### 3.1053 $$\int \frac{\cosh ^3(a+b x)-\sinh ^3(a+b x)}{\cosh ^3(a+b x)+\sinh ^3(a+b x)} \, dx$$

Optimal. Leaf size=47 $-\frac{1}{3 b (\tanh (a+b x)+1)}-\frac{4 \tan ^{-1}\left (\frac{1-2 \tanh (a+b x)}{\sqrt{3}}\right )}{3 \sqrt{3} b}$

[Out]

(-4*ArcTan[(1 - 2*Tanh[a + b*x])/Sqrt[3]])/(3*Sqrt[3]*b) - 1/(3*b*(1 + Tanh[a + b*x]))

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Rubi [A]  time = 0.327863, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {2074, 618, 204} $-\frac{1}{3 b (\tanh (a+b x)+1)}-\frac{4 \tan ^{-1}\left (\frac{1-2 \tanh (a+b x)}{\sqrt{3}}\right )}{3 \sqrt{3} b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-4*ArcTan[(1 - 2*Tanh[a + b*x])/Sqrt[3]])/(3*Sqrt[3]*b) - 1/(3*b*(1 + Tanh[a + b*x]))

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

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])

Rubi steps

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

Mathematica [B]  time = 1.33687, size = 115, normalized size = 2.45 $\frac{(\sinh (a+b x)-\cosh (a+b x)) \left (\cosh (a+b x) \left (8 \sqrt{3} \tan ^{-1}\left (\frac{\text{sech}(b x) (\cosh (2 a+b x)-2 \sinh (2 a+b x))}{\sqrt{3}}\right )+3\right )+\sinh (a+b x) \left (8 \sqrt{3} \tan ^{-1}\left (\frac{\text{sech}(b x) (\cosh (2 a+b x)-2 \sinh (2 a+b x))}{\sqrt{3}}\right )-3\right )\right )}{18 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-Cosh[a + b*x] + Sinh[a + b*x])*((3 + 8*Sqrt[3]*ArcTan[(Sech[b*x]*(Cosh[2*a + b*x] - 2*Sinh[2*a + b*x]))/Sqr
t[3]])*Cosh[a + b*x] + (-3 + 8*Sqrt[3]*ArcTan[(Sech[b*x]*(Cosh[2*a + b*x] - 2*Sinh[2*a + b*x]))/Sqrt[3]])*Sinh
[a + b*x]))/(18*b)

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Maple [C]  time = 0.142, size = 120, normalized size = 2.6 \begin{align*} -{\frac{2}{3\,b} \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-2}}+{\frac{2}{3\,b} \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{\frac{2\,i}{9}}\sqrt{3}}{b}\ln \left ( \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}+ \left ( -i\sqrt{3}-1 \right ) \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) }-{\frac{{\frac{2\,i}{9}}\sqrt{3}}{b}\ln \left ( \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}+ \left ( i\sqrt{3}-1 \right ) \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1 \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/b/(tanh(1/2*b*x+1/2*a)+1)^2+2/3/b/(tanh(1/2*b*x+1/2*a)+1)+2/9*I/b*3^(1/2)*ln(tanh(1/2*b*x+1/2*a)^2+(-I*3^
(1/2)-1)*tanh(1/2*b*x+1/2*a)+1)-2/9*I/b*3^(1/2)*ln(tanh(1/2*b*x+1/2*a)^2+(I*3^(1/2)-1)*tanh(1/2*b*x+1/2*a)+1)

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Maxima [B]  time = 1.57628, size = 126, normalized size = 2.68 \begin{align*} \frac{4 \,{\left (\sqrt{3} \arctan \left (\frac{1}{6} \cdot 3^{\frac{3}{4}} \sqrt{2}{\left (2 \, \sqrt{3} e^{\left (-b x - a\right )} + 3^{\frac{1}{4}} \sqrt{2}\right )}\right ) - \sqrt{3} \arctan \left (\frac{1}{6} \cdot 3^{\frac{3}{4}} \sqrt{2}{\left (2 \, \sqrt{3} e^{\left (-b x - a\right )} - 3^{\frac{1}{4}} \sqrt{2}\right )}\right )\right )}}{9 \, b} - \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{6 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/9*(sqrt(3)*arctan(1/6*3^(3/4)*sqrt(2)*(2*sqrt(3)*e^(-b*x - a) + 3^(1/4)*sqrt(2))) - sqrt(3)*arctan(1/6*3^(3/
4)*sqrt(2)*(2*sqrt(3)*e^(-b*x - a) - 3^(1/4)*sqrt(2))))/b - 1/6*e^(-2*b*x - 2*a)/b

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Fricas [B]  time = 2.17371, size = 370, normalized size = 7.87 \begin{align*} -\frac{8 \,{\left (\sqrt{3} \cosh \left (b x + a\right )^{2} + 2 \, \sqrt{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sqrt{3} \sinh \left (b x + a\right )^{2}\right )} \arctan \left (-\frac{\sqrt{3} \cosh \left (b x + a\right ) + \sqrt{3} \sinh \left (b x + a\right )}{3 \,{\left (\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}}\right ) + 3}{18 \,{\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(8*(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)*arctan(-1
/3*(sqrt(3)*cosh(b*x + a) + sqrt(3)*sinh(b*x + a))/(cosh(b*x + a) - sinh(b*x + a))) + 3)/(b*cosh(b*x + a)^2 +
2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.17791, size = 51, normalized size = 1.09 \begin{align*} \frac{4 \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3} e^{\left (2 \, b x + 2 \, a\right )}\right )}{9 \, b} - \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{6 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/9*sqrt(3)*arctan(1/3*sqrt(3)*e^(2*b*x + 2*a))/b - 1/6*e^(-2*b*x - 2*a)/b