Integrand size = 23, antiderivative size = 105 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos (c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}} \]
-1/2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)+7/4*arctanh(1/2*cos(d*x+c)*a^(1/2 )*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-2*cos(d*x+c)/a/d/(a+a* sin(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.45 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.28 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (3 \cos \left (\frac {1}{2} (c+d x)\right )+2 \cos \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )+(7+7 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) (1+\sin (c+d x))+2 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{2 d (a (1+\sin (c+d x)))^{3/2}} \]
-1/2*((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(3*Cos[(c + d*x)/2] + 2*Cos[(3 *(c + d*x))/2] - 3*Sin[(c + d*x)/2] + (7 + 7*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(1 + Sin[c + d*x]) + 2*Sin[(3*(c + d*x))/2]))/(d*(a*(1 + Sin[c + d*x]))^(3/2))
Time = 0.45 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3237, 27, 3042, 3230, 3042, 3128, 219}
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.
| \(\displaystyle \int \frac {\sin ^2(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{(a \sin (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3237 |
| \(\displaystyle \frac {\int -\frac {3 a-4 a \sin (c+d x)}{2 \sqrt {\sin (c+d x) a+a}}dx}{2 a^2}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {3 a-4 a \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {3 a-4 a \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle -\frac {7 a \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {8 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 a \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {8 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{4 a^2}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle -\frac {\frac {8 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {14 a \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{4 a^2}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {8 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {7 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\) |
-1/2*Cos[c + d*x]/(d*(a + a*Sin[c + d*x])^(3/2)) - ((-7*Sqrt[2]*Sqrt[a]*Ar cTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (8*a *Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/(4*a^2)
3.1.70.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2* m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.71 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.40
| method | result | size |
| default | \(-\frac {\left (-7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sin \left (d x +c \right )+8 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}\, \sin \left (d x +c \right )-7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a +10 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{4 a^{\frac {5}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(147\) |
-1/4/a^(5/2)*(-7*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2 ))*a*sin(d*x+c)+8*(a-a*sin(d*x+c))^(1/2)*a^(1/2)*sin(d*x+c)-7*2^(1/2)*arct anh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a+10*(a-a*sin(d*x+c))^(1/2 )*a^(1/2))*(-a*(sin(d*x+c)-1))^(1/2)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (88) = 176\).
Time = 0.30 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.61 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {7 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (4 \, \cos \left (d x + c\right )^{2} + {\left (4 \, \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 5 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]
1/8*(7*sqrt(2)*(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos (d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)* sin(d*x + c) - cos(d*x + c) - 2)) + 4*(4*cos(d*x + c)^2 + (4*cos(d*x + c) - 1)*sin(d*x + c) + 5*cos(d*x + c) + 1)*sqrt(a*sin(d*x + c) + a))/(a^2*d*c os(d*x + c)^2 - a^2*d*cos(d*x + c) - 2*a^2*d - (a^2*d*cos(d*x + c) + 2*a^2 *d)*sin(d*x + c))
\[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.64 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\frac {7 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {7 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {16 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \]
-1/8*(7*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(- 1/4*pi + 1/2*d*x + 1/2*c))) - 7*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c ) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 16*sqrt(2)*sin(-1/4 *pi + 1/2*d*x + 1/2*c)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 2*s qrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)*a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]