3.533 \(\int \frac{A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac{-A b \sin (x)+A c \cos (x)-b C+B c}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}-\frac{(b B+c C) \tanh ^{-1}\left (\frac{c \cos (x)-b \sin (x)}{\sqrt{b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}} \]

[Out]

-(((b*B + c*C)*ArcTanh[(c*Cos[x] - b*Sin[x])/Sqrt[b^2 + c^2]])/(b^2 + c^2)^(3/2)) - (B*c - b*C + A*c*Cos[x] -
A*b*Sin[x])/((b^2 + c^2)*(b*Cos[x] + c*Sin[x]))

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Rubi [A]  time = 0.0569903, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3153, 3074, 206} \[ -\frac{-A b \sin (x)+A c \cos (x)-b C+B c}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}-\frac{(b B+c C) \tanh ^{-1}\left (\frac{c \cos (x)-b \sin (x)}{\sqrt{b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x])^2,x]

[Out]

-(((b*B + c*C)*ArcTanh[(c*Cos[x] - b*Sin[x])/Sqrt[b^2 + c^2]])/(b^2 + c^2)^(3/2)) - (B*c - b*C + A*c*Cos[x] -
A*b*Sin[x])/((b^2 + c^2)*(b*Cos[x] + c*Sin[x]))

Rule 3153

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A)
*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - b*B - c*C)/(a^2 -
b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[
a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx &=-\frac{B c-b C+A c \cos (x)-A b \sin (x)}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}+\frac{(b B+c C) \int \frac{1}{b \cos (x)+c \sin (x)} \, dx}{b^2+c^2}\\ &=-\frac{B c-b C+A c \cos (x)-A b \sin (x)}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}-\frac{(b B+c C) \operatorname{Subst}\left (\int \frac{1}{b^2+c^2-x^2} \, dx,x,c \cos (x)-b \sin (x)\right )}{b^2+c^2}\\ &=-\frac{(b B+c C) \tanh ^{-1}\left (\frac{c \cos (x)-b \sin (x)}{\sqrt{b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}}-\frac{B c-b C+A c \cos (x)-A b \sin (x)}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.240411, size = 92, normalized size = 1.08 \[ \frac{A \left (b^2+c^2\right ) \sin (x)+b (b C-B c)}{b \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}+\frac{2 (b B+c C) \tanh ^{-1}\left (\frac{b \tan \left (\frac{x}{2}\right )-c}{\sqrt{b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x])^2,x]

[Out]

(2*(b*B + c*C)*ArcTanh[(-c + b*Tan[x/2])/Sqrt[b^2 + c^2]])/(b^2 + c^2)^(3/2) + (b*(-(B*c) + b*C) + A*(b^2 + c^
2)*Sin[x])/(b*(b^2 + c^2)*(b*Cos[x] + c*Sin[x]))

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Maple [A]  time = 0.09, size = 124, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{b \left ( \tan \left ( x/2 \right ) \right ) ^{2}-2\,c\tan \left ( x/2 \right ) -b} \left ( -{\frac{ \left ( A{b}^{2}+A{c}^{2}-B{c}^{2}+Cbc \right ) \tan \left ( x/2 \right ) }{b \left ({b}^{2}+{c}^{2} \right ) }}+{\frac{Bc-bC}{{b}^{2}+{c}^{2}}} \right ) }+2\,{\frac{bB+Cc}{ \left ({b}^{2}+{c}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) -2\,c}{\sqrt{{b}^{2}+{c}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x))^2,x)

[Out]

2*(-(A*b^2+A*c^2-B*c^2+C*b*c)/b/(b^2+c^2)*tan(1/2*x)+(B*c-C*b)/(b^2+c^2))/(b*tan(1/2*x)^2-2*c*tan(1/2*x)-b)+2*
(B*b+C*c)/(b^2+c^2)^(3/2)*arctanh(1/2*(2*b*tan(1/2*x)-2*c)/(b^2+c^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.29003, size = 544, normalized size = 6.4 \begin{align*} \frac{2 \, C b^{3} - 2 \, B b^{2} c + 2 \, C b c^{2} - 2 \, B c^{3} + \sqrt{b^{2} + c^{2}}{\left ({\left (B b^{2} + C b c\right )} \cos \left (x\right ) +{\left (B b c + C c^{2}\right )} \sin \left (x\right )\right )} \log \left (-\frac{2 \, b c \cos \left (x\right ) \sin \left (x\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} - 2 \, b^{2} - c^{2} + 2 \, \sqrt{b^{2} + c^{2}}{\left (c \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, b c \cos \left (x\right ) \sin \left (x\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + c^{2}}\right ) - 2 \,{\left (A b^{2} c + A c^{3}\right )} \cos \left (x\right ) + 2 \,{\left (A b^{3} + A b c^{2}\right )} \sin \left (x\right )}{2 \,{\left ({\left (b^{5} + 2 \, b^{3} c^{2} + b c^{4}\right )} \cos \left (x\right ) +{\left (b^{4} c + 2 \, b^{2} c^{3} + c^{5}\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x))^2,x, algorithm="fricas")

[Out]

1/2*(2*C*b^3 - 2*B*b^2*c + 2*C*b*c^2 - 2*B*c^3 + sqrt(b^2 + c^2)*((B*b^2 + C*b*c)*cos(x) + (B*b*c + C*c^2)*sin
(x))*log(-(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^2 - 2*b^2 - c^2 + 2*sqrt(b^2 + c^2)*(c*cos(x) - b*sin(x)))
/(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^2 + c^2)) - 2*(A*b^2*c + A*c^3)*cos(x) + 2*(A*b^3 + A*b*c^2)*sin(x)
)/((b^5 + 2*b^3*c^2 + b*c^4)*cos(x) + (b^4*c + 2*b^2*c^3 + c^5)*sin(x))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19079, size = 203, normalized size = 2.39 \begin{align*} -\frac{{\left (B b + C c\right )} \log \left (\frac{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, c - 2 \, \sqrt{b^{2} + c^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac{1}{2} \, x\right ) - 2 \, c + 2 \, \sqrt{b^{2} + c^{2}} \right |}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (A b^{2} \tan \left (\frac{1}{2} \, x\right ) + C b c \tan \left (\frac{1}{2} \, x\right ) + A c^{2} \tan \left (\frac{1}{2} \, x\right ) - B c^{2} \tan \left (\frac{1}{2} \, x\right ) + C b^{2} - B b c\right )}}{{\left (b^{3} + b c^{2}\right )}{\left (b \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac{1}{2} \, x\right ) - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(x)+C*sin(x))/(b*cos(x)+c*sin(x))^2,x, algorithm="giac")

[Out]

-(B*b + C*c)*log(abs(2*b*tan(1/2*x) - 2*c - 2*sqrt(b^2 + c^2))/abs(2*b*tan(1/2*x) - 2*c + 2*sqrt(b^2 + c^2)))/
(b^2 + c^2)^(3/2) - 2*(A*b^2*tan(1/2*x) + C*b*c*tan(1/2*x) + A*c^2*tan(1/2*x) - B*c^2*tan(1/2*x) + C*b^2 - B*b
*c)/((b^3 + b*c^2)*(b*tan(1/2*x)^2 - 2*c*tan(1/2*x) - b))