### 3.1052 $$\int \frac{\cosh ^4(a+b x)-\sinh ^4(a+b x)}{\cosh ^4(a+b x)+\sinh ^4(a+b x)} \, dx$$

Optimal. Leaf size=51 $\frac{\tan ^{-1}\left (\sqrt{2} \tanh (a+b x)+1\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (1-\sqrt{2} \tanh (a+b x)\right )}{\sqrt{2} b}$

[Out]

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

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Rubi [A]  time = 0.17189, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.077, Rules used = {1162, 617, 204} $\frac{\tan ^{-1}\left (\sqrt{2} \tanh (a+b x)+1\right )}{\sqrt{2} b}-\frac{\tan ^{-1}\left (1-\sqrt{2} \tanh (a+b x)\right )}{\sqrt{2} b}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]^4 - Sinh[a + b*x]^4)/(Cosh[a + b*x]^4 + Sinh[a + b*x]^4),x]

[Out]

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 ^4(a+b x)-\sinh ^4(a+b x)}{\cosh ^4(a+b x)+\sinh ^4(a+b x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \tanh (a+b x)\right )}{\sqrt{2} b}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \tanh (a+b x)\right )}{\sqrt{2} b}\\ &=-\frac{\tan ^{-1}\left (1-\sqrt{2} \tanh (a+b x)\right )}{\sqrt{2} b}+\frac{\tan ^{-1}\left (1+\sqrt{2} \tanh (a+b x)\right )}{\sqrt{2} b}\\ \end{align*}

Mathematica [A]  time = 0.0357443, size = 25, normalized size = 0.49 $\frac{\tan ^{-1}\left (\frac{\sinh (2 a+2 b x)}{\sqrt{2}}\right )}{\sqrt{2} b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]^4 - Sinh[a + b*x]^4)/(Cosh[a + b*x]^4 + Sinh[a + b*x]^4),x]

[Out]

ArcTan[Sinh[2*a + 2*b*x]/Sqrt[2]]/(Sqrt[2]*b)

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Maple [C]  time = 0.097, size = 138, normalized size = 2.7 \begin{align*}{\frac{{\frac{i}{4}}\sqrt{2}}{b}\ln \left ( -2\,i\sqrt{2} \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{3}+ \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{4}-2\,i\sqrt{2}\tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -2\, \left ( \tanh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1 \right ) }-{\frac{{\frac{i}{4}}\sqrt{2}}{b}\ln \left ( 2\,i\sqrt{2} \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{3}+ \left ( \tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{4}+2\,i\sqrt{2}\tanh \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) -2\, \left ( \tanh \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+1 \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(b*x+a)^4-sinh(b*x+a)^4)/(cosh(b*x+a)^4+sinh(b*x+a)^4),x)

[Out]

1/4*I/b*2^(1/2)*ln(-2*I*2^(1/2)*tanh(1/2*b*x+1/2*a)^3+tanh(1/2*b*x+1/2*a)^4-2*I*2^(1/2)*tanh(1/2*b*x+1/2*a)-2*
tanh(1/2*b*x+1/2*a)^2+1)-1/4*I/b*2^(1/2)*ln(2*I*2^(1/2)*tanh(1/2*b*x+1/2*a)^3+tanh(1/2*b*x+1/2*a)^4+2*I*2^(1/2
)*tanh(1/2*b*x+1/2*a)-2*tanh(1/2*b*x+1/2*a)^2+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, \int \frac{{\left (e^{\left (-b x - a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )} e^{\left (-b x - a\right )}}{6 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1}\,{d x} + 2 \, \int \frac{e^{\left (6 \, b x + 6 \, a\right )}}{e^{\left (8 \, b x + 8 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} + 1}\,{d x} + 2 \, \int \frac{e^{\left (-6 \, b x - 6 \, a\right )}}{6 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-8 \, b x - 8 \, a\right )} + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*integrate((e^(-b*x - a) + e^(-5*b*x - 5*a))*e^(-b*x - a)/(6*e^(-4*b*x - 4*a) + e^(-8*b*x - 8*a) + 1), x) + 2
*integrate(e^(6*b*x + 6*a)/(e^(8*b*x + 8*a) + 6*e^(4*b*x + 4*a) + 1), x) + 2*integrate(e^(-6*b*x - 6*a)/(6*e^(
-4*b*x - 4*a) + e^(-8*b*x - 8*a) + 1), x)

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Fricas [B]  time = 2.06874, size = 559, normalized size = 10.96 \begin{align*} -\frac{\sqrt{2} \arctan \left (-\frac{\sqrt{2} \cosh \left (b x + a\right )^{3} + 3 \, \sqrt{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sqrt{2} \sinh \left (b x + a\right )^{3} +{\left (3 \, \sqrt{2} \cosh \left (b x + a\right )^{2} - 7 \, \sqrt{2}\right )} \sinh \left (b x + a\right ) + 7 \, \sqrt{2} \cosh \left (b x + a\right )}{4 \,{\left (\cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right ) + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - \sinh \left (b x + a\right )^{3}\right )}}\right ) + \sqrt{2} \arctan \left (-\frac{\sqrt{2} \cosh \left (b x + a\right ) + \sqrt{2} \sinh \left (b x + a\right )}{4 \,{\left (\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}}\right )}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(2)*arctan(-1/4*(sqrt(2)*cosh(b*x + a)^3 + 3*sqrt(2)*cosh(b*x + a)*sinh(b*x + a)^2 + sqrt(2)*sinh(b*
x + a)^3 + (3*sqrt(2)*cosh(b*x + a)^2 - 7*sqrt(2))*sinh(b*x + a) + 7*sqrt(2)*cosh(b*x + a))/(cosh(b*x + a)^3 -
3*cosh(b*x + a)^2*sinh(b*x + a) + 3*cosh(b*x + a)*sinh(b*x + a)^2 - sinh(b*x + a)^3)) + sqrt(2)*arctan(-1/4*(
sqrt(2)*cosh(b*x + a) + sqrt(2)*sinh(b*x + a))/(cosh(b*x + a) - sinh(b*x + a))))/b

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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)**4-sinh(b*x+a)**4)/(cosh(b*x+a)**4+sinh(b*x+a)**4),x)

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x