\(\int (1+\coth (x))^{7/2} \, dx\) [70]

Optimal result

Integrand size = 8, antiderivative size = 57 \[ \int (1+\coth (x))^{7/2} \, dx=8 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\coth (x)}}{\sqrt {2}}\right )-8 \sqrt {1+\coth (x)}-\frac {4}{3} (1+\coth (x))^{3/2}-\frac {2}{5} (1+\coth (x))^{5/2} \] Output:

8*2^(1/2)*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))-8*(1+coth(x))^(1/2)-4/3*( 
1+coth(x))^(3/2)-2/5*(1+coth(x))^(5/2)
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int (1+\coth (x))^{7/2} \, dx=8 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\coth (x)}}{\sqrt {2}}\right )-\frac {2}{15} \sqrt {1+\coth (x)} \left (73+16 \coth (x)+3 \coth ^2(x)\right ) \] Input:

Integrate[(1 + Coth[x])^(7/2),x]
 

Output:

8*Sqrt[2]*ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]] - (2*Sqrt[1 + Coth[x]]*(73 + 
16*Coth[x] + 3*Coth[x]^2))/15
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {3042, 3959, 3042, 3959, 3042, 3959, 3042, 3961, 219}

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.

\(\displaystyle \int (\coth (x)+1)^{7/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (1-i \tan \left (\frac {\pi }{2}+i x\right )\right )^{7/2}dx\)

\(\Big \downarrow \) 3959

\(\displaystyle 2 \int (\coth (x)+1)^{5/2}dx-\frac {2}{5} (\coth (x)+1)^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2}{5} (\coth (x)+1)^{5/2}+2 \int \left (1-i \tan \left (i x+\frac {\pi }{2}\right )\right )^{5/2}dx\)

\(\Big \downarrow \) 3959

\(\displaystyle 2 \left (2 \int (\coth (x)+1)^{3/2}dx-\frac {2}{3} (\coth (x)+1)^{3/2}\right )-\frac {2}{5} (\coth (x)+1)^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2}{5} (\coth (x)+1)^{5/2}+2 \left (-\frac {2}{3} (\coth (x)+1)^{3/2}+2 \int \left (1-i \tan \left (i x+\frac {\pi }{2}\right )\right )^{3/2}dx\right )\)

\(\Big \downarrow \) 3959

\(\displaystyle 2 \left (2 \left (2 \int \sqrt {\coth (x)+1}dx-2 \sqrt {\coth (x)+1}\right )-\frac {2}{3} (\coth (x)+1)^{3/2}\right )-\frac {2}{5} (\coth (x)+1)^{5/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2}{5} (\coth (x)+1)^{5/2}+2 \left (-\frac {2}{3} (\coth (x)+1)^{3/2}+2 \left (-2 \sqrt {\coth (x)+1}+2 \int \sqrt {1-i \tan \left (i x+\frac {\pi }{2}\right )}dx\right )\right )\)

\(\Big \downarrow \) 3961

\(\displaystyle 2 \left (2 \left (4 \int \frac {1}{1-\coth (x)}d\sqrt {\coth (x)+1}-2 \sqrt {\coth (x)+1}\right )-\frac {2}{3} (\coth (x)+1)^{3/2}\right )-\frac {2}{5} (\coth (x)+1)^{5/2}\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (2 \left (2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {\coth (x)+1}}{\sqrt {2}}\right )-2 \sqrt {\coth (x)+1}\right )-\frac {2}{3} (\coth (x)+1)^{3/2}\right )-\frac {2}{5} (\coth (x)+1)^{5/2}\)

Input:

Int[(1 + Coth[x])^(7/2),x]
 

Output:

(-2*(1 + Coth[x])^(5/2))/5 + 2*((-2*(1 + Coth[x])^(3/2))/3 + 2*(2*Sqrt[2]* 
ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]] - 2*Sqrt[1 + Coth[x]]))
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3959

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[2*a   Int[(a + b*Tan[c + d* 
x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && GtQ[n 
, 1]
 

rule 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^2, 0]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(8 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\coth \left (x \right )}\, \sqrt {2}}{2}\right )-8 \sqrt {1+\coth \left (x \right )}-\frac {4 \left (1+\coth \left (x \right )\right )^{\frac {3}{2}}}{3}-\frac {2 \left (1+\coth \left (x \right )\right )^{\frac {5}{2}}}{5}\)	\(43\)
default	\(8 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\coth \left (x \right )}\, \sqrt {2}}{2}\right )-8 \sqrt {1+\coth \left (x \right )}-\frac {4 \left (1+\coth \left (x \right )\right )^{\frac {3}{2}}}{3}-\frac {2 \left (1+\coth \left (x \right )\right )^{\frac {5}{2}}}{5}\)	\(43\)

Input:

int((1+coth(x))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

8*2^(1/2)*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))-8*(1+coth(x))^(1/2)-4/3*( 
1+coth(x))^(3/2)-2/5*(1+coth(x))^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (42) = 84\).

Time = 0.09 (sec) , antiderivative size = 338, normalized size of antiderivative = 5.93 \[ \int (1+\coth (x))^{7/2} \, dx=\frac {4 \, {\left (15 \, {\left (\sqrt {2} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \sqrt {2} \cosh \left (x\right )^{2} - \sqrt {2}\right )} \sinh \left (x\right )^{2} - 2 \, \sqrt {2} \cosh \left (x\right )^{2} + 4 \, {\left (\sqrt {2} \cosh \left (x\right )^{3} - \sqrt {2} \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sqrt {2}\right )} \log \left (2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} + \frac {\sqrt {2} {\left (\sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt {2} \sinh \left (x\right )^{3} + {\left (3 \, \sqrt {2} \cosh \left (x\right )^{2} - \sqrt {2}\right )} \sinh \left (x\right ) - \sqrt {2} \cosh \left (x\right )\right )}}{\sqrt {\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 1}} - 1\right ) - \frac {2 \, \sqrt {2} {\left (23 \, \cosh \left (x\right )^{5} + 115 \, \cosh \left (x\right ) \sinh \left (x\right )^{4} + 23 \, \sinh \left (x\right )^{5} + 5 \, {\left (46 \, \cosh \left (x\right )^{2} - 7\right )} \sinh \left (x\right )^{3} - 35 \, \cosh \left (x\right )^{3} + 5 \, {\left (46 \, \cosh \left (x\right )^{3} - 21 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 5 \, {\left (23 \, \cosh \left (x\right )^{4} - 21 \, \cosh \left (x\right )^{2} + 3\right )} \sinh \left (x\right ) + 15 \, \cosh \left (x\right )\right )}}{\sqrt {\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 1}}\right )}}{15 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \, {\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - 2 \, \cosh \left (x\right )^{2} + 4 \, {\left (\cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \] Input:

integrate((1+coth(x))^(7/2),x, algorithm="fricas")
 

Output:

4/15*(15*(sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x 
)^4 + 2*(3*sqrt(2)*cosh(x)^2 - sqrt(2))*sinh(x)^2 - 2*sqrt(2)*cosh(x)^2 + 
4*(sqrt(2)*cosh(x)^3 - sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(2*cosh(x)^2 
 + 4*cosh(x)*sinh(x) + 2*sinh(x)^2 + sqrt(2)*(sqrt(2)*cosh(x)^3 + 3*sqrt(2 
)*cosh(x)*sinh(x)^2 + sqrt(2)*sinh(x)^3 + (3*sqrt(2)*cosh(x)^2 - sqrt(2))* 
sinh(x) - sqrt(2)*cosh(x))/sqrt(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 
- 1) - 1) - 2*sqrt(2)*(23*cosh(x)^5 + 115*cosh(x)*sinh(x)^4 + 23*sinh(x)^5 
 + 5*(46*cosh(x)^2 - 7)*sinh(x)^3 - 35*cosh(x)^3 + 5*(46*cosh(x)^3 - 21*co 
sh(x))*sinh(x)^2 + 5*(23*cosh(x)^4 - 21*cosh(x)^2 + 3)*sinh(x) + 15*cosh(x 
))/sqrt(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1))/(cosh(x)^4 + 4*cos 
h(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 
 4*(cosh(x)^3 - cosh(x))*sinh(x) + 1)
 

Sympy [F(-1)]

Timed out. \[ \int (1+\coth (x))^{7/2} \, dx=\text {Timed out} \] Input:

integrate((1+coth(x))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (1+\coth (x))^{7/2} \, dx=\int { {\left (\coth \left (x\right ) + 1\right )}^{\frac {7}{2}} \,d x } \] Input:

integrate((1+coth(x))^(7/2),x, algorithm="maxima")
 

Output:

integrate((coth(x) + 1)^(7/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (42) = 84\).

Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.81 \[ \int (1+\coth (x))^{7/2} \, dx=-\frac {4}{15} \, \sqrt {2} {\left (\frac {2 \, {\left (45 \, {\left (\sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} + 135 \, {\left (\sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 170 \, {\left (\sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} + 100 \, \sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 100 \, e^{\left (2 \, x\right )} + 23\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} + 1\right )}^{5}} + 15 \, \log \left ({\left | 2 \, \sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 2 \, e^{\left (2 \, x\right )} + 1 \right |}\right )\right )} \mathrm {sgn}\left (e^{\left (2 \, x\right )} - 1\right ) \] Input:

integrate((1+coth(x))^(7/2),x, algorithm="giac")
 

Output:

-4/15*sqrt(2)*(2*(45*(sqrt(e^(4*x) - e^(2*x)) - e^(2*x))^4 + 135*(sqrt(e^( 
4*x) - e^(2*x)) - e^(2*x))^3 + 170*(sqrt(e^(4*x) - e^(2*x)) - e^(2*x))^2 + 
 100*sqrt(e^(4*x) - e^(2*x)) - 100*e^(2*x) + 23)/(sqrt(e^(4*x) - e^(2*x)) 
- e^(2*x) + 1)^5 + 15*log(abs(2*sqrt(e^(4*x) - e^(2*x)) - 2*e^(2*x) + 1))) 
*sgn(e^(2*x) - 1)
 

Mupad [B] (verification not implemented)

Time = 2.50 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int (1+\coth (x))^{7/2} \, dx=-8\,\sqrt {\mathrm {coth}\left (x\right )+1}-\frac {4\,{\left (\mathrm {coth}\left (x\right )+1\right )}^{3/2}}{3}-\frac {2\,{\left (\mathrm {coth}\left (x\right )+1\right )}^{5/2}}{5}-\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {coth}\left (x\right )+1}\,1{}\mathrm {i}}{2}\right )\,8{}\mathrm {i} \] Input:

int((coth(x) + 1)^(7/2),x)
 

Output:

- 2^(1/2)*atan((2^(1/2)*(coth(x) + 1)^(1/2)*1i)/2)*8i - 8*(coth(x) + 1)^(1 
/2) - (4*(coth(x) + 1)^(3/2))/3 - (2*(coth(x) + 1)^(5/2))/5
 

Reduce [F]

\[ \int (1+\coth (x))^{7/2} \, dx=\int \sqrt {\coth \left (x \right )+1}d x +3 \left (\int \sqrt {\coth \left (x \right )+1}\, \coth \left (x \right )d x \right )+\int \sqrt {\coth \left (x \right )+1}\, \coth \left (x \right )^{3}d x +3 \left (\int \sqrt {\coth \left (x \right )+1}\, \coth \left (x \right )^{2}d x \right ) \] Input:

int((1+coth(x))^(7/2),x)
 

Output:

int(sqrt(coth(x) + 1),x) + 3*int(sqrt(coth(x) + 1)*coth(x),x) + int(sqrt(c 
oth(x) + 1)*coth(x)**3,x) + 3*int(sqrt(coth(x) + 1)*coth(x)**2,x)