3.1.12 \(\int \frac {\cos (x)}{a+b \csc (x)} \, dx\) [12]

Optimal. Leaf size=20 \[ -\frac {b \log (b+a \sin (x))}{a^2}+\frac {\sin (x)}{a} \]

[Out]

-b*ln(b+a*sin(x))/a^2+sin(x)/a

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Rubi [A]
time = 0.04, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3957, 2912, 12, 45} \begin {gather*} \frac {\sin (x)}{a}-\frac {b \log (a \sin (x)+b)}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]/(a + b*Csc[x]),x]

[Out]

-((b*Log[b + a*Sin[x]])/a^2) + Sin[x]/a

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\cos (x)}{a+b \csc (x)} \, dx &=\int \frac {\cos (x) \sin (x)}{b+a \sin (x)} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {x}{a (b+x)} \, dx,x,a \sin (x)\right )}{a}\\ &=\frac {\text {Subst}\left (\int \frac {x}{b+x} \, dx,x,a \sin (x)\right )}{a^2}\\ &=\frac {\text {Subst}\left (\int \left (1-\frac {b}{b+x}\right ) \, dx,x,a \sin (x)\right )}{a^2}\\ &=-\frac {b \log (b+a \sin (x))}{a^2}+\frac {\sin (x)}{a}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 19, normalized size = 0.95 \begin {gather*} \frac {-b \log (b+a \sin (x))+a \sin (x)}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]/(a + b*Csc[x]),x]

[Out]

(-(b*Log[b + a*Sin[x]]) + a*Sin[x])/a^2

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Maple [A]
time = 0.07, size = 31, normalized size = 1.55

method	result	size
derivativedivides	\(\frac {1}{a \csc \left (x \right )}+\frac {b \ln \left (\csc \left (x \right )\right )}{a^{2}}-\frac {b \ln \left (a +b \csc \left (x \right )\right )}{a^{2}}\)	\(31\)
default	\(\frac {1}{a \csc \left (x \right )}+\frac {b \ln \left (\csc \left (x \right )\right )}{a^{2}}-\frac {b \ln \left (a +b \csc \left (x \right )\right )}{a^{2}}\)	\(31\)
risch	\(\frac {i b x}{a^{2}}-\frac {i {\mathrm e}^{i x}}{2 a}+\frac {i {\mathrm e}^{-i x}}{2 a}-\frac {b \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a^{2}}\)	\(58\)
norman	\(\frac {2 \tan \left (\frac {x}{2}\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )}+\frac {b \ln \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )}{a^{2}}-\frac {b \ln \left (b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 a \tan \left (\frac {x}{2}\right )+b \right )}{a^{2}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(a+b*csc(x)),x,method=_RETURNVERBOSE)

[Out]

1/a/csc(x)+1/a^2*b*ln(csc(x))-1/a^2*b*ln(a+b*csc(x))

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Maxima [A]
time = 0.26, size = 20, normalized size = 1.00 \begin {gather*} -\frac {b \log \left (a \sin \left (x\right ) + b\right )}{a^{2}} + \frac {\sin \left (x\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(a+b*csc(x)),x, algorithm="maxima")

[Out]

-b*log(a*sin(x) + b)/a^2 + sin(x)/a

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Fricas [A]
time = 3.25, size = 20, normalized size = 1.00 \begin {gather*} -\frac {b \log \left (a \sin \left (x\right ) + b\right ) - a \sin \left (x\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(a+b*csc(x)),x, algorithm="fricas")

[Out]

-(b*log(a*sin(x) + b) - a*sin(x))/a^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(a+b*csc(x)),x)

[Out]

Integral(cos(x)/(a + b*csc(x)), x)

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Giac [A]
time = 0.40, size = 21, normalized size = 1.05 \begin {gather*} -\frac {b \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a^{2}} + \frac {\sin \left (x\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)/(a+b*csc(x)),x, algorithm="giac")

[Out]

-b*log(abs(a*sin(x) + b))/a^2 + sin(x)/a

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Mupad [B]
time = 0.07, size = 20, normalized size = 1.00 \begin {gather*} -\frac {b\,\ln \left (b+a\,\sin \left (x\right )\right )-a\,\sin \left (x\right )}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)/(a + b/sin(x)),x)

[Out]

-(b*log(b + a*sin(x)) - a*sin(x))/a^2

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