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

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Rubi [A]  time = 0.0657701, antiderivative size = 45, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0661455, 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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Maple [A]  time = 0.106, size = 74, normalized size = 1.6 \begin{align*}{\frac{b\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{b\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{a\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

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 = 1.7852, size = 167, normalized size = 3.71 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 0.495281, size = 144, normalized size = 3.2 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 19.2425, size = 299, normalized size = 6.64 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \cos{\left (c \right )}}{\sin{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{\log{\left (\sin{\left (c + d x \right )} \right )}}{b d} & \text{for}\: a = 0 \\- \frac{i d x \sin{\left (c + d x \right )}}{- 2 b d \sin{\left (c + d x \right )} + 2 i b d \cos{\left (c + d x \right )}} - \frac{d x \cos{\left (c + d x \right )}}{- 2 b d \sin{\left (c + d x \right )} + 2 i b d \cos{\left (c + d x \right )}} - \frac{i \cos{\left (c + d x \right )}}{- 2 b d \sin{\left (c + d x \right )} + 2 i b d \cos{\left (c + d x \right )}} & \text{for}\: a = - i b \\- \frac{i d x \sin{\left (c + d x \right )}}{2 b d \sin{\left (c + d x \right )} + 2 i b d \cos{\left (c + d x \right )}} + \frac{d x \cos{\left (c + d x \right )}}{2 b d \sin{\left (c + d x \right )} + 2 i b d \cos{\left (c + d x \right )}} - \frac{i \cos{\left (c + d x \right )}}{2 b d \sin{\left (c + d x \right )} + 2 i b d \cos{\left (c + d x \right )}} & \text{for}\: a = i b \\\frac{x \cos{\left (c \right )}}{a \cos{\left (c \right )} + b \sin{\left (c \right )}} & \text{for}\: d = 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} \end{align*}

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)), (log(sin(c + d*x))/(b*d), Eq(a, 0)), (-I*d*x*
sin(c + d*x)/(-2*b*d*sin(c + d*x) + 2*I*b*d*cos(c + d*x)) - d*x*cos(c + d*x)/(-2*b*d*sin(c + d*x) + 2*I*b*d*co
s(c + d*x)) - I*cos(c + d*x)/(-2*b*d*sin(c + d*x) + 2*I*b*d*cos(c + d*x)), Eq(a, -I*b)), (-I*d*x*sin(c + d*x)/
(2*b*d*sin(c + d*x) + 2*I*b*d*cos(c + d*x)) + d*x*cos(c + d*x)/(2*b*d*sin(c + d*x) + 2*I*b*d*cos(c + d*x)) - I
*cos(c + d*x)/(2*b*d*sin(c + d*x) + 2*I*b*d*cos(c + d*x)), Eq(a, I*b)), (x*cos(c)/(a*cos(c) + b*sin(c)), Eq(d,
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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Giac [A]  time = 1.19466, size = 100, normalized size = 2.22 \begin{align*} \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} \end{align*}

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