3.504 \(\int \text {csch}^2(a+b x) \text {sech}^3(a+b x) \, dx\)

Optimal. Leaf size=49 \[ -\frac {3 \text {csch}(a+b x)}{2 b}-\frac {3 \tan ^{-1}(\sinh (a+b x))}{2 b}+\frac {\text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b} \]

[Out]

-3/2*arctan(sinh(b*x+a))/b-3/2*csch(b*x+a)/b+1/2*csch(b*x+a)*sech(b*x+a)^2/b

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Rubi [A]  time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2621, 288, 321, 207} \[ -\frac {3 \text {csch}(a+b x)}{2 b}-\frac {3 \tan ^{-1}(\sinh (a+b x))}{2 b}+\frac {\text {csch}(a+b x) \text {sech}^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csch[a + b*x]^2*Sech[a + b*x]^3,x]

[Out]

(-3*ArcTan[Sinh[a + b*x]])/(2*b) - (3*Csch[a + b*x])/(2*b) + (Csch[a + b*x]*Sech[a + b*x]^2)/(2*b)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2621

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.59 \[ -\frac {\text {csch}(a+b x) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\sinh ^2(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[a + b*x]^2*Sech[a + b*x]^3,x]

[Out]

-((Csch[a + b*x]*Hypergeometric2F1[-1/2, 2, 1/2, -Sinh[a + b*x]^2])/b)

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fricas [B]  time = 0.41, size = 511, normalized size = 10.43 \[ -\frac {3 \, \cosh \left (b x + a\right )^{5} + 15 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 3 \, \sinh \left (b x + a\right )^{5} + 2 \, {\left (15 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 2 \, \cosh \left (b x + a\right )^{3} + 6 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 3 \, {\left (\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + {\left (15 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{4} + \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + {\left (15 \, \cosh \left (b x + a\right )^{4} + 6 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{5} + 2 \, \cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 2 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) + 3 \, \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{6} + 6 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + b \sinh \left (b x + a\right )^{6} + b \cosh \left (b x + a\right )^{4} + {\left (15 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )^{2} + {\left (15 \, b \cosh \left (b x + a\right )^{4} + 6 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{5} + 2 \, b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^2*sech(b*x+a)^3,x, algorithm="fricas")

[Out]

-(3*cosh(b*x + a)^5 + 15*cosh(b*x + a)*sinh(b*x + a)^4 + 3*sinh(b*x + a)^5 + 2*(15*cosh(b*x + a)^2 + 1)*sinh(b
*x + a)^3 + 2*cosh(b*x + a)^3 + 6*(5*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^2 + 3*(cosh(b*x + a)^6 + 6
*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + (15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^4 + cosh(b*x + a)^4
+ 4*(5*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 + 6*cosh(b*x + a)^2 - 1)*sinh(b*
x + a)^2 - cosh(b*x + a)^2 + 2*(3*cosh(b*x + a)^5 + 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) - 1)*arct
an(cosh(b*x + a) + sinh(b*x + a)) + 3*(5*cosh(b*x + a)^4 + 2*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + 3*cosh(b*x +
a))/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b*x + a)^6 + b*cosh(b*x + a)^4 + (15*b*co
sh(b*x + a)^2 + b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a)^3 - b*cosh(b*x +
a)^2 + (15*b*cosh(b*x + a)^4 + 6*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^5 + 2*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 = 102, normalized size = 2.08 \[ -\frac {3 \, \pi + \frac {4 \, {\left (3 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 8\right )}}{{\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{3} + 4 \, e^{\left (b x + a\right )} - 4 \, e^{\left (-b x - a\right )}} + 6 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^2*sech(b*x+a)^3,x, algorithm="giac")

[Out]

-1/4*(3*pi + 4*(3*(e^(b*x + a) - e^(-b*x - a))^2 + 8)/((e^(b*x + a) - e^(-b*x - a))^3 + 4*e^(b*x + a) - 4*e^(-
b*x - a)) + 6*arctan(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)))/b

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maple [A]  time = 0.34, size = 52, normalized size = 1.06 \[ -\frac {1}{b \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{2}}-\frac {3 \,\mathrm {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2 b}-\frac {3 \arctan \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(b*x+a)^2*sech(b*x+a)^3,x)

[Out]

-1/b/sinh(b*x+a)/cosh(b*x+a)^2-3/2*sech(b*x+a)*tanh(b*x+a)/b-3*arctan(exp(b*x+a))/b

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maxima [B]  time = 0.41, size = 90, normalized size = 1.84 \[ \frac {3 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} - \frac {3 \, e^{\left (-b x - a\right )} + 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + 3 \, e^{\left (-5 \, b x - 5 \, a\right )}}{b {\left (e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (-6 \, b x - 6 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)^2*sech(b*x+a)^3,x, algorithm="maxima")

[Out]

3*arctan(e^(-b*x - a))/b - (3*e^(-b*x - a) + 2*e^(-3*b*x - 3*a) + 3*e^(-5*b*x - 5*a))/(b*(e^(-2*b*x - 2*a) - e
^(-4*b*x - 4*a) - e^(-6*b*x - 6*a) + 1))

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mupad [B]  time = 0.08, size = 107, normalized size = 2.18 \[ \frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(a + b*x)^3*sinh(a + b*x)^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {csch}^{2}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(b*x+a)**2*sech(b*x+a)**3,x)

[Out]

Integral(csch(a + b*x)**2*sech(a + b*x)**3, x)

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