Integrand size = 21, antiderivative size = 45 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{a}-\frac {\cos (c+d x)}{a d}-\frac {\cos (c+d x)}{a d (1+\sin (c+d x))} \] Output:
-x/a-cos(d*x+c)/a/d-cos(d*x+c)/a/d/(1+sin(d*x+c))
Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.89 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right ) (c+d x+\cos (c+d x))+(-2+c+d x+\cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{a d (1+\sin (c+d x))} \] Input:
Integrate[Sin[c + d*x]^2/(a + a*Sin[c + d*x]),x]
Output:
-(((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Cos[(c + d*x)/2]*(c + d*x + Cos[ c + d*x]) + (-2 + c + d*x + Cos[c + d*x])*Sin[(c + d*x)/2]))/(a*d*(1 + Sin [c + d*x])))
Time = 0.36 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3225, 25, 3042, 3214, 3042, 3127}
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.
\(\displaystyle \int \frac {\sin ^2(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3225 |
\(\displaystyle \frac {\int -\frac {\sin (c+d x)}{\sin (c+d x)+1}dx}{a}-\frac {\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sin (c+d x)}{\sin (c+d x)+1}dx}{a}-\frac {\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sin (c+d x)}{\sin (c+d x)+1}dx}{a}-\frac {\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle -\frac {x-\int \frac {1}{\sin (c+d x)+1}dx}{a}-\frac {\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {x-\int \frac {1}{\sin (c+d x)+1}dx}{a}-\frac {\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle -\frac {\cos (c+d x)}{a d}-\frac {\frac {\cos (c+d x)}{d (\sin (c+d x)+1)}+x}{a}\) |
Input:
Int[Sin[c + d*x]^2/(a + a*Sin[c + d*x]),x]
Output:
-(Cos[c + d*x]/(a*d)) - (x + Cos[c + d*x]/(d*(1 + Sin[c + d*x])))/a
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[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 = 0.53 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(\frac {-3-\cos \left (2 d x +2 c \right )-2 d x \cos \left (d x +c \right )+2 \sin \left (d x +c \right )}{2 a d \cos \left (d x +c \right )}\) | \(48\) |
derivativedivides | \(\frac {-\frac {2}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a d}\) | \(54\) |
default | \(\frac {-\frac {2}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a d}\) | \(54\) |
risch | \(-\frac {x}{a}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}-\frac {2}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\) | \(64\) |
norman | \(\frac {-\frac {2}{a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {x}{a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(185\) |
Input:
int(sin(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/2/a/d*(-3-cos(2*d*x+2*c)-2*d*x*cos(d*x+c)+2*sin(d*x+c))/cos(d*x+c)
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {d x + {\left (d x + 2\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + {\left (d x + \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1}{a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d} \] Input:
integrate(sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")
Output:
-(d*x + (d*x + 2)*cos(d*x + c) + cos(d*x + c)^2 + (d*x + cos(d*x + c) - 1) *sin(d*x + c) + 1)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)
Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (32) = 64\).
Time = 1.05 (sec) , antiderivative size = 422, normalized size of antiderivative = 9.38 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} - \frac {d x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {d x}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {2 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} - \frac {4}{a d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \] Input:
integrate(sin(d*x+c)**2/(a+a*sin(d*x+c)),x)
Output:
Piecewise((-d*x*tan(c/2 + d*x/2)**3/(a*d*tan(c/2 + d*x/2)**3 + a*d*tan(c/2 + d*x/2)**2 + a*d*tan(c/2 + d*x/2) + a*d) - d*x*tan(c/2 + d*x/2)**2/(a*d* tan(c/2 + d*x/2)**3 + a*d*tan(c/2 + d*x/2)**2 + a*d*tan(c/2 + d*x/2) + a*d ) - d*x*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**3 + a*d*tan(c/2 + d*x/2)** 2 + a*d*tan(c/2 + d*x/2) + a*d) - d*x/(a*d*tan(c/2 + d*x/2)**3 + a*d*tan(c /2 + d*x/2)**2 + a*d*tan(c/2 + d*x/2) + a*d) - 2*tan(c/2 + d*x/2)**2/(a*d* tan(c/2 + d*x/2)**3 + a*d*tan(c/2 + d*x/2)**2 + a*d*tan(c/2 + d*x/2) + a*d ) - 2*tan(c/2 + d*x/2)/(a*d*tan(c/2 + d*x/2)**3 + a*d*tan(c/2 + d*x/2)**2 + a*d*tan(c/2 + d*x/2) + a*d) - 4/(a*d*tan(c/2 + d*x/2)**3 + a*d*tan(c/2 + d*x/2)**2 + a*d*tan(c/2 + d*x/2) + a*d), Ne(d, 0)), (x*sin(c)**2/(a*sin(c ) + a), True))
Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (45) = 90\).
Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.87 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 2}{a + \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}\right )}}{d} \] Input:
integrate(sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")
Output:
-2*((sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2)/(a + a*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) + arctan(sin(d*x + c)/ (cos(d*x + c) + 1))/a)/d
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.71 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {d x + c}{a} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} a}}{d} \] Input:
integrate(sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
Output:
-((d*x + c)/a + 2*(tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 2)/((ta n(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) + 1)* a))/d
Time = 35.51 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{a}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4}{a\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int(sin(c + d*x)^2/(a + a*sin(c + d*x)),x)
Output:
- x/a - (2*tan(c/2 + (d*x)/2) + 2*tan(c/2 + (d*x)/2)^2 + 4)/(a*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1))
Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.64 \[ \int \frac {\sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-\cos \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\cos \left (d x +c \right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-d x}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \] Input:
int(sin(d*x+c)^2/(a+a*sin(d*x+c)),x)
Output:
( - cos(c + d*x)*tan((c + d*x)/2) - cos(c + d*x) - tan((c + d*x)/2)*d*x + 2*tan((c + d*x)/2) - d*x)/(a*d*(tan((c + d*x)/2) + 1))