\(\int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx\) [230]

Optimal result

Integrand size = 17, antiderivative size = 110 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=\frac {16 b^2 \text {arctanh}(\tanh (a+b x))^{5/2}}{315 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^3}+\frac {8 b \text {arctanh}(\tanh (a+b x))^{5/2}}{63 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}+\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))} \] Output:

16/315*b^2*arctanh(tanh(b*x+a))^(5/2)/x^(5/2)/(b*x-arctanh(tanh(b*x+a)))^3 
+8/63*b*arctanh(tanh(b*x+a))^(5/2)/x^(7/2)/(b*x-arctanh(tanh(b*x+a)))^2+2/ 
9*arctanh(tanh(b*x+a))^(5/2)/x^(9/2)/(b*x-arctanh(tanh(b*x+a)))
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.60 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2} \left (63 b^2 x^2-90 b x \text {arctanh}(\tanh (a+b x))+35 \text {arctanh}(\tanh (a+b x))^2\right )}{315 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))^3} \] Input:

Integrate[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(11/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(5/2)*(63*b^2*x^2 - 90*b*x*ArcTanh[Tanh[a + b*x] 
] + 35*ArcTanh[Tanh[a + b*x]]^2))/(315*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x 
]])^3)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {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.

\(\displaystyle \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx\)

\(\Big \downarrow \) 2602

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

\(\Big \downarrow \) 2602

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

\(\Big \downarrow \) 2598

\(\displaystyle \frac {2 \text {arctanh}(\tanh (a+b x))^{5/2}}{9 x^{9/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \left (\frac {2 \text {arctanh}(\tanh (a+b x))^{5/2}}{7 x^{7/2} (b x-\text {arctanh}(\tanh (a+b x)))}+\frac {4 b \text {arctanh}(\tanh (a+b x))^{5/2}}{35 x^{5/2} (b x-\text {arctanh}(\tanh (a+b x)))^2}\right )}{9 (b x-\text {arctanh}(\tanh (a+b x)))}\)

Input:

Int[ArcTanh[Tanh[a + b*x]]^(3/2)/x^(11/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(5/2))/(9*x^(9/2)*(b*x - ArcTanh[Tanh[a + b*x]]) 
) + (4*b*((4*b*ArcTanh[Tanh[a + b*x]]^(5/2))/(35*x^(5/2)*(b*x - ArcTanh[Ta 
nh[a + b*x]])^2) + (2*ArcTanh[Tanh[a + b*x]]^(5/2))/(7*x^(7/2)*(b*x - ArcT 
anh[Tanh[a + b*x]]))))/(9*(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 = 0.40 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{35 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {5}{2}}}\right )}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(105\)
default	\(-\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {9}{2}}}-\frac {8 b \left (-\frac {\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{7 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) x^{\frac {7}{2}}}+\frac {2 b \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {5}{2}}}{35 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )^{2} x^{\frac {5}{2}}}\right )}{9 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right )}\)	\(105\)

Input:

int(arctanh(tanh(b*x+a))^(3/2)/x^(11/2),x,method=_RETURNVERBOSE)
 

Output:

-2/9/(arctanh(tanh(b*x+a))-b*x)/x^(9/2)*arctanh(tanh(b*x+a))^(5/2)-8/9*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))^(5/2)+2/35*b/(arctanh(tanh(b*x+a))-b*x)^2/x^(5/2)*arctanh(ta 
nh(b*x+a))^(5/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.51 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=-\frac {2 \, {\left (8 \, b^{4} x^{4} - 4 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 50 \, a^{3} b x + 35 \, a^{4}\right )} \sqrt {b x + a}}{315 \, a^{3} x^{\frac {9}{2}}} \] Input:

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(11/2),x, algorithm="fricas")
 

Output:

-2/315*(8*b^4*x^4 - 4*a*b^3*x^3 + 3*a^2*b^2*x^2 + 50*a^3*b*x + 35*a^4)*sqr 
t(b*x + a)/(a^3*x^(9/2))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=\text {Timed out} \] Input:

integrate(atanh(tanh(b*x+a))**(3/2)/x**(11/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.41 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{3} - 12 \, a b^{2} x^{2} + 15 \, a^{2} b x + 35 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {3}{2}}}{315 \, a^{3} x^{\frac {9}{2}}} \] Input:

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(11/2),x, algorithm="maxima")
 

Output:

-2/315*(8*b^3*x^3 - 12*a*b^2*x^2 + 15*a^2*b*x + 35*a^3)*(b*x + a)^(3/2)/(a 
^3*x^(9/2))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.71 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {63 \, \sqrt {2} b^{9}}{a} + 4 \, {\left (\frac {2 \, \sqrt {2} {\left (b x + a\right )} b^{9}}{a^{3}} - \frac {9 \, \sqrt {2} b^{9}}{a^{2}}\right )} {\left (b x + a\right )}\right )} {\left (b x + a\right )}^{\frac {5}{2}} b}{315 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {9}{2}} {\left | b \right |}} \] Input:

integrate(arctanh(tanh(b*x+a))^(3/2)/x^(11/2),x, algorithm="giac")
 

Output:

-1/315*sqrt(2)*(63*sqrt(2)*b^9/a + 4*(2*sqrt(2)*(b*x + a)*b^9/a^3 - 9*sqrt 
(2)*b^9/a^2)*(b*x + a))*(b*x + a)^(5/2)*b/(((b*x + a)*b - a*b)^(9/2)*abs(b 
))
 

Mupad [B] (verification not implemented)

Time = 3.94 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.62 \[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=\frac {\sqrt {\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}-\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{2}}\,\left (\frac {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{9}-\frac {2\,b\,x}{21}+\frac {4\,b^2\,x^2}{105\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {32\,b^3\,x^3}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}+\frac {128\,b^4\,x^4}{315\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3}\right )}{x^{9/2}} \] Input:

int(atanh(tanh(a + b*x))^(3/2)/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)*(log(2/(exp(2*a)*exp(2*b*x) + 1))/9 - log((2* 
exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/9 - (2*b*x)/21 + (4*b^2*x^ 
2)/(105*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*exp(2*a)*exp(2*b*x))/(e 
xp(2*a)*exp(2*b*x) + 1)) + 2*b*x)) + (32*b^3*x^3)/(315*(log(2/(exp(2*a)*ex 
p(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^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)^3)))/x^(9/2)
 

Reduce [F]

\[ \int \frac {\text {arctanh}(\tanh (a+b x))^{3/2}}{x^{11/2}} \, dx=\frac {-14 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, \mathit {atanh} \left (\tanh \left (b x +a \right )\right )-6 \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}\, b x +3 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {\mathit {atanh} \left (\tanh \left (b x +a \right )\right )}}{\mathit {atanh} \left (\tanh \left (b x +a \right )\right ) x^{4}}d x \right ) b^{2} x^{4}}{63 \sqrt {x}\, x^{4}} \] Input:

int(atanh(tanh(b*x+a))^(3/2)/x^(11/2),x)
 

Output:

( - 14*sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x)) - 6*sqrt(atanh(tanh 
(a + b*x)))*b*x + 3*sqrt(x)*int((sqrt(x)*sqrt(atanh(tanh(a + b*x))))/(atan 
h(tanh(a + b*x))*x**4),x)*b**2*x**4)/(63*sqrt(x)*x**4)