Integrand size = 13, antiderivative size = 48 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a}-\frac {3 \arctan (\sinh (x))}{8 a}-\frac {(8-3 \text {sech}(x)) \tanh (x)}{8 a}-\frac {(4-3 \text {sech}(x)) \tanh ^3(x)}{12 a} \]
Time = 0.39 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.25 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {\cosh ^2\left (\frac {x}{2}\right ) \text {sech}(x) \left (6 \left (4 x-3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )\right )+\left (-32+15 \text {sech}(x)+8 \text {sech}^2(x)-6 \text {sech}^3(x)\right ) \tanh (x)\right )}{12 a (1+\text {sech}(x))} \]
(Cosh[x/2]^2*Sech[x]*(6*(4*x - 3*ArcTan[Tanh[x/2]]) + (-32 + 15*Sech[x] + 8*Sech[x]^2 - 6*Sech[x]^3)*Tanh[x]))/(12*a*(1 + Sech[x]))
Time = 0.43 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {3042, 25, 4376, 25, 3042, 4369, 25, 3042, 25, 4369, 2009}
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.
\(\displaystyle \int \frac {\tanh ^6(x)}{a \text {sech}(x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (\frac {\pi }{2}+i x\right )^6}{a+a \csc \left (\frac {\pi }{2}+i x\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (i x+\frac {\pi }{2}\right )^6}{\csc \left (i x+\frac {\pi }{2}\right ) a+a}dx\) |
\(\Big \downarrow \) 4376 |
\(\displaystyle -\frac {\int -\left ((a-a \text {sech}(x)) \tanh ^4(x)\right )dx}{a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int (a-a \text {sech}(x)) \tanh ^4(x)dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cot \left (i x+\frac {\pi }{2}\right )^4 \left (a-a \csc \left (i x+\frac {\pi }{2}\right )\right )dx}{a^2}\) |
\(\Big \downarrow \) 4369 |
\(\displaystyle \frac {-\frac {1}{4} \int -\left ((4 a-3 a \text {sech}(x)) \tanh ^2(x)\right )dx-\frac {1}{12} \tanh ^3(x) (4 a-3 a \text {sech}(x))}{a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{4} \int (4 a-3 a \text {sech}(x)) \tanh ^2(x)dx-\frac {1}{12} \tanh ^3(x) (4 a-3 a \text {sech}(x))}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {1}{12} \tanh ^3(x) (4 a-3 a \text {sech}(x))+\frac {1}{4} \int -\cot \left (i x+\frac {\pi }{2}\right )^2 \left (4 a-3 a \csc \left (i x+\frac {\pi }{2}\right )\right )dx}{a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {1}{12} \tanh ^3(x) (4 a-3 a \text {sech}(x))-\frac {1}{4} \int \cot \left (i x+\frac {\pi }{2}\right )^2 \left (4 a-3 a \csc \left (i x+\frac {\pi }{2}\right )\right )dx}{a^2}\) |
\(\Big \downarrow \) 4369 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int (8 a-3 a \text {sech}(x))dx-\frac {1}{2} \tanh (x) (8 a-3 a \text {sech}(x))\right )-\frac {1}{12} \tanh ^3(x) (4 a-3 a \text {sech}(x))}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} (8 a x-3 a \arctan (\sinh (x)))-\frac {1}{2} \tanh (x) (8 a-3 a \text {sech}(x))\right )-\frac {1}{12} \tanh ^3(x) (4 a-3 a \text {sech}(x))}{a^2}\) |
(-1/12*((4*a - 3*a*Sech[x])*Tanh[x]^3) + ((8*a*x - 3*a*ArcTan[Sinh[x]])/2 - ((8*a - 3*a*Sech[x])*Tanh[x])/2)/4)/a^2
3.2.3.3.1 Defintions of rubi rules used
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-e)*(e*Cot[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc [c + d*x])/(d*m*(m - 1))), x] - Simp[e^2/m Int[(e*Cot[c + d*x])^(m - 2)*( a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m , 1]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n) Int[(e*Cot[c + d*x])^(m + 2* n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a ^2 - b^2, 0] && ILtQ[n, 0]
Time = 0.67 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.56
method | result | size |
default | \(\frac {\frac {2 \left (-\frac {11 \tanh \left (\frac {x}{2}\right )^{7}}{8}-\frac {137 \tanh \left (\frac {x}{2}\right )^{5}}{24}-\frac {71 \tanh \left (\frac {x}{2}\right )^{3}}{24}-\frac {5 \tanh \left (\frac {x}{2}\right )}{8}\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{4}}-\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{4}+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}\) | \(75\) |
risch | \(\frac {x}{a}+\frac {15 \,{\mathrm e}^{7 x}+48 \,{\mathrm e}^{6 x}-9 \,{\mathrm e}^{5 x}+96 \,{\mathrm e}^{4 x}+9 \,{\mathrm e}^{3 x}+80 \,{\mathrm e}^{2 x}-15 \,{\mathrm e}^{x}+32}{12 \left (1+{\mathrm e}^{2 x}\right )^{4} a}+\frac {3 i \ln \left ({\mathrm e}^{x}-i\right )}{8 a}-\frac {3 i \ln \left ({\mathrm e}^{x}+i\right )}{8 a}\) | \(86\) |
64/a*(1/32*(-11/8*tanh(1/2*x)^7-137/24*tanh(1/2*x)^5-71/24*tanh(1/2*x)^3-5 /8*tanh(1/2*x))/(1+tanh(1/2*x)^2)^4-3/256*arctan(tanh(1/2*x))+1/64*ln(tanh (1/2*x)+1)-1/64*ln(tanh(1/2*x)-1))
Leaf count of result is larger than twice the leaf count of optimal. 686 vs. \(2 (42) = 84\).
Time = 0.27 (sec) , antiderivative size = 686, normalized size of antiderivative = 14.29 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\text {Too large to display} \]
1/12*(12*x*cosh(x)^8 + 12*x*sinh(x)^8 + 3*(32*x*cosh(x) + 5)*sinh(x)^7 + 4 8*(x + 1)*cosh(x)^6 + 15*cosh(x)^7 + 3*(112*x*cosh(x)^2 + 16*x + 35*cosh(x ) + 16)*sinh(x)^6 + 3*(224*x*cosh(x)^3 + 96*(x + 1)*cosh(x) + 105*cosh(x)^ 2 - 3)*sinh(x)^5 + 24*(3*x + 4)*cosh(x)^4 - 9*cosh(x)^5 + 3*(280*x*cosh(x) ^4 + 240*(x + 1)*cosh(x)^2 + 175*cosh(x)^3 + 24*x - 15*cosh(x) + 32)*sinh( x)^4 + 3*(224*x*cosh(x)^5 + 320*(x + 1)*cosh(x)^3 + 175*cosh(x)^4 + 32*(3* x + 4)*cosh(x) - 30*cosh(x)^2 + 3)*sinh(x)^3 + 16*(3*x + 5)*cosh(x)^2 + 9* cosh(x)^3 + (336*x*cosh(x)^6 + 720*(x + 1)*cosh(x)^4 + 315*cosh(x)^5 + 144 *(3*x + 4)*cosh(x)^2 - 90*cosh(x)^3 + 48*x + 27*cosh(x) + 80)*sinh(x)^2 - 9*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 + 1)*sinh( x)^6 + 4*cosh(x)^6 + 8*(7*cosh(x)^3 + 3*cosh(x))*sinh(x)^5 + 2*(35*cosh(x) ^4 + 30*cosh(x)^2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 + 10*cosh( x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 + 15*cosh(x)^4 + 9*cosh(x)^2 + 1)*sinh(x)^2 + 4*cosh(x)^2 + 8*(cosh(x)^7 + 3*cosh(x)^5 + 3*cosh(x)^3 + cosh(x))*sinh(x) + 1)*arctan(cosh(x) + sinh(x)) + (96*x*cosh(x)^7 + 288*(x + 1)*cosh(x)^5 + 105*cosh(x)^6 + 96*(3*x + 4)*cosh(x)^3 - 45*cosh(x)^4 + 32*(3*x + 5)*cosh(x) + 27*cosh(x)^2 - 15)*sinh(x) + 12*x - 15*cosh(x) + 32 )/(a*cosh(x)^8 + 8*a*cosh(x)*sinh(x)^7 + a*sinh(x)^8 + 4*a*cosh(x)^6 + 4*( 7*a*cosh(x)^2 + a)*sinh(x)^6 + 8*(7*a*cosh(x)^3 + 3*a*cosh(x))*sinh(x)^5 + 6*a*cosh(x)^4 + 2*(35*a*cosh(x)^4 + 30*a*cosh(x)^2 + 3*a)*sinh(x)^4 + ...
\[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {\int \frac {\tanh ^{6}{\left (x \right )}}{\operatorname {sech}{\left (x \right )} + 1}\, dx}{a} \]
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (42) = 84\).
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.94 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a} + \frac {15 \, e^{\left (-x\right )} - 80 \, e^{\left (-2 \, x\right )} - 9 \, e^{\left (-3 \, x\right )} - 96 \, e^{\left (-4 \, x\right )} + 9 \, e^{\left (-5 \, x\right )} - 48 \, e^{\left (-6 \, x\right )} - 15 \, e^{\left (-7 \, x\right )} - 32}{12 \, {\left (4 \, a e^{\left (-2 \, x\right )} + 6 \, a e^{\left (-4 \, x\right )} + 4 \, a e^{\left (-6 \, x\right )} + a e^{\left (-8 \, x\right )} + a\right )}} + \frac {3 \, \arctan \left (e^{\left (-x\right )}\right )}{4 \, a} \]
x/a + 1/12*(15*e^(-x) - 80*e^(-2*x) - 9*e^(-3*x) - 96*e^(-4*x) + 9*e^(-5*x ) - 48*e^(-6*x) - 15*e^(-7*x) - 32)/(4*a*e^(-2*x) + 6*a*e^(-4*x) + 4*a*e^( -6*x) + a*e^(-8*x) + a) + 3/4*arctan(e^(-x))/a
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a} - \frac {3 \, \arctan \left (e^{x}\right )}{4 \, a} + \frac {15 \, e^{\left (7 \, x\right )} + 48 \, e^{\left (6 \, x\right )} - 9 \, e^{\left (5 \, x\right )} + 96 \, e^{\left (4 \, x\right )} + 9 \, e^{\left (3 \, x\right )} + 80 \, e^{\left (2 \, x\right )} - 15 \, e^{x} + 32}{12 \, a {\left (e^{\left (2 \, x\right )} + 1\right )}^{4}} \]
x/a - 3/4*arctan(e^x)/a + 1/12*(15*e^(7*x) + 48*e^(6*x) - 9*e^(5*x) + 96*e ^(4*x) + 9*e^(3*x) + 80*e^(2*x) - 15*e^x + 32)/(a*(e^(2*x) + 1)^4)
Time = 2.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.98 \[ \int \frac {\tanh ^6(x)}{a+a \text {sech}(x)} \, dx=\frac {\frac {8}{3\,a}+\frac {6\,{\mathrm {e}}^x}{a}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {\frac {4}{a}+\frac {9\,{\mathrm {e}}^x}{2\,a}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {x}{a}+\frac {\frac {4}{a}+\frac {5\,{\mathrm {e}}^x}{4\,a}}{{\mathrm {e}}^{2\,x}+1}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^2}}{a}\right )}{4\,\sqrt {a^2}}-\frac {4\,{\mathrm {e}}^x}{a\,\left (4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1\right )} \]