3.103 \(\int \frac{\tanh ^6(x)}{a+a \text{sech}(x)} \, dx\)

Optimal. Leaf size=48 \[ \frac{x}{a}-\frac{3 \tan ^{-1}(\sinh (x))}{8 a}-\frac{\tanh ^3(x) (4-3 \text{sech}(x))}{12 a}-\frac{\tanh (x) (8-3 \text{sech}(x))}{8 a} \]

[Out]

x/a - (3*ArcTan[Sinh[x]])/(8*a) - ((8 - 3*Sech[x])*Tanh[x])/(8*a) - ((4 - 3*Sech[x])*Tanh[x]^3)/(12*a)

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Rubi [A]  time = 0.0967035, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3888, 3881, 3770} \[ \frac{x}{a}-\frac{3 \tan ^{-1}(\sinh (x))}{8 a}-\frac{\tanh ^3(x) (4-3 \text{sech}(x))}{12 a}-\frac{\tanh (x) (8-3 \text{sech}(x))}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^6/(a + a*Sech[x]),x]

[Out]

x/a - (3*ArcTan[Sinh[x]])/(8*a) - ((8 - 3*Sech[x])*Tanh[x])/(8*a) - ((4 - 3*Sech[x])*Tanh[x]^3)/(12*a)

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[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] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3881

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] - Dist[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]

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{\tanh ^6(x)}{a+a \text{sech}(x)} \, dx &=-\frac{\int (-a+a \text{sech}(x)) \tanh ^4(x) \, dx}{a^2}\\ &=-\frac{(4-3 \text{sech}(x)) \tanh ^3(x)}{12 a}-\frac{\int (-4 a+3 a \text{sech}(x)) \tanh ^2(x) \, dx}{4 a^2}\\ &=-\frac{(8-3 \text{sech}(x)) \tanh (x)}{8 a}-\frac{(4-3 \text{sech}(x)) \tanh ^3(x)}{12 a}-\frac{\int (-8 a+3 a \text{sech}(x)) \, dx}{8 a^2}\\ &=\frac{x}{a}-\frac{(8-3 \text{sech}(x)) \tanh (x)}{8 a}-\frac{(4-3 \text{sech}(x)) \tanh ^3(x)}{12 a}-\frac{3 \int \text{sech}(x) \, dx}{8 a}\\ &=\frac{x}{a}-\frac{3 \tan ^{-1}(\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}\\ \end{align*}

Mathematica [A]  time = 0.117712, size = 60, normalized size = 1.25 \[ \frac{\cosh ^2\left (\frac{x}{2}\right ) \text{sech}(x) \left (6 \left (4 x-3 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )\right )+\tanh (x) \left (-6 \text{sech}^3(x)+8 \text{sech}^2(x)+15 \text{sech}(x)-32\right )\right )}{12 a (\text{sech}(x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^6/(a + a*Sech[x]),x]

[Out]

(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]))

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Maple [B]  time = 0.059, size = 117, normalized size = 2.4 \begin{align*}{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{11}{4\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{7} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}-{\frac{137}{12\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}-{\frac{71}{12\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}-{\frac{5}{4\,a}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-4}}-{\frac{3}{4\,a}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^6/(a+a*sech(x)),x)

[Out]

1/a*ln(tanh(1/2*x)+1)-1/a*ln(tanh(1/2*x)-1)-11/4/a/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^7-137/12/a/(tanh(1/2*x)^2+1
)^4*tanh(1/2*x)^5-71/12/a/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)^3-5/4/a/(tanh(1/2*x)^2+1)^4*tanh(1/2*x)-3/4/a*arctan
(tanh(1/2*x))

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Maxima [B]  time = 1.70188, size = 126, normalized size = 2.62 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+a*sech(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.60885, size = 2288, normalized size = 47.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+a*sech(x)),x, algorithm="fricas")

[Out]

1/12*(12*x*cosh(x)^8 + 12*x*sinh(x)^8 + 3*(32*x*cosh(x) + 5)*sinh(x)^7 + 48*(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*c
osh(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 + 7
20*(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*cos
h(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 + 8*(7*a*cosh(x)^5 + 10*a*cosh(x)^3 + 3*a*cosh(x))*sinh(x)^3 + 4*
a*cosh(x)^2 + 4*(7*a*cosh(x)^6 + 15*a*cosh(x)^4 + 9*a*cosh(x)^2 + a)*sinh(x)^2 + 8*(a*cosh(x)^7 + 3*a*cosh(x)^
5 + 3*a*cosh(x)^3 + a*cosh(x))*sinh(x) + a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tanh ^{6}{\left (x \right )}}{\operatorname{sech}{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**6/(a+a*sech(x)),x)

[Out]

Integral(tanh(x)**6/(sech(x) + 1), x)/a

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Giac [A]  time = 1.15037, size = 93, normalized size = 1.94 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^6/(a+a*sech(x)),x, algorithm="giac")

[Out]

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)