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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0483274, antiderivative size = 49, normalized size of antiderivative = 1., 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 = {2622, 288, 321, 207} \[ -\frac{3 \text{sech}(a+b x)}{2 b}+\frac{3 \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{\text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2622

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

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0349659, size = 68, normalized size = 1.39 \[ -\frac{\text{csch}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{\text{sech}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{\text{sech}(a+b x)}{b}-\frac{3 \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Csch[(a + b*x)/2]^2/(8*b) - (3*Log[Tanh[(a + b*x)/2]])/(2*b) - Sech[(a + b*x)/2]^2/(8*b) - Sech[a + b*x]/b

________________________________________________________________________________________

Maple [A]  time = 0.019, size = 43, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{1}{2\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}-{\frac{3}{2\,\cosh \left ( bx+a \right ) }}+3\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.1438, size = 143, normalized size = 2.92 \begin{align*} \frac{3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} - \frac{3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, 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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.87732, size = 1935, normalized size = 39.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*cosh(b*x + a)^5 + 30*cosh(b*x + a)*sinh(b*x + a)^4 + 6*sinh(b*x + a)^5 + 4*(15*cosh(b*x + a)^2 - 1)*si
nh(b*x + a)^3 - 4*cosh(b*x + a)^3 + 12*(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)*si
nh(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)
*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 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)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 6*(5*cosh
(b*x + a)^4 - 2*cosh(b*x + a)^2 + 1)*sinh(b*x + a) + 6*cosh(b*x + a))/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*s
inh(b*x + a)^5 + b*sinh(b*x + a)^6 - b*cosh(b*x + a)^4 + (15*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^4 + 4*(5*b*c
osh(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
)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{csch}^{3}{\left (a + b x \right )} \operatorname{sech}^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.15404, size = 154, normalized size = 3.14 \begin{align*} \frac{3 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + 2\right )}{4 \, b} - \frac{3 \, \log \left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - 2\right )}{4 \, b} - \frac{3 \,{\left (e^{\left (b x + a\right )} + e^{\left (-b x - a\right )}\right )}^{2} - 8}{{\left ({\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 )}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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