Integrand size = 48, antiderivative size = 110 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {2 \left (a b B-a^2 C-b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac {b (b B-2 a C) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:
2*(B*a*b-C*a^2-C*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/( a-b)^(3/2)/(a+b)^(3/2)/d-b*(B*b-2*C*a)*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x +c))
Time = 0.66 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.97 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 \left (-a b B+a^2 C+b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\frac {b (-b B+2 a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{d} \] Input:
Integrate[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^3,x]
Output:
((-2*(-(a*b*B) + a^2*C + b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a ^2 + b^2]])/(-a^2 + b^2)^(3/2) + (b*(-(b*B) + 2*a*C)*Sin[c + d*x])/((a - b )*(a + b)*(a + b*Cos[c + d*x])))/d
Time = 0.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2014, 3042, 3233, 25, 27, 3042, 3138, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 2014 |
| \(\displaystyle \frac {\int \frac {C \cos (c+d x) b^3+(b B-a C) b^2}{(a+b \cos (c+d x))^2}dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{b^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {-\frac {\int -\frac {b^2 \left (-C a^2+b B a-b^2 C\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {b^2 \left (-C a^2+b B a-b^2 C\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {b^2 \left (a^2 (-C)+a b B-b^2 C\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b^2 \left (a^2 (-C)+a b B-b^2 C\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 b^2 \left (a^2 (-C)+a b B-b^2 C\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {2 b^2 \left (a^2 (-C)+a b B-b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{b^2}\) |
Input:
Int[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + b*Cos [c + d*x])^3,x]
Output:
((2*b^2*(a*b*B - a^2*C - b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt [a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - (b^3*(b*B - 2*a*C)*Sin [c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/b^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S ymbol] :> Simp[1/b^2 Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ [m, -1]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[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]
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] + Simp[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]
Time = 0.53 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.25
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 b \left (B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {2 \left (B a b -a^{2} C -C \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(138\) |
| default | \(\frac {-\frac {2 b \left (B b -2 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {2 \left (B a b -a^{2} C -C \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(138\) |
| risch | \(\frac {2 i \left (B b -2 C a \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{\left (-a^{2}+b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )} b +2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B a b}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2} C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B a b}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2} C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(572\) |
Input:
int((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x ,method=_RETURNVERBOSE)
Output:
1/d*(-2*b*(B*b-2*C*a)/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a -tan(1/2*d*x+1/2*c)^2*b+a+b)+2*(B*a*b-C*a^2-C*b^2)/(a-b)/(a+b)/((a-b)*(a+b ))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.11 (sec) , antiderivative size = 421, normalized size of antiderivative = 3.83 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\left [\frac {{\left (C a^{3} - B a^{2} b + C a b^{2} + {\left (C a^{2} b - B a b^{2} + C b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (2 \, C a^{3} b - B a^{2} b^{2} - 2 \, C a b^{3} + B b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, -\frac {{\left (C a^{3} - B a^{2} b + C a b^{2} + {\left (C a^{2} b - B a b^{2} + C b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, C a^{3} b - B a^{2} b^{2} - 2 \, C a b^{3} + B b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d}\right ] \] Input:
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^3,x, algorithm="fricas")
Output:
[1/2*((C*a^3 - B*a^2*b + C*a*b^2 + (C*a^2*b - B*a*b^2 + C*b^3)*cos(d*x + c ))*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^ 2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 2*(2*C*a^3*b - B*a^2*b^2 - 2*C*a*b^3 + B*b^4)*sin(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c ) + (a^5 - 2*a^3*b^2 + a*b^4)*d), -((C*a^3 - B*a^2*b + C*a*b^2 + (C*a^2*b - B*a*b^2 + C*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*C*a^3*b - B*a^2*b^2 - 2*C*a*b^3 + B*b^4)*sin(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c) + (a^5 - 2 *a^3*b^2 + a*b^4)*d)]
Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((B*a*b-a**2*C+b**2*B*cos(d*x+c)+b**2*C*cos(d*x+c)**2)/(a+b*cos(d *x+c))**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^3,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.17 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.55 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {{\left (C a^{2} - B a b + C b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} - b^{2}\right )}}\right )}}{d} \] Input:
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^3,x, algorithm="giac")
Output:
-2*((C*a^2 - B*a*b + C*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2* b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b ^2)))/(a^2 - b^2)^(3/2) - (2*C*a*b*tan(1/2*d*x + 1/2*c) - B*b^2*tan(1/2*d* x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b) *(a^2 - b^2)))/d
Time = 0.96 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.13 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )}{2\,\sqrt {a+b}\,\sqrt {a-b}}\right )\,\left (C\,a^2-B\,a\,b+C\,b^2\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,b^2-2\,C\,a\,b\right )}{d\,\left (a+b\right )\,\left (a-b\right )\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a+b\right )} \] Input:
int((C*b^2*cos(c + d*x)^2 - C*a^2 + B*a*b + B*b^2*cos(c + d*x))/(a + b*cos (c + d*x))^3,x)
Output:
- (2*atan((tan(c/2 + (d*x)/2)*(2*a - 2*b))/(2*(a + b)^(1/2)*(a - b)^(1/2)) )*(C*a^2 + C*b^2 - B*a*b))/(d*(a + b)^(3/2)*(a - b)^(3/2)) - (2*tan(c/2 + (d*x)/2)*(B*b^2 - 2*C*a*b))/(d*(a + b)*(a - b)*(a + b + tan(c/2 + (d*x)/2) ^2*(a - b)))
Time = 0.16 (sec) , antiderivative size = 455, normalized size of antiderivative = 4.14 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{2} b c +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a \,b^{3}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) b^{3} c -2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} c +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} b^{2}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a \,b^{2} c +2 \sin \left (d x +c \right ) a^{3} b c -\sin \left (d x +c \right ) a^{2} b^{3}-2 \sin \left (d x +c \right ) a \,b^{3} c +\sin \left (d x +c \right ) b^{5}}{d \left (\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:
int((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x )
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr t(a**2 - b**2))*cos(c + d*x)*a**2*b*c + 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a*b**3 - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a* *2 - b**2))*cos(c + d*x)*b**3*c - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/ 2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**3*c + 2*sqrt(a**2 - b**2) *atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**2*b* *2 - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt(a**2 - b**2))*a*b**2*c + 2*sin(c + d*x)*a**3*b*c - sin(c + d*x)*a**2*b* *3 - 2*sin(c + d*x)*a*b**3*c + sin(c + d*x)*b**5)/(d*(cos(c + d*x)*a**4*b - 2*cos(c + d*x)*a**2*b**3 + cos(c + d*x)*b**5 + a**5 - 2*a**3*b**2 + a*b* *4))