Integrand size = 23, antiderivative size = 78 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 \sqrt {a} (a+b)^{3/2} d}-\frac {\cos (c+d x) \sin (c+d x)}{2 (a+b) d \left (a+b \sin ^2(c+d x)\right )} \]
-1/2*cos(d*x+c)*sin(d*x+c)/(a+b)/d/(a+b*sin(d*x+c)^2)+1/2*arctan((a+b)^(1/ 2)*tan(d*x+c)/a^(1/2))/(a+b)^(3/2)/d/a^(1/2)
Time = 11.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {\arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{3/2}}-\frac {\sin (2 (c+d x))}{(a+b) (2 a+b-b \cos (2 (c+d x)))}}{2 d} \]
(ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]/(Sqrt[a]*(a + b)^(3/2)) - Sin[ 2*(c + d*x)]/((a + b)*(2*a + b - b*Cos[2*(c + d*x)])))/(2*d)
Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3652, 27, 3042, 3660, 218}
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.
\(\displaystyle \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2}{\left (a+b \sin (c+d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3652 |
\(\displaystyle \frac {\int \frac {a}{b \sin ^2(c+d x)+a}dx}{2 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{b \sin ^2(c+d x)+a}dx}{2 (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {1}{b \sin (c+d x)^2+a}dx}{2 (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle \frac {\int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 d (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 \sqrt {a} d (a+b)^{3/2}}-\frac {\sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}\) |
ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]]/(2*Sqrt[a]*(a + b)^(3/2)*d) - ( Cos[c + d*x]*Sin[c + d*x])/(2*(a + b)*d*(a + b*Sin[c + d*x]^2))
3.2.2.3.1 Defintions of rubi rules used
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/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x ]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Simp[1/(2* a*(a + b)*(p + 1)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*( p + 1) + b*(2*p + 3)) + 2*(A*b - a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[1/(a + (a + b)*ff^2*x^ 2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]
Time = 0.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {\tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(75\) |
default | \(\frac {-\frac {\tan \left (d x +c \right )}{2 \left (a +b \right ) \left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 \left (a +b \right ) \sqrt {a \left (a +b \right )}}}{d}\) | \(75\) |
risch | \(\frac {i \left (2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}{b \left (a +b \right ) d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right ) d}\) | \(267\) |
1/d*(-1/2/(a+b)*tan(d*x+c)/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)+1/2/(a+b)/(a* (a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (66) = 132\).
Time = 0.28 (sec) , antiderivative size = 419, normalized size of antiderivative = 5.37 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [\frac {4 \, {\left (a^{2} + a b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{8 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} d\right )}}, \frac {2 \, {\left (a^{2} + a b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{4 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} d\right )}}\right ] \]
[1/8*(4*(a^2 + a*b)*cos(d*x + c)*sin(d*x + c) - (b*cos(d*x + c)^2 - a - b) *sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5 *a*b + b^2)*cos(d*x + c)^2 + 4*((2*a + b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(-a^2 - a*b)*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c )^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)))/((a^3*b + 2*a^2* b^2 + a*b^3)*d*cos(d*x + c)^2 - (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d), 1/ 4*(2*(a^2 + a*b)*cos(d*x + c)*sin(d*x + c) - (b*cos(d*x + c)^2 - a - b)*sq rt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a* b)*cos(d*x + c)*sin(d*x + c))))/((a^3*b + 2*a^2*b^2 + a*b^3)*d*cos(d*x + c )^2 - (a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d)]
Timed out. \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {\tan \left (d x + c\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2} + a b} - \frac {\arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a + b\right )}}}{2 \, d} \]
-1/2*(tan(d*x + c)/((a^2 + 2*a*b + b^2)*tan(d*x + c)^2 + a^2 + a*b) - arct an((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*(a + b)))/d
Time = 0.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )}{\sqrt {a^{2} + a b} {\left (a + b\right )}} - \frac {\tan \left (d x + c\right )}{{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )} {\left (a + b\right )}}}{2 \, d} \]
1/2*((pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))/(sqrt(a^2 + a*b)*(a + b)) - tan(d*x + c)/((a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)*(a + b)))/d
Time = 13.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,{\left (2\,a+2\,b\right )}^2}{4\,\sqrt {a}\,{\left (a+b\right )}^{3/2}}\right )}{2\,\sqrt {a}\,d\,{\left (a+b\right )}^{3/2}}-\frac {\mathrm {tan}\left (c+d\,x\right )}{2\,d\,\left (\left (a+b\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )\,\left (a+b\right )} \]