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*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

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Rubi [A]  time = 0.0409798, antiderivative size = 57, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.200731, 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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Maple [A]  time = 0.019, size = 43, normalized size = 0.8 \begin{align*} 8\,{\it Artanh} \left ( 1/2\,\sqrt{1+\tanh \left ( x \right ) }\sqrt{2} \right ) \sqrt{2}-8\,\sqrt{1+\tanh \left ( x \right ) }-{\frac{4}{3} \left ( 1+\tanh \left ( x \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{2}{5} \left ( 1+\tanh \left ( x \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

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 = 1.63342, size = 112, normalized size = 1.96 \begin{align*} -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}}} \end{align*}

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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Fricas [B]  time = 2.34353, size = 1470, normalized size = 25.79 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.2774, size = 189, normalized size = 3.32 \begin{align*} \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 )} \end{align*}

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))