Integrand size = 18, antiderivative size = 76 \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-\frac {b B \text {arctanh}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}}-\frac {B c+A c \cos (x)-A b \sin (x)}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))} \] Output:
-b*B*arctanh((c*cos(x)-b*sin(x))/(b^2+c^2)^(1/2))/(b^2+c^2)^(3/2)-(B*c+A*c *cos(x)-A*b*sin(x))/(b^2+c^2)/(b*cos(x)+c*sin(x))
Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=\frac {2 b B \text {arctanh}\left (\frac {-c+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}}+\frac {-b B c+A \left (b^2+c^2\right ) \sin (x)}{b \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))} \] Input:
Integrate[(A + B*Cos[x])/(b*Cos[x] + c*Sin[x])^2,x]
Output:
(2*b*B*ArcTanh[(-c + b*Tan[x/2])/Sqrt[b^2 + c^2]])/(b^2 + c^2)^(3/2) + (-( b*B*c) + A*(b^2 + c^2)*Sin[x])/(b*(b^2 + c^2)*(b*Cos[x] + c*Sin[x]))
Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 3634, 3042, 3553, 219}
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.
\(\displaystyle \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2}dx\) |
\(\Big \downarrow \) 3634 |
\(\displaystyle \frac {b B \int \frac {1}{b \cos (x)+c \sin (x)}dx}{b^2+c^2}-\frac {-A b \sin (x)+A c \cos (x)+B c}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {b B \int \frac {1}{b \cos (x)+c \sin (x)}dx}{b^2+c^2}-\frac {-A b \sin (x)+A c \cos (x)+B c}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}\) |
\(\Big \downarrow \) 3553 |
\(\displaystyle -\frac {b B \int \frac {1}{b^2+c^2-(c \cos (x)-b \sin (x))^2}d(c \cos (x)-b \sin (x))}{b^2+c^2}-\frac {-A b \sin (x)+A c \cos (x)+B c}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {-A b \sin (x)+A c \cos (x)+B c}{\left (b^2+c^2\right ) (b \cos (x)+c \sin (x))}-\frac {b B \text {arctanh}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{\left (b^2+c^2\right )^{3/2}}\) |
Input:
Int[(A + B*Cos[x])/(b*Cos[x] + c*Sin[x])^2,x]
Output:
-((b*B*ArcTanh[(c*Cos[x] - b*Sin[x])/Sqrt[b^2 + c^2]])/(b^2 + c^2)^(3/2)) - (B*c + A*c*Cos[x] - A*b*Sin[x])/((b^2 + c^2)*(b*Cos[x] + c*Sin[x]))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x _Symbol] :> Simp[-d^(-1) Subst[Int[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/((a_.) + cos[(d_.) + (e_.)*(x_) ]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B + c*A*Co s[d + e*x] + (a*B - b*A)*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x])), x] + Simp[(a*A - b*B)/(a^2 - 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}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B, 0]
Time = 0.38 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.43
method | result | size |
default | \(\frac {-\frac {2 \left (A \,b^{2}+A \,c^{2}-B \,c^{2}\right ) \tan \left (\frac {x}{2}\right )}{b \left (b^{2}+c^{2}\right )}+\frac {2 B c}{b^{2}+c^{2}}}{b \tan \left (\frac {x}{2}\right )^{2}-2 c \tan \left (\frac {x}{2}\right )-b}+\frac {2 b B \,\operatorname {arctanh}\left (\frac {2 b \tan \left (\frac {x}{2}\right )-2 c}{2 \sqrt {b^{2}+c^{2}}}\right )}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\) | \(109\) |
risch | \(-\frac {2 i \left (c A -i A b +B c \,{\mathrm e}^{i x}\right )}{\left (-i b +c \right ) \left (i b +c \right ) \left (c \,{\mathrm e}^{2 i x}+i b \,{\mathrm e}^{2 i x}-c +i b \right )}+\frac {b B \ln \left ({\mathrm e}^{i x}+\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {b B \ln \left ({\mathrm e}^{i x}-\frac {i b^{3}+i b \,c^{2}-b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}}\) | \(174\) |
Input:
int((A+B*cos(x))/(b*cos(x)+c*sin(x))^2,x,method=_RETURNVERBOSE)
Output:
2*(-(A*b^2+A*c^2-B*c^2)/b/(b^2+c^2)*tan(1/2*x)+B*c/(b^2+c^2))/(b*tan(1/2*x )^2-2*c*tan(1/2*x)-b)+2*b*B/(b^2+c^2)^(3/2)*arctanh(1/2*(2*b*tan(1/2*x)-2* c)/(b^2+c^2)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (72) = 144\).
Time = 0.09 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.64 \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-\frac {2 \, B b^{2} c + 2 \, B c^{3} - {\left (B b^{2} \cos \left (x\right ) + B b c \sin \left (x\right )\right )} \sqrt {b^{2} + c^{2}} \log \left (-\frac {2 \, b c \cos \left (x\right ) \sin \left (x\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} - 2 \, b^{2} - c^{2} + 2 \, \sqrt {b^{2} + c^{2}} {\left (c \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, b c \cos \left (x\right ) \sin \left (x\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (x\right )^{2} + c^{2}}\right ) + 2 \, {\left (A b^{2} c + A c^{3}\right )} \cos \left (x\right ) - 2 \, {\left (A b^{3} + A b c^{2}\right )} \sin \left (x\right )}{2 \, {\left ({\left (b^{5} + 2 \, b^{3} c^{2} + b c^{4}\right )} \cos \left (x\right ) + {\left (b^{4} c + 2 \, b^{2} c^{3} + c^{5}\right )} \sin \left (x\right )\right )}} \] Input:
integrate((A+B*cos(x))/(b*cos(x)+c*sin(x))^2,x, algorithm="fricas")
Output:
-1/2*(2*B*b^2*c + 2*B*c^3 - (B*b^2*cos(x) + B*b*c*sin(x))*sqrt(b^2 + c^2)* log(-(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^2 - 2*b^2 - c^2 + 2*sqrt(b^ 2 + c^2)*(c*cos(x) - b*sin(x)))/(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^ 2 + c^2)) + 2*(A*b^2*c + A*c^3)*cos(x) - 2*(A*b^3 + A*b*c^2)*sin(x))/((b^5 + 2*b^3*c^2 + b*c^4)*cos(x) + (b^4*c + 2*b^2*c^3 + c^5)*sin(x))
Timed out. \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(x))/(b*cos(x)+c*sin(x))**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (72) = 144\).
Time = 0.12 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.05 \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-B {\left (\frac {b \log \left (\frac {c - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {b^{2} + c^{2}}}{c - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {b^{2} + c^{2}}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b c + \frac {c^{2} \sin \left (x\right )}{\cos \left (x\right ) + 1}\right )}}{b^{4} + b^{2} c^{2} + \frac {2 \, {\left (b^{3} c + b c^{3}\right )} \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {{\left (b^{4} + b^{2} c^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}}\right )} - \frac {A}{c^{2} \tan \left (x\right ) + b c} \] Input:
integrate((A+B*cos(x))/(b*cos(x)+c*sin(x))^2,x, algorithm="maxima")
Output:
-B*(b*log((c - b*sin(x)/(cos(x) + 1) + sqrt(b^2 + c^2))/(c - b*sin(x)/(cos (x) + 1) - sqrt(b^2 + c^2)))/(b^2 + c^2)^(3/2) + 2*(b*c + c^2*sin(x)/(cos( x) + 1))/(b^4 + b^2*c^2 + 2*(b^3*c + b*c^3)*sin(x)/(cos(x) + 1) - (b^4 + b ^2*c^2)*sin(x)^2/(cos(x) + 1)^2)) - A/(c^2*tan(x) + b*c)
Time = 0.14 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.74 \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-\frac {B b \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, c - 2 \, \sqrt {b^{2} + c^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, c + 2 \, \sqrt {b^{2} + c^{2}} \right |}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (A b^{2} \tan \left (\frac {1}{2} \, x\right ) + A c^{2} \tan \left (\frac {1}{2} \, x\right ) - B c^{2} \tan \left (\frac {1}{2} \, x\right ) - B b c\right )}}{{\left (b^{3} + b c^{2}\right )} {\left (b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac {1}{2} \, x\right ) - b\right )}} \] Input:
integrate((A+B*cos(x))/(b*cos(x)+c*sin(x))^2,x, algorithm="giac")
Output:
-B*b*log(abs(2*b*tan(1/2*x) - 2*c - 2*sqrt(b^2 + c^2))/abs(2*b*tan(1/2*x) - 2*c + 2*sqrt(b^2 + c^2)))/(b^2 + c^2)^(3/2) - 2*(A*b^2*tan(1/2*x) + A*c^ 2*tan(1/2*x) - B*c^2*tan(1/2*x) - B*b*c)/((b^3 + b*c^2)*(b*tan(1/2*x)^2 - 2*c*tan(1/2*x) - b))
Time = 15.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.66 \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=-\frac {\frac {2\,B\,c}{b^2+c^2}-\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^2+A\,c^2-B\,c^2\right )}{b\,\left (b^2+c^2\right )}}{-b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {x}{2}\right )+b}+\frac {B\,b\,\mathrm {atan}\left (\frac {b^2\,c\,1{}\mathrm {i}+c^3\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b^2+c^2\right )\,1{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}}\right )\,2{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}} \] Input:
int((A + B*cos(x))/(b*cos(x) + c*sin(x))^2,x)
Output:
(B*b*atan((b^2*c*1i + c^3*1i - b*tan(x/2)*(b^2 + c^2)*1i)/(b^2 + c^2)^(3/2 ))*2i)/(b^2 + c^2)^(3/2) - ((2*B*c)/(b^2 + c^2) - (2*tan(x/2)*(A*b^2 + A*c ^2 - B*c^2))/(b*(b^2 + c^2)))/(b + 2*c*tan(x/2) - b*tan(x/2)^2)
Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.36 \[ \int \frac {A+B \cos (x)}{(b \cos (x)+c \sin (x))^2} \, dx=\frac {-2 \sqrt {b^{2}+c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -c i}{\sqrt {b^{2}+c^{2}}}\right ) \cos \left (x \right ) b^{3} c i -2 \sqrt {b^{2}+c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -c i}{\sqrt {b^{2}+c^{2}}}\right ) \sin \left (x \right ) b^{2} c^{2} i -\cos \left (x \right ) a \,b^{4}-2 \cos \left (x \right ) a \,b^{2} c^{2}-\cos \left (x \right ) a \,c^{4}-b^{3} c^{2}-b \,c^{4}}{c \left (\cos \left (x \right ) b^{5}+2 \cos \left (x \right ) b^{3} c^{2}+\cos \left (x \right ) b \,c^{4}+\sin \left (x \right ) b^{4} c +2 \sin \left (x \right ) b^{2} c^{3}+\sin \left (x \right ) c^{5}\right )} \] Input:
int((A+B*cos(x))/(b*cos(x)+c*sin(x))^2,x)
Output:
( - 2*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c**2))*cos(x )*b**3*c*i - 2*sqrt(b**2 + c**2)*atan((tan(x/2)*b*i - c*i)/sqrt(b**2 + c** 2))*sin(x)*b**2*c**2*i - cos(x)*a*b**4 - 2*cos(x)*a*b**2*c**2 - cos(x)*a*c **4 - b**3*c**2 - b*c**4)/(c*(cos(x)*b**5 + 2*cos(x)*b**3*c**2 + cos(x)*b* c**4 + sin(x)*b**4*c + 2*sin(x)*b**2*c**3 + sin(x)*c**5))