Integrand size = 23, antiderivative size = 107 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\cos (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}} \]
-1/4*cos(d*x+c)/d/(a+a*sin(d*x+c))^(5/2)+13/16*cos(d*x+c)/a/d/(a+a*sin(d*x +c))^(3/2)-19/32*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^( 1/2))/a^(5/2)/d*2^(1/2)
Result contains complex when optimal does not.
Time = 0.43 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.83 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (8 \sin \left (\frac {1}{2} (c+d x)\right )-4 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-26 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2+13 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3+(19+19 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4\right )}{16 d (a (1+\sin (c+d x)))^{5/2}} \]
((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(8*Sin[(c + d*x)/2] - 4*(Cos[(c + d *x)/2] + Sin[(c + d*x)/2]) - 26*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[( c + d*x)/2])^2 + 13*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (19 + 19*I)* (-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4))/(16*d*(a*(1 + Sin[c + d*x]))^(5/2))
Time = 0.45 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06, 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, 3229, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x)}{(a \sin (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{(a \sin (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3237 |
| \(\displaystyle \frac {\int -\frac {5 a-8 a \sin (c+d x)}{2 (\sin (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {5 a-8 a \sin (c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {5 a-8 a \sin (c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle -\frac {-\frac {19}{4} \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {19}{4} \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle -\frac {\frac {19 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 d}-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {19 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} \sqrt {a} d}-\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}\) |
-1/4*Cos[c + d*x]/(d*(a + a*Sin[c + d*x])^(5/2)) - ((19*ArcTanh[(Sqrt[a]*C os[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(2*Sqrt[2]*Sqrt[a]*d) - (13*a*Cos[c + d*x])/(2*d*(a + a*Sin[c + d*x])^(3/2)))/(8*a^2)
3.1.79.3.1 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[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, 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)]
Leaf count of result is larger than twice the leaf count of optimal. \(192\) vs. \(2(88)=176\).
Time = 0.77 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.80
| method | result | size |
| default | \(-\frac {\left (-19 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )+38 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )+38 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-44 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}+26 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(193\) |
-1/32/a^(9/2)*(-19*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1 /2))*a^2*cos(d*x+c)^2+38*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2 )/a^(1/2))*a^2*sin(d*x+c)+38*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^ (1/2)/a^(1/2))*a^2-44*(a-a*sin(d*x+c))^(1/2)*a^(3/2)+26*(a-a*sin(d*x+c))^( 3/2)*a^(1/2))*(-a*(sin(d*x+c)-1))^(1/2)/(1+sin(d*x+c))/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. 320 vs. \(2 (88) = 176\).
Time = 0.29 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.99 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {19 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (13 \, \cos \left (d x + c\right )^{2} + {\left (13 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) + 9 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]
1/64*(19*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2* cos(d*x + c) - 4)*sin(d*x + c) - 2*cos(d*x + c) - 4)*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*(13*cos(d*x + c)^2 + (13*cos(d*x + c) + 4)*sin(d*x + c) + 9*cos(d*x + c) - 4)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d* x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3* d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))
\[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.36 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\frac {19 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {19 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (13 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]
1/64*(19*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos( -1/4*pi + 1/2*d*x + 1/2*c))) - 19*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2 *c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 2*sqrt(2)*(13*sqr t(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 11*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1 /2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\sin ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]