### 3.111 $$\int \cosh ^4(a+b x) \coth (a+b x) \, dx$$

Optimal. Leaf size=39 $\frac{\sinh ^4(a+b x)}{4 b}+\frac{\sinh ^2(a+b x)}{b}+\frac{\log (\sinh (a+b x))}{b}$

[Out]

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

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Rubi [A]  time = 0.0308386, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2590, 266, 43} $\frac{\sinh ^4(a+b x)}{4 b}+\frac{\sinh ^2(a+b x)}{b}+\frac{\log (\sinh (a+b x))}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]^4*Coth[a + b*x],x]

[Out]

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

Rule 2590

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

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 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])

Rubi steps

\begin{align*} \int \cosh ^4(a+b x) \coth (a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x} \, dx,x,-i \sinh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x} \, dx,x,-\sinh ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-2+\frac{1}{x}+x\right ) \, dx,x,-\sinh ^2(a+b x)\right )}{2 b}\\ &=\frac{\log (\sinh (a+b x))}{b}+\frac{\sinh ^2(a+b x)}{b}+\frac{\sinh ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0323827, size = 35, normalized size = 0.9 $\frac{\sinh ^4(a+b x)+4 \sinh ^2(a+b x)+4 \log (\sinh (a+b x))}{4 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]^4*Coth[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.017, size = 39, normalized size = 1. \begin{align*}{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}{4\,b}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2\,b}}+{\frac{\ln \left ( \sinh \left ( bx+a \right ) \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^4*coth(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.09916, size = 128, normalized size = 3.28 \begin{align*} \frac{{\left (12 \, e^{\left (-2 \, b x - 2 \, a\right )} + 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{64 \, b} + \frac{b x + a}{b} + \frac{12 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}}{64 \, b} + \frac{\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac{\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^4*coth(b*x+a),x, algorithm="maxima")

[Out]

1/64*(12*e^(-2*b*x - 2*a) + 1)*e^(4*b*x + 4*a)/b + (b*x + a)/b + 1/64*(12*e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a))
/b + log(e^(-b*x - a) + 1)/b + log(e^(-b*x - a) - 1)/b

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Fricas [B]  time = 1.90217, size = 1270, normalized size = 32.56 \begin{align*} \frac{\cosh \left (b x + a\right )^{8} + 8 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{7} + \sinh \left (b x + a\right )^{8} + 4 \,{\left (7 \, \cosh \left (b x + a\right )^{2} + 3\right )} \sinh \left (b x + a\right )^{6} - 64 \, b x \cosh \left (b x + a\right )^{4} + 12 \, \cosh \left (b x + a\right )^{6} + 8 \,{\left (7 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{5} + 2 \,{\left (35 \, \cosh \left (b x + a\right )^{4} - 32 \, b x + 90 \, \cosh \left (b x + a\right )^{2}\right )} \sinh \left (b x + a\right )^{4} + 8 \,{\left (7 \, \cosh \left (b x + a\right )^{5} - 32 \, b x \cosh \left (b x + a\right ) + 30 \, \cosh \left (b x + a\right )^{3}\right )} \sinh \left (b x + a\right )^{3} + 4 \,{\left (7 \, \cosh \left (b x + a\right )^{6} - 96 \, b x \cosh \left (b x + a\right )^{2} + 45 \, \cosh \left (b x + a\right )^{4} + 3\right )} \sinh \left (b x + a\right )^{2} + 12 \, \cosh \left (b x + a\right )^{2} + 64 \,{\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4}\right )} \log \left (\frac{2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 8 \,{\left (\cosh \left (b x + a\right )^{7} - 32 \, b x \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right )^{5} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{64 \,{\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, b \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^4*coth(b*x+a),x, algorithm="fricas")

[Out]

1/64*(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 + 3)*sinh(b*x
+ a)^6 - 64*b*x*cosh(b*x + a)^4 + 12*cosh(b*x + a)^6 + 8*(7*cosh(b*x + a)^3 + 9*cosh(b*x + a))*sinh(b*x + a)^
5 + 2*(35*cosh(b*x + a)^4 - 32*b*x + 90*cosh(b*x + a)^2)*sinh(b*x + a)^4 + 8*(7*cosh(b*x + a)^5 - 32*b*x*cosh(
b*x + a) + 30*cosh(b*x + a)^3)*sinh(b*x + a)^3 + 4*(7*cosh(b*x + a)^6 - 96*b*x*cosh(b*x + a)^2 + 45*cosh(b*x +
a)^4 + 3)*sinh(b*x + a)^2 + 12*cosh(b*x + a)^2 + 64*(cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*co
sh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4)*log(2*sinh(b*x + a)/(cosh(b
*x + a) - sinh(b*x + a))) + 8*(cosh(b*x + a)^7 - 32*b*x*cosh(b*x + a)^3 + 9*cosh(b*x + a)^5 + 3*cosh(b*x + a))
*sinh(b*x + a) + 1)/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*cosh(b*x + a)^2*sinh(b*x + a)
^2 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4)

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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(cosh(b*x+a)**4*coth(b*x+a),x)

[Out]

Timed out

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Giac [B]  time = 1.23364, size = 117, normalized size = 3. \begin{align*} -\frac{64 \, b x -{\left (48 \, e^{\left (4 \, b x + 4 \, a\right )} + 12 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )} -{\left (e^{\left (4 \, b x + 16 \, a\right )} + 12 \, e^{\left (2 \, b x + 14 \, a\right )}\right )} e^{\left (-12 \, a\right )} - 64 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{64 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^4*coth(b*x+a),x, algorithm="giac")

[Out]

-1/64*(64*b*x - (48*e^(4*b*x + 4*a) + 12*e^(2*b*x + 2*a) + 1)*e^(-4*b*x - 4*a) - (e^(4*b*x + 16*a) + 12*e^(2*b
*x + 14*a))*e^(-12*a) - 64*log(abs(e^(2*b*x + 2*a) - 1)))/b