\(\int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx\) [129]

Optimal result

Integrand size = 15, antiderivative size = 101 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {16 x^3 \text {arctanh}(\tanh (a+b x))^{9/2}}{63 b^2}+\frac {32 x^2 \text {arctanh}(\tanh (a+b x))^{11/2}}{231 b^3}-\frac {128 x \text {arctanh}(\tanh (a+b x))^{13/2}}{3003 b^4}+\frac {256 \text {arctanh}(\tanh (a+b x))^{15/2}}{45045 b^5} \] Output:

2/7*x^4*arctanh(tanh(b*x+a))^(7/2)/b-16/63*x^3*arctanh(tanh(b*x+a))^(9/2)/ 
b^2+32/231*x^2*arctanh(tanh(b*x+a))^(11/2)/b^3-128/3003*x*arctanh(tanh(b*x 
+a))^(13/2)/b^4+256/45045*arctanh(tanh(b*x+a))^(15/2)/b^5
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \text {arctanh}(\tanh (a+b x))^{7/2} \left (6435 b^4 x^4-5720 b^3 x^3 \text {arctanh}(\tanh (a+b x))+3120 b^2 x^2 \text {arctanh}(\tanh (a+b x))^2-960 b x \text {arctanh}(\tanh (a+b x))^3+128 \text {arctanh}(\tanh (a+b x))^4\right )}{45045 b^5} \] Input:

Integrate[x^4*ArcTanh[Tanh[a + b*x]]^(5/2),x]
 

Output:

(2*ArcTanh[Tanh[a + b*x]]^(7/2)*(6435*b^4*x^4 - 5720*b^3*x^3*ArcTanh[Tanh[ 
a + b*x]] + 3120*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 - 960*b*x*ArcTanh[Tanh[a 
 + b*x]]^3 + 128*ArcTanh[Tanh[a + b*x]]^4))/(45045*b^5)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.24, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2599, 2599, 2599, 2599, 2588, 15}

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.

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

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 \int x^3 \text {arctanh}(\tanh (a+b x))^{7/2}dx}{7 b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \int x^2 \text {arctanh}(\tanh (a+b x))^{9/2}dx}{3 b}\right )}{7 b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {4 \int x \text {arctanh}(\tanh (a+b x))^{11/2}dx}{11 b}\right )}{3 b}\right )}{7 b}\)

\(\Big \downarrow \) 2599

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{13/2}}{13 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{13/2}dx}{13 b}\right )}{11 b}\right )}{3 b}\right )}{7 b}\)

\(\Big \downarrow \) 2588

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{13/2}}{13 b}-\frac {2 \int \text {arctanh}(\tanh (a+b x))^{13/2}d\text {arctanh}(\tanh (a+b x))}{13 b^2}\right )}{11 b}\right )}{3 b}\right )}{7 b}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 x^4 \text {arctanh}(\tanh (a+b x))^{7/2}}{7 b}-\frac {8 \left (\frac {2 x^3 \text {arctanh}(\tanh (a+b x))^{9/2}}{9 b}-\frac {2 \left (\frac {2 x^2 \text {arctanh}(\tanh (a+b x))^{11/2}}{11 b}-\frac {4 \left (\frac {2 x \text {arctanh}(\tanh (a+b x))^{13/2}}{13 b}-\frac {4 \text {arctanh}(\tanh (a+b x))^{15/2}}{195 b^2}\right )}{11 b}\right )}{3 b}\right )}{7 b}\)

Input:

Int[x^4*ArcTanh[Tanh[a + b*x]]^(5/2),x]
 

Output:

(2*x^4*ArcTanh[Tanh[a + b*x]]^(7/2))/(7*b) - (8*((2*x^3*ArcTanh[Tanh[a + b 
*x]]^(9/2))/(9*b) - (2*((2*x^2*ArcTanh[Tanh[a + b*x]]^(11/2))/(11*b) - (4* 
((2*x*ArcTanh[Tanh[a + b*x]]^(13/2))/(13*b) - (4*ArcTanh[Tanh[a + b*x]]^(1 
5/2))/(195*b^2)))/(11*b)))/(3*b)))/(7*b)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 2588

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Simp[1/c   Subst 
[Int[x^m, x], x, u], x]] /; FreeQ[m, x] && PiecewiseLinearQ[u, 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]))
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.52

method	result	size
default	\(\frac {\frac {2 \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {15}{2}}}{15}+\frac {2 \left (-4 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )+4 b x \right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {2 \left (2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}+\left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2}\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {11}{2}}}{11}+\frac {4 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{2} \left (2 b x -2 \,\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right ) \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 \left (b x -\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )\right )^{4} \operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )^{\frac {7}{2}}}{7}}{b^{5}}\)	\(154\)

Input:

int(x^4*arctanh(tanh(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/b^5*(1/15*arctanh(tanh(b*x+a))^(15/2)+1/13*(-4*arctanh(tanh(b*x+a))+4*b* 
x)*arctanh(tanh(b*x+a))^(13/2)+1/11*(2*(b*x-arctanh(tanh(b*x+a)))^2+(2*b*x 
-2*arctanh(tanh(b*x+a)))^2)*arctanh(tanh(b*x+a))^(11/2)+2/9*(b*x-arctanh(t 
anh(b*x+a)))^2*(2*b*x-2*arctanh(tanh(b*x+a)))*arctanh(tanh(b*x+a))^(9/2)+1 
/7*(b*x-arctanh(tanh(b*x+a)))^4*arctanh(tanh(b*x+a))^(7/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \, {\left (3003 \, b^{7} x^{7} + 7161 \, a b^{6} x^{6} + 4473 \, a^{2} b^{5} x^{5} + 35 \, a^{3} b^{4} x^{4} - 40 \, a^{4} b^{3} x^{3} + 48 \, a^{5} b^{2} x^{2} - 64 \, a^{6} b x + 128 \, a^{7}\right )} \sqrt {b x + a}}{45045 \, b^{5}} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

2/45045*(3003*b^7*x^7 + 7161*a*b^6*x^6 + 4473*a^2*b^5*x^5 + 35*a^3*b^4*x^4 
 - 40*a^4*b^3*x^3 + 48*a^5*b^2*x^2 - 64*a^6*b*x + 128*a^7)*sqrt(b*x + a)/b 
^5
 

Sympy [F(-1)]

Timed out. \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(x**4*atanh(tanh(b*x+a))**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {2 \, {\left (3003 \, b^{5} x^{5} + 1155 \, a b^{4} x^{4} - 840 \, a^{2} b^{3} x^{3} + 560 \, a^{3} b^{2} x^{2} - 320 \, a^{4} b x + 128 \, a^{5}\right )} {\left (b x + a\right )}^{\frac {5}{2}}}{45045 \, b^{5}} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

2/45045*(3003*b^5*x^5 + 1155*a*b^4*x^4 - 840*a^2*b^3*x^3 + 560*a^3*b^2*x^2 
 - 320*a^4*b*x + 128*a^5)*(b*x + a)^(5/2)/b^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (81) = 162\).

Time = 0.11 (sec) , antiderivative size = 344, normalized size of antiderivative = 3.41 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\frac {\sqrt {2} {\left (\frac {143 \, \sqrt {2} {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{3}}{b^{4}} + \frac {195 \, \sqrt {2} {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a^{2}}{b^{4}} + \frac {45 \, \sqrt {2} {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )} a}{b^{4}} + \frac {7 \, \sqrt {2} {\left (429 \, {\left (b x + a\right )}^{\frac {15}{2}} - 3465 \, {\left (b x + a\right )}^{\frac {13}{2}} a + 12285 \, {\left (b x + a\right )}^{\frac {11}{2}} a^{2} - 25025 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{3} + 32175 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4} - 27027 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} + 15015 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} - 6435 \, \sqrt {b x + a} a^{7}\right )}}{b^{4}}\right )}}{45045 \, b} \] Input:

integrate(x^4*arctanh(tanh(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

1/45045*sqrt(2)*(143*sqrt(2)*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 
 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4 
)*a^3/b^4 + 195*sqrt(2)*(63*(b*x + a)^(11/2) - 385*(b*x + a)^(9/2)*a + 990 
*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 
 - 693*sqrt(b*x + a)*a^5)*a^2/b^4 + 45*sqrt(2)*(231*(b*x + a)^(13/2) - 163 
8*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b*x + a)^(7/2)*a^3 
 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x + a 
)*a^6)*a/b^4 + 7*sqrt(2)*(429*(b*x + a)^(15/2) - 3465*(b*x + a)^(13/2)*a + 
 12285*(b*x + a)^(11/2)*a^2 - 25025*(b*x + a)^(9/2)*a^3 + 32175*(b*x + a)^ 
(7/2)*a^4 - 27027*(b*x + a)^(5/2)*a^5 + 15015*(b*x + a)^(3/2)*a^6 - 6435*s 
qrt(b*x + a)*a^7)/b^4)/b
 

Mupad [B] (verification not implemented)

Time = 3.42 (sec) , antiderivative size = 2681, normalized size of antiderivative = 26.54 \[ \int x^4 \text {arctanh}(\tanh (a+b x))^{5/2} \, dx=\text {Too large to display} \] Input:

int(x^4*atanh(tanh(a + b*x))^(5/2),x)
 

Output:

(2*b^2*x^7*(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))/15 + (x^5*(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)*((3*b*(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)/2 - (12*(3*b^2*(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) - (28*b^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp( 
2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b*x))/15)*(log(2/(exp(2*a)*exp(2*b* 
x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))/2 + b* 
x))/(13*b)))/(11*b) - (x^6*(3*b^2*(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) - (28*b^2*(log 
(2/(exp(2*a)*exp(2*b*x) + 1))/2 - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*ex 
p(2*b*x) + 1))/2 + b*x))/15)*(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))/(13*b) - (x^4*(l 
og((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)) - log((2*exp( 
2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3/4 - (10*((3*b*(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)/2 - (12*(3*b^2*(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) - (28*...
 

Reduce [F]

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

int(x^4*atanh(tanh(b*x+a))^(5/2),x)
 

Output:

int(sqrt(atanh(tanh(a + b*x)))*atanh(tanh(a + b*x))**2*x**4,x)