Integrand size = 31, antiderivative size = 158 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {4 \cos (c+d x)}{45 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^3(c+d x)}{63 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 d \sqrt {a+a \sin (c+d x)}}+\frac {8 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a d}-\frac {4 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{105 a^2 d} \]
-4/105*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a^2/d-4/45*cos(d*x+c)/d/(a+a*sin( d*x+c))^(1/2)-2/63*cos(d*x+c)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+2/9*co s(d*x+c)*sin(d*x+c)^4/d/(a+a*sin(d*x+c))^(1/2)+8/315*cos(d*x+c)*(a+a*sin(d *x+c))^(1/2)/a/d
Time = 1.38 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (-124+60 \cos (2 (c+d x))-201 \sin (c+d x)+35 \sin (3 (c+d x)))}{630 d \sqrt {a (1+\sin (c+d x))}} \]
((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x) /2])*(-124 + 60*Cos[2*(c + d*x)] - 201*Sin[c + d*x] + 35*Sin[3*(c + d*x)]) )/(630*d*Sqrt[a*(1 + Sin[c + d*x])])
Time = 0.95 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.18, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3358, 3042, 3460, 3042, 3249, 3042, 3238, 27, 3042, 3230, 3042, 3125}
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.
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^2(c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^2}{\sqrt {a \sin (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3358 |
\(\displaystyle \frac {\int \sin ^3(c+d x) (a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin (c+d x)^3 (a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{a^2}\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle \frac {\frac {1}{9} a \int \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{9} a \int \sin (c+d x)^3 \sqrt {\sin (c+d x) a+a}dx+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3249 |
\(\displaystyle \frac {\frac {1}{9} a \left (\frac {6}{7} \int \sin ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{9} a \left (\frac {6}{7} \int \sin (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3238 |
\(\displaystyle \frac {\frac {1}{9} a \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{9} a \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{9} a \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {1}{9} a \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{9} a \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}}{a^2}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}+\frac {1}{9} a \left (\frac {6}{7} \left (\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )}{a^2}\) |
((2*a^2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (a*( (-2*a*Cos[c + d*x]*Sin[c + d*x]^3)/(7*d*Sqrt[a + a*Sin[c + d*x]]) + (6*((- 2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*a*d) + ((-14*a^2*Cos[c + d*x ])/(3*d*Sqrt[a + a*Sin[c + d*x]]) + (4*a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d *x]])/(3*d))/(5*a)))/7))/9)/a^2
3.4.36.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[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[(-Cos[e + f*x])*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*Si n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && ! LtQ[m, -2^(-1)]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 2*n + 1))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/b^2 Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /; Fre eQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.47
method | result | size |
default | \(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{2} \left (35 \left (\sin ^{3}\left (d x +c \right )\right )+30 \left (\sin ^{2}\left (d x +c \right )\right )+24 \sin \left (d x +c \right )+16\right )}{315 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(74\) |
-2/315*(1+sin(d*x+c))*(sin(d*x+c)-1)^2*(35*sin(d*x+c)^3+30*sin(d*x+c)^2+24 *sin(d*x+c)+16)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 \, {\left (35 \, \cos \left (d x + c\right )^{5} + 40 \, \cos \left (d x + c\right )^{4} - 64 \, \cos \left (d x + c\right )^{3} - 82 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{3} - 69 \, \cos \left (d x + c\right )^{2} + 13 \, \cos \left (d x + c\right ) + 26\right )} \sin \left (d x + c\right ) + 13 \, \cos \left (d x + c\right ) + 26\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{315 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]
2/315*(35*cos(d*x + c)^5 + 40*cos(d*x + c)^4 - 64*cos(d*x + c)^3 - 82*cos( d*x + c)^2 - (35*cos(d*x + c)^4 - 5*cos(d*x + c)^3 - 69*cos(d*x + c)^2 + 1 3*cos(d*x + c) + 26)*sin(d*x + c) + 13*cos(d*x + c) + 26)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)
Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{3}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {4 \, \sqrt {2} {\left (280 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 540 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{315 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]
-4/315*sqrt(2)*(280*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 540*sqrt(a) *sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 + 378*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/ 2*c)^5 - 105*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3)/(a*d*sgn(cos(-1/4*p i + 1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]