3.200 \(\int \frac {A+B \sinh (x)}{1+\cosh (x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {A \sinh (x)}{\cosh (x)+1}+B \log (\cosh (x)+1) \]

[Out]

B*ln(1+cosh(x))+A*sinh(x)/(1+cosh(x))

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Rubi [A]  time = 0.08, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4401, 2648, 2667, 31} \[ \frac {A \sinh (x)}{\cosh (x)+1}+B \log (\cosh (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sinh[x])/(1 + Cosh[x]),x]

[Out]

B*Log[1 + Cosh[x]] + (A*Sinh[x])/(1 + Cosh[x])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \frac {A+B \sinh (x)}{1+\cosh (x)} \, dx &=\int \left (\frac {A}{1+\cosh (x)}+\frac {B \sinh (x)}{1+\cosh (x)}\right ) \, dx\\ &=A \int \frac {1}{1+\cosh (x)} \, dx+B \int \frac {\sinh (x)}{1+\cosh (x)} \, dx\\ &=\frac {A \sinh (x)}{1+\cosh (x)}+B \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\cosh (x)\right )\\ &=B \log (1+\cosh (x))+\frac {A \sinh (x)}{1+\cosh (x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 19, normalized size = 1.06 \[ A \tanh \left (\frac {x}{2}\right )+2 B \log \left (\cosh \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sinh[x])/(1 + Cosh[x]),x]

[Out]

2*B*Log[Cosh[x/2]] + A*Tanh[x/2]

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fricas [B]  time = 0.49, size = 46, normalized size = 2.56 \[ -\frac {B x \cosh \relax (x) + B x \sinh \relax (x) + B x - 2 \, {\left (B \cosh \relax (x) + B \sinh \relax (x) + B\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + 2 \, A}{\cosh \relax (x) + \sinh \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(1+cosh(x)),x, algorithm="fricas")

[Out]

-(B*x*cosh(x) + B*x*sinh(x) + B*x - 2*(B*cosh(x) + B*sinh(x) + B)*log(cosh(x) + sinh(x) + 1) + 2*A)/(cosh(x) +
sinh(x) + 1)

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giac [A]  time = 0.12, size = 22, normalized size = 1.22 \[ -B x + 2 \, B \log \left (e^{x} + 1\right ) - \frac {2 \, A}{e^{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(1+cosh(x)),x, algorithm="giac")

[Out]

-B*x + 2*B*log(e^x + 1) - 2*A/(e^x + 1)

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maple [A]  time = 0.05, size = 28, normalized size = 1.56 \[ A \tanh \left (\frac {x}{2}\right )-B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-B \ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sinh(x))/(1+cosh(x)),x)

[Out]

A*tanh(1/2*x)-B*ln(tanh(1/2*x)-1)-B*ln(tanh(1/2*x)+1)

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maxima [A]  time = 0.37, size = 19, normalized size = 1.06 \[ B \log \left (\cosh \relax (x) + 1\right ) + \frac {2 \, A}{e^{\left (-x\right )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(1+cosh(x)),x, algorithm="maxima")

[Out]

B*log(cosh(x) + 1) + 2*A/(e^(-x) + 1)

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mupad [B]  time = 0.06, size = 22, normalized size = 1.22 \[ 2\,B\,\ln \left ({\mathrm {e}}^x+1\right )-\frac {2\,A}{{\mathrm {e}}^x+1}-B\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sinh(x))/(cosh(x) + 1),x)

[Out]

2*B*log(exp(x) + 1) - (2*A)/(exp(x) + 1) - B*x

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sympy [A]  time = 0.34, size = 20, normalized size = 1.11 \[ A \tanh {\left (\frac {x}{2} \right )} + B x - 2 B \log {\left (\tanh {\left (\frac {x}{2} \right )} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(1+cosh(x)),x)

[Out]

A*tanh(x/2) + B*x - 2*B*log(tanh(x/2) + 1)

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