \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{9/2}}dx}{9 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{11 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{11 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{13 (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{13 x^{13/2} (b x-\text {arctanh}(\tanh (a+b x)))}\)

\(\Big \downarrow \) 2598

\(\displaystyle \frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{13 x^{13/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {6 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{11 x^{11/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{7/2}}{63 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}\right )}{11 (b x-\text {arctanh}(\tanh (a+b x)))}\right )}{13 (b x-\text {arctanh}(\tanh (a+b x)))}\)

Input:

Int[ArcTanh[Tanh[a + b*x]]^(5/2)/x^(15/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(7/2))/(13*x^(13/2)*(b*x - ArcTanh[Tanh[a + b*x] 
])) + (6*b*((2*ArcTanh[Tanh[a + b*x]]^(7/2))/(11*x^(11/2)*(b*x - ArcTanh[T 
anh[a + b*x]])) + (4*b*((4*b*ArcTanh[Tanh[a + b*x]]^(7/2))/(63*x^(7/2)*(b* 
x - ArcTanh[Tanh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(7/2))/(9*x^(9/ 
2)*(b*x - ArcTanh[Tanh[a + b*x]]))))/(11*(b*x - ArcTanh[Tanh[a + b*x]])))) 
/(13*(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 = 0.50 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{13 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {13}{2}}}-\frac {12 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}}\right )}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{13 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(151\)
default	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{13 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {13}{2}}}-\frac {12 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {11}{2}}}-\frac {4 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{63 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {7}{2}}}\right )}{11 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\right )}{13 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(151\)

Input:

int(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x,method=_RETURNVERBOSE)
 

Output:

-2/13/(arctanh(tanh(b*x+a))-b*x)/x^(13/2)*arctanh(tanh(b*x+a))^(7/2)-12/13 
*b/(arctanh(tanh(b*x+a))-b*x)*(-1/11/(arctanh(tanh(b*x+a))-b*x)/x^(11/2)*a 
rctanh(tanh(b*x+a))^(7/2)-4/11*b/(arctanh(tanh(b*x+a))-b*x)*(-1/9/(arctanh 
(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(7/2)+2/63*b/(arctanh(tanh 
(b*x+a))-b*x)^2/x^(7/2)*arctanh(tanh(b*x+a))^(7/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.53 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=\frac {2 \, {\left (16 \, b^{6} x^{6} - 8 \, a b^{5} x^{5} + 6 \, a^{2} b^{4} x^{4} - 5 \, a^{3} b^{3} x^{3} - 371 \, a^{4} b^{2} x^{2} - 567 \, a^{5} b x - 231 \, a^{6}\right )} \sqrt {b x + a}}{3003 \, a^{4} x^{\frac {13}{2}}} \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x, algorithm="fricas")
 

Output:

2/3003*(16*b^6*x^6 - 8*a*b^5*x^5 + 6*a^2*b^4*x^4 - 5*a^3*b^3*x^3 - 371*a^4 
*b^2*x^2 - 567*a^5*b*x - 231*a^6)*sqrt(b*x + a)/(a^4*x^(13/2))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=\text {Timed out} \] Input:

integrate(atanh(tanh(b*x+a))**(5/2)/x**(15/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.38 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=\frac {2 \, {\left (16 \, b^{4} x^{4} - 40 \, a b^{3} x^{3} + 70 \, a^{2} b^{2} x^{2} - 105 \, a^{3} b x - 231 \, a^{4}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{3003 \, a^{4} x^{\frac {13}{2}}} \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x, algorithm="maxima")
 

Output:

2/3003*(16*b^4*x^4 - 40*a*b^3*x^3 + 70*a^2*b^2*x^2 - 105*a^3*b*x - 231*a^4 
)*(b*x + a)^(5/2)/(a^4*x^(13/2))
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {429 \, \sqrt {2} b^{13}}{a} - 2 \, {\left (\frac {143 \, \sqrt {2} b^{13}}{a^{2}} + 4 \, {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{13}}{a^{4}} - \frac {13 \, \sqrt {2} b^{13}}{a^{3}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {7}{2}} b}{3003 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {13}{2}} {\left | b \right |}} \] Input:

integrate(arctanh(tanh(b*x+a))^(5/2)/x^(15/2),x, algorithm="giac")
 

Output:

-1/3003*sqrt(2)*(429*sqrt(2)*b^13/a - 2*(143*sqrt(2)*b^13/a^2 + 4*(2*sqrt( 
2)*(b*x + a)*b^13/a^4 - 13*sqrt(2)*b^13/a^3)*(b*x + a))*(b*x + a))*(b*x + 
a)^(7/2)*b/(((b*x + a)*b - a*b)^(13/2)*abs(b))
 

Mupad [B] (verification not implemented)

Time = 3.93 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.79 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {27\,b\,x\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{143}-\frac {106\,b^2\,x^2}{429}-\frac {{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{26}+\frac {20\,b^3\,x^3}{3003\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {16\,b^4\,x^4}{1001\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^5\,x^5}{3003\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}+\frac {512\,b^6\,x^6}{3003\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4}\right )}{x^{13/2}} \] Input:

int(atanh(tanh(a + b*x))^(5/2)/x^(15/2),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)*((27*b*x*(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))/143 - (10 
6*b^2*x^2)/429 - (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)^2/26 + (20*b^3*x^3)/(3003*(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)) + (16*b^4*x^4)/(1001*(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^5*x^5)/(3003*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*ex 
p(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3) + (512*b^6*x^6)/(3003*(lo 
g(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)))/x^(13/2)
 

Reduce [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{5/2}}{x^{15/2}} \, dx=\frac {-66 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )^{2}-30 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right ) b x -10 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, b^{2} x^{2}+5 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{5}}d x \right ) b^{3} x^{6}}{429 \sqrt {x}\, x^{6}} \] Input:

int(atanh(tanh(b*x+a))^(5/2)/x^(15/2),x)
 

Output:

( - 66*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))**2 - 30*sqrt(atanh( 
tanh(a + b*x)))*atanh(tanh(a + b*x))*b*x - 10*sqrt(atanh(tanh(a + b*x)))*b 
**2*x**2 + 5*sqrt(x)*int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atanh(tanh( 
a + b*x))*x**5),x)*b**3*x**6)/(429*sqrt(x)*x**6)