3.101 \(\int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=34 \[ \frac {\text {sech}^5(x)}{5}+\frac {3}{8} \tan ^{-1}(\sinh (x))+\frac {1}{4} \tanh (x) \text {sech}^3(x)+\frac {3}{8} \tanh (x) \text {sech}(x) \]

[Out]

3/8*arctan(sinh(x))+1/5*sech(x)^5+3/8*sech(x)*tanh(x)+1/4*sech(x)^3*tanh(x)

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Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3501, 3768, 3770} \[ \frac {\text {sech}^5(x)}{5}+\frac {3}{8} \tan ^{-1}(\sinh (x))+\frac {1}{4} \tanh (x) \text {sech}^3(x)+\frac {3}{8} \tanh (x) \text {sech}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^7/(1 + Tanh[x]),x]

[Out]

(3*ArcTan[Sinh[x]])/8 + Sech[x]^5/5 + (3*Sech[x]*Tanh[x])/8 + (Sech[x]^3*Tanh[x])/4

Rule 3501

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d^2*
(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(d^2*(m - 2))/(a*(m + n -
1)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2
+ b^2, 0] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx &=\frac {\text {sech}^5(x)}{5}+\int \text {sech}^5(x) \, dx\\ &=\frac {\text {sech}^5(x)}{5}+\frac {1}{4} \text {sech}^3(x) \tanh (x)+\frac {3}{4} \int \text {sech}^3(x) \, dx\\ &=\frac {\text {sech}^5(x)}{5}+\frac {3}{8} \text {sech}(x) \tanh (x)+\frac {1}{4} \text {sech}^3(x) \tanh (x)+\frac {3}{8} \int \text {sech}(x) \, dx\\ &=\frac {3}{8} \tan ^{-1}(\sinh (x))+\frac {\text {sech}^5(x)}{5}+\frac {3}{8} \text {sech}(x) \tanh (x)+\frac {1}{4} \text {sech}^3(x) \tanh (x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 34, normalized size = 1.00 \[ \frac {1}{40} \left (8 \text {sech}^5(x)+30 \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+10 \tanh (x) \text {sech}^3(x)+15 \tanh (x) \text {sech}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^7/(1 + Tanh[x]),x]

[Out]

(30*ArcTan[Tanh[x/2]] + 8*Sech[x]^5 + 15*Sech[x]*Tanh[x] + 10*Sech[x]^3*Tanh[x])/40

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fricas [B]  time = 0.50, size = 670, normalized size = 19.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(15*cosh(x)^9 + 135*cosh(x)*sinh(x)^8 + 15*sinh(x)^9 + 10*(54*cosh(x)^2 + 7)*sinh(x)^7 + 70*cosh(x)^7 + 7
0*(18*cosh(x)^3 + 7*cosh(x))*sinh(x)^6 + 2*(945*cosh(x)^4 + 735*cosh(x)^2 + 64)*sinh(x)^5 + 128*cosh(x)^5 + 10
*(189*cosh(x)^5 + 245*cosh(x)^3 + 64*cosh(x))*sinh(x)^4 + 10*(126*cosh(x)^6 + 245*cosh(x)^4 + 128*cosh(x)^2 -
7)*sinh(x)^3 - 70*cosh(x)^3 + 10*(54*cosh(x)^7 + 147*cosh(x)^5 + 128*cosh(x)^3 - 21*cosh(x))*sinh(x)^2 + 15*(c
osh(x)^10 + 10*cosh(x)*sinh(x)^9 + sinh(x)^10 + 5*(9*cosh(x)^2 + 1)*sinh(x)^8 + 5*cosh(x)^8 + 40*(3*cosh(x)^3
+ cosh(x))*sinh(x)^7 + 10*(21*cosh(x)^4 + 14*cosh(x)^2 + 1)*sinh(x)^6 + 10*cosh(x)^6 + 4*(63*cosh(x)^5 + 70*co
sh(x)^3 + 15*cosh(x))*sinh(x)^5 + 10*(21*cosh(x)^6 + 35*cosh(x)^4 + 15*cosh(x)^2 + 1)*sinh(x)^4 + 10*cosh(x)^4
+ 40*(3*cosh(x)^7 + 7*cosh(x)^5 + 5*cosh(x)^3 + cosh(x))*sinh(x)^3 + 5*(9*cosh(x)^8 + 28*cosh(x)^6 + 30*cosh(
x)^4 + 12*cosh(x)^2 + 1)*sinh(x)^2 + 5*cosh(x)^2 + 10*(cosh(x)^9 + 4*cosh(x)^7 + 6*cosh(x)^5 + 4*cosh(x)^3 + c
osh(x))*sinh(x) + 1)*arctan(cosh(x) + sinh(x)) + 5*(27*cosh(x)^8 + 98*cosh(x)^6 + 128*cosh(x)^4 - 42*cosh(x)^2
- 3)*sinh(x) - 15*cosh(x))/(cosh(x)^10 + 10*cosh(x)*sinh(x)^9 + sinh(x)^10 + 5*(9*cosh(x)^2 + 1)*sinh(x)^8 +
5*cosh(x)^8 + 40*(3*cosh(x)^3 + cosh(x))*sinh(x)^7 + 10*(21*cosh(x)^4 + 14*cosh(x)^2 + 1)*sinh(x)^6 + 10*cosh(
x)^6 + 4*(63*cosh(x)^5 + 70*cosh(x)^3 + 15*cosh(x))*sinh(x)^5 + 10*(21*cosh(x)^6 + 35*cosh(x)^4 + 15*cosh(x)^2
+ 1)*sinh(x)^4 + 10*cosh(x)^4 + 40*(3*cosh(x)^7 + 7*cosh(x)^5 + 5*cosh(x)^3 + cosh(x))*sinh(x)^3 + 5*(9*cosh(
x)^8 + 28*cosh(x)^6 + 30*cosh(x)^4 + 12*cosh(x)^2 + 1)*sinh(x)^2 + 5*cosh(x)^2 + 10*(cosh(x)^9 + 4*cosh(x)^7 +
6*cosh(x)^5 + 4*cosh(x)^3 + cosh(x))*sinh(x) + 1)

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giac [A]  time = 0.12, size = 45, normalized size = 1.32 \[ \frac {15 \, e^{\left (9 \, x\right )} + 70 \, e^{\left (7 \, x\right )} + 128 \, e^{\left (5 \, x\right )} - 70 \, e^{\left (3 \, x\right )} - 15 \, e^{x}}{20 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} + \frac {3}{4} \, \arctan \left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(15*e^(9*x) + 70*e^(7*x) + 128*e^(5*x) - 70*e^(3*x) - 15*e^x)/(e^(2*x) + 1)^5 + 3/4*arctan(e^x)

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maple [B]  time = 0.09, size = 67, normalized size = 1.97 \[ \frac {-\frac {5 \left (\tanh ^{9}\left (\frac {x}{2}\right )\right )}{4}+2 \left (\tanh ^{8}\left (\frac {x}{2}\right )\right )-\frac {\left (\tanh ^{7}\left (\frac {x}{2}\right )\right )}{2}+4 \left (\tanh ^{4}\left (\frac {x}{2}\right )\right )+\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{2}+\frac {5 \tanh \left (\frac {x}{2}\right )}{4}+\frac {2}{5}}{\left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{5}}+\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-5/8*tanh(1/2*x)^9+tanh(1/2*x)^8-1/4*tanh(1/2*x)^7+2*tanh(1/2*x)^4+1/4*tanh(1/2*x)^3+5/8*tanh(1/2*x)+1/5)/(
tanh(1/2*x)^2+1)^5+3/4*arctan(tanh(1/2*x))

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maxima [B]  time = 0.41, size = 73, normalized size = 2.15 \[ \frac {15 \, e^{\left (-x\right )} + 70 \, e^{\left (-3 \, x\right )} + 128 \, e^{\left (-5 \, x\right )} - 70 \, e^{\left (-7 \, x\right )} - 15 \, e^{\left (-9 \, x\right )}}{20 \, {\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} - \frac {3}{4} \, \arctan \left (e^{\left (-x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(15*e^(-x) + 70*e^(-3*x) + 128*e^(-5*x) - 70*e^(-7*x) - 15*e^(-9*x))/(5*e^(-2*x) + 10*e^(-4*x) + 10*e^(-6
*x) + 5*e^(-8*x) + e^(-10*x) + 1) - 3/4*arctan(e^(-x))

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mupad [B]  time = 0.08, size = 137, normalized size = 4.03 \[ \frac {3\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{4}-\frac {32\,{\mathrm {e}}^{3\,x}}{5\,\left (5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{4\,x}+10\,{\mathrm {e}}^{6\,x}+5\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{10\,x}+1\right )}-\frac {12\,{\mathrm {e}}^x}{5\,\left (4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1\right )}+\frac {3\,{\mathrm {e}}^x}{4\,\left ({\mathrm {e}}^{2\,x}+1\right )}+\frac {2\,{\mathrm {e}}^x}{5\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}+\frac {{\mathrm {e}}^x}{2\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^7*(tanh(x) + 1)),x)

[Out]

(3*atan(exp(x)))/4 - (32*exp(3*x))/(5*(5*exp(2*x) + 10*exp(4*x) + 10*exp(6*x) + 5*exp(8*x) + exp(10*x) + 1)) -
(12*exp(x))/(5*(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x) + exp(8*x) + 1)) + (3*exp(x))/(4*(exp(2*x) + 1)) + (2*ex
p(x))/(5*(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1)) + exp(x)/(2*(2*exp(2*x) + exp(4*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)**7/(tanh(x) + 1), x)

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