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

Optimal. Leaf size=23 \[ \frac{A \cos (x)}{1-\sin (x)}-B \log (1-\sin (x)) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.077904, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4401, 2648, 2667, 31} \[ \frac{A \cos (x)}{1-\sin (x)}-B \log (1-\sin (x)) \]

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\\ &=-\left (A \int \frac{1}{-1+\sin (x)} \, dx\right )-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 [A]  time = 0.0491608, size = 46, normalized size = 2. \[ \frac{2 A \sin \left (\frac{x}{2}\right )}{\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )}-2 B \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right ) \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.057, size = 34, normalized size = 1.5 \begin{align*} -2\,{\frac{A}{\tan \left ( x/2 \right ) -1}}-2\,B\ln \left ( \tan \left ( x/2 \right ) -1 \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*cos(x))/(1-sin(x)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.976851, size = 34, normalized size = 1.48 \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)

________________________________________________________________________________________

Fricas [A]  time = 1.96223, size = 127, normalized size = 5.52 \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)

________________________________________________________________________________________

Sympy [B]  time = 1.3663, size = 94, normalized size = 4.09 \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)

________________________________________________________________________________________

Giac [B]  time = 1.08306, size = 62, normalized size = 2.7 \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)