\(\int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\) [303]

Optimal result

Integrand size = 26, antiderivative size = 76 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} c^{3/2} f}+\frac {a \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.41 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {a \sec (e+f x) \left (2 \sqrt {c} (1+\sin (e+f x))-\sqrt {2} \arctan \left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) (-1+\sin (e+f x)) \sqrt {-c (1+\sin (e+f x))}\right )}{2 c^{3/2} f \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^(3/2),x]
 

Output:

(a*Sec[e + f*x]*(2*Sqrt[c]*(1 + Sin[e + f*x]) - Sqrt[2]*ArcTan[Sqrt[-(c*(1 
 + Sin[e + f*x]))]/(Sqrt[2]*Sqrt[c])]*(-1 + Sin[e + f*x])*Sqrt[-(c*(1 + Si 
n[e + f*x]))]))/(2*c^(3/2)*f*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3215, 3042, 3159, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

Int[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^(3/2),x]
 

Output:

a*c*(-(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/ 
(Sqrt[2]*c^(5/2)*f)) + Cos[e + f*x]/(c*f*(c - c*Sin[e + f*x])^(3/2)))
 

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 3159

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[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 = 0.77 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.58

method	result	size
default	\(\frac {a \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{2 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(120\)
parts	\(-\frac {a \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sin \left (f x +e \right )-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(245\)

Input:

int((a+sin(f*x+e)*a)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*a/c^(5/2)*(2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2 
))*c*sin(f*x+e)-2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/ 
2))*c+2*(c*(1+sin(f*x+e)))^(1/2)*c^(1/2))*(c*(1+sin(f*x+e)))^(1/2)/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. 255 vs. \(2 (65) = 130\).

Time = 0.09 (sec) , antiderivative size = 255, normalized size of antiderivative = 3.36 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\frac {\sqrt {2} {\left (a c \cos \left (f x + e\right )^{2} - a c \cos \left (f x + e\right ) - 2 \, a c + {\left (a c \cos \left (f x + e\right ) + 2 \, a c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 4 \, {\left (a \cos \left (f x + e\right ) + a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))**(3/2),x)
 

Output:

a*(Integral(sin(e + f*x)/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sq 
rt(-c*sin(e + f*x) + c)), x) + Integral(1/(-c*sqrt(-c*sin(e + f*x) + c)*si 
n(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x))
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)/(-c*sin(f*x + e) + c)^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),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. \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((a + a*sin(e + f*x))/(c - c*sin(e + f*x))^(3/2),x)
 

Output:

int((a + a*sin(e + f*x))/(c - c*sin(e + f*x))^(3/2), x)
 

Reduce [F]

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

int((a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x)
 

Output:

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