Integrand size = 21, antiderivative size = 74 \[ \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 \sqrt {b} (a+b)^{3/2} d}-\frac {\cos (c+d x)}{2 (a+b) d \left (a+b-b \cos ^2(c+d x)\right )} \]
-1/2*cos(d*x+c)/(a+b)/d/(a+b-b*cos(d*x+c)^2)-1/2*arctanh(cos(d*x+c)*b^(1/2 )/(a+b)^(1/2))/(a+b)^(3/2)/d/b^(1/2)
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.01 \[ \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {\arctan \left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{\sqrt {-a-b} \sqrt {b}}+\frac {\arctan \left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{\sqrt {-a-b} \sqrt {b}}-\frac {2 \cos (c+d x)}{2 a+b-b \cos (2 (c+d x))}}{2 (a+b) d} \]
(ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]]/(Sqrt[-a - b] *Sqrt[b]) + ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]]/(S qrt[-a - b]*Sqrt[b]) - (2*Cos[c + d*x])/(2*a + b - b*Cos[2*(c + d*x)]))/(2 *(a + b)*d)
Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3665, 215, 221}
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.
\(\displaystyle \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{\left (a+b \sin (c+d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1}{\left (-b \cos ^2(c+d x)+a+b\right )^2}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle -\frac {\frac {\int \frac {1}{-b \cos ^2(c+d x)+a+b}d\cos (c+d x)}{2 (a+b)}+\frac {\cos (c+d x)}{2 (a+b) \left (a-b \cos ^2(c+d x)+b\right )}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\frac {\text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 \sqrt {b} (a+b)^{3/2}}+\frac {\cos (c+d x)}{2 (a+b) \left (a-b \cos ^2(c+d x)+b\right )}}{d}\) |
-((ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]]/(2*Sqrt[b]*(a + b)^(3/2)) + Cos[c + d*x]/(2*(a + b)*(a + b - b*Cos[c + d*x]^2)))/d)
3.1.97.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.58 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (d x +c \right )}{2 \left (a +b \right ) \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}-\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{d}\) | \(65\) |
default | \(\frac {-\frac {\cos \left (d x +c \right )}{2 \left (a +b \right ) \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}-\frac {\operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{d}\) | \(65\) |
risch | \(\frac {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}{\left (a +b \right ) d \left (b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}+\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d}\) | \(193\) |
1/d*(-1/2*cos(d*x+c)/(a+b)/(a+b-b*cos(d*x+c)^2)-1/2/(a+b)/((a+b)*b)^(1/2)* arctanh(b*cos(d*x+c)/((a+b)*b)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 282, normalized size of antiderivative = 3.81 \[ \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [\frac {{\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {a b + b^{2}} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a b + b^{2}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) + 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} d\right )}}, \frac {{\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {-a b - b^{2}} \arctan \left (\frac {\sqrt {-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) + {\left (a b + b^{2}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} d\right )}}\right ] \]
[1/4*((b*cos(d*x + c)^2 - a - b)*sqrt(a*b + b^2)*log(-(b*cos(d*x + c)^2 - 2*sqrt(a*b + b^2)*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b)) + 2*(a *b + b^2)*cos(d*x + c))/((a^2*b^2 + 2*a*b^3 + b^4)*d*cos(d*x + c)^2 - (a^3 *b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d), 1/2*((b*cos(d*x + c)^2 - a - b)*sqrt(- a*b - b^2)*arctan(sqrt(-a*b - b^2)*cos(d*x + c)/(a + b)) + (a*b + b^2)*cos (d*x + c))/((a^2*b^2 + 2*a*b^3 + b^4)*d*cos(d*x + c)^2 - (a^3*b + 3*a^2*b^ 2 + 3*a*b^3 + b^4)*d)]
Timed out. \[ \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.32 \[ \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {2 \, \cos \left (d x + c\right )}{{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, a b - b^{2}} + \frac {\log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} {\left (a + b\right )}}}{4 \, d} \]
1/4*(2*cos(d*x + c)/((a*b + b^2)*cos(d*x + c)^2 - a^2 - 2*a*b - b^2) + log ((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + c) + sqrt((a + b)*b)))/(s qrt((a + b)*b)*(a + b)))/d
Time = 0.47 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07 \[ \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\arctan \left (\frac {b \cos \left (d x + c\right )}{\sqrt {-a b - b^{2}}}\right )}{2 \, \sqrt {-a b - b^{2}} {\left (a + b\right )} d} + \frac {\cos \left (d x + c\right )}{2 \, {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} {\left (a + b\right )} d} \]
1/2*arctan(b*cos(d*x + c)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*(a + b)*d) + 1/2*cos(d*x + c)/((b*cos(d*x + c)^2 - a - b)*(a + b)*d)
Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \frac {\sin (c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\cos \left (c+d\,x\right )}{2\,d\,\left (a+b\right )\,\left (-b\,{\cos \left (c+d\,x\right )}^2+a+b\right )}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b}\,\cos \left (c+d\,x\right )}{\sqrt {a+b}}\right )}{2\,\sqrt {b}\,d\,{\left (a+b\right )}^{3/2}} \]