Integrand size = 31, antiderivative size = 92 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {22 a \cos ^3(c+d x)}{105 d (a+a \sin (c+d x))^{3/2}}+\frac {12 \cos ^3(c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 a d} \] Output:
-22/105*a*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)+12/35*cos(d*x+c)^3/d/(a+a* sin(d*x+c))^(1/2)-2/7*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/a/d
Time = 0.64 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (31-15 \cos (2 (c+d x))+24 \sin (c+d x))}{105 d \sqrt {a (1+\sin (c+d x))}} \] Input:
Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/Sqrt[a + a*Sin[c + d*x]],x]
Output:
-1/105*((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(31 - 15*Cos[2*(c + d*x)] + 24*Sin[c + d*x]))/(d*Sqrt[a*(1 + S in[c + d*x])])
Time = 0.78 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.48, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3356, 27, 3042, 3335, 3042, 3153, 3042, 3152}
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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^2(c+d x)}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^2}{\sqrt {a \sin (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3356 |
\(\displaystyle \frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}-\frac {\int -\frac {1}{2} \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a} (4 \sin (c+d x) a+a)dx}{2 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a} (4 \sin (c+d x) a+a)dx}{4 a^2}+\frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \cos (c+d x)^2 \sqrt {\sin (c+d x) a+a} (4 \sin (c+d x) a+a)dx}{4 a^2}+\frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3335 |
\(\displaystyle \frac {\frac {11}{7} a \int \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {8 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{4 a^2}+\frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {11}{7} a \int \cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {8 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{4 a^2}+\frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {\frac {11}{7} a \left (\frac {4}{5} a \int \frac {\cos ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {8 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{4 a^2}+\frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {11}{7} a \left (\frac {4}{5} a \int \frac {\cos (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {8 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{4 a^2}+\frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle \frac {\frac {11}{7} a \left (-\frac {8 a^2 \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {8 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{4 a^2}+\frac {\cos ^3(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\) |
Input:
Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/Sqrt[a + a*Sin[c + d*x]],x]
Output:
Cos[c + d*x]^3/(2*d*Sqrt[a + a*Sin[c + d*x]]) + ((-8*a*Cos[c + d*x]^3*Sqrt [a + a*Sin[c + d*x]])/(7*d) + (11*a*((-8*a^2*Cos[c + d*x]^3)/(15*d*(a + a* Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^3)/(5*d*Sqrt[a + a*Sin[c + d*x]]) ))/7)/(4*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^( p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1))), x] - Simp[1/(a^2*(2* m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*m - b* (2*m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[ a^2 - b^2, 0] && LeQ[m, -2^(-1)] && NeQ[2*m + p + 1, 0]
\[\int \frac {\cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}}{\sqrt {a +a \sin \left (d x +c \right )}}d x\]
Input:
int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x)
Output:
int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x)
Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 3 \, \cos \left (d x + c\right )^{3} - 29 \, \cos \left (d x + c\right )^{2} + {\left (15 \, \cos \left (d x + c\right )^{3} + 18 \, \cos \left (d x + c\right )^{2} - 11 \, \cos \left (d x + c\right ) - 22\right )} \sin \left (d x + c\right ) + 11 \, \cos \left (d x + c\right ) + 22\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="f ricas")
Output:
-2/105*(15*cos(d*x + c)^4 - 3*cos(d*x + c)^3 - 29*cos(d*x + c)^2 + (15*cos (d*x + c)^3 + 18*cos(d*x + c)^2 - 11*cos(d*x + c) - 22)*sin(d*x + c) + 11* cos(d*x + c) + 22)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d* x + c) + a*d)
\[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \] Input:
integrate(cos(d*x+c)**2*sin(d*x+c)**2/(a+a*sin(d*x+c))**(1/2),x)
Output:
Integral(sin(c + d*x)**2*cos(c + d*x)**2/sqrt(a*(sin(c + d*x) + 1)), x)
\[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="m axima")
Output:
integrate(cos(d*x + c)^2*sin(d*x + c)^2/sqrt(a*sin(d*x + c) + a), x)
Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {4 \, \sqrt {2} {\left (60 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 84 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 35 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{105 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="g iac")
Output:
4/105*sqrt(2)*(60*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 84*sqrt(a)*si n(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 35*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c) ^3)/(a*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^2}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \] Input:
int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^(1/2),x)
Output:
int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^(1/2), x)
\[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}}{\sin \left (d x +c \right )+1}d x \right )}{a} \] Input:
int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(1/2),x)
Output:
(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x)**2)/(sin (c + d*x) + 1),x))/a