Integrand size = 23, antiderivative size = 83 \[ \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 b^{3/2} (a+b)^{3/2} d}+\frac {a \cos (c+d x)}{2 b (a+b) d \left (a+b-b \cos ^2(c+d x)\right )} \]
-1/2*(a+2*b)*arctanh(cos(d*x+c)*b^(1/2)/(a+b)^(1/2))/b^(3/2)/(a+b)^(3/2)/d +1/2*a*cos(d*x+c)/b/(a+b)/d/(a+b-b*cos(d*x+c)^2)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.93 \[ \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {(a+2 b) \arctan \left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{\sqrt {-a-b}}+\frac {(a+2 b) \arctan \left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{\sqrt {-a-b}}+\frac {2 a \sqrt {b} \cos (c+d x)}{2 a+b-b \cos (2 (c+d x))}}{2 b^{3/2} (a+b) d} \]
(((a + 2*b)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b]])/S qrt[-a - b] + ((a + 2*b)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d*x)/2])/Sqr t[-a - b]])/Sqrt[-a - b] + (2*a*Sqrt[b]*Cos[c + d*x])/(2*a + b - b*Cos[2*( c + d*x)]))/(2*b^(3/2)*(a + b)*d)
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3665, 298, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3}{\left (a+b \sin (c+d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1-\cos ^2(c+d x)}{\left (-b \cos ^2(c+d x)+a+b\right )^2}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle -\frac {\frac {(a+2 b) \int \frac {1}{-b \cos ^2(c+d x)+a+b}d\cos (c+d x)}{2 b (a+b)}-\frac {a \cos (c+d x)}{2 b (a+b) \left (a-b \cos ^2(c+d x)+b\right )}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\frac {(a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 b^{3/2} (a+b)^{3/2}}-\frac {a \cos (c+d x)}{2 b (a+b) \left (a-b \cos ^2(c+d x)+b\right )}}{d}\) |
-((((a + 2*b)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(2*b^(3/2)*(a + b)^(3/2)) - (a*Cos[c + d*x])/(2*b*(a + b)*(a + b - b*Cos[c + d*x]^2)))/d)
3.1.96.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
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 = 0.72 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {\frac {a \cos \left (d x +c \right )}{2 \left (a +b \right ) b \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}-\frac {\left (a +2 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) b \sqrt {\left (a +b \right ) b}}}{d}\) | \(77\) |
default | \(\frac {\frac {a \cos \left (d x +c \right )}{2 \left (a +b \right ) b \left (a +b -b \left (\cos ^{2}\left (d x +c \right )\right )\right )}-\frac {\left (a +2 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) b \sqrt {\left (a +b \right ) b}}}{d}\) | \(77\) |
risch | \(-\frac {a \left ({\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}\right )}{b \left (a +b \right ) d \left (b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}-\frac {i a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d b}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d}+\frac {i a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d b}+\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d}\) | \(330\) |
1/d*(1/2*a/(a+b)/b*cos(d*x+c)/(a+b-b*cos(d*x+c)^2)-1/2*(a+2*b)/(a+b)/b/((a +b)*b)^(1/2)*arctanh(b*cos(d*x+c)/((a+b)*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (74) = 148\).
Time = 0.28 (sec) , antiderivative size = 327, normalized size of antiderivative = 3.94 \[ \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [\frac {{\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 3 \, a b - 2 \, b^{2}\right )} \sqrt {a b + b^{2}} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a b + b^{2}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) - 2 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} d\right )}}, \frac {{\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 3 \, a b - 2 \, b^{2}\right )} \sqrt {-a b - b^{2}} \arctan \left (\frac {\sqrt {-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) - {\left (a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} d\right )}}\right ] \]
[1/4*(((a*b + 2*b^2)*cos(d*x + c)^2 - a^2 - 3*a*b - 2*b^2)*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^2*b + a*b^2)*cos(d*x + c))/((a^2*b^3 + 2*a*b^4 + b^5)*d*cos(d*x + c)^2 - (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*d), 1/2*(( (a*b + 2*b^2)*cos(d*x + c)^2 - a^2 - 3*a*b - 2*b^2)*sqrt(-a*b - b^2)*arcta n(sqrt(-a*b - b^2)*cos(d*x + c)/(a + b)) - (a^2*b + a*b^2)*cos(d*x + c))/( (a^2*b^3 + 2*a*b^4 + b^5)*d*cos(d*x + c)^2 - (a^3*b^2 + 3*a^2*b^3 + 3*a*b^ 4 + b^5)*d)]
Timed out. \[ \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.34 \[ \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {2 \, a \cos \left (d x + c\right )}{a^{2} b + 2 \, a b^{2} + b^{3} - {\left (a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}} + \frac {{\left (a + 2 \, b\right )} \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} {\left (a b + b^{2}\right )}}}{4 \, d} \]
1/4*(2*a*cos(d*x + c)/(a^2*b + 2*a*b^2 + b^3 - (a*b^2 + b^3)*cos(d*x + c)^ 2) + (a + 2*b)*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x + c) + sq rt((a + b)*b)))/(sqrt((a + b)*b)*(a*b + b^2)))/d
Time = 0.43 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {b \cos \left (d x + c\right )}{\sqrt {-a b - b^{2}}}\right )}{2 \, {\left (a b + b^{2}\right )} \sqrt {-a b - b^{2}} d} - \frac {a \cos \left (d x + c\right )}{2 \, {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} {\left (a b + b^{2}\right )} d} \]
1/2*(a + 2*b)*arctan(b*cos(d*x + c)/sqrt(-a*b - b^2))/((a*b + b^2)*sqrt(-a *b - b^2)*d) - 1/2*a*cos(d*x + c)/((b*cos(d*x + c)^2 - a - b)*(a*b + b^2)* d)
Time = 13.53 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {a\,\cos \left (c+d\,x\right )}{2\,b\,d\,\left (a+b\right )\,\left (-b\,{\cos \left (c+d\,x\right )}^2+a+b\right )}-\frac {\mathrm {atanh}\left (\frac {\sqrt {b}\,\cos \left (c+d\,x\right )}{\sqrt {a+b}}\right )\,\left (a+2\,b\right )}{2\,b^{3/2}\,d\,{\left (a+b\right )}^{3/2}} \]