\(\int x^3 \sqrt {a+a \cos (c+d x)} \, dx\) [143]

Optimal result

Integrand size = 18, antiderivative size = 110 \[ \int x^3 \sqrt {a+a \cos (c+d x)} \, dx=-\frac {96 \sqrt {a+a \cos (c+d x)}}{d^4}+\frac {12 x^2 \sqrt {a+a \cos (c+d x)}}{d^2}-\frac {48 x \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^3}+\frac {2 x^3 \sqrt {a+a \cos (c+d x)} \tan \left (\frac {c}{2}+\frac {d x}{2}\right )}{d} \] Output:

-96*(a+a*cos(d*x+c))^(1/2)/d^4+12*x^2*(a+a*cos(d*x+c))^(1/2)/d^2-48*x*(a+a 
*cos(d*x+c))^(1/2)*tan(1/2*d*x+1/2*c)/d^3+2*x^3*(a+a*cos(d*x+c))^(1/2)*tan 
(1/2*d*x+1/2*c)/d
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.48 \[ \int x^3 \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} \left (6 \left (-8+d^2 x^2\right )+d x \left (-24+d^2 x^2\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d^4} \] Input:

Integrate[x^3*Sqrt[a + a*Cos[c + d*x]],x]
 

Output:

(2*Sqrt[a*(1 + Cos[c + d*x])]*(6*(-8 + d^2*x^2) + d*x*(-24 + d^2*x^2)*Tan[ 
(c + d*x)/2]))/d^4
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3800, 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.

\(\displaystyle \int x^3 \sqrt {a \cos (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {6 \int -x^2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{d}+\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \int x^2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \int x^2 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{d}\right )\)

\(\Big \downarrow \) 3777

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \left (\frac {4 \int x \cos \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{d}-\frac {2 x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \left (\frac {4 \int x \sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{2}\right )dx}{d}-\frac {2 x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}\right )\)

\(\Big \downarrow \) 3777

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \left (\frac {4 \left (\frac {2 \int -\sin \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{d}+\frac {2 x \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}-\frac {2 x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \left (\frac {4 \left (\frac {2 x \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {2 \int \sin \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{d}\right )}{d}-\frac {2 x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \left (\frac {4 \left (\frac {2 x \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {2 \int \sin \left (\frac {c}{2}+\frac {d x}{2}\right )dx}{d}\right )}{d}-\frac {2 x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}\right )\)

\(\Big \downarrow \) 3118

\(\displaystyle \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {a \cos (c+d x)+a} \left (\frac {2 x^3 \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {6 \left (\frac {4 \left (\frac {4 \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^2}+\frac {2 x \sin \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}-\frac {2 x^2 \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d}\right )}{d}\right )\)

Input:

Int[x^3*Sqrt[a + a*Cos[c + d*x]],x]
 

Output:

Sqrt[a + a*Cos[c + d*x]]*Sec[c/2 + (d*x)/2]*((2*x^3*Sin[c/2 + (d*x)/2])/d 
- (6*((-2*x^2*Cos[c/2 + (d*x)/2])/d + (4*((4*Cos[c/2 + (d*x)/2])/d^2 + (2* 
x*Sin[c/2 + (d*x)/2])/d))/d))/d)
 

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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Simp[(2*a)^IntPart[n]*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e 
/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n]))   Int[(c + d*x)^m*Sin[e/2 + a 
*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && 
EqQ[a^2 - b^2, 0] && IntegerQ[n + 1/2] && (GtQ[n, 0] || IGtQ[m, 0])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.10 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.20

method	result	size
risch	\(-\frac {i \sqrt {2}\, \sqrt {a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{2} {\mathrm e}^{-i \left (d x +c \right )}}\, \left (d^{3} x^{3} {\mathrm e}^{i \left (d x +c \right )}+6 i d^{2} x^{2} {\mathrm e}^{i \left (d x +c \right )}-d^{3} x^{3}+6 i d^{2} x^{2}-24 d x \,{\mathrm e}^{i \left (d x +c \right )}-48 i {\mathrm e}^{i \left (d x +c \right )}+24 d x -48 i\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) d^{4}}\)	\(132\)

Input:

int(x^3*(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-I*2^(1/2)*(a*(exp(I*(d*x+c))+1)^2*exp(-I*(d*x+c)))^(1/2)/(exp(I*(d*x+c))+ 
1)*(d^3*x^3*exp(I*(d*x+c))+6*I*d^2*x^2*exp(I*(d*x+c))-d^3*x^3+6*I*d^2*x^2- 
24*d*x*exp(I*(d*x+c))-48*I*exp(I*(d*x+c))+24*d*x-48*I)/d^4
 

Fricas [F(-2)]

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

integrate(x^3*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**3*(a+a*cos(d*x+c))**(1/2),x)
 

Output:

Integral(x**3*sqrt(a*(cos(c + d*x) + 1)), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (94) = 188\).

Time = 0.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.87 \[ \int x^3 \sqrt {a+a \cos (c+d x)} \, dx=-\frac {2 \, {\left (\sqrt {2} \sqrt {a} c^{3} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, {\left (\sqrt {2} {\left (d x + c\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a} c^{2} + 3 \, {\left (\sqrt {2} {\left (d x + c\right )}^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, \sqrt {2} {\left (d x + c\right )} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a} c - {\left (\sqrt {2} {\left (d x + c\right )}^{3} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, \sqrt {2} {\left (d x + c\right )}^{2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, \sqrt {2} {\left (d x + c\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}\right )}}{d^{4}} \] Input:

integrate(x^3*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

-2*(sqrt(2)*sqrt(a)*c^3*sin(1/2*d*x + 1/2*c) - 3*(sqrt(2)*(d*x + c)*sin(1/ 
2*d*x + 1/2*c) + 2*sqrt(2)*cos(1/2*d*x + 1/2*c))*sqrt(a)*c^2 + 3*(sqrt(2)* 
(d*x + c)^2*sin(1/2*d*x + 1/2*c) + 4*sqrt(2)*(d*x + c)*cos(1/2*d*x + 1/2*c 
) - 8*sqrt(2)*sin(1/2*d*x + 1/2*c))*sqrt(a)*c - (sqrt(2)*(d*x + c)^3*sin(1 
/2*d*x + 1/2*c) + 6*sqrt(2)*(d*x + c)^2*cos(1/2*d*x + 1/2*c) - 24*sqrt(2)* 
(d*x + c)*sin(1/2*d*x + 1/2*c) - 48*sqrt(2)*cos(1/2*d*x + 1/2*c))*sqrt(a)) 
/d^4
 

Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.89 \[ \int x^3 \sqrt {a+a \cos (c+d x)} \, dx=2 \, \sqrt {2} \sqrt {a} {\left (\frac {6 \, {\left (d^{2} x^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d^{4}} + \frac {{\left (d^{3} x^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 24 \, d x \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d^{4}}\right )} \] Input:

integrate(x^3*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

2*sqrt(2)*sqrt(a)*(6*(d^2*x^2*sgn(cos(1/2*d*x + 1/2*c)) - 8*sgn(cos(1/2*d* 
x + 1/2*c)))*cos(1/2*d*x + 1/2*c)/d^4 + (d^3*x^3*sgn(cos(1/2*d*x + 1/2*c)) 
 - 24*d*x*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c)/d^4)
 

Mupad [B] (verification not implemented)

Time = 40.99 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75 \[ \int x^3 \sqrt {a+a \cos (c+d x)} \, dx=-\frac {2\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}\,\left (48\,\cos \left (c+d\,x\right )-6\,d^2\,x^2-6\,d^2\,x^2\,\cos \left (c+d\,x\right )-d^3\,x^3\,\sin \left (c+d\,x\right )+24\,d\,x\,\sin \left (c+d\,x\right )+48\right )}{d^4\,\left (\cos \left (c+d\,x\right )+1\right )} \] Input:

int(x^3*(a + a*cos(c + d*x))^(1/2),x)
 

Output:

-(2*(a*(cos(c + d*x) + 1))^(1/2)*(48*cos(c + d*x) - 6*d^2*x^2 - 6*d^2*x^2* 
cos(c + d*x) - d^3*x^3*sin(c + d*x) + 24*d*x*sin(c + d*x) + 48))/(d^4*(cos 
(c + d*x) + 1))
 

Reduce [F]

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

int(x^3*(a+a*cos(d*x+c))^(1/2),x)
 

Output:

sqrt(a)*int(sqrt(cos(c + d*x) + 1)*x**3,x)