\(\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [343]

Optimal result

Integrand size = 31, antiderivative size = 184 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {344 \cos (c+d x)}{105 a d \sqrt {a+a \sin (c+d x)}}-\frac {16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {76 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 a^2 d} \] 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-344/105*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-16/35*cos(d*x+c)*si 
n(d*x+c)^2/a/d/(a+a*sin(d*x+c))^(1/2)+2/7*cos(d*x+c)*sin(d*x+c)^3/a/d/(a+a 
*sin(d*x+c))^(1/2)+76/105*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 = 3.15 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left ((1680+1680 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )-1365 \cos \left (\frac {1}{2} (c+d x)\right )+245 \cos \left (\frac {3}{2} (c+d x)\right )+63 \cos \left (\frac {5}{2} (c+d x)\right )-15 \cos \left (\frac {7}{2} (c+d x)\right )+1365 \sin \left (\frac {1}{2} (c+d x)\right )+245 \sin \left (\frac {3}{2} (c+d x)\right )-63 \sin \left (\frac {5}{2} (c+d x)\right )-15 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{420 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

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

Output:

(Sqrt[a*(1 + Sin[c + d*x])]*((1680 + 1680*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])] - 136 
5*Cos[(c + d*x)/2] + 245*Cos[(3*(c + d*x))/2] + 63*Cos[(5*(c + d*x))/2] - 
15*Cos[(7*(c + d*x))/2] + 1365*Sin[(c + d*x)/2] + 245*Sin[(3*(c + d*x))/2] 
 - 63*Sin[(5*(c + d*x))/2] - 15*Sin[(7*(c + d*x))/2]))/(420*a^2*d*(Cos[(c 
+ d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.13, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.581, Rules used = {3042, 3358, 3042, 3462, 25, 3042, 3462, 27, 3042, 3447, 3042, 3502, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3358

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {2 \int -\frac {\sin ^2(c+d x) \left (3 a^2-4 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{7 a}+\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}}{a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \int \frac {\sin ^2(c+d x) \left (3 a^2-4 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{7 a}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \int \frac {\sin (c+d x)^2 \left (3 a^2-4 a^2 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{7 a}}{a^2}\)

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {2 \int -\frac {\sin (c+d x) \left (16 a^3-19 a^3 \sin (c+d x)\right )}{2 \sqrt {\sin (c+d x) a+a}}dx}{5 a}+\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {\sin (c+d x) \left (16 a^3-19 a^3 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {\sin (c+d x) \left (16 a^3-19 a^3 \sin (c+d x)\right )}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {16 a^3 \sin (c+d x)-19 a^3 \sin ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\int \frac {16 a^3 \sin (c+d x)-19 a^3 \sin (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {2 \int -\frac {19 a^4-86 a^4 \sin (c+d x)}{2 \sqrt {\sin (c+d x) a+a}}dx}{3 a}+\frac {38 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {38 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {19 a^4-86 a^4 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {38 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {19 a^4-86 a^4 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {38 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {105 a^4 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {172 a^4 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {38 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {105 a^4 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {172 a^4 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {38 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {172 a^4 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {210 a^4 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{3 a}}{5 a}\right )}{7 a}}{a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \left (\frac {8 a^2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}-\frac {\frac {38 a^2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {172 a^4 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {105 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{3 a}}{5 a}\right )}{7 a}}{a^2}\)

Input:

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

Output:

((2*a*Cos[c + d*x]*Sin[c + d*x]^3)/(7*d*Sqrt[a + a*Sin[c + d*x]]) - (2*((8 
*a^2*Cos[c + d*x]*Sin[c + d*x]^2)/(5*d*Sqrt[a + a*Sin[c + d*x]]) - ((38*a^ 
2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) - ((-105*Sqrt[2]*a^(7/2)*Ar 
cTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (172 
*a^4*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/(3*a))/(5*a)))/(7*a))/a^2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3358

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]
 

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3462

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m 
 + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (105 a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}+21 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}-35 a^{2} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-105 a^{3} \sqrt {a -a \sin \left (d x +c \right )}\right )}{105 a^{5} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(148\)

Input:

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

Output:

2/105/a^5*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(105*a^(7/2)*2^(1/2)*ar 
ctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-15*(a-a*sin(d*x+c))^(7/2 
)+21*a*(a-a*sin(d*x+c))^(5/2)-35*a^2*(a-a*sin(d*x+c))^(3/2)-105*a^3*(a-a*s 
in(d*x+c))^(1/2))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\frac {105 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 24 \, \cos \left (d x + c\right )^{3} - 92 \, \cos \left (d x + c\right )^{2} + {\left (15 \, \cos \left (d x + c\right )^{3} + 39 \, \cos \left (d x + c\right )^{2} - 53 \, \cos \left (d x + c\right ) - 211\right )} \sin \left (d x + c\right ) + 158 \, \cos \left (d x + c\right ) + 211\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

1/105*(105*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) 
*(cos(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* 
(15*cos(d*x + c)^4 - 24*cos(d*x + c)^3 - 92*cos(d*x + c)^2 + (15*cos(d*x + 
 c)^3 + 39*cos(d*x + c)^2 - 53*cos(d*x + c) - 211)*sin(d*x + c) + 158*cos( 
d*x + c) + 211)*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(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {\frac {105 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {105 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (120 \, a^{\frac {25}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 84 \, a^{\frac {25}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 70 \, a^{\frac {25}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{\frac {25}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{14} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{105 \, d} \] Input:

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

Output:

-1/105*(105*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(c 
os(-1/4*pi + 1/2*d*x + 1/2*c))) - 105*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)*(12 
0*a^(25/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 84*a^(25/2)*sin(-1/4*pi + 1/ 
2*d*x + 1/2*c)^5 + 70*a^(25/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 105*a^(2 
5/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^14*sgn(cos(-1/4*pi + 1/2*d*x + 1/2 
*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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