\(\int \frac {\sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [62]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d}+\frac {4 \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a d} \] 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+4/3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/3*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 = 0.55 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\left ((-6-6 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right )-2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d \sqrt {a (1+\sin (c+d x))}} \] Input:

Integrate[Sin[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

-1/3*(((-6 - 6*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + 
 d*x)/4])] - 2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)*(Cos[(c + d*x)/2] 
+ Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Sin[c + d*x])])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3238, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3238

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {4 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {6 a \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{3 a}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a d}\)

\(\Big \downarrow \) 219

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

Input:

Int[Sin[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(-2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*a*d) + ((-3*Sqrt[2]*Sqrt[a]* 
ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (4 
*a*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/(3*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 3238

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), 
x_Symbol] :> Simp[(-Cos[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*(b*(m + 1) - a*Si 
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  ! 
LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (-3 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )+2 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}\right )}{3 a^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(96\)

Input:

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

Output:

1/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-3*a^(3/2)*2^(1/2)*arctanh(1 
/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+2*(a-a*sin(d*x+c))^(3/2))/a^2/c 
os(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. 209 vs. \(2 (88) = 176\).

Time = 0.09 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.99 \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/6*(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) - 4*(cos 
(d*x + c)^2 + (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)*sqrt(a*s 
in(d*x + c) + a))/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15 \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {\frac {8 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {3 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{6 \, d} \] Input:

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

Output:

-1/6*(8*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3/(sqrt(a)*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)/( 
sqrt(a)*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)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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