Integrand size = 21, antiderivative size = 75 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a x}{b^2}+\frac {2 a^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}-\frac {\cos (c+d x)}{b d} \] Output:
-a*x/b^2+2*a^2*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^2/(a^2-b ^2)^(1/2)/d-cos(d*x+c)/b/d
Time = 0.72 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a (c+d x)-\frac {2 a^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+b \cos (c+d x)}{b^2 d} \] Input:
Integrate[Sin[c + d*x]^2/(a + b*Sin[c + d*x]),x]
Output:
-((a*(c + d*x) - (2*a^2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/ Sqrt[a^2 - b^2] + b*Cos[c + d*x])/(b^2*d))
Time = 0.41 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3225, 25, 27, 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.
\(\displaystyle \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3225 |
\(\displaystyle \frac {\int -\frac {a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {a \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a \int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \int \frac {\sin (c+d x)}{a+b \sin (c+d x)}dx}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle -\frac {a \left (\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \left (\frac {x}{b}-\frac {a \int \frac {1}{a+b \sin (c+d x)}dx}{b}\right )}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -\frac {a \left (\frac {x}{b}-\frac {2 a \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}\right )}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -\frac {a \left (\frac {4 a \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b d}+\frac {x}{b}\right )}{b}-\frac {\cos (c+d x)}{b d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {a \left (\frac {x}{b}-\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b d \sqrt {a^2-b^2}}\right )}{b}-\frac {\cos (c+d x)}{b d}\) |
Input:
Int[Sin[c + d*x]^2/(a + b*Sin[c + d*x]),x]
Output:
-((a*(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)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f _.)*(x_)]), x_Symbol] :> Simp[(-b^2)*(Cos[e + f*x]/(d*f)), x] + Simp[1/d Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20
method | result | size |
derivativedivides | \(\frac {-\frac {2 \left (\frac {b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {2 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{2} \sqrt {a^{2}-b^{2}}}}{d}\) | \(90\) |
default | \(\frac {-\frac {2 \left (\frac {b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}+\frac {2 a^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{b^{2} \sqrt {a^{2}-b^{2}}}}{d}\) | \(90\) |
risch | \(-\frac {a x}{b^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\) | \(186\) |
Input:
int(sin(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
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/b^ 2*a^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 = 0.11 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.77 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} a^{2} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} d x + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac {\sqrt {a^{2} - b^{2}} a^{2} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (a^{3} - a b^{2}\right )} d x + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \] Input:
integrate(sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
[-1/2*(sqrt(-a^2 + b^2)*a^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))*sq rt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 2 *(a^3 - a*b^2)*d*x + 2*(a^2*b - b^3)*cos(d*x + c))/((a^2*b^2 - b^4)*d), -( sqrt(a^2 - b^2)*a^2*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + (a^3 - a*b^2)*d*x + (a^2*b - b^3)*cos(d*x + c))/((a^2*b^2 - b^4)* d)]
Leaf count of result is larger than twice the leaf count of optimal. 1690 vs. \(2 (61) = 122\).
Time = 110.28 (sec) , antiderivative size = 1690, normalized size of antiderivative = 22.53 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)**2/(a+b*sin(d*x+c)),x)
Output:
Piecewise((zoo*x*sin(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-cos(c + d*x)/( b*d), 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* sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) - 2*b*ta n(c/2 + d*x/2)**2/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*ta n(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) - 4*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*sqr t(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)), Eq(a, -sqr t(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*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)) - 4*b/(b**2*d*tan(c/2 + ...
Exception generated. \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(d*x+c)^2/(a+b*sin(d*x+c)),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(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.32 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.32 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2}}{\sqrt {a^{2} - b^{2}} b^{2}} - \frac {{\left (d x + c\right )} a}{b^{2}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b}}{d} \] Input:
integrate(sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
(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^2/(sqrt(a^2 - b^2)*b^2) - (d*x + c)*a/b^2 - 2/ ((tan(1/2*d*x + 1/2*c)^2 + 1)*b))/d
Time = 37.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.69 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\cos \left (c+d\,x\right )}{b\,d}-\frac {a\,x}{b^2}-\frac {a^2\,\mathrm {atan}\left (\frac {\left (-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )\,2{}\mathrm {i}}{b^2\,d\,\sqrt {b^2-a^2}} \] Input:
int(sin(c + d*x)^2/(a + b*sin(c + d*x)),x)
Output:
- cos(c + d*x)/(b*d) - (a*x)/b^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))))*2i)/(b^2*d*(b^2 - a^2)^(1/ 2))
Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.29 \[ \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}-\cos \left (d x +c \right ) a^{2} b +\cos \left (d x +c \right ) b^{3}-a^{3} d x +a \,b^{2} d x}{b^{2} d \left (a^{2}-b^{2}\right )} \] Input:
int(sin(d*x+c)^2/(a+b*sin(d*x+c)),x)
Output:
(2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2 - cos(c + d*x)*a**2*b + cos(c + d*x)*b**3 - a**3*d*x + a*b**2*d*x)/(b**2* d*(a**2 - b**2))