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

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

[Out]

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

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Rubi [A]  time = 0.0683109, 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} \[ B \log (\sin (x)+1)-\frac{A \cos (x)}{\sin (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[x])/(1 + Sin[x]),x]

[Out]

B*Log[1 + Sin[x]] - (A*Cos[x])/(1 + Sin[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 \cos (x)}{1+\sin (x)} \, dx &=\int \left (\frac{A}{1+\sin (x)}+\frac{B \cos (x)}{1+\sin (x)}\right ) \, dx\\ &=A \int \frac{1}{1+\sin (x)} \, dx+B \int \frac{\cos (x)}{1+\sin (x)} \, dx\\ &=-\frac{A \cos (x)}{1+\sin (x)}+B \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\sin (x)\right )\\ &=B \log (1+\sin (x))-\frac{A \cos (x)}{1+\sin (x)}\\ \end{align*}

Mathematica [B]  time = 0.0493335, size = 42, normalized size = 2.21 \[ \frac{2 A \sin \left (\frac{x}{2}\right )}{\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )}+2 B \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*B*Log[Cos[x/2] + Sin[x/2]] + (2*A*Sin[x/2])/(Cos[x/2] + Sin[x/2])

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Maple [A]  time = 0.042, size = 35, normalized size = 1.8 \begin{align*} -B\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) -2\,{\frac{A}{\tan \left ( x/2 \right ) +1}}+2\,B\ln \left ( \tan \left ( x/2 \right ) +1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.995644, size = 32, normalized size = 1.68 \begin{align*} B \log \left (\sin \left (x\right ) + 1\right ) - \frac{2 \, A}{\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.99319, size = 127, normalized size = 6.68 \begin{align*} -\frac{A \cos \left (x\right ) -{\left (B \cos \left (x\right ) + B \sin \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) - A \sin \left (x\right ) + A}{\cos \left (x\right ) + \sin \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.41307, size = 94, normalized size = 4.95 \begin{align*} - \frac{2 A}{\tan{\left (\frac{x}{2} \right )} + 1} + \frac{2 B \log{\left (\tan{\left (\frac{x}{2} \right )} + 1 \right )} \tan{\left (\frac{x}{2} \right )}}{\tan{\left (\frac{x}{2} \right )} + 1} + \frac{2 B \log{\left (\tan{\left (\frac{x}{2} \right )} + 1 \right )}}{\tan{\left (\frac{x}{2} \right )} + 1} - \frac{B \log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + 1 \right )} \tan{\left (\frac{x}{2} \right )}}{\tan{\left (\frac{x}{2} \right )} + 1} - \frac{B \log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + 1 \right )}}{\tan{\left (\frac{x}{2} \right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.11003, size = 58, normalized size = 3.05 \begin{align*} -B \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) + 2 \, B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right ) - \frac{2 \,{\left (B \tan \left (\frac{1}{2} \, x\right ) + A + B\right )}}{\tan \left (\frac{1}{2} \, x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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