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

Optimal. Leaf size=114 \[ \frac {2 (a A-c C) \tan ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}-\frac {-\cos (x) (A c-a C)+A b \sin (x)+b C}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \]

[Out]

2*(A*a-C*c)*arctan((c+(a-b)*tan(1/2*x))/(a^2-b^2-c^2)^(1/2))/(a^2-b^2-c^2)^(3/2)+(-b*C+(A*c-C*a)*cos(x)-A*b*si
n(x))/(a^2-b^2-c^2)/(a+b*cos(x)+c*sin(x))

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Rubi [A]  time = 0.10, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3154, 3124, 618, 204} \[ \frac {2 (a A-c C) \tan ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}-\frac {-\cos (x) (A c-a C)+A b \sin (x)+b C}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 3154

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

Rubi steps

\begin {align*} \int \frac {A+C \sin (x)}{(a+b \cos (x)+c \sin (x))^2} \, dx &=-\frac {b C-(A c-a C) \cos (x)+A b \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}+\frac {(a A-c C) \int \frac {1}{a+b \cos (x)+c \sin (x)} \, dx}{a^2-b^2-c^2}\\ &=-\frac {b C-(A c-a C) \cos (x)+A b \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}+\frac {(2 (a A-c C)) \operatorname {Subst}\left (\int \frac {1}{a+b+2 c x+(a-b) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^2-b^2-c^2}\\ &=-\frac {b C-(A c-a C) \cos (x)+A b \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}-\frac {(4 (a A-c C)) \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 )}{a^2-b^2-c^2}\\ &=\frac {2 (a A-c C) \tan ^{-1}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2}}-\frac {b C-(A c-a C) \cos (x)+A b \sin (x)}{\left (a^2-b^2-c^2\right ) (a+b \cos (x)+c \sin (x))}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 123, normalized size = 1.08 \[ \frac {a^2 (-C)+\sin (x) \left (A \left (b^2+c^2\right )-a c C\right )+a A c+b^2 C}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (x)+c \sin (x))}+\frac {2 (a A-c C) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )+c}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.24, size = 1301, normalized size = 11.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*C*a^4*b - 4*C*a^2*b^3 + 2*C*b^5 + 2*C*b*c^4 - 4*(C*a^2*b - C*b^3)*c^2 - (A*a^2*b^2 - C*a*b^2*c + A*a^2
*c^2 - C*a*c^3 + (A*a*b^3 - C*b^3*c + A*a*b*c^2 - C*b*c^3)*cos(x) + (A*a*b^2*c - C*b^2*c^2 + A*a*c^3 - C*c^4)*
sin(x))*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)*cos(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*a*c^4
- A*c^5 + (A*a^2 - 2*A*b^2)*c^3 - (C*a^3 - C*a*b^2)*c^2 + (A*a^2*b^2 - A*b^4)*c)*cos(x) - 2*(A*a^2*b^3 - A*b^
5 + C*a*b*c^3 - A*b*c^4 + (A*a^2*b - 2*A*b^3)*c^2 - (C*a^3*b - C*a*b^3)*c)*sin(x))/(a^5*b^2 - 2*a^3*b^4 + a*b^
6 + a*c^6 - (2*a^3 - 3*a*b^2)*c^4 + (a^5 - 4*a^3*b^2 + 3*a*b^4)*c^2 + (a^4*b^3 - 2*a^2*b^5 + b^7 + b*c^6 - (2*
a^2*b - 3*b^3)*c^4 + (a^4*b - 4*a^2*b^3 + 3*b^5)*c^2)*cos(x) + (c^7 - (2*a^2 - 3*b^2)*c^5 + (a^4 - 4*a^2*b^2 +
3*b^4)*c^3 + (a^4*b^2 - 2*a^2*b^4 + b^6)*c)*sin(x)), (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + C*b*c^4 - 2*(C*a^2*b -
C*b^3)*c^2 + (A*a^2*b^2 - C*a*b^2*c + A*a^2*c^2 - C*a*c^3 + (A*a*b^3 - C*b^3*c + A*a*b*c^2 - C*b*c^3)*cos(x) +
(A*a*b^2*c - C*b^2*c^2 + A*a*c^3 - C*c^4)*sin(x))*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))) + (C*a*c^4 - A*c
^5 + (A*a^2 - 2*A*b^2)*c^3 - (C*a^3 - C*a*b^2)*c^2 + (A*a^2*b^2 - A*b^4)*c)*cos(x) - (A*a^2*b^3 - A*b^5 + C*a*
b*c^3 - A*b*c^4 + (A*a^2*b - 2*A*b^3)*c^2 - (C*a^3*b - C*a*b^3)*c)*sin(x))/(a^5*b^2 - 2*a^3*b^4 + a*b^6 + a*c^
6 - (2*a^3 - 3*a*b^2)*c^4 + (a^5 - 4*a^3*b^2 + 3*a*b^4)*c^2 + (a^4*b^3 - 2*a^2*b^5 + b^7 + b*c^6 - (2*a^2*b -
3*b^3)*c^4 + (a^4*b - 4*a^2*b^3 + 3*b^5)*c^2)*cos(x) + (c^7 - (2*a^2 - 3*b^2)*c^5 + (a^4 - 4*a^2*b^2 + 3*b^4)*
c^3 + (a^4*b^2 - 2*a^2*b^4 + b^6)*c)*sin(x))]

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giac [A]  time = 0.20, size = 206, normalized size = 1.81 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )} {\left (A a - C c\right )}}{{\left (a^{2} - b^{2} - c^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (A a b \tan \left (\frac {1}{2} \, x\right ) - A b^{2} \tan \left (\frac {1}{2} \, x\right ) + C a c \tan \left (\frac {1}{2} \, x\right ) - C b c \tan \left (\frac {1}{2} \, x\right ) - A c^{2} \tan \left (\frac {1}{2} \, x\right ) + C a^{2} - C b^{2} - A a c\right )}}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3} - a c^{2} + b c^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - b \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, x\right ) + a + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.18, size = 255, normalized size = 2.24 \[ \frac {-\frac {2 \left (a A b -A \,b^{2}-A \,c^{2}+a c C -C b c \right ) \tan \left (\frac {x}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b}+\frac {2 \left (a A c -a^{2} C +b^{2} C \right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b}}{a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 c \tan \left (\frac {x}{2}\right )+a +b}+\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) a A}{\left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}}-\frac {2 \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right ) C c}{\left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c^2+b^2-a^2>0)', see `assume?`
for more details)Is c^2+b^2-a^2 positive or negative?

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mupad [B]  time = 3.19, size = 204, normalized size = 1.79 \[ \frac {2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a-2\,b\right )+\frac {2\,\left (-a^2\,c+b^2\,c+c^3\right )}{-a^2+b^2+c^2}}{2\,\sqrt {-a^2+b^2+c^2}}\right )\,\left (A\,a-C\,c\right )}{{\left (-a^2+b^2+c^2\right )}^{3/2}}-\frac {\frac {2\,\left (-C\,a^2+A\,c\,a+C\,b^2\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^2+C\,b\,c-A\,a\,b+A\,c^2-C\,a\,c\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}}{\left (a-b\right )\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {x}{2}\right )+a+b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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