\(\int x^2 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx\) [94]

Optimal result

Integrand size = 33, antiderivative size = 265 \[ \int x^2 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {2 c x \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {3 c x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f}+\frac {c x \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{2 f^2}+\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {2 c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f^3}-\frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{4 f^3} \] Output:

2*c*x*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2-3/4*c*x^2*sec(f*x+ 
e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f+1/2*c*x*sin(f*x+e)*(a-a 
*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2+1/2*x^2*sec(f*x+e)*(a-a*sin( 
f*x+e))^(1/2)*(c+c*sin(f*x+e))^(5/2)/c/f-2*c*(a-a*sin(f*x+e))^(1/2)*(c+c*s 
in(f*x+e))^(1/2)*tan(f*x+e)/f^3-1/4*c*sin(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c 
+c*sin(f*x+e))^(1/2)*tan(f*x+e)/f^3
 

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.36 \[ \int x^2 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} \left (16 f x-\left (-1+2 f^2 x^2\right ) \cos (2 (e+f x)) \sec (e+f x)+4 f x \sin (e+f x)+8 \left (-2+f^2 x^2\right ) \tan (e+f x)\right )}{8 f^3} \] Input:

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

Output:

(c*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(16*f*x - (-1 + 2*f 
^2*x^2)*Cos[2*(e + f*x)]*Sec[e + f*x] + 4*f*x*Sin[e + f*x] + 8*(-2 + f^2*x 
^2)*Tan[e + f*x]))/(8*f^3)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.57, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {5115, 4922, 3042, 3798, 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.

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

\(\Big \downarrow \) 5115

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

\(\Big \downarrow \) 4922

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3798

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*((x^2*(c + 
c*Sin[e + f*x])^2)/(2*c*f) - ((3*c^2*x^2)/4 - (2*c^2*x*Cos[e + f*x])/f + ( 
2*c^2*Sin[e + f*x])/f^2 - (c^2*x*Cos[e + f*x]*Sin[e + f*x])/(2*f) + (c^2*S 
in[e + f*x]^2)/(4*f^2))/(c*f))
 

Defintions of rubi rules used

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ 
m, 0] || NeQ[a^2 - b^2, 0])
 

rule 4922

Int[Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sin[(c 
_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Simp[(e + f*x)^m*((a + b*Sin[c + d*x 
])^(n + 1)/(b*d*(n + 1))), x] - Simp[f*(m/(b*d*(n + 1)))   Int[(e + f*x)^(m 
 - 1)*(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, 
x] && IGtQ[m, 0] && NeQ[n, -1]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 739 vs. \(2 (233) = 466\).

Time = 0.32 (sec) , antiderivative size = 739, normalized size of antiderivative = 2.79 \[ \int x^2 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*sqrt(a)*sqrt(c)*((pi^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin( 
-1/4*pi + 1/2*f*x + 1/2*e)) - 2*pi*(pi - 2*f*x - 2*e)*c*sgn(cos(-1/4*pi + 
1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + (pi - 2*f*x - 2*e) 
^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e 
)) - 4*pi*c*e*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f* 
x + 1/2*e)) + 4*(pi - 2*f*x - 2*e)*c*e*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) 
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*c*e^2*sgn(cos(-1/4*pi + 1/2*f*x + 
 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c*sgn(cos(-1/4*pi + 1/2*f 
*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(2*f*x + 2*e)/f^2 - 1 
6*(pi*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/ 
2*e)) - (pi - 2*f*x - 2*e)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(- 
1/4*pi + 1/2*f*x + 1/2*e)) - 2*c*e*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn 
(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(f*x + e)/f^2 - 2*(pi*c*sgn(cos(-1/4* 
pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - (pi - 2*f*x - 
 2*e)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/ 
2*e)) - 2*c*e*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f* 
x + 1/2*e)))*sin(2*f*x + 2*e)/f^2 - 4*(pi^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 
1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*pi*(pi - 2*f*x - 2*e)*c*sg 
n(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + (p 
i - 2*f*x - 2*e)^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*p...
 

Mupad [B] (verification not implemented)

Time = 17.37 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.60 \[ \int x^2 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c\,\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (\cos \left (e+f\,x\right )+\cos \left (3\,e+3\,f\,x\right )-16\,\sin \left (2\,e+2\,f\,x\right )+16\,f\,x+16\,f\,x\,\cos \left (2\,e+2\,f\,x\right )+2\,f\,x\,\sin \left (3\,e+3\,f\,x\right )-2\,f^2\,x^2\,\cos \left (e+f\,x\right )+2\,f\,x\,\sin \left (e+f\,x\right )-2\,f^2\,x^2\,\cos \left (3\,e+3\,f\,x\right )+8\,f^2\,x^2\,\sin \left (2\,e+2\,f\,x\right )\right )}{8\,f^3\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

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

Output:

(c*(-a*(sin(e + f*x) - 1))^(1/2)*(c*(sin(e + f*x) + 1))^(1/2)*(cos(e + f*x 
) + cos(3*e + 3*f*x) - 16*sin(2*e + 2*f*x) + 16*f*x + 16*f*x*cos(2*e + 2*f 
*x) + 2*f*x*sin(3*e + 3*f*x) - 2*f^2*x^2*cos(e + f*x) + 2*f*x*sin(e + f*x) 
 - 2*f^2*x^2*cos(3*e + 3*f*x) + 8*f^2*x^2*sin(2*e + 2*f*x)))/(8*f^3*(cos(2 
*e + 2*f*x) + 1))
 

Reduce [F]

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

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

Output:

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