3.20 \(\int \frac{A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx\)

Optimal. Leaf size=187 \[ \frac{\left (2 a^2 A-3 a b B+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{e (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right )^2 (a+b \cos (d+e x))}-\frac{(A b-a B) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}+\frac{C}{2 b e (a+b \cos (d+e x))^2} \]

[Out]

((2*a^2*A + A*b^2 - 3*a*b*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*
e) + C/(2*b*e*(a + b*Cos[d + e*x])^2) - ((A*b - a*B)*Sin[d + e*x])/(2*(a^2 - b^2)*e*(a + b*Cos[d + e*x])^2) -
((3*a*A*b - a^2*B - 2*b^2*B)*Sin[d + e*x])/(2*(a^2 - b^2)^2*e*(a + b*Cos[d + e*x]))

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Rubi [A]  time = 0.284189, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4377, 2754, 12, 2659, 205, 2668, 32} \[ \frac{\left (2 a^2 A-3 a b B+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{e (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right )^2 (a+b \cos (d+e x))}-\frac{(A b-a B) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}+\frac{C}{2 b e (a+b \cos (d+e x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*A + A*b^2 - 3*a*b*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*
e) + C/(2*b*e*(a + b*Cos[d + e*x])^2) - ((A*b - a*B)*Sin[d + e*x])/(2*(a^2 - b^2)*e*(a + b*Cos[d + e*x])^2) -
((3*a*A*b - a^2*B - 2*b^2*B)*Sin[d + e*x])/(2*(a^2 - b^2)^2*e*(a + b*Cos[d + e*x]))

Rule 4377

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] :> With[{e = FreeFactors[Cos[c*(a +
b*x)], x]}, Int[ActivateTrig[u*v], x] + Dist[d, Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[
Cos[c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

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

Rule 2659

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

Rule 205

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx &=C \int \frac{\sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx+\int \frac{A+B \cos (d+e x)}{(a+b \cos (d+e x))^3} \, dx\\ &=-\frac{(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac{\int \frac{-2 (a A-b B)+(A b-a B) \cos (d+e x)}{(a+b \cos (d+e x))^2} \, dx}{2 \left (a^2-b^2\right )}-\frac{C \operatorname{Subst}\left (\int \frac{1}{(a+x)^3} \, dx,x,b \cos (d+e x)\right )}{b e}\\ &=\frac{C}{2 b e (a+b \cos (d+e x))^2}-\frac{(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))}+\frac{\int \frac{2 a^2 A+A b^2-3 a b B}{a+b \cos (d+e x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{C}{2 b e (a+b \cos (d+e x))^2}-\frac{(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))}+\frac{\left (2 a^2 A+A b^2-3 a b B\right ) \int \frac{1}{a+b \cos (d+e x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{C}{2 b e (a+b \cos (d+e x))^2}-\frac{(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))}+\frac{\left (2 a^2 A+A b^2-3 a b B\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{\left (a^2-b^2\right )^2 e}\\ &=\frac{\left (2 a^2 A+A b^2-3 a b B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} e}+\frac{C}{2 b e (a+b \cos (d+e x))^2}-\frac{(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac{\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))}\\ \end{align*}

Mathematica [A]  time = 0.783325, size = 175, normalized size = 0.94 \[ \frac{\frac{C \left (a^2-b^2\right )-b (A b-a B) \sin (d+e x)}{b (a-b) (a+b) (a+b \cos (d+e x))^2}+\frac{\left (a^2 B-3 a A b+2 b^2 B\right ) \sin (d+e x)}{(a-b)^2 (a+b)^2 (a+b \cos (d+e x))}-\frac{2 \left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(2*a^2*A + A*b^2 - 3*a*b*B)*ArcTanh[((a - b)*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + ((
-3*a*A*b + a^2*B + 2*b^2*B)*Sin[d + e*x])/((a - b)^2*(a + b)^2*(a + b*Cos[d + e*x])) + ((a^2 - b^2)*C - b*(A*b
- a*B)*Sin[d + e*x])/((a - b)*b*(a + b)*(a + b*Cos[d + e*x])^2))/(2*e)

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Maple [B]  time = 0.045, size = 994, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,x)

[Out]

-4/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*e*x+1/2*d)^3*A*a*b-1/
e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*e*x+1/2*d)^3*A*b^2+2/e/(
tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*e*x+1/2*d)^3*B*a^2+1/e/(tan
(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*e*x+1/2*d)^3*B*a*b+2/e/(tan(1/
2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*e*x+1/2*d)^3*B*b^2-2/e/(tan(1/2*e
*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2*C/(a-b)*tan(1/2*e*x+1/2*d)^2-4/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e
*x+1/2*d)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*e*x+1/2*d)*A*a*b+1/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/
2*d)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*e*x+1/2*d)*A*b^2+2/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^
2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*e*x+1/2*d)*B*a^2-1/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a
+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*e*x+1/2*d)*B*a*b+2/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2
/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*e*x+1/2*d)*B*b^2-2/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)^2*a*C/
(a^2-2*a*b+b^2)+2/e/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*e*x+1/2*d)/((a+b)*(a-b))^(1/2
))*A*a^2+1/e/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*e*x+1/2*d)/((a+b)*(a-b))^(1/2))*A*b^
2-3/e/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*e*x+1/2*d)/((a+b)*(a-b))^(1/2))*B*a*b

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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(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.86969, size = 1782, normalized size = 9.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,x, algorithm="fricas")

[Out]

[1/4*(2*C*a^6 - 6*C*a^4*b^2 + 6*C*a^2*b^4 - 2*C*b^6 - (2*A*a^4*b - 3*B*a^3*b^2 + A*a^2*b^3 + (2*A*a^2*b^3 - 3*
B*a*b^4 + A*b^5)*cos(e*x + d)^2 + 2*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*cos(e*x + d))*sqrt(-a^2 + b^2)*log((
2*a*b*cos(e*x + d) + (2*a^2 - b^2)*cos(e*x + d)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(e*x + d) + b)*sin(e*x + d) - a^2
+ 2*b^2)/(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x + d) + a^2)) + 2*(2*B*a^5*b - 4*A*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*
b^4 - B*a*b^5 - A*b^6 + (B*a^4*b^2 - 3*A*a^3*b^3 + B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6)*cos(e*x + d))*sin(e*x + d)
)/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*cos(e*x + d)^2 + 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*co
s(e*x + d) + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*e), 1/2*(C*a^6 - 3*C*a^4*b^2 + 3*C*a^2*b^4 - C*b^6 + (2
*A*a^4*b - 3*B*a^3*b^2 + A*a^2*b^3 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cos(e*x + d)^2 + 2*(2*A*a^3*b^2 - 3*B*a
^2*b^3 + A*a*b^4)*cos(e*x + d))*sqrt(a^2 - b^2)*arctan(-(a*cos(e*x + d) + b)/(sqrt(a^2 - b^2)*sin(e*x + d))) +
(2*B*a^5*b - 4*A*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*b^4 - B*a*b^5 - A*b^6 + (B*a^4*b^2 - 3*A*a^3*b^3 + B*a^2*b^4 +
3*A*a*b^5 - 2*B*b^6)*cos(e*x + d))*sin(e*x + d))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*cos(e*x + d)^2 +
2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cos(e*x + d) + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*e)]

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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(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.27839, size = 678, normalized size = 3.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,x, algorithm="giac")

[Out]

((2*A*a^2 - 3*B*a*b + A*b^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))/sqrt(a^2 - b^2)))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (2*B*a^3*tan(1/2*x*e
+ 1/2*d)^3 - 4*A*a^2*b*tan(1/2*x*e + 1/2*d)^3 - B*a^2*b*tan(1/2*x*e + 1/2*d)^3 + 3*A*a*b^2*tan(1/2*x*e + 1/2*d
)^3 + B*a*b^2*tan(1/2*x*e + 1/2*d)^3 + A*b^3*tan(1/2*x*e + 1/2*d)^3 - 2*B*b^3*tan(1/2*x*e + 1/2*d)^3 - 2*C*a^3
*tan(1/2*x*e + 1/2*d)^2 - 2*C*a^2*b*tan(1/2*x*e + 1/2*d)^2 + 2*C*a*b^2*tan(1/2*x*e + 1/2*d)^2 + 2*C*b^3*tan(1/
2*x*e + 1/2*d)^2 + 2*B*a^3*tan(1/2*x*e + 1/2*d) - 4*A*a^2*b*tan(1/2*x*e + 1/2*d) + B*a^2*b*tan(1/2*x*e + 1/2*d
) - 3*A*a*b^2*tan(1/2*x*e + 1/2*d) + B*a*b^2*tan(1/2*x*e + 1/2*d) + A*b^3*tan(1/2*x*e + 1/2*d) + 2*B*b^3*tan(1
/2*x*e + 1/2*d) - 2*C*a^3 - 4*C*a^2*b - 2*C*a*b^2)/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*x*e + 1/2*d)^2 - b*tan(
1/2*x*e + 1/2*d)^2 + a + b)^2))*e^(-1)