Integrand size = 26, antiderivative size = 83 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {a \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {b}{\left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))} \]
-a*arctanh((b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)/d- b/(a^2+b^2)/d/(a*cos(d*x+c)+b*sin(d*x+c))
Time = 0.52 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=\frac {\frac {2 a \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {b}{\left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}}{d} \]
((2*a*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2 ) - b/((a^2 + b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x])))/d
Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, 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.
\(\displaystyle \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3634 |
\(\displaystyle \frac {a \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)}dx}{a^2+b^2}-\frac {b}{d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \int \frac {1}{a \cos (c+d x)+b \sin (c+d x)}dx}{a^2+b^2}-\frac {b}{d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3553 |
\(\displaystyle -\frac {a \int \frac {1}{a^2+b^2-(b \cos (c+d x)-a \sin (c+d x))^2}d(b \cos (c+d x)-a \sin (c+d x))}{d \left (a^2+b^2\right )}-\frac {b}{d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {a \text {arctanh}\left (\frac {b \cos (c+d x)-a \sin (c+d x)}{\sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {b}{d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}\) |
-((a*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/((a^2 + b ^2)^(3/2)*d)) - b/((a^2 + b^2)*d*(a*Cos[c + d*x] + b*Sin[c + d*x]))
3.2.25.3.1 Defintions of rubi rules used
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 = 0.53 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.42
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}+b^{2}\right ) a}-\frac {b}{a^{2}+b^{2}}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}+\frac {2 a \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\) | \(118\) |
default | \(\frac {-\frac {2 \left (-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}+b^{2}\right ) a}-\frac {b}{a^{2}+b^{2}}\right )}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a}+\frac {2 a \,\operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\) | \(118\) |
risch | \(-\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}}{\left (-i a +b \right ) d \left (i a +b \right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{3}+i a \,b^{2}-a^{2} b -b^{3}}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{3}+i a \,b^{2}-a^{2} b -b^{3}}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}\) | \(190\) |
1/d*(-2*(-b^2/(a^2+b^2)/a*tan(1/2*d*x+1/2*c)-b/(a^2+b^2))/(tan(1/2*d*x+1/2 *c)^2*a-2*b*tan(1/2*d*x+1/2*c)-a)+2*a/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tan (1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (79) = 158\).
Time = 0.26 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.59 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {2 \, a^{2} b + 2 \, b^{3} - {\left (a^{2} \cos \left (d x + c\right ) + a b \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right )\right )}} \]
-1/2*(2*a^2*b + 2*b^3 - (a^2*cos(d*x + c) + a*b*sin(d*x + c))*sqrt(a^2 + b ^2)*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2 *a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*c os(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)))/((a^5 + 2*a ^3*b^2 + a*b^4)*d*cos(d*x + c) + (a^4*b + 2*a^2*b^3 + b^5)*d*sin(d*x + c))
Exception generated. \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=\text {Exception raised: AttributeError} \]
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (79) = 158\).
Time = 0.31 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.19 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {\frac {a \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a b + \frac {b^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}}{a^{4} + a^{2} b^{2} + \frac {2 \, {\left (a^{3} b + a b^{3}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (a^{4} + a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}}{d} \]
-(a*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*s in(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) + 2*( a*b + b^2*sin(d*x + c)/(cos(d*x + c) + 1))/(a^4 + a^2*b^2 + 2*(a^3*b + a*b ^3)*sin(d*x + c)/(cos(d*x + c) + 1) - (a^4 + a^2*b^2)*sin(d*x + c)^2/(cos( d*x + c) + 1)^2))/d
Time = 0.33 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.66 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {\frac {a \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b\right )}}{{\left (a^{3} + a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}}{d} \]
-(a*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*ta n(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) - 2*(b^2* tan(1/2*d*x + 1/2*c) + a*b)/((a^3 + a*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b *tan(1/2*d*x + 1/2*c) - a)))/d
Time = 22.66 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.64 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx=-\frac {\frac {2\,b}{a^2+b^2}+\frac {2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,\left (a^2+b^2\right )}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {a\,\mathrm {atan}\left (\frac {a^2\,b\,1{}\mathrm {i}+b^3\,1{}\mathrm {i}-a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+b^2\right )\,1{}\mathrm {i}}{{\left (a^2+b^2\right )}^{3/2}}\right )\,2{}\mathrm {i}}{d\,{\left (a^2+b^2\right )}^{3/2}} \]