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

Optimal. Leaf size=118 \[ \frac{2 (a A-c C) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{e \left (a^2-c^2\right )^{3/2}}+\frac{(A c-a C) \cos (d+e x)}{e \left (a^2-c^2\right ) (a+c \sin (d+e x))}-\frac{B}{c e (a+c \sin (d+e x))} \]

[Out]

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

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Rubi [A]  time = 0.156626, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {4376, 2754, 12, 2660, 618, 204, 2668, 32} \[ \frac{2 (a A-c C) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{e \left (a^2-c^2\right )^{3/2}}+\frac{(A c-a C) \cos (d+e x)}{e \left (a^2-c^2\right ) (a+c \sin (d+e x))}-\frac{B}{c e (a+c \sin (d+e x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4376

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

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 2660

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

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

Mathematica [A]  time = 0.460344, size = 114, normalized size = 0.97 \[ \frac{\frac{B \left (a^2-c^2\right )-c (A c-a C) \cos (d+e x)}{c (c-a) (a+c) (a+c \sin (d+e x))}+\frac{2 (a A-c C) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{\left (a^2-c^2\right )^{3/2}}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.16, size = 426, normalized size = 3.6 \begin{align*} 2\,{\frac{\tan \left ( d/2+1/2\,ex \right ) A{c}^{2}}{e \left ( a \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+2\,c\tan \left ( d/2+1/2\,ex \right ) +a \right ) a \left ({a}^{2}-{c}^{2} \right ) }}+2\,{\frac{a\tan \left ( d/2+1/2\,ex \right ) B}{e \left ( a \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+2\,c\tan \left ( d/2+1/2\,ex \right ) +a \right ) \left ({a}^{2}-{c}^{2} \right ) }}-2\,{\frac{\tan \left ( d/2+1/2\,ex \right ) B{c}^{2}}{e \left ( a \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+2\,c\tan \left ( d/2+1/2\,ex \right ) +a \right ) a \left ({a}^{2}-{c}^{2} \right ) }}-2\,{\frac{c\tan \left ( d/2+1/2\,ex \right ) C}{e \left ( a \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+2\,c\tan \left ( d/2+1/2\,ex \right ) +a \right ) \left ({a}^{2}-{c}^{2} \right ) }}+2\,{\frac{Ac}{e \left ( a \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+2\,c\tan \left ( d/2+1/2\,ex \right ) +a \right ) \left ({a}^{2}-{c}^{2} \right ) }}-2\,{\frac{Ca}{e \left ( a \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+2\,c\tan \left ( d/2+1/2\,ex \right ) +a \right ) \left ({a}^{2}-{c}^{2} \right ) }}+2\,{\frac{aA}{e \left ({a}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( d/2+1/2\,ex \right ) +2\,c}{\sqrt{{a}^{2}-{c}^{2}}}} \right ) }-2\,{\frac{Cc}{e \left ({a}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( d/2+1/2\,ex \right ) +2\,c}{\sqrt{{a}^{2}-{c}^{2}}}} \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+c*sin(e*x+d))^2,x)

[Out]

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

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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+c*sin(e*x+d))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.76902, size = 995, normalized size = 8.43 \begin{align*} \left [-\frac{2 \, B a^{4} - 4 \, B a^{2} c^{2} + 2 \, B c^{4} +{\left (A a^{2} c - C a c^{2} +{\left (A a c^{2} - C c^{3}\right )} \sin \left (e x + d\right )\right )} \sqrt{-a^{2} + c^{2}} \log \left (\frac{{\left (2 \, a^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2} + 2 \,{\left (a \cos \left (e x + d\right ) \sin \left (e x + d\right ) + c \cos \left (e x + d\right )\right )} \sqrt{-a^{2} + c^{2}}}{c^{2} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2}}\right ) + 2 \,{\left (C a^{3} c - A a^{2} c^{2} - C a c^{3} + A c^{4}\right )} \cos \left (e x + d\right )}{2 \,{\left ({\left (a^{4} c^{2} - 2 \, a^{2} c^{4} + c^{6}\right )} e \sin \left (e x + d\right ) +{\left (a^{5} c - 2 \, a^{3} c^{3} + a c^{5}\right )} e\right )}}, -\frac{B a^{4} - 2 \, B a^{2} c^{2} + B c^{4} +{\left (A a^{2} c - C a c^{2} +{\left (A a c^{2} - C c^{3}\right )} \sin \left (e x + d\right )\right )} \sqrt{a^{2} - c^{2}} \arctan \left (-\frac{a \sin \left (e x + d\right ) + c}{\sqrt{a^{2} - c^{2}} \cos \left (e x + d\right )}\right ) +{\left (C a^{3} c - A a^{2} c^{2} - C a c^{3} + A c^{4}\right )} \cos \left (e x + d\right )}{{\left (a^{4} c^{2} - 2 \, a^{2} c^{4} + c^{6}\right )} e \sin \left (e x + d\right ) +{\left (a^{5} c - 2 \, a^{3} c^{3} + a c^{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+c*sin(e*x+d))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*B*a^4 - 4*B*a^2*c^2 + 2*B*c^4 + (A*a^2*c - C*a*c^2 + (A*a*c^2 - C*c^3)*sin(e*x + d))*sqrt(-a^2 + c^2)
*log(((2*a^2 - c^2)*cos(e*x + d)^2 - 2*a*c*sin(e*x + d) - a^2 - c^2 + 2*(a*cos(e*x + d)*sin(e*x + d) + c*cos(e
*x + d))*sqrt(-a^2 + c^2))/(c^2*cos(e*x + d)^2 - 2*a*c*sin(e*x + d) - a^2 - c^2)) + 2*(C*a^3*c - A*a^2*c^2 - C
*a*c^3 + A*c^4)*cos(e*x + d))/((a^4*c^2 - 2*a^2*c^4 + c^6)*e*sin(e*x + d) + (a^5*c - 2*a^3*c^3 + a*c^5)*e), -(
B*a^4 - 2*B*a^2*c^2 + B*c^4 + (A*a^2*c - C*a*c^2 + (A*a*c^2 - C*c^3)*sin(e*x + d))*sqrt(a^2 - c^2)*arctan(-(a*
sin(e*x + d) + c)/(sqrt(a^2 - c^2)*cos(e*x + d))) + (C*a^3*c - A*a^2*c^2 - C*a*c^3 + A*c^4)*cos(e*x + d))/((a^
4*c^2 - 2*a^2*c^4 + c^6)*e*sin(e*x + d) + (a^5*c - 2*a^3*c^3 + a*c^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+c*sin(e*x+d))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.14629, size = 252, normalized size = 2.14 \begin{align*} 2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{x e + d}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + c}{\sqrt{a^{2} - c^{2}}}\right )\right )}{\left (A a - C c\right )}}{{\left (a^{2} - c^{2}\right )}^{\frac{3}{2}}} + \frac{B a^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - C a c \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + A c^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - B c^{2} \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - C a^{2} + A a c}{{\left (a^{3} - a c^{2}\right )}{\left (a \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\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+c*sin(e*x+d))^2,x, algorithm="giac")

[Out]

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