3.534 \(\int \frac{A+B \cos (x)+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx\)

Optimal. Leaf size=129 \[ -\frac{-A b \sin (x)+A c \cos (x)-b C+B c}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac{A \tanh ^{-1}\left (\frac{c \cos (x)-b \sin (x)}{\sqrt{b^2+c^2}}\right )}{2 \left (b^2+c^2\right )^{3/2}}-\frac{c \cos (x) (b B+c C)-b \sin (x) (b B+c C)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))} \]

[Out]

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

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Rubi [A]  time = 0.124475, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3156, 3153, 3074, 206} \[ -\frac{-A b \sin (x)+A c \cos (x)-b C+B c}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac{A \tanh ^{-1}\left (\frac{c \cos (x)-b \sin (x)}{\sqrt{b^2+c^2}}\right )}{2 \left (b^2+c^2\right )^{3/2}}-\frac{c \cos (x) (b B+c C)-b \sin (x) (b B+c C)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x])^3,x]

[Out]

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

Rule 3156

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

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

Mathematica [A]  time = 0.584843, size = 122, normalized size = 0.95 \[ \frac{A b^2 \sin (x)-A b c \cos (x)+b^2 B \sin (2 x)+b^2 C-c \cos (2 x) (b B+c C)+b c C \sin (2 x)+c^2 C}{2 b \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}+\frac{A \tanh ^{-1}\left (\frac{b \tan \left (\frac{x}{2}\right )-c}{\sqrt{b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[x] + C*Sin[x])/(b*Cos[x] + c*Sin[x])^3,x]

[Out]

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

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Maple [A]  time = 0.105, size = 218, normalized size = 1.7 \begin{align*} -2\,{\frac{1}{ \left ( b \left ( \tan \left ( x/2 \right ) \right ) ^{2}-2\,c\tan \left ( x/2 \right ) -b \right ) ^{2}} \left ( -1/2\,{\frac{ \left ( A{b}^{2}+2\,A{c}^{2}-2\,B{b}^{2}-2\,B{c}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{3}}{ \left ({b}^{2}+{c}^{2} \right ) b}}-1/2\,{\frac{ \left ( A{b}^{2}c-2\,A{c}^{3}+2\,B{b}^{2}c+2\,B{c}^{3}+2\,C{b}^{3}+2\,Cb{c}^{2} \right ) \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{ \left ({b}^{2}+{c}^{2} \right ){b}^{2}}}-1/2\,{\frac{ \left ( A{b}^{2}-2\,A{c}^{2}+2\,B{b}^{2}+2\,B{c}^{2} \right ) \tan \left ( x/2 \right ) }{ \left ({b}^{2}+{c}^{2} \right ) b}}+1/2\,{\frac{Ac}{{b}^{2}+{c}^{2}}} \right ) }+{A{\it Artanh} \left ({\frac{1}{2} \left ( 2\,b\tan \left ( x/2 \right ) -2\,c \right ){\frac{1}{\sqrt{{b}^{2}+{c}^{2}}}}} \right ) \left ({b}^{2}+{c}^{2} \right ) ^{-{\frac{3}{2}}}} \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))^3,x)

[Out]

-2*(-1/2*(A*b^2+2*A*c^2-2*B*b^2-2*B*c^2)/(b^2+c^2)/b*tan(1/2*x)^3-1/2*(A*b^2*c-2*A*c^3+2*B*b^2*c+2*B*c^3+2*C*b
^3+2*C*b*c^2)/(b^2+c^2)/b^2*tan(1/2*x)^2-1/2*(A*b^2-2*A*c^2+2*B*b^2+2*B*c^2)/(b^2+c^2)/b*tan(1/2*x)+1/2*A*c/(b
^2+c^2))/(b*tan(1/2*x)^2-2*c*tan(1/2*x)-b)^2+A/(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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.26349, size = 728, normalized size = 5.64 \begin{align*} \frac{2 \, C b^{3} + 2 \, B b^{2} c + 6 \, C b c^{2} - 2 \, B c^{3} - 8 \,{\left (B b^{2} c + C b c^{2}\right )} \cos \left (x\right )^{2} +{\left (2 \, A b c \cos \left (x\right ) \sin \left (x\right ) + A c^{2} +{\left (A b^{2} - A c^{2}\right )} \cos \left (x\right )^{2}\right )} \sqrt{b^{2} + c^{2}} \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} + 2 \,{\left (B b^{3} + C b^{2} c - B b c^{2} - C c^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{4 \,{\left (b^{4} c^{2} + 2 \, b^{2} c^{4} + c^{6} +{\left (b^{6} + b^{4} c^{2} - b^{2} c^{4} - c^{6}\right )} \cos \left (x\right )^{2} + 2 \,{\left (b^{5} c + 2 \, b^{3} c^{3} + b c^{5}\right )} \cos \left (x\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))^3,x, algorithm="fricas")

[Out]

1/4*(2*C*b^3 + 2*B*b^2*c + 6*C*b*c^2 - 2*B*c^3 - 8*(B*b^2*c + C*b*c^2)*cos(x)^2 + (2*A*b*c*cos(x)*sin(x) + A*c
^2 + (A*b^2 - A*c^2)*cos(x)^2)*sqrt(b^2 + c^2)*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 + 2*(B*b^3 + C*b^2*c - B*b*c^2 - C*c^3)*cos(x))*sin(x))/(b^4*c^2 + 2*b^2*c^
4 + c^6 + (b^6 + b^4*c^2 - b^2*c^4 - c^6)*cos(x)^2 + 2*(b^5*c + 2*b^3*c^3 + b*c^5)*cos(x)*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))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.29618, size = 365, normalized size = 2.83 \begin{align*} \frac{A \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 )}{2 \,{\left (b^{2} + c^{2}\right )}^{\frac{3}{2}}} + \frac{A b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} - 2 \, B b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, A b c^{2} \tan \left (\frac{1}{2} \, x\right )^{3} - 2 \, B b c^{2} \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, C b^{3} \tan \left (\frac{1}{2} \, x\right )^{2} + A b^{2} c \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, B b^{2} c \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, C b c^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, A c^{3} \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, B c^{3} \tan \left (\frac{1}{2} \, x\right )^{2} + A b^{3} \tan \left (\frac{1}{2} \, x\right ) + 2 \, B b^{3} \tan \left (\frac{1}{2} \, x\right ) - 2 \, A b c^{2} \tan \left (\frac{1}{2} \, x\right ) + 2 \, B b c^{2} \tan \left (\frac{1}{2} \, x\right ) - A b^{2} c}{{\left (b^{4} + b^{2} c^{2}\right )}{\left (b \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac{1}{2} \, x\right ) - b\right )}^{2}} \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))^3,x, algorithm="giac")

[Out]

1/2*A*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) + (A*b^3*tan(1/2*x)^3 - 2*B*b^3*tan(1/2*x)^3 + 2*A*b*c^2*tan(1/2*x)^3 - 2*B*b*c^2*tan(1/2*x)^3 +
2*C*b^3*tan(1/2*x)^2 + A*b^2*c*tan(1/2*x)^2 + 2*B*b^2*c*tan(1/2*x)^2 + 2*C*b*c^2*tan(1/2*x)^2 - 2*A*c^3*tan(1/
2*x)^2 + 2*B*c^3*tan(1/2*x)^2 + A*b^3*tan(1/2*x) + 2*B*b^3*tan(1/2*x) - 2*A*b*c^2*tan(1/2*x) + 2*B*b*c^2*tan(1
/2*x) - A*b^2*c)/((b^4 + b^2*c^2)*(b*tan(1/2*x)^2 - 2*c*tan(1/2*x) - b)^2)