\(\int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx\) [20]

Optimal result

Integrand size = 31, antiderivative size = 187 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} e}+\frac {C}{2 b e (a+b \cos (d+e x))^2}-\frac {(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))} \]

[Out]

(2*A*a^2+A*b^2-3*B*a*b)*arctan((a-b)^(1/2)*tan(1/2*e*x+1/2*d)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/e+1/2*C/b/e
/(a+b*cos(e*x+d))^2-1/2*(A*b-B*a)*sin(e*x+d)/(a^2-b^2)/e/(a+b*cos(e*x+d))^2-1/2*(3*A*a*b-B*a^2-2*B*b^2)*sin(e*
x+d)/(a^2-b^2)^2/e/(a+b*cos(e*x+d))

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4462, 2833, 12, 2738, 211, 2747, 32} \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\frac {\left (2 a^2 A-3 a b B+A b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{e (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right )^2 (a+b \cos (d+e x))}-\frac {(A b-a B) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}+\frac {C}{2 b e (a+b \cos (d+e x))^2} \]

[In]

Int[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + b*Cos[d + e*x])^3,x]

[Out]

((2*a^2*A + A*b^2 - 3*a*b*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*
e) + C/(2*b*e*(a + b*Cos[d + e*x])^2) - ((A*b - a*B)*Sin[d + e*x])/(2*(a^2 - b^2)*e*(a + b*Cos[d + e*x])^2) -
((3*a*A*b - a^2*B - 2*b^2*B)*Sin[d + e*x])/(2*(a^2 - b^2)^2*e*(a + b*Cos[d + e*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

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 2747

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 2833

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 4462

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*} \text {integral}& = C \int \frac {\sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx+\int \frac {A+B \cos (d+e x)}{(a+b \cos (d+e x))^3} \, dx \\ & = -\frac {(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac {\int \frac {-2 (a A-b B)+(A b-a B) \cos (d+e x)}{(a+b \cos (d+e x))^2} \, dx}{2 \left (a^2-b^2\right )}-\frac {C \text {Subst}\left (\int \frac {1}{(a+x)^3} \, dx,x,b \cos (d+e x)\right )}{b e} \\ & = \frac {C}{2 b e (a+b \cos (d+e x))^2}-\frac {(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))}+\frac {\int \frac {2 a^2 A+A b^2-3 a b B}{a+b \cos (d+e x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = \frac {C}{2 b e (a+b \cos (d+e x))^2}-\frac {(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \int \frac {1}{a+b \cos (d+e x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = \frac {C}{2 b e (a+b \cos (d+e x))^2}-\frac {(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \text {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 )^2 e} \\ & = \frac {\left (2 a^2 A+A b^2-3 a b B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} e}+\frac {C}{2 b e (a+b \cos (d+e x))^2}-\frac {(A b-a B) \sin (d+e x)}{2 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (d+e x)}{2 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.94 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\frac {-\frac {2 \left (2 a^2 A+A b^2-3 a b B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {\left (-3 a A b+a^2 B+2 b^2 B\right ) \sin (d+e x)}{(a-b)^2 (a+b)^2 (a+b \cos (d+e x))}+\frac {\left (a^2-b^2\right ) C-b (A b-a B) \sin (d+e x)}{(a-b) b (a+b) (a+b \cos (d+e x))^2}}{2 e} \]

[In]

Integrate[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + b*Cos[d + e*x])^3,x]

[Out]

((-2*(2*a^2*A + A*b^2 - 3*a*b*B)*ArcTanh[((a - b)*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + ((
-3*a*A*b + a^2*B + 2*b^2*B)*Sin[d + e*x])/((a - b)^2*(a + b)^2*(a + b*Cos[d + e*x])) + ((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])^2))/(2*e)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.44

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (4 A a b +A \,b^{2}-2 B \,a^{2}-B a b -2 B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 C \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{a -b}-\frac {\left (4 A a b -A \,b^{2}-2 B \,a^{2}+B a b -2 B \,b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 C a}{a^{2}-2 a b +b^{2}}}{{\left (a \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 A \,a^{2}+A \,b^{2}-3 B a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{e}\)	\(270\)
default	\(\frac {\frac {-\frac {\left (4 A a b +A \,b^{2}-2 B \,a^{2}-B a b -2 B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 C \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{a -b}-\frac {\left (4 A a b -A \,b^{2}-2 B \,a^{2}+B a b -2 B \,b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 C a}{a^{2}-2 a b +b^{2}}}{{\left (a \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )+a +b \right )}^{2}}+\frac {\left (2 A \,a^{2}+A \,b^{2}-3 B a b \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{e}\)	\(270\)
risch	\(\frac {4 i B \,a^{3} b \,{\mathrm e}^{i \left (e x +d \right )}+2 i B \,a^{4} {\mathrm e}^{2 i \left (e x +d \right )}-10 i A \,a^{2} b^{2} {\mathrm e}^{i \left (e x +d \right )}-2 i A \,a^{2} b^{2} {\mathrm e}^{3 i \left (e x +d \right )}+i A \,b^{4} {\mathrm e}^{i \left (e x +d \right )}+3 i B a \,b^{3} {\mathrm e}^{3 i \left (e x +d \right )}-3 i A a \,b^{3} {\mathrm e}^{2 i \left (e x +d \right )}+5 i B a \,b^{3} {\mathrm e}^{i \left (e x +d \right )}-6 i A \,a^{3} b \,{\mathrm e}^{2 i \left (e x +d \right )}+2 i B \,b^{4} {\mathrm e}^{2 i \left (e x +d \right )}-3 i A a \,b^{3}-i A \,b^{4} {\mathrm e}^{3 i \left (e x +d \right )}+2 C \,a^{4} {\mathrm e}^{2 i \left (e x +d \right )}-4 C \,a^{2} b^{2} {\mathrm e}^{2 i \left (e x +d \right )}+2 C \,b^{4} {\mathrm e}^{2 i \left (e x +d \right )}+i B \,a^{2} b^{2}+2 i B \,b^{4}+5 i B \,a^{2} b^{2} {\mathrm e}^{2 i \left (e x +d \right )}}{b \left (a^{2}-b^{2}\right )^{2} e \left (b \,{\mathrm e}^{2 i \left (e x +d \right )}+2 a \,{\mathrm e}^{i \left (e x +d \right )}+b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A \,a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A \,b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} e}+\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B a b}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A \,a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A \,b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} e}-\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B a b}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} e}\)	\(828\)

[In]

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (171) = 342\).

Time = 0.31 (sec) , antiderivative size = 830, normalized size of antiderivative = 4.44 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\left [\frac {2 \, C a^{6} - 6 \, C a^{4} b^{2} + 6 \, C a^{2} b^{4} - 2 \, C b^{6} - {\left (2 \, A a^{4} b - 3 \, B a^{3} b^{2} + A a^{2} b^{3} + {\left (2 \, A a^{2} b^{3} - 3 \, B a b^{4} + A b^{5}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (2 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + A a b^{4}\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 (2 \, B a^{5} b - 4 \, A a^{4} b^{2} - B a^{3} b^{3} + 5 \, A a^{2} b^{4} - B a b^{5} - A b^{6} + {\left (B a^{4} b^{2} - 3 \, A a^{3} b^{3} + B a^{2} b^{4} + 3 \, A a b^{5} - 2 \, B b^{6}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{4 \, {\left ({\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} e \cos \left (e x + d\right )^{2} + 2 \, {\left (a^{7} b^{2} - 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} - a b^{8}\right )} e \cos \left (e x + d\right ) + {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} e\right )}}, \frac {C a^{6} - 3 \, C a^{4} b^{2} + 3 \, C a^{2} b^{4} - C b^{6} + {\left (2 \, A a^{4} b - 3 \, B a^{3} b^{2} + A a^{2} b^{3} + {\left (2 \, A a^{2} b^{3} - 3 \, B a b^{4} + A b^{5}\right )} \cos \left (e x + d\right )^{2} + 2 \, {\left (2 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + A a b^{4}\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 (2 \, B a^{5} b - 4 \, A a^{4} b^{2} - B a^{3} b^{3} + 5 \, A a^{2} b^{4} - B a b^{5} - A b^{6} + {\left (B a^{4} b^{2} - 3 \, A a^{3} b^{3} + B a^{2} b^{4} + 3 \, A a b^{5} - 2 \, B b^{6}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{2 \, {\left ({\left (a^{6} b^{3} - 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} - b^{9}\right )} e \cos \left (e x + d\right )^{2} + 2 \, {\left (a^{7} b^{2} - 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} - a b^{8}\right )} e \cos \left (e x + d\right ) + {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} e\right )}}\right ] \]

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,x, algorithm="fricas")

[Out]

[1/4*(2*C*a^6 - 6*C*a^4*b^2 + 6*C*a^2*b^4 - 2*C*b^6 - (2*A*a^4*b - 3*B*a^3*b^2 + A*a^2*b^3 + (2*A*a^2*b^3 - 3*
B*a*b^4 + A*b^5)*cos(e*x + d)^2 + 2*(2*A*a^3*b^2 - 3*B*a^2*b^3 + A*a*b^4)*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*(2*B*a^5*b - 4*A*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*
b^4 - B*a*b^5 - A*b^6 + (B*a^4*b^2 - 3*A*a^3*b^3 + B*a^2*b^4 + 3*A*a*b^5 - 2*B*b^6)*cos(e*x + d))*sin(e*x + d)
)/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*cos(e*x + d)^2 + 2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*co
s(e*x + d) + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*e), 1/2*(C*a^6 - 3*C*a^4*b^2 + 3*C*a^2*b^4 - C*b^6 + (2
*A*a^4*b - 3*B*a^3*b^2 + A*a^2*b^3 + (2*A*a^2*b^3 - 3*B*a*b^4 + A*b^5)*cos(e*x + d)^2 + 2*(2*A*a^3*b^2 - 3*B*a
^2*b^3 + A*a*b^4)*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))) +
(2*B*a^5*b - 4*A*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*b^4 - B*a*b^5 - A*b^6 + (B*a^4*b^2 - 3*A*a^3*b^3 + B*a^2*b^4 +
3*A*a*b^5 - 2*B*b^6)*cos(e*x + d))*sin(e*x + d))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*e*cos(e*x + d)^2 +
2*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*e*cos(e*x + d) + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*e)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,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 de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 480 vs. \(2 (171) = 342\).

Time = 0.37 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.57 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\frac {\frac {{\left (2 \, A a^{2} - 3 \, B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {e x + 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} \, e x + \frac {1}{2} \, d\right ) - b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, B a^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 4 \, A a^{2} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 3 \, A a b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + B a b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + A b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 2 \, B b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 2 \, C a^{2} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, C a b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, C b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 4 \, A a^{2} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + B a^{2} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 3 \, A a b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + B a b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + A b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 2 \, B b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 2 \, C a^{3} - 4 \, C a^{2} b - 2 \, C a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + a + b\right )}^{2}}}{e} \]

[In]

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.58 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.51 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,A\,a^2-3\,B\,a\,b+A\,b^2\right )}{e\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,C\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{a-b}+\frac {2\,C\,a}{{\left (a-b\right )}^2}-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (2\,B\,a^2-A\,b^2+2\,B\,b^2-4\,A\,a\,b+B\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (A\,b^2+2\,B\,a^2+2\,B\,b^2-4\,A\,a\,b-B\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{e\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]

[In]

int((A + B*cos(d + e*x) + C*sin(d + e*x))/(a + b*cos(d + e*x))^3,x)

[Out]

(atan((tan(d/2 + (e*x)/2)*(2*a - 2*b)*(a^2 - 2*a*b + b^2))/(2*(a + b)^(1/2)*(a - b)^(5/2)))*(2*A*a^2 + A*b^2 -
3*B*a*b))/(e*(a + b)^(5/2)*(a - b)^(5/2)) - ((2*C*tan(d/2 + (e*x)/2)^2)/(a - b) + (2*C*a)/(a - b)^2 - (tan(d/
2 + (e*x)/2)^3*(2*B*a^2 - A*b^2 + 2*B*b^2 - 4*A*a*b + B*a*b))/((a + b)^2*(a - b)) - (tan(d/2 + (e*x)/2)*(A*b^2
+ 2*B*a^2 + 2*B*b^2 - 4*A*a*b - B*a*b))/((a + b)*(a^2 - 2*a*b + b^2)))/(e*(2*a*b + tan(d/2 + (e*x)/2)^2*(2*a^
2 - 2*b^2) + tan(d/2 + (e*x)/2)^4*(a^2 - 2*a*b + b^2) + a^2 + b^2))