3.819 \(\int \frac {1}{(\text {sech}^2(x)-\tanh ^2(x))^3} \, dx\)

Optimal. Leaf size=54 \[ -x+\frac {7 \tanh ^{-1}\left (\sqrt {2} \tanh (x)\right )}{4 \sqrt {2}}-\frac {\tanh (x)}{4 \left (1-2 \tanh ^2(x)\right )}+\frac {\tanh (x)}{2 \left (1-2 \tanh ^2(x)\right )^2} \]

[Out]

-x+7/8*arctanh(2^(1/2)*tanh(x))*2^(1/2)+1/2*tanh(x)/(1-2*tanh(x)^2)^2-1/4*tanh(x)/(1-2*tanh(x)^2)

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Rubi [A]  time = 0.06, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {414, 527, 522, 206} \[ -x+\frac {7 \tanh ^{-1}\left (\sqrt {2} \tanh (x)\right )}{4 \sqrt {2}}-\frac {\tanh (x)}{4 \left (1-2 \tanh ^2(x)\right )}+\frac {\tanh (x)}{2 \left (1-2 \tanh ^2(x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sech[x]^2 - Tanh[x]^2)^(-3),x]

[Out]

-x + (7*ArcTanh[Sqrt[2]*Tanh[x]])/(4*Sqrt[2]) + Tanh[x]/(2*(1 - 2*Tanh[x]^2)^2) - Tanh[x]/(4*(1 - 2*Tanh[x]^2)
)

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])

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {1}{\left (\text {sech}^2(x)-\tanh ^2(x)\right )^3} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1-2 x^2\right )^3 \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\frac {\tanh (x)}{2 \left (1-2 \tanh ^2(x)\right )^2}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {2-6 x^2}{\left (1-2 x^2\right )^2 \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\frac {\tanh (x)}{2 \left (1-2 \tanh ^2(x)\right )^2}-\frac {\tanh (x)}{4 \left (1-2 \tanh ^2(x)\right )}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {6+2 x^2}{\left (1-2 x^2\right ) \left (1-x^2\right )} \, dx,x,\tanh (x)\right )\\ &=\frac {\tanh (x)}{2 \left (1-2 \tanh ^2(x)\right )^2}-\frac {\tanh (x)}{4 \left (1-2 \tanh ^2(x)\right )}+\frac {7}{4} \operatorname {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\tanh (x)\right )-\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (x)\right )\\ &=-x+\frac {7 \tanh ^{-1}\left (\sqrt {2} \tanh (x)\right )}{4 \sqrt {2}}+\frac {\tanh (x)}{2 \left (1-2 \tanh ^2(x)\right )^2}-\frac {\tanh (x)}{4 \left (1-2 \tanh ^2(x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 66, normalized size = 1.22 \[ \frac {-76 x-2 \sinh (2 x)+3 \sinh (4 x)+48 x \cosh (2 x)-4 x \cosh (4 x)+7 \sqrt {2} (\cosh (2 x)-3)^2 \tanh ^{-1}\left (\sqrt {2} \tanh (x)\right )}{8 (\cosh (2 x)-3)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sech[x]^2 - Tanh[x]^2)^(-3),x]

[Out]

(-76*x + 7*Sqrt[2]*ArcTanh[Sqrt[2]*Tanh[x]]*(-3 + Cosh[2*x])^2 + 48*x*Cosh[2*x] - 4*x*Cosh[4*x] - 2*Sinh[2*x]
+ 3*Sinh[4*x])/(8*(-3 + Cosh[2*x])^2)

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fricas [B]  time = 0.45, size = 717, normalized size = 13.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)^2-tanh(x)^2)^3,x, algorithm="fricas")

[Out]

-1/16*(16*x*cosh(x)^8 + 128*x*cosh(x)*sinh(x)^7 + 16*x*sinh(x)^8 - 8*(24*x + 17)*cosh(x)^6 + 8*(56*x*cosh(x)^2
- 24*x - 17)*sinh(x)^6 + 16*(56*x*cosh(x)^3 - 3*(24*x + 17)*cosh(x))*sinh(x)^5 + 152*(4*x + 3)*cosh(x)^4 + 8*
(140*x*cosh(x)^4 - 15*(24*x + 17)*cosh(x)^2 + 76*x + 57)*sinh(x)^4 + 32*(28*x*cosh(x)^5 - 5*(24*x + 17)*cosh(x
)^3 + 19*(4*x + 3)*cosh(x))*sinh(x)^3 - 8*(24*x + 19)*cosh(x)^2 + 8*(56*x*cosh(x)^6 - 15*(24*x + 17)*cosh(x)^4
+ 114*(4*x + 3)*cosh(x)^2 - 24*x - 19)*sinh(x)^2 - 7*(sqrt(2)*cosh(x)^8 + 8*sqrt(2)*cosh(x)*sinh(x)^7 + sqrt(
2)*sinh(x)^8 + 4*(7*sqrt(2)*cosh(x)^2 - 3*sqrt(2))*sinh(x)^6 - 12*sqrt(2)*cosh(x)^6 + 8*(7*sqrt(2)*cosh(x)^3 -
9*sqrt(2)*cosh(x))*sinh(x)^5 + 2*(35*sqrt(2)*cosh(x)^4 - 90*sqrt(2)*cosh(x)^2 + 19*sqrt(2))*sinh(x)^4 + 38*sq
rt(2)*cosh(x)^4 + 8*(7*sqrt(2)*cosh(x)^5 - 30*sqrt(2)*cosh(x)^3 + 19*sqrt(2)*cosh(x))*sinh(x)^3 + 4*(7*sqrt(2)
*cosh(x)^6 - 45*sqrt(2)*cosh(x)^4 + 57*sqrt(2)*cosh(x)^2 - 3*sqrt(2))*sinh(x)^2 - 12*sqrt(2)*cosh(x)^2 + 8*(sq
rt(2)*cosh(x)^7 - 9*sqrt(2)*cosh(x)^5 + 19*sqrt(2)*cosh(x)^3 - 3*sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(-(3*(
2*sqrt(2) - 3)*cosh(x)^2 - 4*(3*sqrt(2) - 4)*cosh(x)*sinh(x) + 3*(2*sqrt(2) - 3)*sinh(x)^2 - 2*sqrt(2) + 3)/(c
osh(x)^2 + sinh(x)^2 - 3)) + 16*(8*x*cosh(x)^7 - 3*(24*x + 17)*cosh(x)^5 + 38*(4*x + 3)*cosh(x)^3 - (24*x + 19
)*cosh(x))*sinh(x) + 16*x + 24)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 - 3)*sinh(x)^6 -
12*cosh(x)^6 + 8*(7*cosh(x)^3 - 9*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 - 90*cosh(x)^2 + 19)*sinh(x)^4 + 38*co
sh(x)^4 + 8*(7*cosh(x)^5 - 30*cosh(x)^3 + 19*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 45*cosh(x)^4 + 57*cosh(x)^2
- 3)*sinh(x)^2 - 12*cosh(x)^2 + 8*(cosh(x)^7 - 9*cosh(x)^5 + 19*cosh(x)^3 - 3*cosh(x))*sinh(x) + 1)

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giac [A]  time = 0.12, size = 77, normalized size = 1.43 \[ -\frac {7}{16} \, \sqrt {2} \log \left (\frac {{\left | -4 \, \sqrt {2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}{{\left | 4 \, \sqrt {2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}\right ) - x + \frac {17 \, e^{\left (6 \, x\right )} - 57 \, e^{\left (4 \, x\right )} + 19 \, e^{\left (2 \, x\right )} - 3}{2 \, {\left (e^{\left (4 \, x\right )} - 6 \, e^{\left (2 \, x\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)^2-tanh(x)^2)^3,x, algorithm="giac")

[Out]

-7/16*sqrt(2)*log(abs(-4*sqrt(2) + 2*e^(2*x) - 6)/abs(4*sqrt(2) + 2*e^(2*x) - 6)) - x + 1/2*(17*e^(6*x) - 57*e
^(4*x) + 19*e^(2*x) - 3)/(e^(4*x) - 6*e^(2*x) + 1)^2

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maple [B]  time = 0.25, size = 140, normalized size = 2.59 \[ \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {2 \left (-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{8}-\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}-\frac {5 \tanh \left (\frac {x}{2}\right )}{8}-\frac {1}{8}\right )}{\left (\tanh ^{2}\left (\frac {x}{2}\right )-2 \tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {7 \sqrt {2}\, \arctanh \left (\frac {\left (2 \tanh \left (\frac {x}{2}\right )-2\right ) \sqrt {2}}{4}\right )}{8}-\frac {2 \left (-\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{8}+\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}-\frac {5 \tanh \left (\frac {x}{2}\right )}{8}+\frac {1}{8}\right )}{\left (\tanh ^{2}\left (\frac {x}{2}\right )+2 \tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {7 \sqrt {2}\, \arctanh \left (\frac {\left (2 \tanh \left (\frac {x}{2}\right )+2\right ) \sqrt {2}}{4}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sech(x)^2-tanh(x)^2)^3,x)

[Out]

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

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maxima [B]  time = 0.44, size = 114, normalized size = 2.11 \[ \frac {7}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - e^{\left (-x\right )} + 1}{\sqrt {2} + e^{\left (-x\right )} - 1}\right ) - \frac {7}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - e^{\left (-x\right )} - 1}{\sqrt {2} + e^{\left (-x\right )} + 1}\right ) - x + \frac {19 \, e^{\left (-2 \, x\right )} - 57 \, e^{\left (-4 \, x\right )} + 17 \, e^{\left (-6 \, x\right )} - 3}{2 \, {\left (12 \, e^{\left (-2 \, x\right )} - 38 \, e^{\left (-4 \, x\right )} + 12 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)^2-tanh(x)^2)^3,x, algorithm="maxima")

[Out]

7/16*sqrt(2)*log(-(sqrt(2) - e^(-x) + 1)/(sqrt(2) + e^(-x) - 1)) - 7/16*sqrt(2)*log(-(sqrt(2) - e^(-x) - 1)/(s
qrt(2) + e^(-x) + 1)) - x + 1/2*(19*e^(-2*x) - 57*e^(-4*x) + 17*e^(-6*x) - 3)/(12*e^(-2*x) - 38*e^(-4*x) + 12*
e^(-6*x) - e^(-8*x) - 1)

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mupad [B]  time = 0.07, size = 114, normalized size = 2.11 \[ \frac {136\,{\mathrm {e}}^{2\,x}-24}{38\,{\mathrm {e}}^{4\,x}-12\,{\mathrm {e}}^{2\,x}-12\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1}-x-\frac {7\,\sqrt {2}\,\ln \left (7\,{\mathrm {e}}^{2\,x}-\frac {7\,\sqrt {2}\,\left (12\,{\mathrm {e}}^{2\,x}-4\right )}{16}\right )}{16}+\frac {7\,\sqrt {2}\,\ln \left (7\,{\mathrm {e}}^{2\,x}+\frac {7\,\sqrt {2}\,\left (12\,{\mathrm {e}}^{2\,x}-4\right )}{16}\right )}{16}+\frac {\frac {17\,{\mathrm {e}}^{2\,x}}{2}+\frac {45}{2}}{{\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/cosh(x)^2 - tanh(x)^2)^3,x)

[Out]

(136*exp(2*x) - 24)/(38*exp(4*x) - 12*exp(2*x) - 12*exp(6*x) + exp(8*x) + 1) - x - (7*2^(1/2)*log(7*exp(2*x) -
(7*2^(1/2)*(12*exp(2*x) - 4))/16))/16 + (7*2^(1/2)*log(7*exp(2*x) + (7*2^(1/2)*(12*exp(2*x) - 4))/16))/16 + (
(17*exp(2*x))/2 + 45/2)/(exp(4*x) - 6*exp(2*x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- \tanh {\relax (x )} + \operatorname {sech}{\relax (x )}\right )^{3} \left (\tanh {\relax (x )} + \operatorname {sech}{\relax (x )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)**2-tanh(x)**2)**3,x)

[Out]

Integral(1/((-tanh(x) + sech(x))**3*(tanh(x) + sech(x))**3), x)

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