Integrand size = 21, antiderivative size = 70 \[ \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a x}{b^2}-\frac {2 \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {\cos (c+d x)}{b d} \]
a*x/b^2+cos(d*x+c)/b/d-2*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))* (a^2-b^2)^(1/2)/b^2/d
Leaf count is larger than twice the leaf count of optimal. \(361\) vs. \(2(70)=140\).
Time = 0.82 (sec) , antiderivative size = 361, normalized size of antiderivative = 5.16 \[ \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\cos (c+d x) \left (2 (a-b) \text {arctanh}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}}{\sqrt {a+b} \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {a+b} \left (-2 \sqrt {a-b} \text {arctanh}\left (\frac {\sqrt {\frac {b (1+\sin (c+d x))}{-a+b}}}{\sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}}}\right ) \sqrt {1-\sin (c+d x)}+\sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \left (2 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}}{\sqrt {2} \sqrt {b}}\right )+\sqrt {a-b} \sqrt {1-\sin (c+d x)} \sqrt {\frac {b (1+\sin (c+d x))}{-a+b}}\right )\right )\right )}{\sqrt {a-b} b \sqrt {a+b} d \sqrt {1-\sin (c+d x)} \sqrt {-\frac {b (-1+\sin (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sin (c+d x))}{a-b}}} \]
(Cos[c + d*x]*(2*(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))])]*Sqrt[ 1 - Sin[c + d*x]] + Sqrt[a + b]*(-2*Sqrt[a - b]*ArcTanh[Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)]/Sqrt[-((b*(-1 + Sin[c + d*x]))/(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])] + Sqrt[a - b]*Sqrt[1 - Sin[c + d*x]]*Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)])) ))/(Sqrt[a - b]*b*Sqrt[a + b]*d*Sqrt[1 - Sin[c + d*x]]*Sqrt[-((b*(-1 + Sin [c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])
Time = 0.41 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3174, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3174 |
\(\displaystyle \frac {\int \frac {b+a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}+\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {b+a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}+\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}+\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}}{b}+\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {2 \left (a^2-b^2\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}+\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\frac {4 \left (a^2-b^2\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {a x}{b}}{b}+\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {a x}{b}-\frac {2 \sqrt {a^2-b^2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d}}{b}+\frac {\cos (c+d x)}{b d}\) |
((a*x)/b - (2*Sqrt[a^2 - b^2]*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[ a^2 - b^2])])/(b*d))/b + Cos[c + d*x]/(b*d)
3.5.33.3.1 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 + p))), x] + Simp[g^2*((p - 1)/(b*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; F reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[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 = 0.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.37
method | result | size |
derivativedivides | \(\frac {\frac {\frac {2 b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {2 \left (-a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{2} \sqrt {a^{2}-b^{2}}}}{d}\) | \(96\) |
default | \(\frac {\frac {\frac {2 b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}+\frac {2 \left (-a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{2} \sqrt {a^{2}-b^{2}}}}{d}\) | \(96\) |
risch | \(\frac {a x}{b^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}\) | \(143\) |
1/d*(2/b^2*(b/(1+tan(1/2*d*x+1/2*c)^2)+a*arctan(tan(1/2*d*x+1/2*c)))+2*(-a ^2+b^2)/b^2/(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)))
Time = 0.29 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.06 \[ \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {2 \, a d x + 2 \, b \cos \left (d x + c\right ) + \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \, b^{2} d}, \frac {a d x + b \cos \left (d x + c\right ) + \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{b^{2} d}\right ] \]
[1/2*(2*a*d*x + 2*b*cos(d*x + c) + 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)))/(b^2*d), (a*d*x + b*cos(d*x + c) + sqrt(a^2 - b^2)*a rctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))))/(b^2*d)]
Leaf count of result is larger than twice the leaf count of optimal. 1923 vs. \(2 (58) = 116\).
Time = 119.15 (sec) , antiderivative size = 1923, normalized size of antiderivative = 27.47 \[ \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
Piecewise((zoo*x*cos(c)**2/sin(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((log( tan(c/2 + d*x/2))*tan(c/2 + d*x/2)**2/(d*tan(c/2 + d*x/2)**2 + d) + log(ta n(c/2 + d*x/2))/(d*tan(c/2 + d*x/2)**2 + d) + 2/(d*tan(c/2 + d*x/2)**2 + d ))/b, Eq(a, 0)), (b*d*x*tan(c/2 + d*x/2)**3/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x /2)) + b*d*x*tan(c/2 + d*x/2)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*s qrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) + 2*b/(b* *2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b *d*sqrt(b**2)*tan(c/2 + d*x/2)) - d*x*sqrt(b**2)*tan(c/2 + d*x/2)**2/(b**2 *d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d *sqrt(b**2)*tan(c/2 + d*x/2)) - d*x*sqrt(b**2)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) - 2*sqrt(b**2)*tan(c/2 + d*x/2)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2 *d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) , Eq(a, -sqrt(b**2))), (b*d*x*tan(c/2 + d*x/2)**3/(b**2*d*tan(c/2 + d*x/2) **2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2)) + b*d*x*tan(c/2 + d*x/2)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2)) + 2 *b/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)* *3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2)) + d*x*sqrt(b**2)*tan(c/2 + d*x/2)...
Exception generated. \[ \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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 = 0.53 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} a}{b^{2}} - \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} \sqrt {a^{2} - b^{2}}}{b^{2}} + \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b}}{d} \]
((d*x + c)*a/b^2 - 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)))*sqrt(a^2 - b^2)/b^2 + 2/((tan( 1/2*d*x + 1/2*c)^2 + 1)*b))/d
Time = 4.77 (sec) , antiderivative size = 318, normalized size of antiderivative = 4.54 \[ \int \frac {\cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,\mathrm {atan}\left (\frac {64\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^2-\frac {64\,a^4}{b^2}}+\frac {64\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^4-64\,a^2\,b^2}\right )}{b^2\,d}+\frac {2\,\mathrm {atanh}\left (\frac {64\,a^2\,\sqrt {b^2-a^2}}{64\,a^2\,b-\frac {64\,a^4}{b}-128\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+128\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {128\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{64\,a^2-\frac {64\,a^4}{b^2}-\frac {128\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}+128\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {64\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{64\,a^4+128\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b-64\,a^2\,b^2-128\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^3}\right )\,\sqrt {b^2-a^2}}{b^2\,d} \]
2/(b*d*(tan(c/2 + (d*x)/2)^2 + 1)) + (2*a*atan((64*a^2*tan(c/2 + (d*x)/2)) /(64*a^2 - (64*a^4)/b^2) + (64*a^4*tan(c/2 + (d*x)/2))/(64*a^4 - 64*a^2*b^ 2)))/(b^2*d) + (2*atanh((64*a^2*(b^2 - a^2)^(1/2))/(64*a^2*b - (64*a^4)/b - 128*a^3*tan(c/2 + (d*x)/2) + 128*a*b^2*tan(c/2 + (d*x)/2)) + (128*a*tan( c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(64*a^2 - (64*a^4)/b^2 - (128*a^3*tan(c/ 2 + (d*x)/2))/b + 128*a*b*tan(c/2 + (d*x)/2)) + (64*a^3*tan(c/2 + (d*x)/2) *(b^2 - a^2)^(1/2))/(64*a^4 - 64*a^2*b^2 - 128*a*b^3*tan(c/2 + (d*x)/2) + 128*a^3*b*tan(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2))/(b^2*d)