Integrand size = 21, antiderivative size = 84 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {x}{b^2}+\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))} \]
-x/b^2-cos(d*x+c)/b/d/(a+b*sin(d*x+c))+2*a*arctan((b+a*tan(1/2*d*x+1/2*c)) /(a^2-b^2)^(1/2))/b^2/d/(a^2-b^2)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(414\) vs. \(2(84)=168\).
Time = 2.02 (sec) , antiderivative size = 414, normalized size of antiderivative = 4.93 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\cos (c+d x) \left (-2 a (a-b) \text {arctanh}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}}{\sqrt {a+b} \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}}}\right ) \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x))+\sqrt {a+b} \left (2 a \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {\frac {b (1+\sin (c+d x))}{-a+b}}}{\sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}}}\right ) \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x))+(-a+b) \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \left (\sqrt {a-b} (a+b) \sqrt {1-\sin (c+d x)} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}+2 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}}{\sqrt {2} \sqrt {b}}\right ) (a+b \sin (c+d x))\right )\right )\right )}{(a-b)^{3/2} b (a+b)^{3/2} d \sqrt {1-\sin (c+d x)} \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}} (a+b \sin (c+d x))} \]
(Cos[c + d*x]*(-2*a*(a - b)*ArcTanh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d* x]))/(a - b))])/(Sqrt[a + b]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))])]*Sq rt[1 - Sin[c + d*x]]*(a + b*Sin[c + d*x]) + Sqrt[a + b]*(2*a*Sqrt[a - b]*A rcTanh[Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)]/Sqrt[-((b*(-1 + Sin[c + d*x]) )/(a + b))]]*Sqrt[1 - Sin[c + d*x]]*(a + b*Sin[c + d*x]) + (-a + b)*Sqrt[- ((b*(-1 + Sin[c + d*x]))/(a + b))]*(Sqrt[a - b]*(a + b)*Sqrt[1 - Sin[c + d *x]]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))] + 2*Sqrt[b]*ArcSinh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(Sqrt[2]*Sqrt[b])]*(a + b*Sin [c + d*x])))))/((a - b)^(3/2)*b*(a + b)^(3/2)*d*Sqrt[1 - Sin[c + d*x]]*Sqr t[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b ))]*(a + b*Sin[c + d*x]))
Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3172, 3042, 3214, 3042, 3139, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2}{(a+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3172 |
\(\displaystyle -\frac {\int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle -\frac {\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -\frac {\frac {x}{b}-\frac {2 a \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -\frac {\frac {4 a \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x}{b}}{b}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\frac {x}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}}{b}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\) |
-((x/b - (2*a*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(b *Sqrt[a^2 - b^2]*d))/b) - Cos[c + d*x]/(b*d*(a + b*Sin[c + d*x]))
3.5.46.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I ntegersQ[2*m, 2*p]
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 = 0.75 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44
method | result | size |
derivativedivides | \(\frac {\frac {\frac {2 \left (-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b \right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\) | \(121\) |
default | \(\frac {\frac {\frac {2 \left (-\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-b \right )}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{b^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\) | \(121\) |
risch | \(-\frac {x}{b^{2}}-\frac {2 \left (i b +a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{b^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\) | \(201\) |
1/d*(2/b^2*((-b^2/a*tan(1/2*d*x+1/2*c)-b)/(a*tan(1/2*d*x+1/2*c)^2+2*b*tan( 1/2*d*x+1/2*c)+a)+a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b )/(a^2-b^2)^(1/2)))-2/b^2*arctan(tan(1/2*d*x+1/2*c)))
Time = 0.30 (sec) , antiderivative size = 388, normalized size of antiderivative = 4.62 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [-\frac {2 \, {\left (a^{2} b - b^{3}\right )} d x \sin \left (d x + c\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} d x + {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{3} - b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d\right )}}, -\frac {{\left (a^{2} b - b^{3}\right )} d x \sin \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} d x + {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{{\left (a^{2} b^{3} - b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d}\right ] \]
[-1/2*(2*(a^2*b - b^3)*d*x*sin(d*x + c) + 2*(a^3 - a*b^2)*d*x + (a*b*sin(d *x + c) + a^2)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b* sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c) )*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 2*(a^2*b - b^3)*cos(d*x + c))/((a^2*b^3 - b^5)*d*sin(d*x + c) + (a^3*b^ 2 - a*b^4)*d), -((a^2*b - b^3)*d*x*sin(d*x + c) + (a^3 - a*b^2)*d*x + (a*b *sin(d*x + c) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^ 2 - b^2)*cos(d*x + c))) + (a^2*b - b^3)*cos(d*x + c))/((a^2*b^3 - b^5)*d*s in(d*x + c) + (a^3*b^2 - a*b^4)*d)]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{\sqrt {a^{2} - b^{2}} b^{2}} - \frac {d x + c}{b^{2}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a b}}{d} \]
(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2* c) + b)/sqrt(a^2 - b^2)))*a/(sqrt(a^2 - b^2)*b^2) - (d*x + c)/b^2 - 2*(b*t an(1/2*d*x + 1/2*c) + a)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/ 2*c) + a)*a*b))/d
Time = 5.10 (sec) , antiderivative size = 329, normalized size of antiderivative = 3.92 \[ \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {b^2\,\sin \left (c+d\,x\right )+\frac {\left (2\,a^3\,\mathrm {atan}\left (\frac {\left (-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )-a^2\,\sqrt {b^2-a^2}\,\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}}-\frac {b\,\left (a\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+a\,\cos \left (c+d\,x\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}-2\,a^2\,\mathrm {atan}\left (\frac {\left (-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )\,\sin \left (c+d\,x\right )+a\,\sin \left (c+d\,x\right )\,\sqrt {b^2-a^2}\,\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}}}{a\,b^2\,d\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \]
-(b^2*sin(c + d*x) + ((2*a^3*atan(((2*b^2*sin(c/2 + (d*x)/2) - a^2*sin(c/2 + (d*x)/2) + a*b*cos(c/2 + (d*x)/2))*1i)/((b^2 - a^2)^(1/2)*(a*cos(c/2 + (d*x)/2) + 2*b*sin(c/2 + (d*x)/2)))) - a^2*(b^2 - a^2)^(1/2)*(c + d*x)*1i) *1i)/(b^2 - a^2)^(1/2) - (b*(a*(b^2 - a^2)^(1/2)*1i + a*cos(c + d*x)*(b^2 - a^2)^(1/2)*1i - 2*a^2*atan(((2*b^2*sin(c/2 + (d*x)/2) - a^2*sin(c/2 + (d *x)/2) + a*b*cos(c/2 + (d*x)/2))*1i)/((b^2 - a^2)^(1/2)*(a*cos(c/2 + (d*x) /2) + 2*b*sin(c/2 + (d*x)/2))))*sin(c + d*x) + a*sin(c + d*x)*(b^2 - a^2)^ (1/2)*(c + d*x)*1i)*1i)/(b^2 - a^2)^(1/2))/(a*b^2*d*(a + b*sin(c + d*x)))