3.10 \(\int \frac {\sin (x)}{a \cos (x)+b \sin (x)} \, dx\)

Optimal. Leaf size=35 \[ \frac {b x}{a^2+b^2}-\frac {a \log (a \cos (x)+b \sin (x))}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3097, 3133} \[ \frac {b x}{a^2+b^2}-\frac {a \log (a \cos (x)+b \sin (x))}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(a*Cos[x] + b*Sin[x]),x]

[Out]

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

Rule 3097

Int[sin[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(b*x)/(a^2 + b^2), x] - Dist[a/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

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

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Mathematica [C]  time = 0.06, size = 47, normalized size = 1.34 \[ \frac {2 x (b-i a)-a \log \left ((a \cos (x)+b \sin (x))^2\right )+2 i a \tan ^{-1}(\tan (x))}{2 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(a*Cos[x] + b*Sin[x]),x]

[Out]

(2*((-I)*a + b)*x + (2*I)*a*ArcTan[Tan[x]] - a*Log[(a*Cos[x] + b*Sin[x])^2])/(2*(a^2 + b^2))

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fricas [A]  time = 0.48, size = 46, normalized size = 1.31 \[ \frac {2 \, b x - a \log \left (2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x - a*log(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2))/(a^2 + b^2)

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giac [A]  time = 1.46, size = 55, normalized size = 1.57 \[ -\frac {a b \log \left ({\left | b \tan \relax (x) + a \right |}\right )}{a^{2} b + b^{3}} + \frac {b x}{a^{2} + b^{2}} + \frac {a \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b*log(abs(b*tan(x) + a))/(a^2*b + b^3) + b*x/(a^2 + b^2) + 1/2*a*log(tan(x)^2 + 1)/(a^2 + b^2)

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maple [A]  time = 0.46, size = 54, normalized size = 1.54 \[ -\frac {a \ln \left (a +b \tan \relax (x )\right )}{a^{2}+b^{2}}+\frac {a \ln \left (1+\tan ^{2}\relax (x )\right )}{2 a^{2}+2 b^{2}}+\frac {b \arctan \left (\tan \relax (x )\right )}{a^{2}+b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a*cos(x)+b*sin(x)),x)

[Out]

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

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maxima [B]  time = 0.43, size = 88, normalized size = 2.51 \[ \frac {2 \, b \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2} + b^{2}} - \frac {a \log \left (-a - \frac {2 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{2} + b^{2}} + \frac {a \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{a^{2} + b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*arctan(sin(x)/(cos(x) + 1))/(a^2 + b^2) - a*log(-a - 2*b*sin(x)/(cos(x) + 1) + a*sin(x)^2/(cos(x) + 1)^2)/
(a^2 + b^2) + a*log(sin(x)^2/(cos(x) + 1)^2 + 1)/(a^2 + b^2)

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mupad [B]  time = 2.14, size = 970, normalized size = 27.71 \[ -\frac {a\,\ln \left (a\,\cos \relax (x)+b\,\sin \relax (x)\right )}{a^2+b^2}-\frac {2\,b\,\mathrm {atan}\left (\frac {\left (a^4+2\,a^2\,b^2+b^4\right )\,\left (\mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {\left (4\,a^4-13\,a^2\,b^2+b^4\right )\,\left (\frac {b\,\left (64\,a\,b^2+\frac {a\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}-\frac {b^3\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^3}+\frac {a\,\left (\frac {b\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}-\frac {6\,a\,b\,\left (2\,a^2-b^2\right )\,\left (64\,a^2+\frac {a\,\left (64\,a\,b^2+\frac {a\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}-\frac {b\,\left (\frac {b\,\left (32\,a^2\,b^2-64\,a^4+\frac {a\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}-\frac {a\,b^2\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^3}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}\right )-\frac {\left (4\,a^4-13\,a^2\,b^2+b^4\right )\,\left (\frac {b\,\left (32\,a^2\,b-\frac {a\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {b^3\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^3}-\frac {a\,\left (\frac {b\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}+\frac {6\,a\,b\,\left (2\,a^2-b^2\right )\,\left (\frac {a\,\left (32\,a^2\,b-\frac {a\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {b\,\left (\frac {b\,\left (64\,a^3\,b-32\,a\,b^3+\frac {a\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{a^2+b^2}\right )}{a^2+b^2}+\frac {a\,b\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^2}\right )}{a^2+b^2}+\frac {a\,b^2\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^3}\right )}{{\left (4\,a^4+5\,a^2\,b^2+b^4\right )}^2}\right )}{32\,a\,b}\right )}{a^2+b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a*cos(x) + b*sin(x)),x)

[Out]

- (a*log(a*cos(x) + b*sin(x)))/(a^2 + b^2) - (2*b*atan(((a^4 + b^4 + 2*a^2*b^2)*(tan(x/2)*(((4*a^4 + b^4 - 13*
a^2*b^2)*((b*(64*a*b^2 + (a*(32*a^2*b^2 - 64*a^4 + (a*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)))/(a^2 + b^2)))/(a^
2 + b^2) - (b^3*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)^3 + (a*((b*(32*a^2*b^2 - 64*a^4 + (a*(96*a*b^4 + 96*a^3*b
^2))/(a^2 + b^2)))/(a^2 + b^2) + (a*b*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)^2))/(a^2 + b^2)))/(4*a^4 + b^4 + 5*
a^2*b^2)^2 - (6*a*b*(2*a^2 - b^2)*(64*a^2 + (a*(64*a*b^2 + (a*(32*a^2*b^2 - 64*a^4 + (a*(96*a*b^4 + 96*a^3*b^2
))/(a^2 + b^2)))/(a^2 + b^2)))/(a^2 + b^2) - (b*((b*(32*a^2*b^2 - 64*a^4 + (a*(96*a*b^4 + 96*a^3*b^2))/(a^2 +
b^2)))/(a^2 + b^2) + (a*b*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)^2))/(a^2 + b^2) - (a*b^2*(96*a*b^4 + 96*a^3*b^2
))/(a^2 + b^2)^3))/(4*a^4 + b^4 + 5*a^2*b^2)^2) - ((4*a^4 + b^4 - 13*a^2*b^2)*((b*(32*a^2*b - (a*(64*a^3*b - 3
2*a*b^3 + (a*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)))/(a^2 + b^2)))/(a^2 + b^2) + (b^3*(96*a^4*b + 96*a^2*b^3))/
(a^2 + b^2)^3 - (a*((b*(64*a^3*b - 32*a*b^3 + (a*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)))/(a^2 + b^2) + (a*b*(96
*a^4*b + 96*a^2*b^3))/(a^2 + b^2)^2))/(a^2 + b^2)))/(4*a^4 + b^4 + 5*a^2*b^2)^2 + (6*a*b*(2*a^2 - b^2)*((a*(32
*a^2*b - (a*(64*a^3*b - 32*a*b^3 + (a*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)))/(a^2 + b^2)))/(a^2 + b^2) + (b*((
b*(64*a^3*b - 32*a*b^3 + (a*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)))/(a^2 + b^2) + (a*b*(96*a^4*b + 96*a^2*b^3))
/(a^2 + b^2)^2))/(a^2 + b^2) + (a*b^2*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)^3))/(4*a^4 + b^4 + 5*a^2*b^2)^2))/(
32*a*b)))/(a^2 + b^2)

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sympy [A]  time = 0.77, size = 173, normalized size = 4.94 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\- \frac {i x \sin {\relax (x )}}{- 2 i b \sin {\relax (x )} - 2 b \cos {\relax (x )}} - \frac {x \cos {\relax (x )}}{- 2 i b \sin {\relax (x )} - 2 b \cos {\relax (x )}} + \frac {i \cos {\relax (x )}}{- 2 i b \sin {\relax (x )} - 2 b \cos {\relax (x )}} & \text {for}\: a = - i b \\\frac {i x \sin {\relax (x )}}{2 i b \sin {\relax (x )} - 2 b \cos {\relax (x )}} - \frac {x \cos {\relax (x )}}{2 i b \sin {\relax (x )} - 2 b \cos {\relax (x )}} - \frac {i \cos {\relax (x )}}{2 i b \sin {\relax (x )} - 2 b \cos {\relax (x )}} & \text {for}\: a = i b \\- \frac {\log {\left (\cos {\relax (x )} \right )}}{a} & \text {for}\: b = 0 \\- \frac {a \log {\left (\frac {a \cos {\relax (x )}}{b} + \sin {\relax (x )} \right )}}{a^{2} + b^{2}} + \frac {b x}{a^{2} + b^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (-I*x*sin(x)/(-2*I*b*sin(x) - 2*b*cos(x)) - x*cos(x)/(-2*I*b*sin(x) -
2*b*cos(x)) + I*cos(x)/(-2*I*b*sin(x) - 2*b*cos(x)), Eq(a, -I*b)), (I*x*sin(x)/(2*I*b*sin(x) - 2*b*cos(x)) - x
*cos(x)/(2*I*b*sin(x) - 2*b*cos(x)) - I*cos(x)/(2*I*b*sin(x) - 2*b*cos(x)), Eq(a, I*b)), (-log(cos(x))/a, Eq(b
, 0)), (-a*log(a*cos(x)/b + sin(x))/(a**2 + b**2) + b*x/(a**2 + b**2), True))

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