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

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Rubi [A]  time = 0.15, antiderivative size = 84, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.25, 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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fricas [A]  time = 1.00, size = 346, normalized size = 4.12 \[ \left [\frac {2 \, {\left (C a^{2} - C c^{2}\right )} e x + {\left (C a - A c\right )} \sqrt {-a^{2} + c^{2}} \log \left (\frac {{\left (2 \, a^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2} + 2 \, {\left (a \cos \left (e x + d\right ) \sin \left (e x + d\right ) + c \cos \left (e x + d\right )\right )} \sqrt {-a^{2} + c^{2}}}{c^{2} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2}}\right ) + {\left (B a^{2} - B c^{2}\right )} \log \left (-c^{2} \cos \left (e x + d\right )^{2} + 2 \, a c \sin \left (e x + d\right ) + a^{2} + c^{2}\right )}{2 \, {\left (a^{2} c - c^{3}\right )} e}, \frac {2 \, {\left (C a^{2} - C c^{2}\right )} e x + 2 \, {\left (C a - A c\right )} \sqrt {a^{2} - c^{2}} \arctan \left (-\frac {a \sin \left (e x + d\right ) + c}{\sqrt {a^{2} - c^{2}} \cos \left (e x + d\right )}\right ) + {\left (B a^{2} - B c^{2}\right )} \log \left (-c^{2} \cos \left (e x + d\right )^{2} + 2 \, a c \sin \left (e x + d\right ) + a^{2} + c^{2}\right )}{2 \, {\left (a^{2} c - c^{3}\right )} e}\right ] \]

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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giac [A]  time = 0.18, size = 141, normalized size = 1.68 \[ {\left (\frac {{\left (x e + d\right )} C}{c} + \frac {B \log \left (a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a\right )}{c} - \frac {B \log \left (\tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 1\right )}{c} - \frac {2 \, {\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + c}{\sqrt {a^{2} - c^{2}}}\right )\right )} {\left (C a - A c\right )}}{\sqrt {a^{2} - c^{2}} c}\right )} e^{\left (-1\right )} \]

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)

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maple [B]  time = 0.25, size = 178, normalized size = 2.12 \[ \frac {B \ln \left (a \left (\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )+2 c \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+a \right )}{e c}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right ) A}{e \sqrt {a^{2}-c^{2}}}-\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d}{2}+\frac {e x}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right ) C a}{e c \sqrt {a^{2}-c^{2}}}-\frac {B \ln \left (1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{e c}+\frac {2 C \arctan \left (\tan \left (\frac {d}{2}+\frac {e x}{2}\right )\right )}{e c} \]

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 = 0.00, 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+c*sin(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*c^2-4*a^2>0)', see `assume?`
for more details)Is 4*c^2-4*a^2 positive or negative?

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mupad [B]  time = 9.63, size = 1143, normalized size = 13.61 \[ \frac {\ln \left (32\,B^3\,a^2-32\,A\,B^2\,a^2+32\,A\,C^2\,a^2+32\,B\,C^2\,a^2+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (c\,A^2\,B-2\,c\,A\,B^2-2\,a\,A\,B\,C-2\,c\,A\,C^2+c\,B^3+2\,a\,B^2\,C+2\,c\,B\,C^2+2\,a\,C^3\right )-32\,A^2\,C\,a\,c+32\,B^2\,C\,a\,c-\frac {\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,C^2\,a^2\,c-32\,B^2\,a^2\,c-128\,B\,C\,a^3+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-A^2\,c^2+4\,A\,B\,c^2+2\,A\,C\,a\,c+2\,B^2\,a^2-3\,B^2\,c^2-4\,B\,C\,a\,c-2\,C^2\,a^2+2\,C^2\,c^2\right )+64\,A\,B\,a^2\,c+64\,B\,C\,a\,c^2+\frac {\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,A\,a^2\,c^2+32\,B\,a^2\,c^2-32\,C\,a\,c^3+32\,a\,c^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,c-2\,C\,a+B\,c\right )+\frac {32\,a\,c\,\left (-2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,a^2+a\,c+3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,c^2\right )\,\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )}{a^2-c^2}\right )}{c\,\left (a^2-c^2\right )}\right )}{c\,\left (a^2-c^2\right )}\right )\,\left (B\,a^2-B\,c^2+A\,c\,\sqrt {c^2-a^2}-C\,a\,\sqrt {c^2-a^2}\right )}{c\,e\,\left (a^2-c^2\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{c\,e}-\frac {\ln \left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )}{c\,e}+\frac {\ln \left (32\,B^3\,a^2-32\,A\,B^2\,a^2+32\,A\,C^2\,a^2+32\,B\,C^2\,a^2+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (c\,A^2\,B-2\,c\,A\,B^2-2\,a\,A\,B\,C-2\,c\,A\,C^2+c\,B^3+2\,a\,B^2\,C+2\,c\,B\,C^2+2\,a\,C^3\right )-32\,A^2\,C\,a\,c+32\,B^2\,C\,a\,c-\frac {\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,C^2\,a^2\,c-32\,B^2\,a^2\,c-128\,B\,C\,a^3+32\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-A^2\,c^2+4\,A\,B\,c^2+2\,A\,C\,a\,c+2\,B^2\,a^2-3\,B^2\,c^2-4\,B\,C\,a\,c-2\,C^2\,a^2+2\,C^2\,c^2\right )+64\,A\,B\,a^2\,c+64\,B\,C\,a\,c^2+\frac {\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )\,\left (32\,A\,a^2\,c^2+32\,B\,a^2\,c^2-32\,C\,a\,c^3+32\,a\,c^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,c-2\,C\,a+B\,c\right )+\frac {32\,a\,c\,\left (-2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,a^2+a\,c+3\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,c^2\right )\,\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )}{a^2-c^2}\right )}{c\,\left (a^2-c^2\right )}\right )}{c\,\left (a^2-c^2\right )}\right )\,\left (B\,a^2-B\,c^2-A\,c\,\sqrt {c^2-a^2}+C\,a\,\sqrt {c^2-a^2}\right )}{c\,e\,\left (a^2-c^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 + c*sin(d + e*x)),x)

[Out]

(log(32*B^3*a^2 - 32*A*B^2*a^2 + 32*A*C^2*a^2 + 32*B*C^2*a^2 + 32*a*tan(d/2 + (e*x)/2)*(2*C^3*a + B^3*c - 2*A*
B^2*c + A^2*B*c + 2*B^2*C*a - 2*A*C^2*c + 2*B*C^2*c - 2*A*B*C*a) - 32*A^2*C*a*c + 32*B^2*C*a*c - ((B*a^2 - B*c
^2 + A*c*(c^2 - a^2)^(1/2) - C*a*(c^2 - a^2)^(1/2))*(32*C^2*a^2*c - 32*B^2*a^2*c - 128*B*C*a^3 + 32*a*tan(d/2
+ (e*x)/2)*(2*B^2*a^2 - A^2*c^2 - 2*C^2*a^2 - 3*B^2*c^2 + 2*C^2*c^2 + 4*A*B*c^2 + 2*A*C*a*c - 4*B*C*a*c) + 64*
A*B*a^2*c + 64*B*C*a*c^2 + ((B*a^2 - B*c^2 + A*c*(c^2 - a^2)^(1/2) - C*a*(c^2 - a^2)^(1/2))*(32*A*a^2*c^2 + 32
*B*a^2*c^2 - 32*C*a*c^3 + 32*a*c^2*tan(d/2 + (e*x)/2)*(2*A*c - 2*C*a + B*c) + (32*a*c*(a*c - 2*a^2*tan(d/2 + (
e*x)/2) + 3*c^2*tan(d/2 + (e*x)/2))*(B*a^2 - B*c^2 + A*c*(c^2 - a^2)^(1/2) - C*a*(c^2 - a^2)^(1/2)))/(a^2 - c^
2)))/(c*(a^2 - c^2))))/(c*(a^2 - c^2)))*(B*a^2 - B*c^2 + A*c*(c^2 - a^2)^(1/2) - C*a*(c^2 - a^2)^(1/2)))/(c*e*
(a^2 - c^2)) - (log(tan(d/2 + (e*x)/2) + 1i)*(B - C*1i))/(c*e) - (log(tan(d/2 + (e*x)/2) - 1i)*(B + C*1i))/(c*
e) + (log(32*B^3*a^2 - 32*A*B^2*a^2 + 32*A*C^2*a^2 + 32*B*C^2*a^2 + 32*a*tan(d/2 + (e*x)/2)*(2*C^3*a + B^3*c -
2*A*B^2*c + A^2*B*c + 2*B^2*C*a - 2*A*C^2*c + 2*B*C^2*c - 2*A*B*C*a) - 32*A^2*C*a*c + 32*B^2*C*a*c - ((B*a^2
- B*c^2 - A*c*(c^2 - a^2)^(1/2) + C*a*(c^2 - a^2)^(1/2))*(32*C^2*a^2*c - 32*B^2*a^2*c - 128*B*C*a^3 + 32*a*tan
(d/2 + (e*x)/2)*(2*B^2*a^2 - A^2*c^2 - 2*C^2*a^2 - 3*B^2*c^2 + 2*C^2*c^2 + 4*A*B*c^2 + 2*A*C*a*c - 4*B*C*a*c)
+ 64*A*B*a^2*c + 64*B*C*a*c^2 + ((B*a^2 - B*c^2 - A*c*(c^2 - a^2)^(1/2) + C*a*(c^2 - a^2)^(1/2))*(32*A*a^2*c^2
+ 32*B*a^2*c^2 - 32*C*a*c^3 + 32*a*c^2*tan(d/2 + (e*x)/2)*(2*A*c - 2*C*a + B*c) + (32*a*c*(a*c - 2*a^2*tan(d/
2 + (e*x)/2) + 3*c^2*tan(d/2 + (e*x)/2))*(B*a^2 - B*c^2 - A*c*(c^2 - a^2)^(1/2) + C*a*(c^2 - a^2)^(1/2)))/(a^2
- c^2)))/(c*(a^2 - c^2))))/(c*(a^2 - c^2)))*(B*a^2 - B*c^2 - A*c*(c^2 - a^2)^(1/2) + C*a*(c^2 - a^2)^(1/2)))/
(c*e*(a^2 - c^2))

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sympy [A]  time = 29.52, size = 1110, normalized size = 13.21 \[ \text {result too large to display} \]

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)), ((A*log(tan(d/2 + e*x/2))/
e - B*log(tan(d/2 + e*x/2)**2 + 1)/e + B*log(tan(d/2 + e*x/2))/e + C*x)/c, Eq(a, 0)), (2*A/(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(tan(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*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/(c*e*tan(d/2 + e*x/2) - c*e), Eq(a, -c)), (-2*A/(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*lo
g(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*ta
n(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/(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*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*s
qrt(-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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