\(\int \frac {A+B \cos (x)}{1+\sin (x)} \, dx\) [2]

Optimal result

Integrand size = 13, antiderivative size = 19 \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=B \log (1+\sin (x))-\frac {A \cos (x)}{1+\sin (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4486, 2727, 2746, 31} \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=B \log (\sin (x)+1)-\frac {A \cos (x)}{\sin (x)+1} \]

[In]

Int[(A + B*Cos[x])/(1 + Sin[x]),x]

[Out]

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

Rule 31

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

Rule 2727

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 2746

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 4486

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

Rubi steps \begin{align*} \text {integral}& = \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 \text {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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(42\) vs. \(2(19)=38\).

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.21 \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=2 B \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {2 A \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )} \]

[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] (verified)

Time = 0.74 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
parts	\(-\frac {2 A}{\tan \left (\frac {x}{2}\right )+1}+B \ln \left (1+\sin \left (x \right )\right )\)	\(20\)
risch	\(-i x B -\frac {2 A}{{\mathrm e}^{i x}+i}+2 B \ln \left ({\mathrm e}^{i x}+i\right )\)	\(32\)
default	\(-\frac {2 A}{\tan \left (\frac {x}{2}\right )+1}+2 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )-B \ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )\)	\(35\)
parallelrisch	\(-B \ln \left (\frac {2}{\cos \left (x \right )+1}\right )+2 B \ln \left (-\cot \left (x \right )+\csc \left (x \right )+1\right )-A \left (\sec \left (x \right )-\tan \left (x \right )+1\right )\)	\(37\)
norman	\(\frac {-2 A \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 A}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right ) \left (\tan \left (\frac {x}{2}\right )+1\right )}+2 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )-B \ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.00 \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=-\frac {A \cos \left (x\right ) - {\left (B \cos \left (x\right ) + B \sin \left (x\right ) + B\right )} \log \left (\sin \left (x\right ) + 1\right ) - A \sin \left (x\right ) + A}{\cos \left (x\right ) + \sin \left (x\right ) + 1} \]

[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] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (17) = 34\).

Time = 0.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 4.95 \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=- \frac {2 A}{\tan {\left (\frac {x}{2} \right )} + 1} + \frac {2 B \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} + \frac {2 B \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} - \frac {B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )} \tan {\left (\frac {x}{2} \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} - \frac {B \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + 1 \right )}}{\tan {\left (\frac {x}{2} \right )} + 1} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=B \log \left (\sin \left (x\right ) + 1\right ) - \frac {2 \, A}{\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (19) = 38\).

Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.26 \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=-B \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) + 2 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) - \frac {2 \, {\left (B \tan \left (\frac {1}{2} \, x\right ) + A + B\right )}}{\tan \left (\frac {1}{2} \, x\right ) + 1} \]

[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)

Mupad [B] (verification not implemented)

Time = 14.97 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.79 \[ \int \frac {A+B \cos (x)}{1+\sin (x)} \, dx=2\,B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )-\frac {2\,A}{\mathrm {tan}\left (\frac {x}{2}\right )+1}-B\,\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right ) \]

[In]

int((A + B*cos(x))/(sin(x) + 1),x)

[Out]

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