3.2 \(\int \frac{A+B \sin (x)}{1+\cos (x)} \, dx\)

Optimal. Leaf size=19 \[ \frac{A \sin (x)}{\cos (x)+1}-B \log (\cos (x)+1) \]

[Out]

-(B*Log[1 + Cos[x]]) + (A*Sin[x])/(1 + Cos[x])

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Rubi [A]  time = 0.0677288, antiderivative size = 19, normalized size of antiderivative = 1., 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} \[ \frac{A \sin (x)}{\cos (x)+1}-B \log (\cos (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[x])/(1 + Cos[x]),x]

[Out]

-(B*Log[1 + Cos[x]]) + (A*Sin[x])/(1 + Cos[x])

Rule 4401

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

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 31

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

Rubi steps

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

Mathematica [A]  time = 0.0318927, size = 19, normalized size = 1. \[ A \tan \left (\frac{x}{2}\right )-2 B \log \left (\cos \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[x])/(1 + Cos[x]),x]

[Out]

-2*B*Log[Cos[x/2]] + A*Tan[x/2]

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Maple [A]  time = 0.012, size = 19, normalized size = 1. \begin{align*} A\tan \left ({\frac{x}{2}} \right ) +B\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(x))/(1+cos(x)),x)

[Out]

A*tan(1/2*x)+B*ln(tan(1/2*x)^2+1)

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Maxima [A]  time = 0.937644, size = 26, normalized size = 1.37 \begin{align*} -B \log \left (\cos \left (x\right ) + 1\right ) + \frac{A \sin \left (x\right )}{\cos \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1+cos(x)),x, algorithm="maxima")

[Out]

-B*log(cos(x) + 1) + A*sin(x)/(cos(x) + 1)

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Fricas [A]  time = 1.92543, size = 88, normalized size = 4.63 \begin{align*} -\frac{{\left (B \cos \left (x\right ) + B\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - A \sin \left (x\right )}{\cos \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1+cos(x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.396981, size = 17, normalized size = 0.89 \begin{align*} A \tan{\left (\frac{x}{2} \right )} + B \log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1+cos(x)),x)

[Out]

A*tan(x/2) + B*log(tan(x/2)**2 + 1)

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Giac [A]  time = 1.14587, size = 24, normalized size = 1.26 \begin{align*} B \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) + A \tan \left (\frac{1}{2} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(x))/(1+cos(x)),x, algorithm="giac")

[Out]

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