### 3.120 $$\int \cosh ^3(a+b x) \text{csch}^{3+n}(a+b x) \, dx$$

Optimal. Leaf size=37 $-\frac{\text{csch}^n(a+b x)}{b n}-\frac{\text{csch}^{n+2}(a+b x)}{b (n+2)}$

[Out]

-(Csch[a + b*x]^n/(b*n)) - Csch[a + b*x]^(2 + n)/(b*(2 + n))

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Rubi [A]  time = 0.0421384, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.105, Rules used = {2621, 14} $-\frac{\text{csch}^n(a+b x)}{b n}-\frac{\text{csch}^{n+2}(a+b x)}{b (n+2)}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]^3*Csch[a + b*x]^(3 + n),x]

[Out]

-(Csch[a + b*x]^n/(b*n)) - Csch[a + b*x]^(2 + n)/(b*(2 + n))

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])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0657785, size = 34, normalized size = 0.92 $-\frac{\text{csch}^n(a+b x) \left (n \text{csch}^2(a+b x)+n+2\right )}{b n (n+2)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]^3*Csch[a + b*x]^(3 + n),x]

[Out]

-((Csch[a + b*x]^n*(2 + n + n*Csch[a + b*x]^2))/(b*n*(2 + n)))

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Maple [C]  time = 0.152, size = 499, normalized size = 13.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(coth(b*x+a)^3*csch(b*x+a)^n,x)

[Out]

-(n*exp(4*b*x+4*a)+2*exp(4*b*x+4*a)+2*n*exp(2*b*x+2*a)-4*exp(2*b*x+2*a)+n+2)/b/n/(n+2)/(exp(2*b*x+2*a)-1)^2*ex
p(-1/2*n*(I*csgn(I/(exp(b*x+a)-1)/(1+exp(b*x+a)))^3*Pi-I*csgn(I/(exp(b*x+a)-1)/(1+exp(b*x+a)))^2*csgn(I/(exp(b
*x+a)-1))*Pi-I*csgn(I/(exp(b*x+a)-1)/(1+exp(b*x+a)))^2*csgn(I/(1+exp(b*x+a)))*Pi+I*csgn(I/(exp(b*x+a)-1)/(1+ex
p(b*x+a)))*csgn(I/(exp(b*x+a)-1))*csgn(I/(1+exp(b*x+a)))*Pi-I*csgn(I/(exp(b*x+a)-1)/(1+exp(b*x+a)))*csgn(I*exp
(b*x+a)/(1+exp(b*x+a))/(exp(b*x+a)-1))^2*Pi+I*csgn(I/(exp(b*x+a)-1)/(1+exp(b*x+a)))*csgn(I*exp(b*x+a)/(1+exp(b
*x+a))/(exp(b*x+a)-1))*csgn(I*exp(b*x+a))*Pi+I*csgn(I*exp(b*x+a)/(1+exp(b*x+a))/(exp(b*x+a)-1))^3*Pi-I*csgn(I*
exp(b*x+a)/(1+exp(b*x+a))/(exp(b*x+a)-1))^2*csgn(I*exp(b*x+a))*Pi-2*ln(2)-2*ln(exp(b*x+a))+2*ln(exp(b*x+a)-1)+
2*ln(1+exp(b*x+a))))

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Maxima [B]  time = 1.91423, size = 559, normalized size = 15.11 \begin{align*} -\frac{2^{n} n e^{\left (-{\left (b x + a\right )} n - n \log \left (e^{\left (-b x - a\right )} + 1\right ) - n \log \left (-e^{\left (-b x - a\right )} + 1\right )\right )}}{{\left (n^{2} - 2 \,{\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} +{\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} - \frac{{\left (2^{n + 1} n - 2^{n + 2}\right )} e^{\left (-{\left (b x + a\right )} n - 2 \, b x - n \log \left (e^{\left (-b x - a\right )} + 1\right ) - n \log \left (-e^{\left (-b x - a\right )} + 1\right ) - 2 \, a\right )}}{{\left (n^{2} - 2 \,{\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} +{\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} - \frac{{\left (2^{n} n + 2^{n + 1}\right )} e^{\left (-{\left (b x + a\right )} n - 4 \, b x - n \log \left (e^{\left (-b x - a\right )} + 1\right ) - n \log \left (-e^{\left (-b x - a\right )} + 1\right ) - 4 \, a\right )}}{{\left (n^{2} - 2 \,{\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} +{\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} - \frac{2^{n + 1} e^{\left (-{\left (b x + a\right )} n - n \log \left (e^{\left (-b x - a\right )} + 1\right ) - n \log \left (-e^{\left (-b x - a\right )} + 1\right )\right )}}{{\left (n^{2} - 2 \,{\left (n^{2} + 2 \, n\right )} e^{\left (-2 \, b x - 2 \, a\right )} +{\left (n^{2} + 2 \, n\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 2 \, n\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.98283, size = 603, normalized size = 16.3 \begin{align*} \frac{{\left ({\left (n + 2\right )} \cosh \left (b x + a\right )^{2} +{\left (n + 2\right )} \sinh \left (b x + a\right )^{2} + n - 2\right )} \cosh \left (n \log \left (\frac{2 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}\right )\right ) +{\left ({\left (n + 2\right )} \cosh \left (b x + a\right )^{2} +{\left (n + 2\right )} \sinh \left (b x + a\right )^{2} + n - 2\right )} \sinh \left (n \log \left (\frac{2 \,{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}\right )\right )}{b n^{2} -{\left (b n^{2} + 2 \, b n\right )} \cosh \left (b x + a\right )^{2} -{\left (b n^{2} + 2 \, b n\right )} \sinh \left (b x + a\right )^{2} + 2 \, b n} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((n + 2)*cosh(b*x + a)^2 + (n + 2)*sinh(b*x + a)^2 + n - 2)*cosh(n*log(2*(cosh(b*x + a) + sinh(b*x + a))/(cos
h(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1))) + ((n + 2)*cosh(b*x + a)^2 + (n + 2)*sin
h(b*x + a)^2 + n - 2)*sinh(n*log(2*(cosh(b*x + a) + sinh(b*x + a))/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x
+ a) + sinh(b*x + a)^2 - 1))))/(b*n^2 - (b*n^2 + 2*b*n)*cosh(b*x + a)^2 - (b*n^2 + 2*b*n)*sinh(b*x + a)^2 + 2
*b*n)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(b*x+a)**3*csch(b*x+a)**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{csch}\left (b x + a\right )^{n} \coth \left (b x + a\right )^{3}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(csch(b*x + a)^n*coth(b*x + a)^3, x)