\(\int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [345]

Optimal result

Integrand size = 29, antiderivative size = 108 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {10 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d} \] 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-10/3*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)+2/3*cos(d*x+c)*(a+a*si 
n(d*x+c))^(1/2)/a^2/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left ((12+12 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )-9 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )+9 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

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

Output:

(Sqrt[a*(1 + Sin[c + d*x])]*((12 + 12*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(- 
1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c + d*x)/4] - Sin[(2*c + d*x)/4])] - 9*Cos[( 
c + d*x)/2] + Cos[(3*(c + d*x))/2] + 9*Sin[(c + d*x)/2] + Sin[(3*(c + d*x) 
)/2]))/(3*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3337

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

\(\displaystyle \frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a^2 d}-\frac {6 a \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {10 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a^2 d}-\frac {6 a \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {10 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a^2}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a^2 d}-\frac {\frac {10 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {12 a \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{3 a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a^2 d}-\frac {\frac {10 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {6 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{3 a^2}\)

Input:

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

Output:

(2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*a^2*d) - ((-6*Sqrt[2]*Sqrt[a] 
*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + ( 
10*a*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/(3*a^2)
 

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]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (3 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-3 \sqrt {a -a \sin \left (d x +c \right )}\, a \right )}{3 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(112\)

Input:

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

Output:

2/3/a^3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*a^(3/2)*2^(1/2)*arctan 
h(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-(a-a*sin(d*x+c))^(3/2)-3*(a- 
a*sin(d*x+c))^(1/2)*a)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (91) = 182\).

Time = 0.09 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.99 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} + 2 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right ) - 5\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

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

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(cos(d*x + c)^2*sin(d*x + c)/(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 = 1.33 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (2 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{3 \, d} \] Input:

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

Output:

-1/3*(3*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))) - 3*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)*(2*a^(9/2 
)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 3*a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2 
*c))/(a^6*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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