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

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Rubi [A]  time = 0.03, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 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]

Rule 2637

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

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Mathematica [A]  time = 0.01, 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

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fricas [B]  time = 0.48, size = 258, normalized size = 5.16 \[ -\frac {\cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + 4 \, \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) + 6 \, \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 4 \, \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} + \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} - {\left (\cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + 4 \, \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) + 6 \, \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 4 \, \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} + \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} - 1\right )} \log \left (\cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) - 1}{2 \, {\left (b \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 2 \, b \cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) + b \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2}\right )}} \]

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)

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giac [B]  time = 0.47, size = 94, normalized size = 1.88 \[ \frac {1}{2} \, {\left (\frac {e^{\left (b x + a\right )}}{b} - \frac {e^{\left (-b x - a\right )}}{b}\right )} \log \left (\frac {1}{4} \, {\left (e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} + e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )}\right )} {\left (e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - e^{\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a\right )}\right )}\right ) - \frac {e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}}{2 \, b} \]

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

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maple [A]  time = 0.89, size = 32, normalized size = 0.64 \[ \frac {\ln \left (\frac {\sinh \left (b x +a \right )}{2}\right ) \sinh \left (b x +a \right )}{b}-\frac {\sinh \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.51, size = 112, normalized size = 2.24 \[ \frac {\log \left (\cosh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \sinh \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right ) \sinh \left (b x + a\right )}{b} - \frac {b {\left (\frac {2 \, {\left (b x + a\right )}}{b} + \frac {e^{\left (b x + a\right )}}{b} - \frac {e^{\left (-b x - a\right )}}{b}\right )} - b {\left (\frac {2 \, {\left (b x + a\right )}}{b} - \frac {e^{\left (b x + a\right )}}{b} + \frac {e^{\left (-b x - a\right )}}{b}\right )}}{4 \, b} \]

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

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mupad [B]  time = 0.51, size = 31, normalized size = 0.62 \[ \frac {\ln \left (\frac {\mathrm {sinh}\left (a+b\,x\right )}{2}\right )\,\mathrm {sinh}\left (a+b\,x\right )}{b}-\frac {\mathrm {sinh}\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(sinh(a + b*x)/2)*sinh(a + b*x))/b - sinh(a + b*x)/b

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (\sinh {\left (\frac {a}{2} + \frac {b x}{2} \right )} \cosh {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \cosh {\left (a + b x \right )}\, dx \]

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)

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