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

Optimal. Leaf size=84 \[ \frac{2 (A c-a C) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{c e \sqrt{a^2-c^2}}+\frac{B \log (a+c \sin (d+e x))}{c e}+\frac{C x}{c} \]

[Out]

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

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Rubi [A]  time = 0.151861, antiderivative size = 84, 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 = {4376, 2735, 2660, 618, 204, 2668, 31} \[ \frac{2 (A c-a C) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{c e \sqrt{a^2-c^2}}+\frac{B \log (a+c \sin (d+e x))}{c e}+\frac{C x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2735

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

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.262016, size = 80, normalized size = 0.95 \[ \frac{\frac{2 (A c-a C) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-c^2}}\right )}{\sqrt{a^2-c^2}}+B \log (a+c \sin (d+e x))+C (d+e x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.059, size = 178, normalized size = 2.1 \begin{align*}{\frac{B}{ce}\ln \left ( a \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2}+2\,c\tan \left ( d/2+1/2\,ex \right ) +a \right ) }+2\,{\frac{A}{e\sqrt{{a}^{2}-{c}^{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{Ca}{ce\sqrt{{a}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( d/2+1/2\,ex \right ) +2\,c}{\sqrt{{a}^{2}-{c}^{2}}}} \right ) }-{\frac{B}{ce}\ln \left ( 1+ \left ( \tan \left ({\frac{d}{2}}+{\frac{ex}{2}} \right ) \right ) ^{2} \right ) }+2\,{\frac{C\arctan \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) }{ce}} \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)),x)

[Out]

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

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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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.63804, size = 763, normalized size = 9.08 \begin{align*} \left [\frac{2 \,{\left (C a^{2} - C c^{2}\right )} e x +{\left (C a - A c\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 ) +{\left (B a^{2} - B c^{2}\right )} \log \left (-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 (a^{2} c - c^{3}\right )} e}, \frac{2 \,{\left (C a^{2} - C c^{2}\right )} e x + 2 \,{\left (C a - A c\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 (B a^{2} - B c^{2}\right )} \log \left (-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 (a^{2} c - c^{3}\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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 52.2945, size = 1151, normalized size = 13.7 \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+c*sin(e*x+d)),x)

[Out]

Piecewise((zoo*x*(A + B*cos(d) + C*sin(d))/sin(d), Eq(a, 0) & Eq(c, 0) & Eq(e, 0)), (2*A*tan(d/2 + e*x/2)/(c*e
*tan(d/2 + e*x/2) - c*e) + 2*B*log(tan(d/2 + e*x/2) - 1)*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) - c*e) - 2*B*l
og(tan(d/2 + e*x/2) - 1)/(c*e*tan(d/2 + e*x/2) - c*e) - B*log(tan(d/2 + e*x/2)**2 + 1)*tan(d/2 + e*x/2)/(c*e*t
an(d/2 + e*x/2) - c*e) + B*log(tan(d/2 + e*x/2)**2 + 1)/(c*e*tan(d/2 + e*x/2) - c*e) + C*e*x*tan(d/2 + e*x/2)/
(c*e*tan(d/2 + e*x/2) - c*e) - C*e*x/(c*e*tan(d/2 + e*x/2) - c*e) + 2*C*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2)
- c*e), Eq(a, -c)), (2*A*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) + c*e) + 2*B*log(tan(d/2 + e*x/2) + 1)*tan(d/
2 + e*x/2)/(c*e*tan(d/2 + e*x/2) + c*e) + 2*B*log(tan(d/2 + e*x/2) + 1)/(c*e*tan(d/2 + e*x/2) + c*e) - B*log(t
an(d/2 + e*x/2)**2 + 1)*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) + c*e) - B*log(tan(d/2 + e*x/2)**2 + 1)/(c*e*ta
n(d/2 + e*x/2) + c*e) + C*e*x*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) + c*e) + C*e*x/(c*e*tan(d/2 + e*x/2) + c*
e) - 2*C*tan(d/2 + e*x/2)/(c*e*tan(d/2 + e*x/2) + c*e), Eq(a, c)), ((A*x + B*sin(d + e*x)/e - C*cos(d + e*x)/e
)/a, Eq(c, 0)), (x*(A + B*cos(d) + C*sin(d))/(a + c*sin(d)), Eq(e, 0)), ((A*log(tan(d/2 + e*x/2))/e - B*log(ta
n(d/2 + e*x/2)**2 + 1)/e + B*log(tan(d/2 + e*x/2))/e + C*x)/c, Eq(a, 0)), (-A*c*sqrt(-a**2 + c**2)*log(tan(d/2
+ e*x/2) + c/a - sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) + A*c*sqrt(-a**2 + c**2)*log(tan(d/2 + e*x/2) + c/
a + sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) - B*a**2*log(tan(d/2 + e*x/2)**2 + 1)/(a**2*c*e - c**3*e) + B*a*
*2*log(tan(d/2 + e*x/2) + c/a - sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) + B*a**2*log(tan(d/2 + e*x/2) + c/a
+ sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) + B*c**2*log(tan(d/2 + e*x/2)**2 + 1)/(a**2*c*e - c**3*e) - B*c**2
*log(tan(d/2 + e*x/2) + c/a - sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) - B*c**2*log(tan(d/2 + e*x/2) + c/a +
sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) + C*a**2*e*x/(a**2*c*e - c**3*e) + C*a*sqrt(-a**2 + c**2)*log(tan(d/
2 + e*x/2) + c/a - sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) - C*a*sqrt(-a**2 + c**2)*log(tan(d/2 + e*x/2) + c
/a + sqrt(-a**2 + c**2)/a)/(a**2*c*e - c**3*e) - C*c**2*e*x/(a**2*c*e - c**3*e), True))

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Giac [A]  time = 1.13396, size = 190, normalized size = 2.26 \begin{align*}{\left (\frac{{\left (x e + d\right )} C}{c} + \frac{B \log \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 )}{c} - \frac{B \log \left (\tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + 1\right )}{c} - \frac{2 \,{\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 (C a - A c\right )}}{\sqrt{a^{2} - c^{2}} c}\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)),x, algorithm="giac")

[Out]

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