3.114 \(\int \frac {\cos (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=45 \[ \frac {b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac {a x}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3098, 3133} \[ \frac {b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac {a x}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

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

Rule 3098

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(a*x)/(a^2 + b^2), x] + Dist[b/(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 {\cos (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx &=\frac {a x}{a^2+b^2}+\frac {b \int \frac {b \cos (c+d x)-a \sin (c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2}\\ &=\frac {a x}{a^2+b^2}+\frac {b \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 41, normalized size = 0.91 \[ \frac {b \log (a \cos (c+d x)+b \sin (c+d x))+a (c+d x)}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 61, normalized size = 1.36 \[ \frac {2 \, a d x + b \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )}{2 \, {\left (a^{2} + b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 74, normalized size = 1.64 \[ \frac {\frac {2 \, b^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}} + \frac {2 \, {\left (d x + c\right )} a}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.15, size = 74, normalized size = 1.64 \[ \frac {b \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )}-\frac {b \ln \left (\tan ^{2}\left (d x +c \right )+1\right )}{2 d \left (a^{2}+b^{2}\right )}+\frac {a \arctan \left (\tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.60, size = 124, normalized size = 2.76 \[ \frac {\frac {2 \, a \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2} + b^{2}} + \frac {b \log \left (-a - \frac {2 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{2} + b^{2}} - \frac {b \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{2} + b^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.16, size = 1069, normalized size = 23.76 \[ \frac {b\,\ln \left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^2+b^2\right )}-\frac {2\,a\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {\left (a^4-13\,a^2\,b^2+4\,b^4\right )\,\left (\frac {a^3\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^3}+\frac {a\,\left (96\,a\,b^2-32\,a^3+\frac {b\,\left (32\,a\,b^3+128\,a^3\,b-\frac {b\,\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 {a\,\left (32\,a\,b^3+128\,a^3\,b-\frac {b\,\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 (a^4+5\,a^2\,b^2+4\,b^4\right )}^2}-\frac {6\,a\,b\,\left (a^2-2\,b^2\right )\,\left (32\,a\,b-\frac {b\,\left (96\,a\,b^2-32\,a^3+\frac {b\,\left (32\,a\,b^3+128\,a^3\,b-\frac {b\,\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 {a\,\left (\frac {a\,\left (32\,a\,b^3+128\,a^3\,b-\frac {b\,\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^2\,b\,\left (96\,a^3\,b^2+96\,a\,b^4\right )}{{\left (a^2+b^2\right )}^3}\right )}{{\left (a^4+5\,a^2\,b^2+4\,b^4\right )}^2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{32\,a^2}+\frac {\left (a^4-13\,a^2\,b^2+4\,b^4\right )\,\left (\frac {a\,\left (32\,a^2\,b-\frac {b\,\left (64\,a^2\,b^2-32\,a^4+\frac {b\,\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 {a^3\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^3}-\frac {b\,\left (\frac {a\,\left (64\,a^2\,b^2-32\,a^4+\frac {b\,\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 (a^4+2\,a^2\,b^2+b^4\right )}{32\,a^2\,{\left (a^4+5\,a^2\,b^2+4\,b^4\right )}^2}+\frac {3\,b\,\left (a^2-2\,b^2\right )\,\left (\frac {b\,\left (32\,a^2\,b-\frac {b\,\left (64\,a^2\,b^2-32\,a^4+\frac {b\,\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 {a\,\left (\frac {a\,\left (64\,a^2\,b^2-32\,a^4+\frac {b\,\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^2\,b\,\left (96\,a^4\,b+96\,a^2\,b^3\right )}{{\left (a^2+b^2\right )}^3}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{16\,a\,{\left (a^4+5\,a^2\,b^2+4\,b^4\right )}^2}\right )}{d\,\left (a^2+b^2\right )}-\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d\,\left (a^2+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*log(a + 2*b*tan(c/2 + (d*x)/2) - a*tan(c/2 + (d*x)/2)^2))/(d*(a^2 + b^2)) - (2*a*atan((tan(c/2 + (d*x)/2)*(
((a^4 + 4*b^4 - 13*a^2*b^2)*((a^3*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)^3 + (a*(96*a*b^2 - 32*a^3 + (b*(32*a*b^
3 + 128*a^3*b - (b*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)))/(a^2 + b^2)))/(a^2 + b^2) + (b*((a*(32*a*b^3 + 128*a
^3*b - (b*(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^4 + 4*b^4 + 5*a^2*b^2)^2 - (6*a*b*(a^2 - 2*b^2)*(32*a*b - (b*(96*a*b^2 - 32*a^3 + (b*(32*a*b^3
+ 128*a^3*b - (b*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)))/(a^2 + b^2)))/(a^2 + b^2) + (a*((a*(32*a*b^3 + 128*a^
3*b - (b*(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^2*b*(96*a*b^4 + 96*a^3*b^2))/(a^2 + b^2)^3))/(a^4 + 4*b^4 + 5*a^2*b^2)^2)*(a^4 + b^4 + 2*a^2*b^
2))/(32*a^2) + ((a^4 + 4*b^4 - 13*a^2*b^2)*((a*(32*a^2*b - (b*(64*a^2*b^2 - 32*a^4 + (b*(96*a^4*b + 96*a^2*b^3
))/(a^2 + b^2)))/(a^2 + b^2)))/(a^2 + b^2) + (a^3*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)^3 - (b*((a*(64*a^2*b^2
- 32*a^4 + (b*(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^4 + b^4 + 2*a^2*b^2))/(32*a^2*(a^4 + 4*b^4 + 5*a^2*b^2)^2) + (3*b*(a^2 - 2*b^2)*((b*(32*a^2
*b - (b*(64*a^2*b^2 - 32*a^4 + (b*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)))/(a^2 + b^2)))/(a^2 + b^2) + (a*((a*(6
4*a^2*b^2 - 32*a^4 + (b*(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^2*b*(96*a^4*b + 96*a^2*b^3))/(a^2 + b^2)^3)*(a^4 + b^4 + 2*a^2*b^2))/(16*a*(a^4
+ 4*b^4 + 5*a^2*b^2)^2)))/(d*(a^2 + b^2)) - (b*log(tan(c/2 + (d*x)/2)^2 + 1))/(d*(a^2 + b^2))

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sympy [A]  time = 1.75, size = 296, normalized size = 6.58 \[ \begin {cases} \frac {\tilde {\infty } x \cos {\relax (c )}}{\sin {\relax (c )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\- \frac {d x \sin {\left (c + d x \right )}}{2 i b d \sin {\left (c + d x \right )} + 2 b d \cos {\left (c + d x \right )}} + \frac {i d x \cos {\left (c + d x \right )}}{2 i b d \sin {\left (c + d x \right )} + 2 b d \cos {\left (c + d x \right )}} - \frac {\cos {\left (c + d x \right )}}{2 i b d \sin {\left (c + d x \right )} + 2 b d \cos {\left (c + d x \right )}} & \text {for}\: a = - i b \\- \frac {d x \sin {\left (c + d x \right )}}{- 2 i b d \sin {\left (c + d x \right )} + 2 b d \cos {\left (c + d x \right )}} - \frac {i d x \cos {\left (c + d x \right )}}{- 2 i b d \sin {\left (c + d x \right )} + 2 b d \cos {\left (c + d x \right )}} - \frac {\cos {\left (c + d x \right )}}{- 2 i b d \sin {\left (c + d x \right )} + 2 b d \cos {\left (c + d x \right )}} & \text {for}\: a = i b \\\frac {x \cos {\relax (c )}}{a \cos {\relax (c )} + b \sin {\relax (c )}} & \text {for}\: d = 0 \\\frac {\log {\left (\sin {\left (c + d x \right )} \right )}}{b d} & \text {for}\: a = 0 \\\frac {a d x}{a^{2} d + b^{2} d} + \frac {b \log {\left (\cos {\left (c + d x \right )} + \frac {b \sin {\left (c + d x \right )}}{a} \right )}}{a^{2} d + b^{2} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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