\(\int \frac {(a+a \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx\) [310]

Optimal result

Integrand size = 28, antiderivative size = 115 \[ \int \frac {(a+a \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {4 \sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \] Output:

4*2^(1/2)*a^2*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/2 
))/c^(1/2)/f-2/3*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)-4*a^2*cos(f*x 
+e)/f/(c-c*sin(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 8.78 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.13 \[ \int \frac {(a+a \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((24+24 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )+15 \cos \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {3}{2} (e+f x)\right )+15 \sin \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {3}{2} (e+f x)\right )\right )}{3 f \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^2/Sqrt[c - c*Sin[e + f*x]],x]
 

Output:

-1/3*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*((24 + 24*I)*(-1)^(1/4)*Ar 
cTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])] + 15*Cos[(e + f*x)/2] 
- Cos[(3*(e + f*x))/2] + 15*Sin[(e + f*x)/2] + Sin[(3*(e + f*x))/2]))/(f*S 
qrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 3215, 3042, 3158, 3042, 3158, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3158

\(\displaystyle a^2 c^2 \left (\frac {2 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {2 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3158

\(\displaystyle a^2 c^2 \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 3128

\(\displaystyle a^2 c^2 \left (\frac {2 \left (-\frac {4 \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle a^2 c^2 \left (\frac {2 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )\)

Input:

Int[(a + a*Sin[e + f*x])^2/Sqrt[c - c*Sin[e + f*x]],x]
 

Output:

a^2*c^2*((-2*Cos[e + f*x]^3)/(3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (2*((2*S 
qrt[2]*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]) 
/(c^(3/2)*f) - (2*Cos[e + f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])))/c)
 

Defintions of rubi rules used

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 3158

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

rule 3215

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97

method	result	size
default	\(-\frac {2 \left (-1+\sin \left (f x +e \right )\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (6 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-\left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}-6 c \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(112\)
parts	\(-\frac {a^{2} \left (-1+\sin \left (f x +e \right )\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\sqrt {c}\, \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a^{2} \left (-1+\sin \left (f x +e \right )\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (-3 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+2 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{2} \left (-1+\sin \left (f x +e \right )\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(270\)

Input:

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

Output:

-2/3*(-1+sin(f*x+e))*(c*(1+sin(f*x+e)))^(1/2)*a^2*(6*c^(3/2)*2^(1/2)*arcta 
nh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))-(c*(1+sin(f*x+e)))^(3/2)- 
6*c*(c*(1+sin(f*x+e)))^(1/2))/c^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (100) = 200\).

Time = 0.09 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.07 \[ \int \frac {(a+a \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, {\left (\frac {3 \, \sqrt {2} {\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c \sin \left (f x + e\right ) + a^{2} c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + {\left (a^{2} \cos \left (f x + e\right )^{2} - 7 \, a^{2} \cos \left (f x + e\right ) - 8 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right ) + 8 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{3 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \] Input:

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

Output:

2/3*(3*sqrt(2)*(a^2*c*cos(f*x + e) - a^2*c*sin(f*x + e) + a^2*c)*log(-(cos 
(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x 
+ e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2)/ 
(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqr 
t(c) + (a^2*cos(f*x + e)^2 - 7*a^2*cos(f*x + e) - 8*a^2 - (a^2*cos(f*x + e 
) + 8*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c* 
f*sin(f*x + e) + c*f)
 

Sympy [F]

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

integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(1/2),x)
 

Output:

a**2*(Integral(2*sin(e + f*x)/sqrt(-c*sin(e + f*x) + c), x) + Integral(sin 
(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x) + Integral(1/sqrt(-c*sin(e + f* 
x) + c), x))
 

Maxima [F]

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

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

Output:

integrate((a*sin(f*x + e) + a)^2/sqrt(-c*sin(f*x + e) + c), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^2}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(1/2),x)
 

Output:

int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(1/2), x)
 

Reduce [F]

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

int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(1/2),x)
 

Output:

(sqrt(c)*a**2*( - int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x) - 1),x) - in 
t((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x) - 1),x) - 2*in 
t((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) - 1),x)))/c