\(\int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx\) [93]

Optimal result

Integrand size = 33, antiderivative size = 393 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {6 c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^4}+\frac {3 c x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}+\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{8 f^3}-\frac {3 c x^3 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f}-\frac {3 c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{8 f^4}+\frac {3 c x^2 \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f^2}+\frac {x^3 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {6 c x \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f^3}-\frac {3 c x \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{4 f^3} \] Output:

-6*c*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^4+3*c*x^2*(a-a*sin(f* 
x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2+3/8*c*x*sec(f*x+e)*(a-a*sin(f*x+e)) 
^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^3-3/4*c*x^3*sec(f*x+e)*(a-a*sin(f*x+e))^(1 
/2)*(c+c*sin(f*x+e))^(1/2)/f-3/8*c*sin(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c* 
sin(f*x+e))^(1/2)/f^4+3/4*c*x^2*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^3*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+ 
e))^(5/2)/c/f-6*c*x*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)*tan(f*x+ 
e)/f^3-3/4*c*x*sin(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)*ta 
n(f*x+e)/f^3
 

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.29 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} \left (-f x \left (-3+2 f^2 x^2\right ) \cos (2 (e+f x)) \sec (e+f x)+\left (-3+6 f^2 x^2\right ) \sin (e+f x)+8 \left (-6+3 f^2 x^2+f x \left (-6+f^2 x^2\right ) \tan (e+f x)\right )\right )}{8 f^4} \] Input:

Integrate[x^3*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]]*(-(f*x*(-3 + 2*f^2* 
x^2)*Cos[2*(e + f*x)]*Sec[e + f*x]) + (-3 + 6*f^2*x^2)*Sin[e + f*x] + 8*(- 
6 + 3*f^2*x^2 + f*x*(-6 + f^2*x^2)*Tan[e + f*x])))/(8*f^4)
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.52, 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.

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

\(\Big \downarrow \) 5115

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

\(\Big \downarrow \) 4922

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3798

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

\(\Big \downarrow \) 2009

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

Input:

Int[x^3*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^3*(c + 
c*Sin[e + f*x])^2)/(2*c*f) - (3*(-1/4*(c^2*x)/f^2 + (c^2*x^3)/2 + (4*c^2*C 
os[e + f*x])/f^3 - (2*c^2*x^2*Cos[e + f*x])/f + (4*c^2*x*Sin[e + f*x])/f^2 
 + (c^2*Cos[e + f*x]*Sin[e + f*x])/(4*f^3) - (c^2*x^2*Cos[e + f*x]*Sin[e + 
 f*x])/(2*f) + (c^2*x*Sin[e + f*x]^2)/(2*f^2)))/(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]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

integrate(x^3*(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(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(x^3*(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^3, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1487 vs. \(2 (345) = 690\).

Time = 0.45 (sec) , antiderivative size = 1487, normalized size of antiderivative = 3.78 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/32*sqrt(a)*sqrt(c)*((pi^3*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)) - 3*pi^2*(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)) + 3*pi*(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)) - (pi - 2*f*x - 2*e)^3*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)) - 6*pi^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)) + 12*pi*(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)) - 
 6*(pi - 2*f*x - 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)) + 12*pi*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)) - 12*(pi - 2*f*x - 2*e)*c*e^2*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 8*c*e^ 
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*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)) + 6*(pi - 2*f*x - 2*e)*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(si 
n(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*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(2*f*x + 2*e)/f^3 - 24*(pi^2*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)) - 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*p 
i + 1/2*f*x + 1/2*e)) + (pi - 2*f*x - 2*e)^2*c*sgn(cos(-1/4*pi + 1/2*f*...
 

Mupad [B] (verification not implemented)

Time = 17.94 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.55 \[ \int x^3 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {c\,\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (3\,\sin \left (e+f\,x\right )+96\,\cos \left (2\,e+2\,f\,x\right )+3\,\sin \left (3\,e+3\,f\,x\right )-48\,f^2\,x^2-6\,f\,x\,\cos \left (3\,e+3\,f\,x\right )+96\,f\,x\,\sin \left (2\,e+2\,f\,x\right )+4\,f^3\,x^3\,\cos \left (e+f\,x\right )-6\,f^2\,x^2\,\sin \left (e+f\,x\right )-6\,f\,x\,\cos \left (e+f\,x\right )-48\,f^2\,x^2\,\cos \left (2\,e+2\,f\,x\right )+4\,f^3\,x^3\,\cos \left (3\,e+3\,f\,x\right )-6\,f^2\,x^2\,\sin \left (3\,e+3\,f\,x\right )-16\,f^3\,x^3\,\sin \left (2\,e+2\,f\,x\right )+96\right )}{16\,f^4\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

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

Output:

-(c*(-a*(sin(e + f*x) - 1))^(1/2)*(c*(sin(e + f*x) + 1))^(1/2)*(3*sin(e + 
f*x) + 96*cos(2*e + 2*f*x) + 3*sin(3*e + 3*f*x) - 48*f^2*x^2 - 6*f*x*cos(3 
*e + 3*f*x) + 96*f*x*sin(2*e + 2*f*x) + 4*f^3*x^3*cos(e + f*x) - 6*f^2*x^2 
*sin(e + f*x) - 6*f*x*cos(e + f*x) - 48*f^2*x^2*cos(2*e + 2*f*x) + 4*f^3*x 
^3*cos(3*e + 3*f*x) - 6*f^2*x^2*sin(3*e + 3*f*x) - 16*f^3*x^3*sin(2*e + 2* 
f*x) + 96))/(16*f^4*(cos(2*e + 2*f*x) + 1))
 

Reduce [F]

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

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