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

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 19, 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 = {4486, 2727, 2746, 31} \begin {gather*} B \log (\sin (x)+1)-\frac {A \cos (x)}{\sin (x)+1} \end {gather*}

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 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*} \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 \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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(42\) vs. \(2(19)=38\).
time = 0.04, size = 42, normalized size = 2.21 \begin {gather*} 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 )} \end {gather*}

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 = 0.19, size = 35, normalized size = 1.84

method	result	size
risch	\(-i B x -\frac {2 A}{{\mathrm e}^{i x}+i}+2 B \ln \left ({\mathrm e}^{i x}+i\right )\)	\(32\)
default	\(-B \ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )-\frac {2 A}{\tan \left (\frac {x}{2}\right )+1}+2 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )\)	\(35\)
norman	\(\frac {-2 A \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 A}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right ) \left (\tan \left (\frac {x}{2}\right )+1\right )}+2 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )-B \ln \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 24, normalized size = 1.26 \begin {gather*} B \log \left (\sin \left (x\right ) + 1\right ) - \frac {2 \, A}{\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1} \end {gather*}

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 = 0.38, size = 38, normalized size = 2.00 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (17) = 34\).
time = 0.28, size = 94, normalized size = 4.95 \begin {gather*} - \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (19) = 38\).
time = 0.42, size = 43, normalized size = 2.26 \begin {gather*} -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 {gather*}

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)

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Mupad [B]
time = 15.15, size = 34, normalized size = 1.79 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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