\(\int x^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx\) [87]

Optimal result

Integrand size = 33, antiderivative size = 155 \[ \int x^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=-\frac {6 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^4}+\frac {3 x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {6 x \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f^3}+\frac {x^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f} \] Output:

-6*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^4+3*x^2*(a-a*sin(f*x+e) 
)^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2-6*x*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x 
+e))^(1/2)*tan(f*x+e)/f^3+x^3*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2 
)*tan(f*x+e)/f
 

Mathematica [A] (verified)

Time = 2.81 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.39 \[ \int x^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\frac {\sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} \left (-6+3 f^2 x^2+f x \left (-6+f^2 x^2\right ) \tan (e+f x)\right )}{f^4} \] Input:

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

Output:

(Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(-6 + 3*f^2*x^2 + f*x 
*(-6 + f^2*x^2)*Tan[e + f*x]))/f^4
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5115, 3042, 3777, 25, 3042, 3777, 3042, 3777, 25, 3042, 3118}

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.

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

\(\Big \downarrow \) 5115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3118

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

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]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.01 \[ \int x^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=-\sqrt {a} \sqrt {c} {\left (\frac {3 \, {\left (f^{2} x^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (f x + e\right )}{f^{4}} + \frac {{\left (f^{3} x^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 6 \, f x \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (f x + e\right )}{f^{4}}\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 16.53 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.72 \[ \int x^3 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=-\frac {\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (6\,\cos \left (2\,e+2\,f\,x\right )-3\,f^2\,x^2+6\,f\,x\,\sin \left (2\,e+2\,f\,x\right )-3\,f^2\,x^2\,\cos \left (2\,e+2\,f\,x\right )-f^3\,x^3\,\sin \left (2\,e+2\,f\,x\right )+6\right )}{f^4\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

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

Output:

-((-a*(sin(e + f*x) - 1))^(1/2)*(c*(sin(e + f*x) + 1))^(1/2)*(6*cos(2*e + 
2*f*x) - 3*f^2*x^2 + 6*f*x*sin(2*e + 2*f*x) - 3*f^2*x^2*cos(2*e + 2*f*x) - 
 f^3*x^3*sin(2*e + 2*f*x) + 6))/(f^4*(cos(2*e + 2*f*x) + 1))
 

Reduce [F]

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

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

Output:

sqrt(c)*sqrt(a)*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*x**3, 
x)