3.65 \(\int \frac{\cosh ^{\frac{5}{3}}(a+b x)}{\sinh ^{\frac{5}{3}}(a+b x)} \, dx\)

Optimal. Leaf size=155 \[ -\frac{3 \cosh ^{\frac{2}{3}}(a+b x)}{2 b \sinh ^{\frac{2}{3}}(a+b x)}-\frac{\log \left (1-\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{\log \left (\frac{\sinh ^{\frac{4}{3}}(a+b x)}{\cosh ^{\frac{4}{3}}(a+b x)}+\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}+1\right )}{4 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}+1}{\sqrt{3}}\right )}{2 b} \]

[Out]

-(Sqrt[3]*ArcTan[(1 + (2*Sinh[a + b*x]^(2/3))/Cosh[a + b*x]^(2/3))/Sqrt[3]])/(2*b) - Log[1 - Sinh[a + b*x]^(2/
3)/Cosh[a + b*x]^(2/3)]/(2*b) + Log[1 + Sinh[a + b*x]^(2/3)/Cosh[a + b*x]^(2/3) + Sinh[a + b*x]^(4/3)/Cosh[a +
 b*x]^(4/3)]/(4*b) - (3*Cosh[a + b*x]^(2/3))/(2*b*Sinh[a + b*x]^(2/3))

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Rubi [A]  time = 0.124612, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2567, 2574, 275, 292, 31, 634, 618, 204, 628} \[ -\frac{3 \cosh ^{\frac{2}{3}}(a+b x)}{2 b \sinh ^{\frac{2}{3}}(a+b x)}-\frac{\log \left (1-\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{\log \left (\frac{\sinh ^{\frac{4}{3}}(a+b x)}{\cosh ^{\frac{4}{3}}(a+b x)}+\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}+1\right )}{4 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}+1}{\sqrt{3}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^(5/3)/Sinh[a + b*x]^(5/3),x]

[Out]

-(Sqrt[3]*ArcTan[(1 + (2*Sinh[a + b*x]^(2/3))/Cosh[a + b*x]^(2/3))/Sqrt[3]])/(2*b) - Log[1 - Sinh[a + b*x]^(2/
3)/Cosh[a + b*x]^(2/3)]/(2*b) + Log[1 + Sinh[a + b*x]^(2/3)/Cosh[a + b*x]^(2/3) + Sinh[a + b*x]^(4/3)/Cosh[a +
 b*x]^(4/3)]/(4*b) - (3*Cosh[a + b*x]^(2/3))/(2*b*Sinh[a + b*x]^(2/3))

Rule 2567

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[e +
 f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
 + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [C]  time = 0.0349135, size = 59, normalized size = 0.38 \[ -\frac{3 \cosh ^2(a+b x)^{2/3} \, _2F_1\left (-\frac{1}{3},-\frac{1}{3};\frac{2}{3};-\sinh ^2(a+b x)\right )}{2 b \sinh ^{\frac{2}{3}}(a+b x) \cosh ^{\frac{4}{3}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^(5/3)/Sinh[a + b*x]^(5/3),x]

[Out]

(-3*(Cosh[a + b*x]^2)^(2/3)*Hypergeometric2F1[-1/3, -1/3, 2/3, -Sinh[a + b*x]^2])/(2*b*Cosh[a + b*x]^(4/3)*Sin
h[a + b*x]^(2/3))

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Maple [F]  time = 0.038, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cosh \left ( bx+a \right ) \right ) ^{{\frac{5}{3}}} \left ( \sinh \left ( bx+a \right ) \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^(5/3)/sinh(b*x+a)^(5/3),x)

[Out]

int(cosh(b*x+a)^(5/3)/sinh(b*x+a)^(5/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (b x + a\right )^{\frac{5}{3}}}{\sinh \left (b x + a\right )^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^(5/3)/sinh(b*x+a)^(5/3),x, algorithm="maxima")

[Out]

integrate(cosh(b*x + a)^(5/3)/sinh(b*x + a)^(5/3), x)

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Fricas [B]  time = 1.92521, size = 2222, normalized size = 14.34 \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)^(5/3)/sinh(b*x+a)^(5/3),x, algorithm="fricas")

[Out]

-1/4*(2*(sqrt(3)*cosh(b*x + a)^2 + 2*sqrt(3)*cosh(b*x + a)*sinh(b*x + a) + sqrt(3)*sinh(b*x + a)^2 - sqrt(3))*
arctan(1/3*(sqrt(3)*cosh(b*x + a)^2 + 2*sqrt(3)*cosh(b*x + a)*sinh(b*x + a) + sqrt(3)*sinh(b*x + a)^2 + 4*(sqr
t(3)*cosh(b*x + a) + sqrt(3)*sinh(b*x + a))*cosh(b*x + a)^(1/3)*sinh(b*x + a)^(2/3) + sqrt(3))/(cosh(b*x + a)^
2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) - (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) +
 sinh(b*x + a)^2 - 1)*log((cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x
 + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 2*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh
(b*x + a)^3 + (3*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - cosh(b*x + a))*cosh(b*x + a)^(2/3)*sinh(b*x + a)^(1/3) +
 2*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 + (3*cosh(b*x + a)^2 + 1)*sinh(b*x + a
) + cosh(b*x + a))*cosh(b*x + a)^(1/3)*sinh(b*x + a)^(2/3) + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a)
 + 1)/(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*
x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)) + 2*(cosh(b*x + a)^2 +
2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*log(-(cosh(b*x + a)^2 - 2*(cosh(b*x + a) + sinh(b*x + a))
*cosh(b*x + a)^(1/3)*sinh(b*x + a)^(2/3) + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)/(cosh(b*x + a)
^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) + 12*(cosh(b*x + a) + sinh(b*x + a))*cosh(b*x + a)^
(2/3)*sinh(b*x + a)^(1/3))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 - b)

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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)**(5/3)/sinh(b*x+a)**(5/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (b x + a\right )^{\frac{5}{3}}}{\sinh \left (b x + a\right )^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^(5/3)/sinh(b*x+a)^(5/3),x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)^(5/3)/sinh(b*x + a)^(5/3), x)