Integrand size = 48, antiderivative size = 61 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=C x+\frac {2 (b B-2 a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} d} \]
C*x+2*(B*b-2*C*a)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/d/(a- b)^(1/2)/(a+b)^(1/2)
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {C (c+d x)}{d}+\frac {2 (-b B+2 a C) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2} d} \]
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])^2,x]
(C*(c + d*x))/d + (2*(-(b*B) + 2*a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/S qrt[-a^2 + b^2]])/(Sqrt[-a^2 + b^2]*d)
Time = 0.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2014, 3042, 3214, 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.
\(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 2014 |
\(\displaystyle \frac {\int \frac {C \cos (c+d x) b^3+(b B-a C) b^2}{a+b \cos (c+d x)}dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {b^2 (b B-2 a C) \int \frac {1}{a+b \cos (c+d x)}dx+b^2 C x}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b^2 (b B-2 a C) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+b^2 C x}{b^2}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {2 b^2 (b B-2 a C) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}+b^2 C x}{b^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {2 b^2 (b B-2 a C) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}+b^2 C x}{b^2}\) |
(b^2*C*x + (2*b^2*(b*B - 2*a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt [a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d))/b^2
3.11.10.3.1 Defintions of rubi rules used
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_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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]
Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 \left (B b -2 C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(67\) |
default | \(\frac {2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 \left (B b -2 C a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(67\) |
risch | \(C x -\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C a}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C a}{\sqrt {-a^{2}+b^{2}}\, d}\) | \(288\) |
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))^2,x ,method=_RETURNVERBOSE)
1/d*(2*C*arctan(tan(1/2*d*x+1/2*c))+2*(B*b-2*C*a)/((a-b)*(a+b))^(1/2)*arct an((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.29 (sec) , antiderivative size = 240, normalized size of antiderivative = 3.93 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [\frac {2 \, {\left (C a^{2} - C b^{2}\right )} d x + {\left (2 \, C a - B b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, {\left (a^{2} - b^{2}\right )} d}, \frac {{\left (C a^{2} - C b^{2}\right )} d x - {\left (2 \, C a - B b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{{\left (a^{2} - b^{2}\right )} d}\right ] \]
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 ))^2,x, algorithm="fricas")
[1/2*(2*(C*a^2 - C*b^2)*d*x + (2*C*a - B*b)*sqrt(-a^2 + b^2)*log((2*a*b*co s(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)))/((a^2 - b^2)*d), ((C*a^2 - C*b^2)*d*x - (2*C*a - B*b)*sqrt(a ^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))))/(( a^2 - b^2)*d)]
Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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 ))^2,x, algorithm="maxima")
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
Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (53) = 106\).
Time = 0.32 (sec) , antiderivative size = 318, normalized size of antiderivative = 5.21 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} B b^{2} {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} C {\left (a + b\right )} {\left | a - b \right |} {\left | b \right |} + \sqrt {a^{2} - b^{2}} B b {\left | a - b \right |} {\left | b \right |} - \sqrt {a^{2} - b^{2}} {\left (3 \, a b - b^{2}\right )} C {\left | a - b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} - \frac {{\left (3 \, C a b - B b^{2} - C b^{2} - C a {\left | b \right |} + B b {\left | b \right |} - C b {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{b^{2} - a {\left | b \right |}}}{d} \]
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 ))^2,x, algorithm="giac")
((sqrt(a^2 - b^2)*B*b^2*abs(a - b) - sqrt(a^2 - b^2)*C*(a + b)*abs(a - b)* abs(b) + sqrt(a^2 - b^2)*B*b*abs(a - b)*abs(b) - sqrt(a^2 - b^2)*(3*a*b - b^2)*C*abs(a - b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)* tan(1/2*d*x + 1/2*c)/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)) - (3*C*a*b - B*b^2 - C*b^2 - C*a*abs(b) + B*b*abs(b) - C*b*abs(b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2*c)/sqrt((2*a - sqrt (-4*(a + b)*(a - b) + 4*a^2))/(a - b))))/(b^2 - a*abs(b)))/d
Time = 2.85 (sec) , antiderivative size = 248, normalized size of antiderivative = 4.07 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2\,C\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B^2\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,B\,C\,a\,b+3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,C^2\,a^2+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,C^2\,b^2}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B^2\,b^2-4\,B\,C\,a\,b+3\,C^2\,a^2+C^2\,b^2\right )}\right )}{d}-\frac {2\,B\,b\,\mathrm {atanh}\left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )}{d\,\sqrt {b^2-a^2}}+\frac {4\,C\,a\,\mathrm {atanh}\left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )}{d\,\sqrt {b^2-a^2}} \]
(2*C*atan((B^2*b^2*sin(c/2 + (d*x)/2) + 3*C^2*a^2*sin(c/2 + (d*x)/2) + C^2 *b^2*sin(c/2 + (d*x)/2) - 4*B*C*a*b*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2 )*(B^2*b^2 + 3*C^2*a^2 + C^2*b^2 - 4*B*C*a*b))))/d - (2*B*b*atanh((a*sin(c /2 + (d*x)/2) - b*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2 ))))/(d*(b^2 - a^2)^(1/2)) + (4*C*a*atanh((a*sin(c/2 + (d*x)/2) - b*sin(c/ 2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))))/(d*(b^2 - a^2)^(1/2 ))