### 3.221 $$\int \cosh (a+b x) \log (\cosh (\frac{a}{2}+\frac{b x}{2}) \sinh (\frac{a}{2}+\frac{b x}{2})) \, dx$$

Optimal. Leaf size=50 $\frac{\sinh (a+b x) \log \left (\sinh \left (\frac{a}{2}+\frac{b x}{2}\right ) \cosh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}-\frac{\sinh (a+b x)}{b}$

[Out]

-(Sinh[a + b*x]/b) + (Log[Cosh[a/2 + (b*x)/2]*Sinh[a/2 + (b*x)/2]]*Sinh[a + b*x])/b

________________________________________________________________________________________

Rubi [A]  time = 0.0286319, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 35, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.057, Rules used = {2637, 2554} $\frac{\sinh (a+b x) \log \left (\sinh \left (\frac{a}{2}+\frac{b x}{2}\right ) \cosh \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}-\frac{\sinh (a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sinh[a + b*x]/b) + (Log[Cosh[a/2 + (b*x)/2]*Sinh[a/2 + (b*x)/2]]*Sinh[a + b*x])/b

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin{align*} \int \cosh (a+b x) \log \left (\cosh \left (\frac{a}{2}+\frac{b x}{2}\right ) \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )\right ) \, dx &=\frac{\log \left (\cosh \left (\frac{a}{2}+\frac{b x}{2}\right ) \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )\right ) \sinh (a+b x)}{b}-\int \cosh (a+b x) \, dx\\ &=-\frac{\sinh (a+b x)}{b}+\frac{\log \left (\cosh \left (\frac{a}{2}+\frac{b x}{2}\right ) \sinh \left (\frac{a}{2}+\frac{b x}{2}\right )\right ) \sinh (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0105279, size = 33, normalized size = 0.66 $\frac{\sinh (a+b x) \log \left (\frac{1}{2} \sinh (a+b x)\right )}{b}-\frac{\sinh (a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sinh[a + b*x]/b) + (Log[Sinh[a + b*x]/2]*Sinh[a + b*x])/b

________________________________________________________________________________________

Maple [A]  time = 0.034, size = 32, normalized size = 0.6 \begin{align*}{\frac{\sinh \left ( bx+a \right ) }{b}\ln \left ({\frac{\sinh \left ( bx+a \right ) }{2}} \right ) }-{\frac{\sinh \left ( bx+a \right ) }{b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*ln(cosh(1/2*a+1/2*b*x)*sinh(1/2*a+1/2*b*x)),x)

[Out]

ln(1/2*sinh(b*x+a))/b*sinh(b*x+a)-sinh(b*x+a)/b

________________________________________________________________________________________

Maxima [B]  time = 1.0706, size = 151, normalized size = 3.02 \begin{align*} \frac{\log \left (\cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )\right ) \sinh \left (b x + a\right )}{b} - \frac{b{\left (\frac{2 \,{\left (b x + a\right )}}{b} + \frac{e^{\left (b x + a\right )}}{b} - \frac{e^{\left (-b x - a\right )}}{b}\right )} - b{\left (\frac{2 \,{\left (b x + a\right )}}{b} - \frac{e^{\left (b x + a\right )}}{b} + \frac{e^{\left (-b x - a\right )}}{b}\right )}}{4 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*log(cosh(1/2*a+1/2*b*x)*sinh(1/2*a+1/2*b*x)),x, algorithm="maxima")

[Out]

log(cosh(1/2*b*x + 1/2*a)*sinh(1/2*b*x + 1/2*a))*sinh(b*x + a)/b - 1/4*(b*(2*(b*x + a)/b + e^(b*x + a)/b - e^(
-b*x - a)/b) - b*(2*(b*x + a)/b - e^(b*x + a)/b + e^(-b*x - a)/b))/b

________________________________________________________________________________________

Fricas [B]  time = 1.91609, size = 782, normalized size = 15.64 \begin{align*} -\frac{\cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{4} + 4 \, \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{3} \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) + 6 \, \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2} \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2} + 4 \, \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{3} + \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{4} -{\left (\cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{4} + 4 \, \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{3} \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) + 6 \, \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2} \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2} + 4 \, \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{3} + \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{4} - 1\right )} \log \left (\cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )\right ) - 1}{2 \,{\left (b \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2} + 2 \, b \cosh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) + b \sinh \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*log(cosh(1/2*a+1/2*b*x)*sinh(1/2*a+1/2*b*x)),x, algorithm="fricas")

[Out]

-1/2*(cosh(1/2*b*x + 1/2*a)^4 + 4*cosh(1/2*b*x + 1/2*a)^3*sinh(1/2*b*x + 1/2*a) + 6*cosh(1/2*b*x + 1/2*a)^2*si
nh(1/2*b*x + 1/2*a)^2 + 4*cosh(1/2*b*x + 1/2*a)*sinh(1/2*b*x + 1/2*a)^3 + sinh(1/2*b*x + 1/2*a)^4 - (cosh(1/2*
b*x + 1/2*a)^4 + 4*cosh(1/2*b*x + 1/2*a)^3*sinh(1/2*b*x + 1/2*a) + 6*cosh(1/2*b*x + 1/2*a)^2*sinh(1/2*b*x + 1/
2*a)^2 + 4*cosh(1/2*b*x + 1/2*a)*sinh(1/2*b*x + 1/2*a)^3 + sinh(1/2*b*x + 1/2*a)^4 - 1)*log(cosh(1/2*b*x + 1/2
*a)*sinh(1/2*b*x + 1/2*a)) - 1)/(b*cosh(1/2*b*x + 1/2*a)^2 + 2*b*cosh(1/2*b*x + 1/2*a)*sinh(1/2*b*x + 1/2*a) +
b*sinh(1/2*b*x + 1/2*a)^2)

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*ln(cosh(1/2*a+1/2*b*x)*sinh(1/2*a+1/2*b*x)),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.34414, size = 127, normalized size = 2.54 \begin{align*} \frac{1}{2} \,{\left (\frac{e^{\left (b x + a\right )}}{b} - \frac{e^{\left (-b x - a\right )}}{b}\right )} \log \left (\frac{1}{4} \,{\left (e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )} + e^{\left (-\frac{1}{2} \, b x - \frac{1}{2} \, a\right )}\right )}{\left (e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )} - e^{\left (-\frac{1}{2} \, b x - \frac{1}{2} \, a\right )}\right )}\right ) - \frac{e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*log(cosh(1/2*a+1/2*b*x)*sinh(1/2*a+1/2*b*x)),x, algorithm="giac")

[Out]

1/2*(e^(b*x + a)/b - e^(-b*x - a)/b)*log(1/4*(e^(1/2*b*x + 1/2*a) + e^(-1/2*b*x - 1/2*a))*(e^(1/2*b*x + 1/2*a)
- e^(-1/2*b*x - 1/2*a))) - 1/2*(e^(b*x + a) - e^(-b*x - a))/b