3.19 \(\int \frac{A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^2} \, dx\)

Optimal. Leaf size=120 \[ -\frac{(A b-a B) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}+\frac{2 (a A-b B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{e (a-b)^{3/2} (a+b)^{3/2}}+\frac{C}{b e (a+b \cos (d+e x))} \]

[Out]

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

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Rubi [A]  time = 0.171644, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{(A b-a B) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}+\frac{2 (a A-b B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{e (a-b)^{3/2} (a+b)^{3/2}}+\frac{C}{b e (a+b \cos (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.404844, size = 115, normalized size = 0.96 \[ \frac{\frac{C \left (a^2-b^2\right )-b (A b-a B) \sin (d+e x)}{b (a-b) (a+b) (a+b \cos (d+e x))}+\frac{2 (a A-b B) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.04, size = 279, normalized size = 2.3 \begin{align*} -2\,{\frac{\tan \left ( 1/2\,ex+d/2 \right ) Ab}{e \left ( \left ( \tan \left ( 1/2\,ex+d/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,ex+d/2 \right ) \right ) ^{2}b+a+b \right ) \left ({a}^{2}-{b}^{2} \right ) }}+2\,{\frac{\tan \left ( 1/2\,ex+d/2 \right ) Ba}{e \left ( \left ( \tan \left ( 1/2\,ex+d/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,ex+d/2 \right ) \right ) ^{2}b+a+b \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{C}{e \left ( \left ( \tan \left ( 1/2\,ex+d/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,ex+d/2 \right ) \right ) ^{2}b+a+b \right ) \left ( a-b \right ) }}+2\,{\frac{Aa}{e \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,ex+d/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{Bb}{e \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,ex+d/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \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))^2,x)

[Out]

-2/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1/2*d)^2*b+a+b)/(a^2-b^2)*tan(1/2*e*x+1/2*d)*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)/(a^2-b^2)*tan(1/2*e*x+1/2*d)*B*a-2/e/(tan(1/2*e*x+1/2*d)^2*a-tan(1/2*e*x+1
/2*d)^2*b+a+b)*C/(a-b)+2/e/(a+b)/(a-b)/((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/e/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*e*x+1/2*d)/((a+b)*(a-b))^(1/2))*B*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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61975, size = 969, normalized size = 8.07 \begin{align*} \left [\frac{2 \, C a^{4} - 4 \, C a^{2} b^{2} + 2 \, C b^{4} -{\left (A a^{2} b - B a b^{2} +{\left (A a b^{2} - B b^{3}\right )} \cos \left (e x + d\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (e x + d\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (e x + d\right ) + b\right )} \sin \left (e x + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}}\right ) + 2 \,{\left (B a^{3} b - A a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sin \left (e x + d\right )}{2 \,{\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} e \cos \left (e x + d\right ) +{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} e\right )}}, \frac{C a^{4} - 2 \, C a^{2} b^{2} + C b^{4} +{\left (A a^{2} b - B a b^{2} +{\left (A a b^{2} - B b^{3}\right )} \cos \left (e x + d\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (e x + d\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (e x + d\right )}\right ) +{\left (B a^{3} b - A a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sin \left (e x + d\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} e \cos \left (e x + d\right ) +{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} e}\right ] \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))^2,x, algorithm="fricas")

[Out]

[1/2*(2*C*a^4 - 4*C*a^2*b^2 + 2*C*b^4 - (A*a^2*b - B*a*b^2 + (A*a*b^2 - B*b^3)*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*(B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4)*si
n(e*x + d))/((a^4*b^2 - 2*a^2*b^4 + b^6)*e*cos(e*x + d) + (a^5*b - 2*a^3*b^3 + a*b^5)*e), (C*a^4 - 2*C*a^2*b^2
+ C*b^4 + (A*a^2*b - B*a*b^2 + (A*a*b^2 - B*b^3)*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))) + (B*a^3*b - A*a^2*b^2 - B*a*b^3 + A*b^4)*sin(e*x + d))/((a^4*b^2 - 2*a^2*b^4 +
b^6)*e*cos(e*x + d) + (a^5*b - 2*a^3*b^3 + a*b^5)*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.22492, size = 234, normalized size = 1.95 \begin{align*} -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 )}{\sqrt{a^{2} - b^{2}}}\right )\right )}{\left (A a - B b\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} - \frac{B a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - A b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - C a - C b}{{\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} + a + b\right )}{\left (a^{2} - b^{2}\right )}}\right )} e^{\left (-1\right )} \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))^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))/sqrt(a^2 - b^2)))*(A*a - B*b)/(a^2 - b^2)^(3/2) - (B*a*tan(1/2*x*e + 1/2*d) - A*b*tan(1/2*x*e + 1/2*d) - C
*a - C*b)/((a*tan(1/2*x*e + 1/2*d)^2 - b*tan(1/2*x*e + 1/2*d)^2 + a + b)*(a^2 - b^2)))*e^(-1)