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

Optimal. Leaf size=87 \[ \frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b e \sqrt {a-b} \sqrt {a+b}}-\frac {C \log (a+b \cos (d+e x))}{b e}+\frac {B x}{b} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4377, 2735, 2659, 205, 2668, 31} \[ \frac {2 (A b-a B) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{b e \sqrt {a-b} \sqrt {a+b}}-\frac {C \log (a+b \cos (d+e x))}{b e}+\frac {B x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 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 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 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 82, normalized size = 0.94 \[ \frac {\frac {2 (a B-A b) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}-C \log (a+b \cos (d+e x))+B (d+e x)}{b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 326, normalized size = 3.75 \[ \left [\frac {2 \, {\left (B a^{2} - B b^{2}\right )} e x + {\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (e x + d\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (e x + d\right ) + b\right )} \sin \left (e x + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}}\right ) - {\left (C a^{2} - C b^{2}\right )} \log \left (b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} e}, \frac {2 \, {\left (B a^{2} - B b^{2}\right )} e x - 2 \, {\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (e x + d\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (e x + d\right )}\right ) - {\left (C a^{2} - C b^{2}\right )} \log \left (b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} e}\right ] \]

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)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.45, size = 459, normalized size = 5.28 \[ -{\left (\frac {C {\left (a + b\right )} {\left (a - b\right )}^{2} \log \left (\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + \frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{2 \, {\left (a - b\right )}}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} + \frac {{\left (\sqrt {a^{2} - b^{2}} B {\left (2 \, a - b\right )} {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} A b {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} A {\left | a - b \right |} {\left | b \right |} + \sqrt {a^{2} - b^{2}} B {\left | a - b \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )}{\sqrt {\frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} + \frac {{\left (2 \, B a - A b - B b + A {\left | b \right |} - B {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )}{\sqrt {\frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{b^{2} - a {\left | b \right |}} + \frac {{\left (C a - C b\right )} \log \left (\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + \frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{2 \, {\left (a - b\right )}}\right )}{b^{2} - a {\left | b \right |}}\right )} e^{\left (-1\right )} \]

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)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.18, size = 226, normalized size = 2.60 \[ -\frac {\ln \left (a \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right ) b +a +b \right ) a C}{e b \left (a -b \right )}+\frac {\ln \left (a \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-\left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right ) b +a +b \right ) C}{e \left (a -b \right )}+\frac {2 \arctan \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{e \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \arctan \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) a B}{e b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {C \ln \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )+1\right )}{e b}+\frac {2 B \arctan \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{e b} \]

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)),x)

[Out]

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

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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((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d)),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(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 4.16, size = 886, normalized size = 10.18 \[ -\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{b\,e}+\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )}{b\,e}-\frac {\ln \left (A^2\,b^3+B^2\,b^3-4\,C^2\,a^3+4\,C^2\,b^3+A^2\,a\,b^2+B^2\,a\,b^2+4\,C^2\,a\,b^2-4\,C^2\,a^2\,b-2\,A\,B\,a\,b^2-2\,A\,B\,a^2\,b+A^2\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}+B^2\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}-4\,C^2\,a^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}+4\,C^2\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}-4\,A\,C\,b^2\,\sqrt {b^2-a^2}+4\,B\,C\,a^2\,\sqrt {b^2-a^2}-4\,A\,C\,b^3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )-4\,B\,C\,a^3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )-4\,A\,C\,a\,b\,\sqrt {b^2-a^2}+4\,B\,C\,a\,b\,\sqrt {b^2-a^2}+4\,A\,C\,a^2\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+4\,B\,C\,a\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )-2\,A\,B\,a\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}\right )\,\left (C\,a^2-C\,b^2+A\,b\,\sqrt {b^2-a^2}-B\,a\,\sqrt {b^2-a^2}\right )}{b\,e\,\left (a^2-b^2\right )}-\frac {\ln \left (A^2\,b^3+B^2\,b^3-4\,C^2\,a^3+4\,C^2\,b^3+A^2\,a\,b^2+B^2\,a\,b^2+4\,C^2\,a\,b^2-4\,C^2\,a^2\,b-2\,A\,B\,a\,b^2-2\,A\,B\,a^2\,b-A^2\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}-B^2\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}+4\,C^2\,a^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}-4\,C^2\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}+4\,A\,C\,b^2\,\sqrt {b^2-a^2}-4\,B\,C\,a^2\,\sqrt {b^2-a^2}-4\,A\,C\,b^3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )-4\,B\,C\,a^3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+4\,A\,C\,a\,b\,\sqrt {b^2-a^2}-4\,B\,C\,a\,b\,\sqrt {b^2-a^2}+4\,A\,C\,a^2\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+4\,B\,C\,a\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+2\,A\,B\,a\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\sqrt {b^2-a^2}\right )\,\left (C\,a^2-C\,b^2-A\,b\,\sqrt {b^2-a^2}+B\,a\,\sqrt {b^2-a^2}\right )}{b\,e\,\left (a^2-b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 26.56, size = 672, normalized size = 7.72 \[ \begin {cases} \frac {\tilde {\infty } x \left (A + B \cos {\relax (d )} + C \sin {\relax (d )}\right )}{\cos {\relax (d )}} & \text {for}\: a = 0 \wedge b = 0 \wedge e = 0 \\\frac {A \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{b e} + \frac {B x}{b} - \frac {B \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}}{b e} + \frac {C \log {\left (\tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{b e} & \text {for}\: a = b \\\frac {A}{b e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}} + \frac {B x}{b} + \frac {B}{b e \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )}} + \frac {C \log {\left (\tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{b e} - \frac {2 C \log {\left (\tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{b e} & \text {for}\: a = - b \\\frac {A x + \frac {B \sin {\left (d + e x \right )}}{e} - \frac {C \cos {\left (d + e x \right )}}{e}}{a} & \text {for}\: b = 0 \\\frac {x \left (A + B \cos {\relax (d )} + C \sin {\relax (d )}\right )}{a + b \cos {\relax (d )}} & \text {for}\: e = 0 \\- \frac {A b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {A b \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {B a e x}{a b e + b^{2} e} + \frac {B a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} - \frac {B a \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {B b e x}{a b e + b^{2} e} - \frac {C a \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} - \frac {C a \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {C a \log {\left (\tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{a b e + b^{2} e} - \frac {C b \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} - \frac {C b \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {d}{2} + \frac {e x}{2} \right )} \right )}}{a b e + b^{2} e} + \frac {C b \log {\left (\tan ^{2}{\left (\frac {d}{2} + \frac {e x}{2} \right )} + 1 \right )}}{a b e + b^{2} e} & \text {otherwise} \end {cases} \]

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)),x)

[Out]

Piecewise((zoo*x*(A + B*cos(d) + C*sin(d))/cos(d), Eq(a, 0) & Eq(b, 0) & Eq(e, 0)), (A*tan(d/2 + e*x/2)/(b*e)
+ B*x/b - B*tan(d/2 + e*x/2)/(b*e) + C*log(tan(d/2 + e*x/2)**2 + 1)/(b*e), Eq(a, b)), (A/(b*e*tan(d/2 + e*x/2)
) + B*x/b + B/(b*e*tan(d/2 + e*x/2)) + C*log(tan(d/2 + e*x/2)**2 + 1)/(b*e) - 2*C*log(tan(d/2 + e*x/2))/(b*e),
Eq(a, -b)), ((A*x + B*sin(d + e*x)/e - C*cos(d + e*x)/e)/a, Eq(b, 0)), (x*(A + B*cos(d) + C*sin(d))/(a + b*co
s(d)), Eq(e, 0)), (-A*b*sqrt(-a/(a - b) - b/(a - b))*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2))/(a*
b*e + b**2*e) + A*b*sqrt(-a/(a - b) - b/(a - b))*log(sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e +
b**2*e) + B*a*e*x/(a*b*e + b**2*e) + B*a*sqrt(-a/(a - b) - b/(a - b))*log(-sqrt(-a/(a - b) - b/(a - b)) + tan
(d/2 + e*x/2))/(a*b*e + b**2*e) - B*a*sqrt(-a/(a - b) - b/(a - b))*log(sqrt(-a/(a - b) - b/(a - b)) + tan(d/2
+ e*x/2))/(a*b*e + b**2*e) + B*b*e*x/(a*b*e + b**2*e) - C*a*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/
2))/(a*b*e + b**2*e) - C*a*log(sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b**2*e) + C*a*log(tan
(d/2 + e*x/2)**2 + 1)/(a*b*e + b**2*e) - C*b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b*
*2*e) - C*b*log(sqrt(-a/(a - b) - b/(a - b)) + tan(d/2 + e*x/2))/(a*b*e + b**2*e) + C*b*log(tan(d/2 + e*x/2)**
2 + 1)/(a*b*e + b**2*e), True))

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