3.2.92 \(\int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))} \, dx\) [192]

3.2.92.1 Optimal result

Integrand size = 15, antiderivative size = 116 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 x^{5/2}}{5 b}+\frac {2 x^{3/2} (b x-\text {arctanh}(\tanh (a+b x)))}{3 b^2}+\frac {2 \sqrt {x} (b x-\text {arctanh}(\tanh (a+b x)))^2}{b^3}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x-\text {arctanh}(\tanh (a+b x))}}\right ) (b x-\text {arctanh}(\tanh (a+b x)))^{5/2}}{b^{7/2}} \]

output

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

3.2.92.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2 \left (23 b^{5/2} x^{5/2}-35 b^{3/2} x^{3/2} \text {arctanh}(\tanh (a+b x))+15 \sqrt {b} \sqrt {x} \text {arctanh}(\tanh (a+b x))^2-15 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {-b x+\text {arctanh}(\tanh (a+b x))}}\right ) (-b x+\text {arctanh}(\tanh (a+b x)))^{5/2}\right )}{15 b^{7/2}} \]

input

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

output

(2*(23*b^(5/2)*x^(5/2) - 35*b^(3/2)*x^(3/2)*ArcTanh[Tanh[a + b*x]] + 15*Sq 
rt[b]*Sqrt[x]*ArcTanh[Tanh[a + b*x]]^2 - 15*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[ 
-(b*x) + ArcTanh[Tanh[a + b*x]]]]*(-(b*x) + ArcTanh[Tanh[a + b*x]])^(5/2)) 
)/(15*b^(7/2))
 

3.2.92.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2590, 2590, 2590, 2593}

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.

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2590

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

\(\Big \downarrow \) 2593

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

input

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

output

(2*x^(5/2))/(5*b) + (((2*x^(3/2))/(3*b) + (((2*Sqrt[x])/b - (2*ArcTanh[(Sq 
rt[b]*Sqrt[x])/Sqrt[b*x - ArcTanh[Tanh[a + b*x]]]]*Sqrt[b*x - ArcTanh[Tanh 
[a + b*x]]])/b^(3/2))*(b*x - ArcTanh[Tanh[a + b*x]]))/b)*(b*x - ArcTanh[Ta 
nh[a + b*x]]))/b
 

3.2.92.3.1 Defintions of rubi rules used

rule 2590

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

rule 2593

Int[1/((u_)*Sqrt[v_]), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simpli 
fy[D[v, x]]}, Simp[-2*(ArcTanh[Sqrt[v]/Rt[-(b*u - a*v)/a, 2]]/(a*Rt[-(b*u - 
 a*v)/a, 2])), x] /; NeQ[b*u - a*v, 0] && NegQ[(b*u - a*v)/a]] /; Piecewise 
LinearQ[u, v, x]
 

3.2.92.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(200\) vs. \(2(96)=192\).

Time = 0.12 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(\frac {\frac {2 b^{2} x^{\frac {5}{2}}}{5}-\frac {2 b \,x^{\frac {3}{2}} a}{3}-\frac {2 b \,x^{\frac {3}{2}} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3}+2 a^{2} \sqrt {x}+4 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}+2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{b^{3}}+\frac {2 \left (-a^{3}-3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{b^{3} \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\)	\(201\)
default	\(\frac {\frac {2 b^{2} x^{\frac {5}{2}}}{5}-\frac {2 b \,x^{\frac {3}{2}} a}{3}-\frac {2 b \,x^{\frac {3}{2}} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )}{3}+2 a^{2} \sqrt {x}+4 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right ) \sqrt {x}+2 \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} \sqrt {x}}{b^{3}}+\frac {2 \left (-a^{3}-3 a^{2} \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )-3 a \left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}-\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}\right ) \arctan \left (\frac {b \sqrt {x}}{\sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\right )}{b^{3} \sqrt {\left (\operatorname {arctanh}\left (\tanh \left (b x +a \right )\right )-b x \right ) b}}\)	\(201\)

input

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

output

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

3.2.92.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.14 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))} \, dx=\left [\frac {15 \, a^{2} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, a^{2} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, b^{3}}\right ] \]

input

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

output

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

3.2.92.6 Sympy [F]

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

input

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

output

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

3.2.92.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.47 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, b^{2} x^{\frac {5}{2}} - 5 \, a b x^{\frac {3}{2}} + 15 \, a^{2} \sqrt {x}\right )}}{15 \, b^{3}} \]

input

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

output

-2*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*b^2*x^(5/2) - 
 5*a*b*x^(3/2) + 15*a^2*sqrt(x))/b^3
 

3.2.92.8 Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.51 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))} \, dx=-\frac {2 \, a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (3 \, b^{4} x^{\frac {5}{2}} - 5 \, a b^{3} x^{\frac {3}{2}} + 15 \, a^{2} b^{2} \sqrt {x}\right )}}{15 \, b^{5}} \]

input

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

output

-2*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*b^4*x^(5/2) - 
 5*a*b^3*x^(3/2) + 15*a^2*b^2*sqrt(x))/b^5
 

3.2.92.9 Mupad [B] (verification not implemented)

Time = 4.50 (sec) , antiderivative size = 415, normalized size of antiderivative = 3.58 \[ \int \frac {x^{5/2}}{\text {arctanh}(\tanh (a+b x))} \, dx=\frac {2\,x^{5/2}}{5\,b}+\frac {x^{3/2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{3\,b^2}+\frac {\sqrt {x}\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{2\,b^3}+\frac {\sqrt {2}\,\ln \left (\frac {16\,b^{15/2}\,\left (\sqrt {2}\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )-4\,\sqrt {b}\,\sqrt {x}\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}+2\,\sqrt {2}\,b\,x\right )}{\left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\sqrt {\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x}}\right )\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^{5/2}}{8\,b^{7/2}} \]

input

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

output

(2*x^(5/2))/(5*b) + (x^(3/2)*(log(2/(exp(2*a)*exp(2*b*x) + 1)) - log((2*ex 
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(3*b^2) + (x^(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)^2)/(2*b^3) + (2^(1/2)*log((16*b^(15/2)*(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) - 4*b^(1/2)*x^(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)^( 
1/2) + 2*2^(1/2)*b*x))/((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) 
+ 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))*(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)^(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)^(5/2))/(8*b^(7/2))