3.2.83 \(\int \frac {x \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx\) [183]

3.2.83.1 Optimal result

Integrand size = 31, antiderivative size = 171 \[ \int \frac {x \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=-\frac {a}{c f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}-\frac {a x \sec (e+f x)}{c f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a \sin (e+f x)}{c f^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}+\frac {a x \tan (e+f x)}{c f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}} \]

output

-a/c/f^2/(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)-a*x*sec(f*x+e)/c/f/ 
(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)+a*sin(f*x+e)/c/f^2/(a-a*sin( 
f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2)+a*x*tan(f*x+e)/c/f/(a-a*sin(f*x+e))^( 
1/2)/(c+c*sin(f*x+e))^(1/2)
 

3.2.83.2 Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.88 \[ \int \frac {x \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=-\frac {\left ((-1+f x) \cos \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}+f x\right )+\sin \left (\frac {e}{2}\right )+f x \sin \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}+f x\right )\right ) \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)}}{c^2 f^2 \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

input

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

output

-((((-1 + f*x)*Cos[e/2] + Cos[e/2 + f*x] + Sin[e/2] + f*x*Sin[e/2] - Sin[e 
/2 + f*x])*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]])/(c^2*f^2*( 
Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/ 
2] + Sin[(e + f*x)/2])^3))
 

3.2.83.3 Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.54, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {5115, 7292, 27, 7293, 2009}

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.

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

\(\Big \downarrow \) 5115

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \cos (e+f x) \left (\frac {\tan (e+f x)}{f^2}-\frac {\sec (e+f x)}{f^2}-\frac {x \sec ^2(e+f x)}{f}+\frac {x \tan (e+f x) \sec (e+f x)}{f}\right )}{c \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}\)

input

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

output

(a*Cos[e + f*x]*(-(Sec[e + f*x]/f^2) - (x*Sec[e + f*x]^2)/f + Tan[e + f*x] 
/f^2 + (x*Sec[e + f*x]*Tan[e + f*x])/f))/(c*Sqrt[a - a*Sin[e + f*x]]*Sqrt[ 
c + c*Sin[e + f*x]])
 

3.2.83.3.1 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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5115

Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)* 
((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m] 
*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPa 
rt[m]/Cos[e + f*x]^(2*FracPart[m]))   Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c 
 + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && 
 EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ[2*m] && 
IGeQ[n - m, 0]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.2.83.4 Maple [F]

\[\int \frac {x \sqrt {a -\sin \left (f x +e \right ) a}}{\left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

input

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

output

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

3.2.83.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.42 \[ \int \frac {x \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=-\frac {{\left (f x + \cos \left (f x + e\right )\right )} \sqrt {-a \sin \left (f x + e\right ) + a} \sqrt {c \sin \left (f x + e\right ) + c}}{c^{2} f^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + c^{2} f^{2} \cos \left (f x + e\right )} \]

input

integrate(x*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

output

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

3.2.83.6 Sympy [F]

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

input

integrate(x*(a-a*sin(f*x+e))**(1/2)/(c+c*sin(f*x+e))**(3/2),x)
 

output

Integral(x*sqrt(-a*(sin(e + f*x) - 1))/(c*(sin(e + f*x) + 1))**(3/2), x)
 

3.2.83.7 Maxima [F]

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

input

integrate(x*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

output

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

3.2.83.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (155) = 310\).

Time = 0.38 (sec) , antiderivative size = 446, normalized size of antiderivative = 2.61 \[ \int \frac {x \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (\pi \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} - {\left (\pi - 2 \, f x - 2 \, e\right )} \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} - 2 \, \sqrt {c} e \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} + 2 \, \pi \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} - 2 \, {\left (\pi - 2 \, f x - 2 \, e\right )} \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} - 4 \, \sqrt {c} e \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} + 8 \, \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{3} + \pi \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - {\left (\pi - 2 \, f x - 2 \, e\right )} \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, \sqrt {c} e \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 8 \, \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )\right )} \sqrt {a}}{4 \, {\left (\sqrt {2} c^{2} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{4} - 2 \, \sqrt {2} c^{2} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (-\frac {1}{8} \, \pi + \frac {1}{4} \, f x + \frac {1}{4} \, e\right )^{2} + \sqrt {2} c^{2} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} f} \]

input

integrate(x*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

output

1/4*sqrt(2)*(pi*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 
1/4*f*x + 1/4*e)^4 - (pi - 2*f*x - 2*e)*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x 
+ 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^4 - 2*sqrt(c)*e*sgn(sin(-1/4*pi + 
 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^4 + 2*pi*sqrt(c)*sgn(sin 
(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 - 2*(pi - 2* 
f*x - 2*e)*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f 
*x + 1/4*e)^2 - 4*sqrt(c)*e*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*p 
i + 1/4*f*x + 1/4*e)^2 + 8*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*tan 
(-1/8*pi + 1/4*f*x + 1/4*e)^3 + pi*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2 
*e)) - (pi - 2*f*x - 2*e)*sqrt(c)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2* 
sqrt(c)*e*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 8*sqrt(c)*sgn(sin(-1/4*pi 
+ 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e))*sqrt(a)/((sqrt(2)*c^2* 
f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^4 - 2 
*sqrt(2)*c^2*f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 
 1/4*e)^2 + sqrt(2)*c^2*f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*f)
 

3.2.83.9 Mupad [B] (verification not implemented)

Time = 28.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.51 \[ \int \frac {x \sqrt {a-a \sin (e+f x)}}{(c+c \sin (e+f x))^{3/2}} \, dx=-\frac {\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (\cos \left (2\,e+2\,f\,x\right )+2\,f\,x\,\cos \left (e+f\,x\right )+1-\cos \left (e+f\,x\right )\,2{}\mathrm {i}-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{c\,f^2\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}} \]

input

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

output

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