3.540 \(\int \frac{A+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx\)

Optimal. Leaf size=116 \[ \frac{2 \left (A \left (b^2+c^2\right )-a c C\right ) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (b^2+c^2\right ) \sqrt{a^2-b^2-c^2}}-\frac{b C \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac{c C x}{b^2+c^2} \]

[Out]

(c*C*x)/(b^2 + c^2) + (2*(A*(b^2 + c^2) - a*c*C)*ArcTan[(c + (a - b)*Tan[x/2])/Sqrt[a^2 - b^2 - c^2]])/(Sqrt[a
^2 - b^2 - c^2]*(b^2 + c^2)) - (b*C*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.107879, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3137, 3124, 618, 204} \[ \frac{2 \left (A \left (b^2+c^2\right )-a c C\right ) \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{\left (b^2+c^2\right ) \sqrt{a^2-b^2-c^2}}-\frac{b C \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac{c C x}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*C*x)/(b^2 + c^2) + (2*(A*(b^2 + c^2) - a*c*C)*ArcTan[(c + (a - b)*Tan[x/2])/Sqrt[a^2 - b^2 - c^2]])/(Sqrt[a
^2 - b^2 - c^2]*(b^2 + c^2)) - (b*C*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)

Rule 3137

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

Rule 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{A+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx &=\frac{c C x}{b^2+c^2}-\frac{b C \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\left (A-\frac{a c C}{b^2+c^2}\right ) \int \frac{1}{a+b \cos (x)+c \sin (x)} \, dx\\ &=\frac{c C x}{b^2+c^2}-\frac{b C \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\left (2 \left (A-\frac{a c C}{b^2+c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+2 c x+(a-b) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=\frac{c C x}{b^2+c^2}-\frac{b C \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}-\left (4 \left (A-\frac{a c C}{b^2+c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2-c^2\right )-x^2} \, dx,x,2 c+2 (a-b) \tan \left (\frac{x}{2}\right )\right )\\ &=\frac{c C x}{b^2+c^2}+\frac{2 \left (A-\frac{a c C}{b^2+c^2}\right ) \tan ^{-1}\left (\frac{c+(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2-c^2}}\right )}{\sqrt{a^2-b^2-c^2}}-\frac{b C \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}\\ \end{align*}

Mathematica [A]  time = 0.277802, size = 96, normalized size = 0.83 \[ \frac{C (c x-b \log (a+b \cos (x)+c \sin (x)))-\frac{2 \left (A \left (b^2+c^2\right )-a c C\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )+c}{\sqrt{-a^2+b^2+c^2}}\right )}{\sqrt{-a^2+b^2+c^2}}}{b^2+c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.06, size = 542, normalized size = 4.7 \begin{align*} -{\frac{abC}{ \left ({b}^{2}+{c}^{2} \right ) \left ( a-b \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}-b \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,c\tan \left ( x/2 \right ) +a+b \right ) }+{\frac{{b}^{2}C}{ \left ({b}^{2}+{c}^{2} \right ) \left ( a-b \right ) }\ln \left ( a \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}-b \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,c\tan \left ( x/2 \right ) +a+b \right ) }+2\,{\frac{A{b}^{2}}{ \left ({b}^{2}+{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }+2\,{\frac{A{c}^{2}}{ \left ({b}^{2}+{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }-2\,{\frac{acC}{ \left ({b}^{2}+{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }-2\,{\frac{Cbc}{ \left ({b}^{2}+{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }+2\,{\frac{abcC}{ \left ({b}^{2}+{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}} \left ( a-b \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }-2\,{\frac{c{b}^{2}C}{ \left ({b}^{2}+{c}^{2} \right ) \sqrt{{a}^{2}-{b}^{2}-{c}^{2}} \left ( a-b \right ) }\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( x/2 \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) }+{\frac{bC}{{b}^{2}+{c}^{2}}\ln \left ( 1+ \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2} \right ) }+2\,{\frac{Cc\arctan \left ( \tan \left ( x/2 \right ) \right ) }{{b}^{2}+{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.85079, size = 1374, normalized size = 11.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((A*b^2 - C*a*c + A*c^2)*sqrt(-a^2 + b^2 + c^2)*log((a^2*b^2 - 2*b^4 - c^4 - (a^2 + 3*b^2)*c^2 - (2*a^2*b
^2 - b^4 - 2*a^2*c^2 + c^4)*cos(x)^2 - 2*(a*b^3 + a*b*c^2)*cos(x) - 2*(a*b^2*c + a*c^3 - (b*c^3 - (2*a^2*b - b
^3)*c)*cos(x))*sin(x) - 2*(2*a*b*c*cos(x)^2 - a*b*c + (b^2*c + c^3)*cos(x) - (b^3 + b*c^2 + (a*b^2 - a*c^2)*co
s(x))*sin(x))*sqrt(-a^2 + b^2 + c^2))/(2*a*b*cos(x) + (b^2 - c^2)*cos(x)^2 + a^2 + c^2 + 2*(b*c*cos(x) + a*c)*
sin(x))) - 2*(C*c^3 - (C*a^2 - C*b^2)*c)*x - (C*a^2*b - C*b^3 - C*b*c^2)*log(2*a*b*cos(x) + (b^2 - c^2)*cos(x)
^2 + a^2 + c^2 + 2*(b*c*cos(x) + a*c)*sin(x)))/(a^2*b^2 - b^4 - c^4 + (a^2 - 2*b^2)*c^2), 1/2*(2*(A*b^2 - C*a*
c + A*c^2)*sqrt(a^2 - b^2 - c^2)*arctan(-(a*b*cos(x) + a*c*sin(x) + b^2 + c^2)*sqrt(a^2 - b^2 - c^2)/((c^3 - (
a^2 - b^2)*c)*cos(x) + (a^2*b - b^3 - b*c^2)*sin(x))) - 2*(C*c^3 - (C*a^2 - C*b^2)*c)*x - (C*a^2*b - C*b^3 - C
*b*c^2)*log(2*a*b*cos(x) + (b^2 - c^2)*cos(x)^2 + a^2 + c^2 + 2*(b*c*cos(x) + a*c)*sin(x)))/(a^2*b^2 - b^4 - c
^4 + (a^2 - 2*b^2)*c^2)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.15184, size = 239, normalized size = 2.06 \begin{align*} \frac{C c x}{b^{2} + c^{2}} - \frac{C b \log \left (-a \tan \left (\frac{1}{2} \, x\right )^{2} + b \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac{1}{2} \, x\right ) - a - b\right )}{b^{2} + c^{2}} + \frac{C b \log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}{b^{2} + c^{2}} - \frac{2 \,{\left (A b^{2} - C a c + A c^{2}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right ) + c}{\sqrt{a^{2} - b^{2} - c^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2} - c^{2}}{\left (b^{2} + c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C*c*x/(b^2 + c^2) - C*b*log(-a*tan(1/2*x)^2 + b*tan(1/2*x)^2 - 2*c*tan(1/2*x) - a - b)/(b^2 + c^2) + C*b*log(t
an(1/2*x)^2 + 1)/(b^2 + c^2) - 2*(A*b^2 - C*a*c + A*c^2)*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(
a*tan(1/2*x) - b*tan(1/2*x) + c)/sqrt(a^2 - b^2 - c^2)))/(sqrt(a^2 - b^2 - c^2)*(b^2 + c^2))