Integrand size = 23, antiderivative size = 131 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx=\frac {(b B+c C) x}{b^2+c^2}+\frac {2 \left (A \left (b^2+c^2\right )-a (b B+c C)\right ) \arctan \left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\sqrt {a^2-b^2-c^2} \left (b^2+c^2\right )}+\frac {(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2} \] Output:
(B*b+C*c)*x/(b^2+c^2)+2*(A*(b^2+c^2)-a*(B*b+C*c))*arctan((c+(a-b)*tan(1/2* x))/(a^2-b^2-c^2)^(1/2))/(a^2-b^2-c^2)^(1/2)/(b^2+c^2)+(B*c-C*b)*ln(a+b*co s(x)+c*sin(x))/(b^2+c^2)
Time = 0.46 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.84 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx=\frac {(b B+c C) x+\frac {2 \left (-A \left (b^2+c^2\right )+a (b B+c C)\right ) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\sqrt {-a^2+b^2+c^2}}+(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2} \] Input:
Integrate[(A + B*Cos[x] + C*Sin[x])/(a + b*Cos[x] + c*Sin[x]),x]
Output:
((b*B + c*C)*x + (2*(-(A*(b^2 + c^2)) + a*(b*B + c*C))*ArcTanh[(c + (a - b )*Tan[x/2])/Sqrt[-a^2 + b^2 + c^2]])/Sqrt[-a^2 + b^2 + c^2] + (B*c - b*C)* Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)
Time = 0.40 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3615, 3042, 3603, 1083, 217}
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.
\(\displaystyle \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)}dx\) |
\(\Big \downarrow \) 3615 |
\(\displaystyle \left (A-\frac {a (b B+c C)}{b^2+c^2}\right ) \int \frac {1}{a+b \cos (x)+c \sin (x)}dx+\frac {(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac {x (b B+c C)}{b^2+c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (A-\frac {a (b B+c C)}{b^2+c^2}\right ) \int \frac {1}{a+b \cos (x)+c \sin (x)}dx+\frac {(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac {x (b B+c C)}{b^2+c^2}\) |
\(\Big \downarrow \) 3603 |
\(\displaystyle 2 \left (A-\frac {a (b B+c C)}{b^2+c^2}\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {x}{2}\right )+2 c \tan \left (\frac {x}{2}\right )+a+b}d\tan \left (\frac {x}{2}\right )+\frac {(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac {x (b B+c C)}{b^2+c^2}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -4 \left (A-\frac {a (b B+c C)}{b^2+c^2}\right ) \int \frac {1}{-\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2-c^2\right )}d\left (2 c+2 (a-b) \tan \left (\frac {x}{2}\right )\right )+\frac {(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac {x (b B+c C)}{b^2+c^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 \left (A-\frac {a (b B+c C)}{b^2+c^2}\right ) \arctan \left (\frac {2 (a-b) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^2-b^2-c^2}}\right )}{\sqrt {a^2-b^2-c^2}}+\frac {(B c-b C) \log (a+b \cos (x)+c \sin (x))}{b^2+c^2}+\frac {x (b B+c C)}{b^2+c^2}\) |
Input:
Int[(A + B*Cos[x] + C*Sin[x])/(a + b*Cos[x] + c*Sin[x]),x]
Output:
((b*B + c*C)*x)/(b^2 + c^2) + (2*(A - (a*(b*B + c*C))/(b^2 + c^2))*ArcTan[ (2*c + 2*(a - b)*Tan[x/2])/(2*Sqrt[a^2 - b^2 - c^2])])/Sqrt[a^2 - b^2 - c^ 2] + ((B*c - b*C)*Log[a + b*Cos[x] + c*Sin[x]])/(b^2 + c^2)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f /e) Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) /2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x _Symbol] :> Simp[(b*B + c*C)*(x/(b^2 + c^2)), x] + (Simp[(c*B - b*C)*(Log[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/(e*(b^2 + c^2))), x] + Simp[(A*(b^2 + c ^2) - a*(b*B + c*C))/(b^2 + c^2) Int[1/(a + b*Cos[d + e*x] + c*Sin[d + e* x]), x], x]) /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2 + c^2, 0] && NeQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]
Time = 0.42 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.73
method | result | size |
default | \(\frac {\left (-B c +b C \right ) \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )+2 \left (B b +C c \right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{b^{2}+c^{2}}+\frac {\frac {2 \left (a B c -b B c -a b C +b^{2} C \right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2} a -b \tan \left (\frac {x}{2}\right )^{2}+2 c \tan \left (\frac {x}{2}\right )+a +b \right )}{2 a -2 b}+\frac {2 \left (A \,b^{2}+A \,c^{2}-a b B +B \,c^{2}-a c C -C b c -\frac {\left (a B c -b B c -a b C +b^{2} C \right ) c}{a -b}\right ) \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {x}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\sqrt {a^{2}-b^{2}-c^{2}}}}{b^{2}+c^{2}}\) | \(227\) |
risch | \(\text {Expression too large to display}\) | \(10609\) |
Input:
int((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x,method=_RETURNVERBOSE)
Output:
2/(b^2+c^2)*(1/2*(-B*c+C*b)*ln(1+tan(1/2*x)^2)+(B*b+C*c)*arctan(tan(1/2*x) ))+2/(b^2+c^2)*(1/2*(B*a*c-B*b*c-C*a*b+C*b^2)/(a-b)*ln(tan(1/2*x)^2*a-b*ta n(1/2*x)^2+2*c*tan(1/2*x)+a+b)+(A*b^2+A*c^2-a*b*B+B*c^2-a*c*C-C*b*c-(B*a*c -B*b*c-C*a*b+C*b^2)*c/(a-b))/(a^2-b^2-c^2)^(1/2)*arctan(1/2*(2*(a-b)*tan(1 /2*x)+2*c)/(a^2-b^2-c^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (126) = 252\).
Time = 0.15 (sec) , antiderivative size = 713, normalized size of antiderivative = 5.44 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx =\text {Too large to display} \] Input:
integrate((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x, algorithm="fricas ")
Output:
[1/2*((B*a*b - A*b^2 + C*a*c - A*c^2)*sqrt(-a^2 + b^2 + c^2)*log(-(a^2*b^2 - 2*b^4 - c^4 - (a^2 + 3*b^2)*c^2 - (2*a^2*b^2 - b^4 - 2*a^2*c^2 + c^4)*c os(x)^2 - 2*(a*b^3 + a*b*c^2)*cos(x) - 2*(a*b^2*c + a*c^3 - (b*c^3 - (2*a^ 2*b - b^3)*c)*cos(x))*sin(x) + 2*(2*a*b*c*cos(x)^2 - a*b*c + (b^2*c + c^3) *cos(x) - (b^3 + b*c^2 + (a*b^2 - a*c^2)*cos(x))*sin(x))*sqrt(-a^2 + b^2 + c^2))/(2*a*b*cos(x) + (b^2 - c^2)*cos(x)^2 + a^2 + c^2 + 2*(b*c*cos(x) + a*c)*sin(x))) + 2*(B*a^2*b - B*b^3 - B*b*c^2 - C*c^3 + (C*a^2 - C*b^2)*c)* x - (C*a^2*b - C*b^3 - C*b*c^2 + B*c^3 - (B*a^2 - B*b^2)*c)*log(2*a*b*cos( x) + (b^2 - c^2)*cos(x)^2 + a^2 + c^2 + 2*(b*c*cos(x) + a*c)*sin(x)))/(a^2 *b^2 - b^4 - c^4 + (a^2 - 2*b^2)*c^2), -1/2*(2*(B*a*b - A*b^2 + C*a*c - A* c^2)*sqrt(a^2 - b^2 - c^2)*arctan(-(a*b*cos(x) + a*c*sin(x) + b^2 + c^2)*s qrt(a^2 - b^2 - c^2)/((c^3 - (a^2 - b^2)*c)*cos(x) + (a^2*b - b^3 - b*c^2) *sin(x))) - 2*(B*a^2*b - B*b^3 - B*b*c^2 - C*c^3 + (C*a^2 - C*b^2)*c)*x + (C*a^2*b - C*b^3 - C*b*c^2 + B*c^3 - (B*a^2 - B*b^2)*c)*log(2*a*b*cos(x) + (b^2 - c^2)*cos(x)^2 + a^2 + c^2 + 2*(b*c*cos(x) + a*c)*sin(x)))/(a^2*b^2 - b^4 - c^4 + (a^2 - 2*b^2)*c^2)]
Timed out. \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x)
Output:
Timed out
Exception generated. \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),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(c^2+b^2-a^2>0)', see `assume?` f or more de
Time = 0.31 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.52 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx=\frac {{\left (B b + C c\right )} x}{b^{2} + c^{2}} - \frac {{\left (C b - B c\right )} \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac {1}{2} \, x\right ) - a - b\right )}{b^{2} + c^{2}} + \frac {{\left (C b - B c\right )} \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}{b^{2} + c^{2}} + \frac {2 \, {\left (B a b - A b^{2} + C a c - A c^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2} - c^{2}} {\left (b^{2} + c^{2}\right )}} \] Input:
integrate((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x, algorithm="giac")
Output:
(B*b + C*c)*x/(b^2 + c^2) - (C*b - B*c)*log(-a*tan(1/2*x)^2 + b*tan(1/2*x) ^2 - 2*c*tan(1/2*x) - a - b)/(b^2 + c^2) + (C*b - B*c)*log(tan(1/2*x)^2 + 1)/(b^2 + c^2) + 2*(B*a*b - A*b^2 + C*a*c - A*c^2)*(pi*floor(1/2*x/pi + 1/ 2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x) + c)/sqrt(a^2 - b^2 - c^2)))/(sqrt(a^2 - b^2 - c^2)*(b^2 + c^2))
Time = 81.56 (sec) , antiderivative size = 2711, normalized size of antiderivative = 20.69 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx=\text {Too large to display} \] Input:
int((A + B*cos(x) + C*sin(x))/(a + b*cos(x) + c*sin(x)),x)
Output:
(log(tan(x/2) - 1i)*(B + C*1i))/(b*1i - c) - (log(tan(x/2) + 1i)*(B - C*1i ))/(b*1i + c) + (log(32*B^3*a^2 - 32*A*B^2*a^2 + 32*A*B^2*b^2 + 32*A*C^2*a ^2 - 32*A^2*B*b^2 + 32*A*C^2*b^2 + 32*B*C^2*a^2 + 32*B*C^2*b^2 + 32*tan(x/ 2)*(a - b)*(2*C^3*a + B^3*c - 2*C^3*b - 2*A*B^2*c + A^2*B*c + A^2*C*b + 2* B^2*C*a - 2*A*C^2*c - B^2*C*b + 2*B*C^2*c - 2*A*B*C*a) - 32*B^3*a*b + 32*A ^2*B*a*b - 64*A*C^2*a*b - 64*B*C^2*a*b - 32*A^2*C*a*c + 32*A^2*C*b*c + 32* B^2*C*a*c - 32*B^2*C*b*c - ((B*c^3 - C*b^3 - A*b^2*(b^2 - a^2 + c^2)^(1/2) - B*a^2*c + C*a^2*b - A*c^2*(b^2 - a^2 + c^2)^(1/2) + B*b^2*c - C*b*c^2 + B*a*b*(b^2 - a^2 + c^2)^(1/2) + C*a*c*(b^2 - a^2 + c^2)^(1/2))*(64*A^2*b^ 2*c - 32*B^2*a^2*c + 32*B^2*b^2*c + 32*C^2*a^2*c + 32*C^2*b^2*c + 32*tan(x /2)*(a - b)*(A^2*b^2 + 2*B^2*a^2 - A^2*c^2 + B^2*b^2 - 2*C^2*a^2 - 3*B^2*c ^2 + 2*C^2*c^2 + 4*A*B*c^2 - 2*B^2*a*b + 2*C^2*a*b - 2*A*B*a*b + 2*A*C*a*c - 4*A*C*b*c - 4*B*C*a*c + 6*B*C*b*c) + 64*A*C*b^3 - 128*B*C*a^3 - 64*B*C* b^3 + 64*A*B*a^2*c - 128*A*C*a*b^2 + 64*A*C*a^2*b - 64*A*B*b^2*c + 192*B*C *a^2*b + 64*B*C*a*c^2 - 64*B*C*b*c^2 - 64*A^2*a*b*c - 64*C^2*a*b*c - ((B*c ^3 - C*b^3 - A*b^2*(b^2 - a^2 + c^2)^(1/2) - B*a^2*c + C*a^2*b - A*c^2*(b^ 2 - a^2 + c^2)^(1/2) + B*b^2*c - C*b*c^2 + B*a*b*(b^2 - a^2 + c^2)^(1/2) + C*a*c*(b^2 - a^2 + c^2)^(1/2))*(32*A*b^4 + 32*B*b^4 - 32*tan(x/2)*(a - b) *(2*A*c^3 + B*c^3 - 2*C*b^3 + 2*A*b^2*c + 2*C*a*b^2 + 4*B*b^2*c - 2*C*a*c^ 2 + C*b*c^2 - 2*A*a*b*c - 4*B*a*b*c) + 32*A*a^2*b^2 - 32*A*a^2*c^2 + 32...
Time = 0.16 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.01 \[ \int \frac {A+B \cos (x)+C \sin (x)}{a+b \cos (x)+c \sin (x)} \, dx=x \] Input:
int((A+B*cos(x)+C*sin(x))/(a+b*cos(x)+c*sin(x)),x)
Output:
x