\(\int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [61]

Optimal result

Integrand size = 23, antiderivative size = 139 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}-\frac {28 \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^2(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 a d} \] Output:

2^(1/2)*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^( 
1/2)/d-28/15*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/5*cos(d*x+c)*sin(d*x+c) 
^2/d/(a+a*sin(d*x+c))^(1/2)+2/15*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left ((-60-60 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right )-60 \cos \left (\frac {1}{2} (c+d x)\right )+5 \cos \left (\frac {3}{2} (c+d x)\right )+3 \cos \left (\frac {5}{2} (c+d x)\right )+60 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d \sqrt {a (1+\sin (c+d x))}} \] Input:

Integrate[Sin[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*((-60 - 60*I)*(-1)^(3/4)*ArcTanh[(1 
/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])] - 60*Cos[(c + d*x)/2] + 5*Co 
s[(3*(c + d*x))/2] + 3*Cos[(5*(c + d*x))/2] + 60*Sin[(c + d*x)/2] + 5*Sin[ 
(3*(c + d*x))/2] - 3*Sin[(5*(c + d*x))/2]))/(30*d*Sqrt[a*(1 + Sin[c + d*x] 
)])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.11, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3257, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3257

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {2 \int -\frac {a^2-14 a^2 \sin (c+d x)}{2 \sqrt {\sin (c+d x) a+a}}dx}{3 a}+\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {a^2-14 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{5 a}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {a^2-14 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{5 a}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {15 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {28 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{5 a}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {15 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {28 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{5 a}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {28 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {30 a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{3 a}}{5 a}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {28 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {15 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{3 a}}{5 a}-\frac {2 \sin ^2(c+d x) \cos (c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\)

Input:

Int[Sin[c + d*x]^3/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(-2*Cos[c + d*x]*Sin[c + d*x]^2)/(5*d*Sqrt[a + a*Sin[c + d*x]]) + ((2*Cos[ 
c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) - ((-15*Sqrt[2]*a^(3/2)*ArcTanh[( 
Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (28*a^2*Cos 
[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/(3*a))/(5*a)
 

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 3257

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(b*(2*n - 1)) 
   Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - 
 b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n - 3))*Sin[e + f*x], 
x], 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, 1] && IntegerQ[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 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 = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (15 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-6 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}+10 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-30 \sqrt {a -a \sin \left (d x +c \right )}\, a^{2}\right )}{15 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(130\)

Input:

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

Output:

1/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(15*a^(5/2)*2^(1/2)*arctanh( 
1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-6*(a-a*sin(d*x+c))^(5/2)+10*a* 
(a-a*sin(d*x+c))^(3/2)-30*(a-a*sin(d*x+c))^(1/2)*a^2)/a^3/cos(d*x+c)/(a+a* 
sin(d*x+c))^(1/2)/d
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.68 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 17\right )} \sin \left (d x + c\right ) - 16 \, \cos \left (d x + c\right ) - 17\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \] Input:

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/30*(15*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) + 4*(3 
*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - cos(d*x + c) - 17 
)*sin(d*x + c) - 16*cos(d*x + c) - 17)*sqrt(a*sin(d*x + c) + a))/(a*d*cos( 
d*x + c) + a*d*sin(d*x + c) + a*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^3(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\frac {15 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {15 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (12 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{30 \, d} \] Input:

integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

-1/30*(15*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos 
(-1/4*pi + 1/2*d*x + 1/2*c))) - 15*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/ 
2*c) + 1)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 4*sqrt(2)*(12*a^ 
(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 10*a^(9/2)*sin(-1/4*pi + 1/2*d*x 
+ 1/2*c)^3 + 15*a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^5*sgn(cos(-1/4* 
pi + 1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*sin(c + d*x)**3)/(sin(c + d*x) + 1),x 
))/a