### 3.60 $$\int \frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)} \, dx$$

Optimal. Leaf size=218 $-\frac{\log \left (\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}-\frac{\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}+1\right )}{4 b}+\frac{\log \left (\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}+\frac{\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}+1\right )}{4 b}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}}{\sqrt{3}}\right )}{2 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}+1}{\sqrt{3}}\right )}{2 b}+\frac{\tanh ^{-1}\left (\frac{\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}\right )}{b}$

[Out]

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

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Rubi [A]  time = 0.223059, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {2574, 296, 634, 618, 204, 628, 206} $-\frac{\log \left (\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}-\frac{\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}+1\right )}{4 b}+\frac{\log \left (\frac{\sinh ^{\frac{2}{3}}(a+b x)}{\cosh ^{\frac{2}{3}}(a+b x)}+\frac{\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}+1\right )}{4 b}+\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}}{\sqrt{3}}\right )}{2 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}+1}{\sqrt{3}}\right )}{2 b}+\frac{\tanh ^{-1}\left (\frac{\sqrt [3]{\sinh (a+b x)}}{\sqrt [3]{\cosh (a+b x)}}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[a + b*x]^(2/3)/Cosh[a + b*x]^(2/3),x]

[Out]

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

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 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

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]

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

Mathematica [C]  time = 0.0399421, size = 59, normalized size = 0.27 $\frac{3 \sinh ^{\frac{5}{3}}(a+b x) \cosh ^2(a+b x)^{5/6} \, _2F_1\left (\frac{5}{6},\frac{5}{6};\frac{11}{6};-\sinh ^2(a+b x)\right )}{5 b \cosh ^{\frac{5}{3}}(a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[a + b*x]^(2/3)/Cosh[a + b*x]^(2/3),x]

[Out]

(3*(Cosh[a + b*x]^2)^(5/6)*Hypergeometric2F1[5/6, 5/6, 11/6, -Sinh[a + b*x]^2]*Sinh[a + b*x]^(5/3))/(5*b*Cosh[
a + b*x]^(5/3))

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Maple [F]  time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sinh \left ( bx+a \right ) \right ) ^{{\frac{2}{3}}} \left ( \cosh \left ( bx+a \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^(2/3)/cosh(b*x+a)^(2/3),x)

[Out]

int(sinh(b*x+a)^(2/3)/cosh(b*x+a)^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x + a\right )^{\frac{2}{3}}}{\cosh \left (b x + a\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(2/3)/cosh(b*x+a)^(2/3),x, algorithm="maxima")

[Out]

integrate(sinh(b*x + a)^(2/3)/cosh(b*x + a)^(2/3), x)

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Fricas [B]  time = 2.21773, size = 2232, normalized size = 10.24 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(2/3)/cosh(b*x+a)^(2/3),x, algorithm="fricas")

[Out]

1/4*(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*(sqrt(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)) + 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*(sqrt(3)*cosh(b*x + a) + sqr
t(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)) + log((cosh(b*x + a)^2 + 2*(cosh(b*x + a) + sinh(b*x + a))*cosh(b*x + a)^(2/
3)*sinh(b*x + a)^(1/3) + 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)) - log((cosh(b*x + a)^2 + 2*(cosh(b*x + a) + sinh(b*x + a))*cosh(b*x + a)^(2/3)*sinh(b*x + a)^(1/3) - 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)) + 2*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)) - 2*log(-(c
osh(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)*s
inh(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)))/
b

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)**(2/3)/cosh(b*x+a)**(2/3),x)

[Out]

Integral(sinh(a + b*x)**(2/3)/cosh(a + b*x)**(2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x + a\right )^{\frac{2}{3}}}{\cosh \left (b x + a\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^(2/3)/cosh(b*x+a)^(2/3),x, algorithm="giac")

[Out]

integrate(sinh(b*x + a)^(2/3)/cosh(b*x + a)^(2/3), x)