3.400 \(\int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx\)

Optimal. Leaf size=121 \[ \frac {2 a \tan ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{e \left (a^2-b^2-c^2\right )^{3/2}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{e \left (a^2-b^2-c^2\right ) (a+b \cos (d+e x)+c \sin (d+e x))} \]

[Out]

2*a*arctan((c+(a-b)*tan(1/2*e*x+1/2*d))/(a^2-b^2-c^2)^(1/2))/(a^2-b^2-c^2)^(3/2)/e+(c*cos(e*x+d)-b*sin(e*x+d))
/(a^2-b^2-c^2)/e/(a+b*cos(e*x+d)+c*sin(e*x+d))

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Rubi [A]  time = 0.11, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3129, 12, 3124, 618, 204} \[ \frac {2 a \tan ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )+c}{\sqrt {a^2-b^2-c^2}}\right )}{e \left (a^2-b^2-c^2\right )^{3/2}}+\frac {c \cos (d+e x)-b \sin (d+e x)}{e \left (a^2-b^2-c^2\right ) (a+b \cos (d+e x)+c \sin (d+e x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(-2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3129

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[((-(c*Cos[d
+ e*x]) + b*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*(n + 1) - b*(n + 2)*Cos[d + e*x] - c*(n + 2)*Sin[d + e*x])*(a + b*C
os[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n
, -1] && NeQ[n, -3/2]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 116, normalized size = 0.96 \[ \frac {\frac {a c+\left (b^2+c^2\right ) \sin (d+e x)}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (d+e x)+c \sin (d+e x))}+\frac {2 a \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )+c}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}}}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(-2),x]

[Out]

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

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fricas [B]  time = 1.04, size = 819, normalized size = 6.77 \[ \left [\frac {{\left (a b \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2} + c^{2}} \log \left (\frac {a^{2} b^{2} - 2 \, b^{4} - c^{4} - {\left (a^{2} + 3 \, b^{2}\right )} c^{2} - {\left (2 \, a^{2} b^{2} - b^{4} - 2 \, a^{2} c^{2} + c^{4}\right )} \cos \left (e x + d\right )^{2} - 2 \, {\left (a b^{3} + a b c^{2}\right )} \cos \left (e x + d\right ) - 2 \, {\left (a b^{2} c + a c^{3} - {\left (b c^{3} - {\left (2 \, a^{2} b - b^{3}\right )} c\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right ) - 2 \, {\left (2 \, a b c \cos \left (e x + d\right )^{2} - a b c + {\left (b^{2} c + c^{3}\right )} \cos \left (e x + d\right ) - {\left (b^{3} + b c^{2} + {\left (a b^{2} - a c^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \sqrt {-a^{2} + b^{2} + c^{2}}}{2 \, a b \cos \left (e x + d\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} + a^{2} + c^{2} + 2 \, {\left (b c \cos \left (e x + d\right ) + a c\right )} \sin \left (e x + d\right )}\right ) - 2 \, {\left (c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} \cos \left (e x + d\right ) - 2 \, {\left (a^{2} b - b^{3} - b c^{2}\right )} \sin \left (e x + d\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + b c^{4} - 2 \, {\left (a^{2} b - b^{3}\right )} c^{2}\right )} e \cos \left (e x + d\right ) + {\left (c^{5} - 2 \, {\left (a^{2} - b^{2}\right )} c^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} c\right )} e \sin \left (e x + d\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + a c^{4} - 2 \, {\left (a^{3} - a b^{2}\right )} c^{2}\right )} e\right )}}, \frac {{\left (a b \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2} - c^{2}} \arctan \left (-\frac {{\left (a b \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + b^{2} + c^{2}\right )} \sqrt {a^{2} - b^{2} - c^{2}}}{{\left (c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} \cos \left (e x + d\right ) + {\left (a^{2} b - b^{3} - b c^{2}\right )} \sin \left (e x + d\right )}\right ) - {\left (c^{3} - {\left (a^{2} - b^{2}\right )} c\right )} \cos \left (e x + d\right ) - {\left (a^{2} b - b^{3} - b c^{2}\right )} \sin \left (e x + d\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} + b c^{4} - 2 \, {\left (a^{2} b - b^{3}\right )} c^{2}\right )} e \cos \left (e x + d\right ) + {\left (c^{5} - 2 \, {\left (a^{2} - b^{2}\right )} c^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} c\right )} e \sin \left (e x + d\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + a c^{4} - 2 \, {\left (a^{3} - a b^{2}\right )} c^{2}\right )} e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^2,x, algorithm="fricas")

[Out]

[1/2*((a*b*cos(e*x + d) + a*c*sin(e*x + d) + a^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(e*x + d)^2 - 2*(a*b^3 + a*b*c^2)*cos(e*x + d) - 2*(a*b^2*c
+ a*c^3 - (b*c^3 - (2*a^2*b - b^3)*c)*cos(e*x + d))*sin(e*x + d) - 2*(2*a*b*c*cos(e*x + d)^2 - a*b*c + (b^2*c
+ c^3)*cos(e*x + d) - (b^3 + b*c^2 + (a*b^2 - a*c^2)*cos(e*x + d))*sin(e*x + d))*sqrt(-a^2 + b^2 + c^2))/(2*a
*b*cos(e*x + d) + (b^2 - c^2)*cos(e*x + d)^2 + a^2 + c^2 + 2*(b*c*cos(e*x + d) + a*c)*sin(e*x + d))) - 2*(c^3
- (a^2 - b^2)*c)*cos(e*x + d) - 2*(a^2*b - b^3 - b*c^2)*sin(e*x + d))/((a^4*b - 2*a^2*b^3 + b^5 + b*c^4 - 2*(a
^2*b - b^3)*c^2)*e*cos(e*x + d) + (c^5 - 2*(a^2 - b^2)*c^3 + (a^4 - 2*a^2*b^2 + b^4)*c)*e*sin(e*x + d) + (a^5
- 2*a^3*b^2 + a*b^4 + a*c^4 - 2*(a^3 - a*b^2)*c^2)*e), ((a*b*cos(e*x + d) + a*c*sin(e*x + d) + a^2)*sqrt(a^2 -
b^2 - c^2)*arctan(-(a*b*cos(e*x + d) + a*c*sin(e*x + d) + b^2 + c^2)*sqrt(a^2 - b^2 - c^2)/((c^3 - (a^2 - b^2
)*c)*cos(e*x + d) + (a^2*b - b^3 - b*c^2)*sin(e*x + d))) - (c^3 - (a^2 - b^2)*c)*cos(e*x + d) - (a^2*b - b^3 -
b*c^2)*sin(e*x + d))/((a^4*b - 2*a^2*b^3 + b^5 + b*c^4 - 2*(a^2*b - b^3)*c^2)*e*cos(e*x + d) + (c^5 - 2*(a^2
- b^2)*c^3 + (a^4 - 2*a^2*b^2 + b^4)*c)*e*sin(e*x + d) + (a^5 - 2*a^3*b^2 + a*b^4 + a*c^4 - 2*(a^3 - a*b^2)*c^
2)*e)]

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giac [A]  time = 0.16, size = 222, normalized size = 1.83 \[ -2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )} a}{{\left (a^{2} - b^{2} - c^{2}\right )}^{\frac {3}{2}}} + \frac {a b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - c^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) - a c}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3} - a c^{2} + b c^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a + b\right )}}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^2,x, algorithm="giac")

[Out]

-2*((pi*floor(1/2*(x*e + d)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x*e + 1/2*d) - b*tan(1/2*x*e + 1/2*
d) + c)/sqrt(a^2 - b^2 - c^2)))*a/(a^2 - b^2 - c^2)^(3/2) + (a*b*tan(1/2*x*e + 1/2*d) - b^2*tan(1/2*x*e + 1/2*
d) - c^2*tan(1/2*x*e + 1/2*d) - a*c)/((a^3 - a^2*b - a*b^2 + b^3 - a*c^2 + b*c^2)*(a*tan(1/2*x*e + 1/2*d)^2 -
b*tan(1/2*x*e + 1/2*d)^2 + 2*c*tan(1/2*x*e + 1/2*d) + a + b)))*e^(-1)

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maple [B]  time = 0.50, size = 424, normalized size = 3.50 \[ -\frac {2 \tan \left (\frac {d}{2}+\frac {e x}{2}\right ) a b}{e \left (a \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )+2 c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+a +b \right ) \left (a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b \right )}+\frac {2 \tan \left (\frac {d}{2}+\frac {e x}{2}\right ) b^{2}}{e \left (a \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )+2 c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+a +b \right ) \left (a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b \right )}+\frac {2 \tan \left (\frac {d}{2}+\frac {e x}{2}\right ) c^{2}}{e \left (a \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )+2 c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+a +b \right ) \left (a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b \right )}+\frac {2 a c}{e \left (a \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )+2 c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+a +b \right ) \left (a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+c^{2} b \right )}+\frac {2 a \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{e \left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^2,x)

[Out]

-2/e/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a+b)/(a^3-a^2*b-a*b^2-a*c^2+b^3+b*c
^2)*tan(1/2*d+1/2*e*x)*a*b+2/e/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a+b)/(a^3
-a^2*b-a*b^2-a*c^2+b^3+b*c^2)*tan(1/2*d+1/2*e*x)*b^2+2/e/(a*tan(1/2*d+1/2*e*x)^2-b*tan(1/2*d+1/2*e*x)^2+2*c*ta
n(1/2*d+1/2*e*x)+a+b)/(a^3-a^2*b-a*b^2-a*c^2+b^3+b*c^2)*tan(1/2*d+1/2*e*x)*c^2+2/e/(a*tan(1/2*d+1/2*e*x)^2-b*t
an(1/2*d+1/2*e*x)^2+2*c*tan(1/2*d+1/2*e*x)+a+b)*a*c/(a^3-a^2*b-a*b^2-a*c^2+b^3+b*c^2)+2/e*a/(a^2-b^2-c^2)^(3/2
)*arctan(1/2*(2*(a-b)*tan(1/2*d+1/2*e*x)+2*c)/(a^2-b^2-c^2)^(1/2))

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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(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^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 = 3.06, size = 195, normalized size = 1.61 \[ \frac {2\,a\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )+\frac {2\,\left (-a^2\,c+b^2\,c+c^3\right )}{-a^2+b^2+c^2}}{2\,\sqrt {-a^2+b^2+c^2}}\right )}{e\,{\left (-a^2+b^2+c^2\right )}^{3/2}}-\frac {\frac {2\,a\,c}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (b^2-a\,b+c^2\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}}{e\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(d + e*x) + c*sin(d + e*x))^2,x)

[Out]

(2*a*atanh((tan(d/2 + (e*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))))/(e*(b^2 - a^2 + c^2)^(3/2)) - ((2*a*c)/((a - b)*(b^2 - a^2 + c^2)) + (2*tan(d/2 + (e*x)/2)*(b^2 - a*
b + c^2))/((a - b)*(b^2 - a^2 + c^2)))/(e*(a + b + tan(d/2 + (e*x)/2)^2*(a - b) + 2*c*tan(d/2 + (e*x)/2)))

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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(1/(a+b*cos(e*x+d)+c*sin(e*x+d))**2,x)

[Out]

Timed out

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