Integrand size = 8, antiderivative size = 57 \[ \int (1+\tanh (x))^{7/2} \, dx=8 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-8 \sqrt {1+\tanh (x)}-\frac {4}{3} (1+\tanh (x))^{3/2}-\frac {2}{5} (1+\tanh (x))^{5/2} \] Output:
8*2^(1/2)*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))-8*(1+tanh(x))^(1/2)-4/3*( 1+tanh(x))^(3/2)-2/5*(1+tanh(x))^(5/2)
Time = 0.38 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int (1+\tanh (x))^{7/2} \, dx=8 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-\frac {2}{15} \sqrt {1+\tanh (x)} \left (73+16 \tanh (x)+3 \tanh ^2(x)\right ) \] Input:
Integrate[(1 + Tanh[x])^(7/2),x]
Output:
8*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - (2*Sqrt[1 + Tanh[x]]*(73 + 16*Tanh[x] + 3*Tanh[x]^2))/15
Time = 0.47 (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.
\(\displaystyle \int (\tanh (x)+1)^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (1-i \tan (i x))^{7/2}dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 \int (\tanh (x)+1)^{5/2}dx-\frac {2}{5} (\tanh (x)+1)^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2}{5} (\tanh (x)+1)^{5/2}+2 \int (1-i \tan (i x))^{5/2}dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 \left (2 \int (\tanh (x)+1)^{3/2}dx-\frac {2}{3} (\tanh (x)+1)^{3/2}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2}{5} (\tanh (x)+1)^{5/2}+2 \left (-\frac {2}{3} (\tanh (x)+1)^{3/2}+2 \int (1-i \tan (i x))^{3/2}dx\right )\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 \left (2 \left (2 \int \sqrt {\tanh (x)+1}dx-2 \sqrt {\tanh (x)+1}\right )-\frac {2}{3} (\tanh (x)+1)^{3/2}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2}{5} (\tanh (x)+1)^{5/2}+2 \left (-\frac {2}{3} (\tanh (x)+1)^{3/2}+2 \left (-2 \sqrt {\tanh (x)+1}+2 \int \sqrt {1-i \tan (i x)}dx\right )\right )\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle 2 \left (2 \left (4 \int \frac {1}{1-\tanh (x)}d\sqrt {\tanh (x)+1}-2 \sqrt {\tanh (x)+1}\right )-\frac {2}{3} (\tanh (x)+1)^{3/2}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (2 \left (2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-2 \sqrt {\tanh (x)+1}\right )-\frac {2}{3} (\tanh (x)+1)^{3/2}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}\) |
Input:
Int[(1 + Tanh[x])^(7/2),x]
Output:
(-2*(1 + Tanh[x])^(5/2))/5 + 2*((-2*(1 + Tanh[x])^(3/2))/3 + 2*(2*Sqrt[2]* ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - 2*Sqrt[1 + Tanh[x]]))
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])
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]
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]
Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(8 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right )-8 \sqrt {1+\tanh \left (x \right )}-\frac {4 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}{3}-\frac {2 \left (1+\tanh \left (x \right )\right )^{\frac {5}{2}}}{5}\) | \(43\) |
default | \(8 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right )-8 \sqrt {1+\tanh \left (x \right )}-\frac {4 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}{3}-\frac {2 \left (1+\tanh \left (x \right )\right )^{\frac {5}{2}}}{5}\) | \(43\) |
Input:
int((1+tanh(x))^(7/2),x,method=_RETURNVERBOSE)
Output:
8*2^(1/2)*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))-8*(1+tanh(x))^(1/2)-4/3*( 1+tanh(x))^(3/2)-2/5*(1+tanh(x))^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 331 vs. \(2 (42) = 84\).
Time = 0.10 (sec) , antiderivative size = 331, normalized size of antiderivative = 5.81 \[ \int (1+\tanh (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+tanh(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*c osh(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*co sh(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)
\[ \int (1+\tanh (x))^{7/2} \, dx=\int \left (\tanh {\left (x \right )} + 1\right )^{\frac {7}{2}}\, dx \] Input:
integrate((1+tanh(x))**(7/2),x)
Output:
Integral((tanh(x) + 1)**(7/2), x)
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.46 \[ \int (1+\tanh (x))^{7/2} \, dx=-4 \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) - \frac {8 \, \sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}} - \frac {8 \, \sqrt {2}}{3 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {3}{2}}} - \frac {8 \, \sqrt {2}}{5 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} \] Input:
integrate((1+tanh(x))^(7/2),x, algorithm="maxima")
Output:
-4*sqrt(2)*log(-(sqrt(2) - sqrt(2)/sqrt(e^(-2*x) + 1))/(sqrt(2) + sqrt(2)/ sqrt(e^(-2*x) + 1))) - 8*sqrt(2)/sqrt(e^(-2*x) + 1) - 8/3*sqrt(2)/(e^(-2*x ) + 1)^(3/2) - 8/5*sqrt(2)/(e^(-2*x) + 1)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (42) = 84\).
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.46 \[ \int (1+\tanh (x))^{7/2} \, dx=\frac {4}{15} \, \sqrt {2} {\left (\frac {2 \, {\left (45 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} - 135 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 170 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 100 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 100 \, e^{\left (2 \, x\right )} + 23\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1\right )}^{5}} - 15 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \] Input:
integrate((1+tanh(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(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))
Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int (1+\tanh (x))^{7/2} \, dx=-8\,\sqrt {\mathrm {tanh}\left (x\right )+1}-\frac {4\,{\left (\mathrm {tanh}\left (x\right )+1\right )}^{3/2}}{3}-\frac {2\,{\left (\mathrm {tanh}\left (x\right )+1\right )}^{5/2}}{5}-\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}\,1{}\mathrm {i}}{2}\right )\,8{}\mathrm {i} \] Input:
int((tanh(x) + 1)^(7/2),x)
Output:
- 2^(1/2)*atan((2^(1/2)*(tanh(x) + 1)^(1/2)*1i)/2)*8i - 8*(tanh(x) + 1)^(1 /2) - (4*(tanh(x) + 1)^(3/2))/3 - (2*(tanh(x) + 1)^(5/2))/5
\[ \int (1+\tanh (x))^{7/2} \, dx=\int \sqrt {\tanh \left (x \right )+1}d x +\int \sqrt {\tanh \left (x \right )+1}\, \tanh \left (x \right )^{3}d x +3 \left (\int \sqrt {\tanh \left (x \right )+1}\, \tanh \left (x \right )^{2}d x \right )+3 \left (\int \sqrt {\tanh \left (x \right )+1}\, \tanh \left (x \right )d x \right ) \] Input:
int((1+tanh(x))^(7/2),x)
Output:
int(sqrt(tanh(x) + 1),x) + int(sqrt(tanh(x) + 1)*tanh(x)**3,x) + 3*int(sqr t(tanh(x) + 1)*tanh(x)**2,x) + 3*int(sqrt(tanh(x) + 1)*tanh(x),x)