\(\int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [344]

Optimal result

Integrand size = 31, antiderivative size = 140 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {18 \cos (c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a+a \sin (c+d x)}}-\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{5 a^2 d} \] Output:

-2*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/ 
a^(3/2)/d+18/5*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-2/5*cos(d*x+c)^3/a/d/ 
(a+a*sin(d*x+c))^(1/2)-4/5*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left ((40+40 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right )+30 \cos \left (\frac {1}{2} (c+d x)\right )-5 \cos \left (\frac {3}{2} (c+d x)\right )-\cos \left (\frac {5}{2} (c+d x)\right )-30 \sin \left (\frac {1}{2} (c+d x)\right )-5 \sin \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {5}{2} (c+d x)\right )\right )}{10 d (a (1+\sin (c+d x)))^{3/2}} \] Input:

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]
 

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*((40 + 40*I)*(-1)^(3/4)*ArcTanh[( 
1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])] + 30*Cos[(c + d*x)/2] - 5*C 
os[(3*(c + d*x))/2] - Cos[(5*(c + d*x))/2] - 30*Sin[(c + d*x)/2] - 5*Sin[( 
3*(c + d*x))/2] + Sin[(5*(c + d*x))/2]))/(10*d*(a*(1 + Sin[c + d*x]))^(3/2 
))
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3357, 27, 3042, 3337, 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.

\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^2(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^2}{(a \sin (c+d x)+a)^{3/2}}dx\)

\(\Big \downarrow \) 3357

\(\displaystyle \frac {2 \int -\frac {\cos ^2(c+d x) (6 \sin (c+d x) a+a)}{2 (\sin (c+d x) a+a)^{3/2}}dx}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\cos ^2(c+d x) (6 \sin (c+d x) a+a)}{(\sin (c+d x) a+a)^{3/2}}dx}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {\cos (c+d x)^2 (6 \sin (c+d x) a+a)}{(\sin (c+d x) a+a)^{3/2}}dx}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3337

\(\displaystyle -\frac {\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{a d}-\frac {2 \int \frac {3 \left (a^2-9 a^2 \sin (c+d x)\right )}{2 \sqrt {\sin (c+d x) a+a}}dx}{3 a^2}}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{a d}-\frac {\int \frac {a^2-9 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{a^2}}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{a d}-\frac {\int \frac {a^2-9 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{a^2}}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3230

\(\displaystyle -\frac {\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{a d}-\frac {10 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {18 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a^2}}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{a d}-\frac {10 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {18 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{a^2}}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{a d}-\frac {\frac {18 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {20 a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{a^2}}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{a d}-\frac {\frac {18 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {10 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{a^2}}{5 a}-\frac {2 \cos ^3(c+d x)}{5 a d \sqrt {a \sin (c+d x)+a}}\)

Input:

Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]
 

Output:

(-2*Cos[c + d*x]^3)/(5*a*d*Sqrt[a + a*Sin[c + d*x]]) - ((4*Cos[c + d*x]*Sq 
rt[a + a*Sin[c + d*x]])/(a*d) - ((-10*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[a]*Cos 
[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (18*a^2*Cos[c + d*x])/ 
(d*Sqrt[a + a*Sin[c + d*x]]))/a^2)/(5*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

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]
 

rule 3230

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)]
 

rule 3337

Int[cos[(e_.) + (f_.)*(x_)]^2*((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 + 2)/(b^2*f*(m + 3))), x] - Simp[1/(b^2*(m + 3))   I 
nt[(a + b*Sin[e + f*x])^(m + 1)*(b*d*(m + 2) - a*c*(m + 3) + (b*c*(m + 3) - 
 a*d*(m + 4))*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[ 
a^2 - b^2, 0] && GeQ[m, -3/2] && LtQ[m, 0]
 

rule 3357

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^( 
p + 1))*((a + b*Sin[e + f*x])^(m + 1)/(b*f*g*(m + p + 2))), x] + Simp[1/(b* 
(m + p + 2))   Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m*(b*(m + 1) - a 
*(p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^ 
2 - b^2, 0] && NeQ[m + p + 2, 0]
 

Maple [F]

\[\int \frac {\cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

Input:

int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x)
 

Output:

int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.69 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\frac {5 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - 2 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 9\right )} \sin \left (d x + c\right ) - 7 \, \cos \left (d x + c\right ) - 9\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="f 
ricas")
 

Output:

1/5*(5*sqrt(2)*(a*cos(d*x + c) + a*sin(d*x + c) + a)*log(-(cos(d*x + c)^2 
- (cos(d*x + c) - 2)*sin(d*x + c) - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*(co 
s(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x + c) 
^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt(a) - 2*(cos 
(d*x + c)^3 + 3*cos(d*x + c)^2 - (cos(d*x + c)^2 - 2*cos(d*x + c) - 9)*sin 
(d*x + c) - 7*cos(d*x + c) - 9)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + 
 c) + a^2*d*sin(d*x + c) + a^2*d)
 

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cos(d*x+c)**2*sin(d*x+c)**2/(a+a*sin(d*x+c))**(3/2),x)
 

Output:

Integral(sin(c + d*x)**2*cos(c + d*x)**2/(a*(sin(c + d*x) + 1))**(3/2), x)
 

Maxima [F]

\[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="m 
axima")
 

Output:

integrate(cos(d*x + c)^2*sin(d*x + c)^2/(a*sin(d*x + c) + a)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\frac {5 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {5 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (4 \, a^{\frac {17}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5 \, a^{\frac {17}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{5 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="g 
iac")
 

Output:

1/5*(5*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))) - 5*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))) - 2*sqrt(2)*(4*a^(17/2 
)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 5*a^(17/2)*sin(-1/4*pi + 1/2*d*x + 1/ 
2*c))/(a^10*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^(3/2),x)
 

Output:

int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:

int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),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)**2 + 2*sin(c + d*x) + 1),x))/a**2