3.50 \(\int (1+\tanh (x))^{7/2} \, dx\)

Optimal. Leaf size=57 \[ 8 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}-\frac {4}{3} (\tanh (x)+1)^{3/2}-8 \sqrt {\tanh (x)+1} \]

[Out]

8*arctanh(1/2*(1+tanh(x))^(1/2)*2^(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)

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3478, 3480, 206} \[ 8 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}-\frac {4}{3} (\tanh (x)+1)^{3/2}-8 \sqrt {\tanh (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tanh[x])^(7/2),x]

[Out]

8*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - 8*Sqrt[1 + Tanh[x]] - (4*(1 + Tanh[x])^(3/2))/3 - (2*(1 + Tanh[
x])^(5/2))/5

Rule 206

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

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[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] && G
tQ[n, 1]

Rule 3480

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

Rubi steps

\begin {align*} \int (1+\tanh (x))^{7/2} \, dx &=-\frac {2}{5} (1+\tanh (x))^{5/2}+2 \int (1+\tanh (x))^{5/2} \, dx\\ &=-\frac {4}{3} (1+\tanh (x))^{3/2}-\frac {2}{5} (1+\tanh (x))^{5/2}+4 \int (1+\tanh (x))^{3/2} \, dx\\ &=-8 \sqrt {1+\tanh (x)}-\frac {4}{3} (1+\tanh (x))^{3/2}-\frac {2}{5} (1+\tanh (x))^{5/2}+8 \int \sqrt {1+\tanh (x)} \, dx\\ &=-8 \sqrt {1+\tanh (x)}-\frac {4}{3} (1+\tanh (x))^{3/2}-\frac {2}{5} (1+\tanh (x))^{5/2}+16 \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1+\tanh (x)}\right )\\ &=8 \sqrt {2} \tanh ^{-1}\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}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 65, normalized size = 1.14 \[ \frac {\cosh ^3(x) \left (8 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right ) (\tanh (x)+1)^3-\frac {2}{15} (\tanh (x)+1)^{7/2} \left (16 \tanh (x)-3 \text {sech}^2(x)+76\right )\right )}{(\sinh (x)+\cosh (x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tanh[x])^(7/2),x]

[Out]

(Cosh[x]^3*(8*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]*(1 + Tanh[x])^3 - (2*(1 + Tanh[x])^(7/2)*(76 - 3*Sech
[x]^2 + 16*Tanh[x]))/15))/(Cosh[x] + Sinh[x])^3

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fricas [B]  time = 0.70, size = 434, normalized size = 7.61 \[ -\frac {4 \, {\left (2 \, \sqrt {2} {\left (23 \, \sqrt {2} \cosh \relax (x)^{5} + 115 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{4} + 23 \, \sqrt {2} \sinh \relax (x)^{5} + 5 \, {\left (46 \, \sqrt {2} \cosh \relax (x)^{2} + 7 \, \sqrt {2}\right )} \sinh \relax (x)^{3} + 35 \, \sqrt {2} \cosh \relax (x)^{3} + 5 \, {\left (46 \, \sqrt {2} \cosh \relax (x)^{3} + 21 \, \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x)^{2} + 5 \, {\left (23 \, \sqrt {2} \cosh \relax (x)^{4} + 21 \, \sqrt {2} \cosh \relax (x)^{2} + 3 \, \sqrt {2}\right )} \sinh \relax (x) + 15 \, \sqrt {2} \cosh \relax (x)\right )} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} - 15 \, {\left (\sqrt {2} \cosh \relax (x)^{6} + 6 \, \sqrt {2} \cosh \relax (x) \sinh \relax (x)^{5} + \sqrt {2} \sinh \relax (x)^{6} + 3 \, {\left (5 \, \sqrt {2} \cosh \relax (x)^{2} + \sqrt {2}\right )} \sinh \relax (x)^{4} + 3 \, \sqrt {2} \cosh \relax (x)^{4} + 4 \, {\left (5 \, \sqrt {2} \cosh \relax (x)^{3} + 3 \, \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 3 \, {\left (5 \, \sqrt {2} \cosh \relax (x)^{4} + 6 \, \sqrt {2} \cosh \relax (x)^{2} + \sqrt {2}\right )} \sinh \relax (x)^{2} + 3 \, \sqrt {2} \cosh \relax (x)^{2} + 6 \, {\left (\sqrt {2} \cosh \relax (x)^{5} + 2 \, \sqrt {2} \cosh \relax (x)^{3} + \sqrt {2} \cosh \relax (x)\right )} \sinh \relax (x) + \sqrt {2}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}} {\left (\cosh \relax (x) + \sinh \relax (x)\right )} - 2 \, \cosh \relax (x)^{2} - 4 \, \cosh \relax (x) \sinh \relax (x) - 2 \, \sinh \relax (x)^{2} - 1\right )\right )}}{15 \, {\left (\cosh \relax (x)^{6} + 6 \, \cosh \relax (x) \sinh \relax (x)^{5} + \sinh \relax (x)^{6} + 3 \, {\left (5 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{4} + 3 \, \cosh \relax (x)^{4} + 4 \, {\left (5 \, \cosh \relax (x)^{3} + 3 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 3 \, {\left (5 \, \cosh \relax (x)^{4} + 6 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + 3 \, \cosh \relax (x)^{2} + 6 \, {\left (\cosh \relax (x)^{5} + 2 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*(2*sqrt(2)*(23*sqrt(2)*cosh(x)^5 + 115*sqrt(2)*cosh(x)*sinh(x)^4 + 23*sqrt(2)*sinh(x)^5 + 5*(46*sqrt(2)*
cosh(x)^2 + 7*sqrt(2))*sinh(x)^3 + 35*sqrt(2)*cosh(x)^3 + 5*(46*sqrt(2)*cosh(x)^3 + 21*sqrt(2)*cosh(x))*sinh(x
)^2 + 5*(23*sqrt(2)*cosh(x)^4 + 21*sqrt(2)*cosh(x)^2 + 3*sqrt(2))*sinh(x) + 15*sqrt(2)*cosh(x))*sqrt(cosh(x)/(
cosh(x) - sinh(x))) - 15*(sqrt(2)*cosh(x)^6 + 6*sqrt(2)*cosh(x)*sinh(x)^5 + sqrt(2)*sinh(x)^6 + 3*(5*sqrt(2)*c
osh(x)^2 + sqrt(2))*sinh(x)^4 + 3*sqrt(2)*cosh(x)^4 + 4*(5*sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x))*sinh(x)^3 +
3*(5*sqrt(2)*cosh(x)^4 + 6*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^2 + 3*sqrt(2)*cosh(x)^2 + 6*(sqrt(2)*cosh(x)^5
+ 2*sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(-2*sqrt(2)*sqrt(cosh(x)/(cosh(x) - sinh(x)))*
(cosh(x) + sinh(x)) - 2*cosh(x)^2 - 4*cosh(x)*sinh(x) - 2*sinh(x)^2 - 1))/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + s
inh(x)^6 + 3*(5*cosh(x)^2 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(5*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^
4 + 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 + 2*cosh(x)^3 + cosh(x))*sinh(x) + 1)

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giac [B]  time = 0.15, size = 140, normalized size = 2.46 \[ \frac {4}{15} \, \sqrt {2} {\left (\frac {2 \, {\left (45 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} - 135 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 170 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 100 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 100 \, e^{\left (2 \, x\right )} + 23\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1\right )}^{5}} - 15 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*(s
qrt(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))

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maple [A]  time = 0.05, size = 43, normalized size = 0.75 \[ 8 \arctanh \left (\frac {\sqrt {1+\tanh \relax (x )}\, \sqrt {2}}{2}\right ) \sqrt {2}-8 \sqrt {1+\tanh \relax (x )}-\frac {4 \left (1+\tanh \relax (x )\right )^{\frac {3}{2}}}{3}-\frac {2 \left (1+\tanh \relax (x )\right )^{\frac {5}{2}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*arctanh(1/2*(1+tanh(x))^(1/2)*2^(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)

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maxima [A]  time = 0.44, size = 83, normalized size = 1.46 \[ -4 \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) - \frac {8 \, \sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}} - \frac {8 \, \sqrt {2}}{3 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {3}{2}}} - \frac {8 \, \sqrt {2}}{5 \, {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)/sqr
t(e^(-2*x) + 1) - 8/3*sqrt(2)/(e^(-2*x) + 1)^(3/2) - 8/5*sqrt(2)/(e^(-2*x) + 1)^(5/2)

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mupad [B]  time = 0.19, size = 44, normalized size = 0.77 \[ -8\,\sqrt {\mathrm {tanh}\relax (x)+1}-\frac {4\,{\left (\mathrm {tanh}\relax (x)+1\right )}^{3/2}}{3}-\frac {2\,{\left (\mathrm {tanh}\relax (x)+1\right )}^{5/2}}{5}-\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\relax (x)+1}\,1{}\mathrm {i}}{2}\right )\,8{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x) + 1)^(7/2),x)

[Out]

- 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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\tanh {\relax (x )} + 1\right )^{\frac {7}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((tanh(x) + 1)**(7/2), x)

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