3.44 \(\int \text {csch}^5(a+b x) \text {sech}(a+b x) \, dx\)

Optimal. Leaf size=39 \[ -\frac {\coth ^4(a+b x)}{4 b}+\frac {\coth ^2(a+b x)}{b}+\frac {\log (\tanh (a+b x))}{b} \]

[Out]

coth(b*x+a)^2/b-1/4*coth(b*x+a)^4/b+ln(tanh(b*x+a))/b

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2620, 266, 43} \[ -\frac {\coth ^4(a+b x)}{4 b}+\frac {\coth ^2(a+b x)}{b}+\frac {\log (\tanh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[a + b*x]^5*Sech[a + b*x],x]

[Out]

Coth[a + b*x]^2/b - Coth[a + b*x]^4/(4*b) + Log[Tanh[a + b*x]]/b

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rubi steps

\begin {align*} \int \text {csch}^5(a+b x) \text {sech}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^5} \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(1+x)^2}{x^3} \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{x^3}+\frac {2}{x^2}+\frac {1}{x}\right ) \, dx,x,-\tanh ^2(a+b x)\right )}{2 b}\\ &=\frac {\coth ^2(a+b x)}{b}-\frac {\coth ^4(a+b x)}{4 b}+\frac {\log (\tanh (a+b x))}{b}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 46, normalized size = 1.18 \[ \frac {-\text {csch}^4(a+b x)+2 \text {csch}^2(a+b x)+4 \log (\sinh (a+b x))-4 \log (\cosh (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[a + b*x]^5*Sech[a + b*x],x]

[Out]

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

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fricas [B]  time = 0.49, size = 1082, normalized size = 27.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a),x, algorithm="fricas")

[Out]

(2*cosh(b*x + a)^6 + 12*cosh(b*x + a)*sinh(b*x + a)^5 + 2*sinh(b*x + a)^6 + 2*(15*cosh(b*x + a)^2 - 4)*sinh(b*
x + a)^4 - 8*cosh(b*x + a)^4 + 8*(5*cosh(b*x + a)^3 - 4*cosh(b*x + a))*sinh(b*x + a)^3 + 2*(15*cosh(b*x + a)^4
- 24*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 - (cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x +
a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3
- 3*cosh(b*x + a))*sinh(b*x + a)^5 + 2*(35*cosh(b*x + a)^4 - 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh
(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 - 10*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a
)^6 - 15*cosh(b*x + a)^4 + 9*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 - 3
*cosh(b*x + a)^5 + 3*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(2*cosh(b*x + a)/(cosh(b*x + a) -
sinh(b*x + a))) + (cosh(b*x + a)^8 + 8*cosh(b*x + a)*sinh(b*x + a)^7 + sinh(b*x + a)^8 + 4*(7*cosh(b*x + a)^2
- 1)*sinh(b*x + a)^6 - 4*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^5 + 2*(35*cos
h(b*x + a)^4 - 30*cosh(b*x + a)^2 + 3)*sinh(b*x + a)^4 + 6*cosh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 - 10*cosh(b*
x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 15*cosh(b*x + a)^4 + 9*cosh(b*x + a)^2 -
1)*sinh(b*x + a)^2 - 4*cosh(b*x + a)^2 + 8*(cosh(b*x + a)^7 - 3*cosh(b*x + a)^5 + 3*cosh(b*x + a)^3 - cosh(b*x
+ a))*sinh(b*x + a) + 1)*log(2*sinh(b*x + a)/(cosh(b*x + a) - sinh(b*x + a))) + 4*(3*cosh(b*x + a)^5 - 8*cosh
(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^8 + 8*b*cosh(b*x + a)*sinh(b*x + a)^7 + b*sinh(b*
x + a)^8 - 4*b*cosh(b*x + a)^6 + 4*(7*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^6 + 8*(7*b*cosh(b*x + a)^3 - 3*b*co
sh(b*x + a))*sinh(b*x + a)^5 + 6*b*cosh(b*x + a)^4 + 2*(35*b*cosh(b*x + a)^4 - 30*b*cosh(b*x + a)^2 + 3*b)*sin
h(b*x + a)^4 + 8*(7*b*cosh(b*x + a)^5 - 10*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^3 - 4*b*cosh(b
*x + a)^2 + 4*(7*b*cosh(b*x + a)^6 - 15*b*cosh(b*x + a)^4 + 9*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 8*(b*co
sh(b*x + a)^7 - 3*b*cosh(b*x + a)^5 + 3*b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)

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giac [B]  time = 0.15, size = 122, normalized size = 3.13 \[ -\frac {\frac {3 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}\right )}^{2} - 20 \, e^{\left (2 \, b x + 2 \, a\right )} - 20 \, e^{\left (-2 \, b x - 2 \, a\right )} + 44}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}^{2}} + 2 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} + 2\right ) - 2 \, \log \left (e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )} - 2\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a),x, algorithm="giac")

[Out]

-1/4*((3*(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a))^2 - 20*e^(2*b*x + 2*a) - 20*e^(-2*b*x - 2*a) + 44)/(e^(2*b*x + 2
*a) + e^(-2*b*x - 2*a) - 2)^2 + 2*log(e^(2*b*x + 2*a) + e^(-2*b*x - 2*a) + 2) - 2*log(e^(2*b*x + 2*a) + e^(-2*
b*x - 2*a) - 2))/b

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maple [A]  time = 0.17, size = 39, normalized size = 1.00 \[ -\frac {1}{4 b \sinh \left (b x +a \right )^{4}}+\frac {1}{2 b \sinh \left (b x +a \right )^{2}}+\frac {\ln \left (\tanh \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)^5*sech(b*x+a),x)

[Out]

-1/4/b/sinh(b*x+a)^4+1/2/b/sinh(b*x+a)^2+ln(tanh(b*x+a))/b

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maxima [B]  time = 0.47, size = 133, normalized size = 3.41 \[ \frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac {\log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )}{b} - \frac {2 \, {\left (e^{\left (-2 \, b x - 2 \, a\right )} - 4 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}}{b {\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} - 6 \, e^{\left (-4 \, b x - 4 \, a\right )} + 4 \, e^{\left (-6 \, b x - 6 \, a\right )} - e^{\left (-8 \, b x - 8 \, a\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^5*sech(b*x+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 169, normalized size = 4.33 \[ \frac {2}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {8}{b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1\right )}-\frac {4}{b\,\left (6\,{\mathrm {e}}^{4\,a+4\,b\,x}-4\,{\mathrm {e}}^{2\,a+2\,b\,x}-4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(a + b*x)*sinh(a + b*x)^5),x)

[Out]

2/(b*(exp(2*a + 2*b*x) - 1)) - (2*atan((exp(2*a)*exp(2*b*x)*(-b^2)^(1/2))/b))/(-b^2)^(1/2) - 2/(b*(exp(4*a + 4
*b*x) - 2*exp(2*a + 2*b*x) + 1)) - 8/(b*(3*exp(2*a + 2*b*x) - 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) - 1)) - 4/
(b*(6*exp(4*a + 4*b*x) - 4*exp(2*a + 2*b*x) - 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{5}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**5*sech(b*x+a),x)

[Out]

Integral(csch(a + b*x)**5*sech(a + b*x), x)

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