3.58 \(\int \frac {\text {csch}^3(x)}{a+a \text {sech}(x)} \, dx\)

Optimal. Leaf size=46 \[ \frac {\text {csch}^4(x)}{4 a}+\frac {\tanh ^{-1}(\cosh (x))}{8 a}-\frac {\coth (x) \text {csch}^3(x)}{4 a}-\frac {\coth (x) \text {csch}(x)}{8 a} \]

[Out]

1/8*arctanh(cosh(x))/a-1/8*coth(x)*csch(x)/a-1/4*coth(x)*csch(x)^3/a+1/4*csch(x)^4/a

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Rubi [A]  time = 0.19, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3872, 2835, 2606, 30, 2611, 3768, 3770} \[ \frac {\text {csch}^4(x)}{4 a}+\frac {\tanh ^{-1}(\cosh (x))}{8 a}-\frac {\coth (x) \text {csch}^3(x)}{4 a}-\frac {\coth (x) \text {csch}(x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^3/(a + a*Sech[x]),x]

[Out]

ArcTanh[Cosh[x]]/(8*a) - (Coth[x]*Csch[x])/(8*a) - (Coth[x]*Csch[x]^3)/(4*a) + Csch[x]^4/(4*a)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2611

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

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

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]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\text {csch}^3(x)}{a+a \text {sech}(x)} \, dx &=-\int \frac {\coth (x) \text {csch}^2(x)}{-a-a \cosh (x)} \, dx\\ &=\frac {\int \coth ^2(x) \text {csch}^3(x) \, dx}{a}-\frac {\int \coth (x) \text {csch}^4(x) \, dx}{a}\\ &=-\frac {\coth (x) \text {csch}^3(x)}{4 a}+\frac {\int \text {csch}^3(x) \, dx}{4 a}+\frac {\operatorname {Subst}\left (\int x^3 \, dx,x,-i \text {csch}(x)\right )}{a}\\ &=-\frac {\coth (x) \text {csch}(x)}{8 a}-\frac {\coth (x) \text {csch}^3(x)}{4 a}+\frac {\text {csch}^4(x)}{4 a}-\frac {\int \text {csch}(x) \, dx}{8 a}\\ &=\frac {\tanh ^{-1}(\cosh (x))}{8 a}-\frac {\coth (x) \text {csch}(x)}{8 a}-\frac {\coth (x) \text {csch}^3(x)}{4 a}+\frac {\text {csch}^4(x)}{4 a}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 59, normalized size = 1.28 \[ \frac {\cosh ^2\left (\frac {x}{2}\right ) \text {sech}(x) \left (-2 \text {csch}^2\left (\frac {x}{2}\right )+\text {sech}^4\left (\frac {x}{2}\right )-4 \log \left (\sinh \left (\frac {x}{2}\right )\right )+4 \log \left (\cosh \left (\frac {x}{2}\right )\right )\right )}{16 (a \text {sech}(x)+a)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^3/(a + a*Sech[x]),x]

[Out]

(Cosh[x/2]^2*(-2*Csch[x/2]^2 + 4*Log[Cosh[x/2]] - 4*Log[Sinh[x/2]] + Sech[x/2]^4)*Sech[x])/(16*(a + a*Sech[x])
)

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fricas [B]  time = 0.40, size = 630, normalized size = 13.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+a*sech(x)),x, algorithm="fricas")

[Out]

-1/8*(2*cosh(x)^5 + 2*(5*cosh(x) + 2)*sinh(x)^4 + 2*sinh(x)^5 + 4*cosh(x)^4 + 4*(5*cosh(x)^2 + 4*cosh(x) + 5)*
sinh(x)^3 + 20*cosh(x)^3 + 4*(5*cosh(x)^3 + 6*cosh(x)^2 + 15*cosh(x) + 1)*sinh(x)^2 + 4*cosh(x)^2 - (cosh(x)^6
+ 2*(3*cosh(x) + 1)*sinh(x)^5 + sinh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cosh(x) - 1)*sinh(x)^4 - cosh(x)
^4 + 4*(5*cosh(x)^3 + 5*cosh(x)^2 - cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*cosh(x)^4 + 20*cosh(x)^3 - 6*co
sh(x)^2 - 12*cosh(x) - 1)*sinh(x)^2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 - 2*cosh(x)^3 - 6*cosh(x)^2 - c
osh(x) + 1)*sinh(x) + 2*cosh(x) + 1)*log(cosh(x) + sinh(x) + 1) + (cosh(x)^6 + 2*(3*cosh(x) + 1)*sinh(x)^5 + s
inh(x)^6 + 2*cosh(x)^5 + (15*cosh(x)^2 + 10*cosh(x) - 1)*sinh(x)^4 - cosh(x)^4 + 4*(5*cosh(x)^3 + 5*cosh(x)^2
- cosh(x) - 1)*sinh(x)^3 - 4*cosh(x)^3 + (15*cosh(x)^4 + 20*cosh(x)^3 - 6*cosh(x)^2 - 12*cosh(x) - 1)*sinh(x)^
2 - cosh(x)^2 + 2*(3*cosh(x)^5 + 5*cosh(x)^4 - 2*cosh(x)^3 - 6*cosh(x)^2 - cosh(x) + 1)*sinh(x) + 2*cosh(x) +
1)*log(cosh(x) + sinh(x) - 1) + 2*(5*cosh(x)^4 + 8*cosh(x)^3 + 30*cosh(x)^2 + 4*cosh(x) + 1)*sinh(x) + 2*cosh(
x))/(a*cosh(x)^6 + a*sinh(x)^6 + 2*a*cosh(x)^5 + 2*(3*a*cosh(x) + a)*sinh(x)^5 - a*cosh(x)^4 + (15*a*cosh(x)^2
+ 10*a*cosh(x) - a)*sinh(x)^4 - 4*a*cosh(x)^3 + 4*(5*a*cosh(x)^3 + 5*a*cosh(x)^2 - a*cosh(x) - a)*sinh(x)^3 -
a*cosh(x)^2 + (15*a*cosh(x)^4 + 20*a*cosh(x)^3 - 6*a*cosh(x)^2 - 12*a*cosh(x) - a)*sinh(x)^2 + 2*a*cosh(x) +
2*(3*a*cosh(x)^5 + 5*a*cosh(x)^4 - 2*a*cosh(x)^3 - 6*a*cosh(x)^2 - a*cosh(x) + a)*sinh(x) + a)

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giac [B]  time = 0.11, size = 90, normalized size = 1.96 \[ \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{16 \, a} - \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{16 \, a} + \frac {e^{\left (-x\right )} + e^{x} - 6}{16 \, a {\left (e^{\left (-x\right )} + e^{x} - 2\right )}} - \frac {3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 12 \, e^{\left (-x\right )} + 12 \, e^{x} - 4}{32 \, a {\left (e^{\left (-x\right )} + e^{x} + 2\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+a*sech(x)),x, algorithm="giac")

[Out]

1/16*log(e^(-x) + e^x + 2)/a - 1/16*log(e^(-x) + e^x - 2)/a + 1/16*(e^(-x) + e^x - 6)/(a*(e^(-x) + e^x - 2)) -
1/32*(3*(e^(-x) + e^x)^2 + 12*e^(-x) + 12*e^x - 4)/(a*(e^(-x) + e^x + 2)^2)

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maple [A]  time = 0.15, size = 45, normalized size = 0.98 \[ \frac {\tanh ^{4}\left (\frac {x}{2}\right )}{32 a}-\frac {\tanh ^{2}\left (\frac {x}{2}\right )}{16 a}-\frac {1}{16 a \tanh \left (\frac {x}{2}\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{8 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)^3/(a+a*sech(x)),x)

[Out]

1/32/a*tanh(1/2*x)^4-1/16/a*tanh(1/2*x)^2-1/16/a/tanh(1/2*x)^2-1/8/a*ln(tanh(1/2*x))

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maxima [B]  time = 0.32, size = 99, normalized size = 2.15 \[ -\frac {e^{\left (-x\right )} + 2 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 2 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )}}{4 \, {\left (2 \, a e^{\left (-x\right )} - a e^{\left (-2 \, x\right )} - 4 \, a e^{\left (-3 \, x\right )} - a e^{\left (-4 \, x\right )} + 2 \, a e^{\left (-5 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} + \frac {\log \left (e^{\left (-x\right )} + 1\right )}{8 \, a} - \frac {\log \left (e^{\left (-x\right )} - 1\right )}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^3/(a+a*sech(x)),x, algorithm="maxima")

[Out]

-1/4*(e^(-x) + 2*e^(-2*x) + 10*e^(-3*x) + 2*e^(-4*x) + e^(-5*x))/(2*a*e^(-x) - a*e^(-2*x) - 4*a*e^(-3*x) - a*e
^(-4*x) + 2*a*e^(-5*x) + a*e^(-6*x) + a) + 1/8*log(e^(-x) + 1)/a - 1/8*log(e^(-x) - 1)/a

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mupad [B]  time = 1.35, size = 121, normalized size = 2.63 \[ \frac {1}{2\,a\,\left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )}-\frac {1}{4\,a\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )}+\frac {1}{2\,a\,\left (6\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^x+1\right )}-\frac {1}{4\,a\,\left ({\mathrm {e}}^x-1\right )}+\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {-a^2}}{a}\right )}{4\,\sqrt {-a^2}}-\frac {1}{a\,\left (3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^3*(a + a/cosh(x))),x)

[Out]

1/(2*a*(exp(2*x) + 2*exp(x) + 1)) - 1/(4*a*(exp(2*x) - 2*exp(x) + 1)) + 1/(2*a*(6*exp(2*x) + 4*exp(3*x) + exp(
4*x) + 4*exp(x) + 1)) - 1/(4*a*(exp(x) - 1)) + atan((exp(x)*(-a^2)^(1/2))/a)/(4*(-a^2)^(1/2)) - 1/(a*(3*exp(2*
x) + exp(3*x) + 3*exp(x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)**3/(a+a*sech(x)),x)

[Out]

Integral(csch(x)**3/(sech(x) + 1), x)/a

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