### 3.110 $$\int \cosh ^3(a+b x) \coth ^3(a+b x) \, dx$$

Optimal. Leaf size=66 $\frac{5 \cosh ^3(a+b x)}{6 b}+\frac{5 \cosh (a+b x)}{2 b}-\frac{\cosh ^3(a+b x) \coth ^2(a+b x)}{2 b}-\frac{5 \tanh ^{-1}(\cosh (a+b x))}{2 b}$

[Out]

(-5*ArcTanh[Cosh[a + b*x]])/(2*b) + (5*Cosh[a + b*x])/(2*b) + (5*Cosh[a + b*x]^3)/(6*b) - (Cosh[a + b*x]^3*Cot
h[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.0482733, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {2592, 288, 302, 206} $\frac{5 \cosh ^3(a+b x)}{6 b}+\frac{5 \cosh (a+b x)}{2 b}-\frac{\cosh ^3(a+b x) \coth ^2(a+b x)}{2 b}-\frac{5 \tanh ^{-1}(\cosh (a+b x))}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[a + b*x]^3*Coth[a + b*x]^3,x]

[Out]

(-5*ArcTanh[Cosh[a + b*x]])/(2*b) + (5*Cosh[a + b*x])/(2*b) + (5*Cosh[a + b*x]^3)/(6*b) - (Cosh[a + b*x]^3*Cot
h[a + b*x]^2)/(2*b)

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

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

Rubi steps

\begin{align*} \int \cosh ^3(a+b x) \coth ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\cosh (a+b x)\right )}{b}\\ &=-\frac{\cosh ^3(a+b x) \coth ^2(a+b x)}{2 b}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cosh (a+b x)\right )}{2 b}\\ &=-\frac{\cosh ^3(a+b x) \coth ^2(a+b x)}{2 b}-\frac{5 \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cosh (a+b x)\right )}{2 b}\\ &=\frac{5 \cosh (a+b x)}{2 b}+\frac{5 \cosh ^3(a+b x)}{6 b}-\frac{\cosh ^3(a+b x) \coth ^2(a+b x)}{2 b}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cosh (a+b x)\right )}{2 b}\\ &=-\frac{5 \tanh ^{-1}(\cosh (a+b x))}{2 b}+\frac{5 \cosh (a+b x)}{2 b}+\frac{5 \cosh ^3(a+b x)}{6 b}-\frac{\cosh ^3(a+b x) \coth ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.040455, size = 85, normalized size = 1.29 $\frac{9 \cosh (a+b x)}{4 b}+\frac{\cosh (3 (a+b x))}{12 b}-\frac{\text{csch}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{\text{sech}^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{5 \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[a + b*x]^3*Coth[a + b*x]^3,x]

[Out]

(9*Cosh[a + b*x])/(4*b) + Cosh[3*(a + b*x)]/(12*b) - Csch[(a + b*x)/2]^2/(8*b) + (5*Log[Tanh[(a + b*x)/2]])/(2
*b) - Sech[(a + b*x)/2]^2/(8*b)

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Maple [A]  time = 0.021, size = 81, normalized size = 1.2 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{5}}{3\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{5\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{3\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}-5\,{\frac{\cosh \left ( bx+a \right ) }{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{5\,{\rm csch} \left (bx+a\right ){\rm coth} \left (bx+a\right )}{2}}-5\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/3/sinh(b*x+a)^2*cosh(b*x+a)^5+5/3/sinh(b*x+a)^2*cosh(b*x+a)^3-5/sinh(b*x+a)^2*cosh(b*x+a)+5/2*csch(b*x+
a)*coth(b*x+a)-5*arctanh(exp(b*x+a)))

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Maxima [B]  time = 1.04491, size = 180, normalized size = 2.73 \begin{align*} \frac{27 \, e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}}{24 \, b} - \frac{5 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac{5 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} + \frac{25 \, e^{\left (-2 \, b x - 2 \, a\right )} - 77 \, e^{\left (-4 \, b x - 4 \, a\right )} + 3 \, e^{\left (-6 \, b x - 6 \, a\right )} + 1}{24 \, b{\left (e^{\left (-3 \, b x - 3 \, a\right )} - 2 \, e^{\left (-5 \, b x - 5 \, a\right )} + e^{\left (-7 \, b x - 7 \, a\right )}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(27*e^(-b*x - a) + e^(-3*b*x - 3*a))/b - 5/2*log(e^(-b*x - a) + 1)/b + 5/2*log(e^(-b*x - a) - 1)/b + 1/24
*(25*e^(-2*b*x - 2*a) - 77*e^(-4*b*x - 4*a) + 3*e^(-6*b*x - 6*a) + 1)/(b*(e^(-3*b*x - 3*a) - 2*e^(-5*b*x - 5*a
) + e^(-7*b*x - 7*a)))

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Fricas [B]  time = 1.97856, size = 3027, normalized size = 45.86 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(cosh(b*x + a)^10 + 10*cosh(b*x + a)*sinh(b*x + a)^9 + sinh(b*x + a)^10 + 5*(9*cosh(b*x + a)^2 + 5)*sinh(
b*x + a)^8 + 25*cosh(b*x + a)^8 + 40*(3*cosh(b*x + a)^3 + 5*cosh(b*x + a))*sinh(b*x + a)^7 + 10*(21*cosh(b*x +
a)^4 + 70*cosh(b*x + a)^2 - 5)*sinh(b*x + a)^6 - 50*cosh(b*x + a)^6 + 4*(63*cosh(b*x + a)^5 + 350*cosh(b*x +
a)^3 - 75*cosh(b*x + a))*sinh(b*x + a)^5 + 10*(21*cosh(b*x + a)^6 + 175*cosh(b*x + a)^4 - 75*cosh(b*x + a)^2 -
5)*sinh(b*x + a)^4 - 50*cosh(b*x + a)^4 + 40*(3*cosh(b*x + a)^7 + 35*cosh(b*x + a)^5 - 25*cosh(b*x + a)^3 - 5
*cosh(b*x + a))*sinh(b*x + a)^3 + 5*(9*cosh(b*x + a)^8 + 140*cosh(b*x + a)^6 - 150*cosh(b*x + a)^4 - 60*cosh(b
*x + a)^2 + 5)*sinh(b*x + a)^2 + 25*cosh(b*x + a)^2 - 60*(cosh(b*x + a)^7 + 7*cosh(b*x + a)*sinh(b*x + a)^6 +
sinh(b*x + a)^7 + (21*cosh(b*x + a)^2 - 2)*sinh(b*x + a)^5 - 2*cosh(b*x + a)^5 + 5*(7*cosh(b*x + a)^3 - 2*cosh
(b*x + a))*sinh(b*x + a)^4 + (35*cosh(b*x + a)^4 - 20*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + cosh(b*x + a)^3 +
(21*cosh(b*x + a)^5 - 20*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 + (7*cosh(b*x + a)^6 - 10*cosh(b*
x + a)^4 + 3*cosh(b*x + a)^2)*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 60*(cosh(b*x + a)^7 + 7*
cosh(b*x + a)*sinh(b*x + a)^6 + sinh(b*x + a)^7 + (21*cosh(b*x + a)^2 - 2)*sinh(b*x + a)^5 - 2*cosh(b*x + a)^5
+ 5*(7*cosh(b*x + a)^3 - 2*cosh(b*x + a))*sinh(b*x + a)^4 + (35*cosh(b*x + a)^4 - 20*cosh(b*x + a)^2 + 1)*sin
h(b*x + a)^3 + cosh(b*x + a)^3 + (21*cosh(b*x + a)^5 - 20*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 +
(7*cosh(b*x + a)^6 - 10*cosh(b*x + a)^4 + 3*cosh(b*x + a)^2)*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a)
- 1) + 10*(cosh(b*x + a)^9 + 20*cosh(b*x + a)^7 - 30*cosh(b*x + a)^5 - 20*cosh(b*x + a)^3 + 5*cosh(b*x + a))*
sinh(b*x + a) + 1)/(b*cosh(b*x + a)^7 + 7*b*cosh(b*x + a)*sinh(b*x + a)^6 + b*sinh(b*x + a)^7 - 2*b*cosh(b*x +
a)^5 + (21*b*cosh(b*x + a)^2 - 2*b)*sinh(b*x + a)^5 + 5*(7*b*cosh(b*x + a)^3 - 2*b*cosh(b*x + a))*sinh(b*x +
a)^4 + b*cosh(b*x + a)^3 + (35*b*cosh(b*x + a)^4 - 20*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^3 + (21*b*cosh(b*x
+ a)^5 - 20*b*cosh(b*x + a)^3 + 3*b*cosh(b*x + a))*sinh(b*x + a)^2 + (7*b*cosh(b*x + a)^6 - 10*b*cosh(b*x + a)
^4 + 3*b*cosh(b*x + a)^2)*sinh(b*x + a))

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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)**3*coth(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.32919, size = 146, normalized size = 2.21 \begin{align*} \frac{{\left (27 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} +{\left (e^{\left (3 \, b x + 30 \, a\right )} + 27 \, e^{\left (b x + 28 \, a\right )}\right )} e^{\left (-27 \, a\right )} - \frac{24 \,{\left (e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 60 \, \log \left (e^{\left (b x + a\right )} + 1\right ) + 60 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{24 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((27*e^(2*b*x + 2*a) + 1)*e^(-3*b*x - 3*a) + (e^(3*b*x + 30*a) + 27*e^(b*x + 28*a))*e^(-27*a) - 24*(e^(3*
b*x + 3*a) + e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^2 - 60*log(e^(b*x + a) + 1) + 60*log(abs(e^(b*x + a) - 1)))/b