Integrand size = 11, antiderivative size = 34 \[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\frac {3}{8} \arctan (\sinh (x))+\frac {\text {sech}^5(x)}{5}+\frac {3}{8} \text {sech}(x) \tanh (x)+\frac {1}{4} \text {sech}^3(x) \tanh (x) \] Output:
3/8*arctan(sinh(x))+1/5*sech(x)^5+3/8*sech(x)*tanh(x)+1/4*sech(x)^3*tanh(x )
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\frac {1}{40} \left (30 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+8 \text {sech}^5(x)+15 \text {sech}(x) \tanh (x)+10 \text {sech}^3(x) \tanh (x)\right ) \] Input:
Integrate[Sech[x]^7/(1 + Tanh[x]),x]
Output:
(30*ArcTan[Tanh[x/2]] + 8*Sech[x]^5 + 15*Sech[x]*Tanh[x] + 10*Sech[x]^3*Ta nh[x])/40
Time = 0.39 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {3042, 3982, 3042, 4255, 3042, 4255, 3042, 4257}
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.
\(\displaystyle \int \frac {\text {sech}^7(x)}{\tanh (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (i x)^7}{1-i \tan (i x)}dx\) |
\(\Big \downarrow \) 3982 |
\(\displaystyle \int \text {sech}^5(x)dx+\frac {\text {sech}^5(x)}{5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\text {sech}^5(x)}{5}+\int \csc \left (i x+\frac {\pi }{2}\right )^5dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {3}{4} \int \text {sech}^3(x)dx+\frac {\text {sech}^5(x)}{5}+\frac {1}{4} \tanh (x) \text {sech}^3(x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \int \csc \left (i x+\frac {\pi }{2}\right )^3dx+\frac {\text {sech}^5(x)}{5}+\frac {1}{4} \tanh (x) \text {sech}^3(x)\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {3}{4} \left (\frac {\int \text {sech}(x)dx}{2}+\frac {1}{2} \tanh (x) \text {sech}(x)\right )+\frac {\text {sech}^5(x)}{5}+\frac {1}{4} \tanh (x) \text {sech}^3(x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \tanh (x) \text {sech}(x)+\frac {1}{2} \int \csc \left (i x+\frac {\pi }{2}\right )dx\right )+\frac {\text {sech}^5(x)}{5}+\frac {1}{4} \tanh (x) \text {sech}^3(x)\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \arctan (\sinh (x))+\frac {1}{2} \tanh (x) \text {sech}(x)\right )+\frac {\text {sech}^5(x)}{5}+\frac {1}{4} \tanh (x) \text {sech}^3(x)\) |
Input:
Int[Sech[x]^7/(1 + Tanh[x]),x]
Output:
Sech[x]^5/5 + (Sech[x]^3*Tanh[x])/4 + (3*(ArcTan[Sinh[x]]/2 + (Sech[x]*Tan h[x])/2))/4
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] + Simp[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] && !IL tQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
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] + Simp[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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(66\) vs. \(2(26)=52\).
Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.97
\[\frac {-\frac {5 \tanh \left (\frac {x}{2}\right )^{9}}{4}+2 \tanh \left (\frac {x}{2}\right )^{8}-\frac {\tanh \left (\frac {x}{2}\right )^{7}}{2}+4 \tanh \left (\frac {x}{2}\right )^{4}+\frac {\tanh \left (\frac {x}{2}\right )^{3}}{2}+\frac {5 \tanh \left (\frac {x}{2}\right )}{4}+\frac {2}{5}}{\left (\tanh \left (\frac {x}{2}\right )^{2}+1\right )^{5}}+\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{4}\]
Input:
int(sech(x)^7/(1+tanh(x)),x)
Output:
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))
Leaf count of result is larger than twice the leaf count of optimal. 670 vs. \(2 (26) = 52\).
Time = 0.08 (sec) , antiderivative size = 670, normalized size of antiderivative = 19.71 \[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\text {Too large to display} \] Input:
integrate(sech(x)^7/(1+tanh(x)),x, algorithm="fricas")
Output:
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 + 70*(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*cos h(x)^7 + 147*cosh(x)^5 + 128*cosh(x)^3 - 21*cosh(x))*sinh(x)^2 + 15*(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 + 1 2*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)*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*c osh(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)^...
\[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\int \frac {\operatorname {sech}^{7}{\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \] Input:
integrate(sech(x)**7/(1+tanh(x)),x)
Output:
Integral(sech(x)**7/(tanh(x) + 1), x)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (26) = 52\).
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.15 \[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\frac {15 \, e^{\left (-x\right )} + 70 \, e^{\left (-3 \, x\right )} + 128 \, e^{\left (-5 \, x\right )} - 70 \, e^{\left (-7 \, x\right )} - 15 \, e^{\left (-9 \, x\right )}}{20 \, {\left (5 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} + 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} + 1\right )}} - \frac {3}{4} \, \arctan \left (e^{\left (-x\right )}\right ) \] Input:
integrate(sech(x)^7/(1+tanh(x)),x, algorithm="maxima")
Output:
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))
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32 \[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\frac {15 \, e^{\left (9 \, x\right )} + 70 \, e^{\left (7 \, x\right )} + 128 \, e^{\left (5 \, x\right )} - 70 \, e^{\left (3 \, x\right )} - 15 \, e^{x}}{20 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} + \frac {3}{4} \, \arctan \left (e^{x}\right ) \] Input:
integrate(sech(x)^7/(1+tanh(x)),x, algorithm="giac")
Output:
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)
Time = 2.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 4.03 \[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\frac {3\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{4}-\frac {32\,{\mathrm {e}}^{3\,x}}{5\,\left (5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{4\,x}+10\,{\mathrm {e}}^{6\,x}+5\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{10\,x}+1\right )}-\frac {12\,{\mathrm {e}}^x}{5\,\left (4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1\right )}+\frac {3\,{\mathrm {e}}^x}{4\,\left ({\mathrm {e}}^{2\,x}+1\right )}+\frac {2\,{\mathrm {e}}^x}{5\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}+\frac {{\mathrm {e}}^x}{2\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )} \] Input:
int(1/(cosh(x)^7*(tanh(x) + 1)),x)
Output:
(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*exp( 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))
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 3.97 \[ \int \frac {\text {sech}^7(x)}{1+\tanh (x)} \, dx=\frac {15 e^{10 x} \mathit {atan} \left (e^{x}\right )+75 e^{8 x} \mathit {atan} \left (e^{x}\right )+150 e^{6 x} \mathit {atan} \left (e^{x}\right )+150 e^{4 x} \mathit {atan} \left (e^{x}\right )+75 e^{2 x} \mathit {atan} \left (e^{x}\right )+15 \mathit {atan} \left (e^{x}\right )+15 e^{9 x}+70 e^{7 x}+128 e^{5 x}-70 e^{3 x}-15 e^{x}}{20 e^{10 x}+100 e^{8 x}+200 e^{6 x}+200 e^{4 x}+100 e^{2 x}+20} \] Input:
int(sech(x)^7/(1+tanh(x)),x)
Output:
(15*e**(10*x)*atan(e**x) + 75*e**(8*x)*atan(e**x) + 150*e**(6*x)*atan(e**x ) + 150*e**(4*x)*atan(e**x) + 75*e**(2*x)*atan(e**x) + 15*atan(e**x) + 15* e**(9*x) + 70*e**(7*x) + 128*e**(5*x) - 70*e**(3*x) - 15*e**x)/(20*(e**(10 *x) + 5*e**(8*x) + 10*e**(6*x) + 10*e**(4*x) + 5*e**(2*x) + 1))