\(\int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx\) [171]

Optimal result

Integrand size = 18, antiderivative size = 262 \[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {4 i x^2 \arctan \left (e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cos (c+d x)}}+\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {8 i x \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {16 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,-i e^{\frac {1}{2} i (c+d x)}\right )}{d^3 \sqrt {a+a \cos (c+d x)}}+\frac {16 \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,i e^{\frac {1}{2} i (c+d x)}\right )}{d^3 \sqrt {a+a \cos (c+d x)}} \] Output:

-4*I*x^2*arctan(exp(1/2*I*(d*x+c)))*cos(1/2*d*x+1/2*c)/d/(a+a*cos(d*x+c))^ 
(1/2)+8*I*x*cos(1/2*d*x+1/2*c)*polylog(2,-I*exp(1/2*I*(d*x+c)))/d^2/(a+a*c 
os(d*x+c))^(1/2)-8*I*x*cos(1/2*d*x+1/2*c)*polylog(2,I*exp(1/2*I*(d*x+c)))/ 
d^2/(a+a*cos(d*x+c))^(1/2)-16*cos(1/2*d*x+1/2*c)*polylog(3,-I*exp(1/2*I*(d 
*x+c)))/d^3/(a+a*cos(d*x+c))^(1/2)+16*cos(1/2*d*x+1/2*c)*polylog(3,I*exp(1 
/2*I*(d*x+c)))/d^3/(a+a*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.56 \[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-i d^2 x^2 \arctan \left (e^{\frac {1}{2} i (c+d x)}\right )+2 i d x \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )-2 i d x \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )-4 \operatorname {PolyLog}\left (3,-i e^{\frac {1}{2} i (c+d x)}\right )+4 \operatorname {PolyLog}\left (3,i e^{\frac {1}{2} i (c+d x)}\right )\right )}{d^3 \sqrt {a (1+\cos (c+d x))}} \] Input:

Integrate[x^2/Sqrt[a + a*Cos[c + d*x]],x]
 

Output:

(4*Cos[(c + d*x)/2]*((-I)*d^2*x^2*ArcTan[E^((I/2)*(c + d*x))] + (2*I)*d*x* 
PolyLog[2, (-I)*E^((I/2)*(c + d*x))] - (2*I)*d*x*PolyLog[2, I*E^((I/2)*(c 
+ d*x))] - 4*PolyLog[3, (-I)*E^((I/2)*(c + d*x))] + 4*PolyLog[3, I*E^((I/2 
)*(c + d*x))]))/(d^3*Sqrt[a*(1 + Cos[c + d*x])])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.64, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 3800, 3042, 4669, 3011, 2720, 7143}

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.

\(\displaystyle \int \frac {x^2}{\sqrt {a \cos (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

\(\displaystyle \frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {4 \int x \log \left (1-i e^{\frac {1}{2} i (c+d x)}\right )dx}{d}+\frac {4 \int x \log \left (1+i e^{\frac {1}{2} i (c+d x)}\right )dx}{d}-\frac {4 i x^2 \arctan \left (e^{\frac {1}{2} i (c+d x)}\right )}{d}\right )}{\sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )}{d}-\frac {2 i \int \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )dx}{d}\right )}{d}-\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )}{d}-\frac {2 i \int \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )dx}{d}\right )}{d}-\frac {4 i x^2 \arctan \left (e^{\frac {1}{2} i (c+d x)}\right )}{d}\right )}{\sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )}{d}-\frac {4 \int e^{-\frac {1}{2} i (c+d x)} \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )de^{\frac {1}{2} i (c+d x)}}{d^2}\right )}{d}-\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )}{d}-\frac {4 \int e^{-\frac {1}{2} i (c+d x)} \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )de^{\frac {1}{2} i (c+d x)}}{d^2}\right )}{d}-\frac {4 i x^2 \arctan \left (e^{\frac {1}{2} i (c+d x)}\right )}{d}\right )}{\sqrt {a \cos (c+d x)+a}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {4 i x^2 \arctan \left (e^{\frac {1}{2} i (c+d x)}\right )}{d}+\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )}{d}-\frac {4 \operatorname {PolyLog}\left (3,-i e^{\frac {1}{2} i (c+d x)}\right )}{d^2}\right )}{d}-\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )}{d}-\frac {4 \operatorname {PolyLog}\left (3,i e^{\frac {1}{2} i (c+d x)}\right )}{d^2}\right )}{d}\right )}{\sqrt {a \cos (c+d x)+a}}\)

Input:

Int[x^2/Sqrt[a + a*Cos[c + d*x]],x]
 

Output:

(Cos[c/2 + (d*x)/2]*(((-4*I)*x^2*ArcTan[E^((I/2)*(c + d*x))])/d + (4*(((2* 
I)*x*PolyLog[2, (-I)*E^((I/2)*(c + d*x))])/d - (4*PolyLog[3, (-I)*E^((I/2) 
*(c + d*x))])/d^2))/d - (4*(((2*I)*x*PolyLog[2, I*E^((I/2)*(c + d*x))])/d 
- (4*PolyLog[3, I*E^((I/2)*(c + d*x))])/d^2))/d))/Sqrt[a + a*Cos[c + d*x]]
 

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

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])
 

rule 4669

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x 
))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [F]

\[\int \frac {x^{2}}{\sqrt {a +a \cos \left (d x +c \right )}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {x^{2}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:

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

Output:

integral(x^2/sqrt(a*cos(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {x^{2}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \] Input:

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

Output:

Integral(x**2/sqrt(a*(cos(c + d*x) + 1)), x)
 

Maxima [F]

\[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {x^{2}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:

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

Output:

-2*(sqrt(2)*d^2*x^2*sin(1/2*d*x + 1/2*c) - 4*sqrt(2)*d*x*cos(1/2*d*x + 1/2 
*c) + (sqrt(2)*d^2*x^2*sin(1/2*d*x + 1/2*c) - 4*sqrt(2)*d*x*cos(1/2*d*x + 
1/2*c))*cos(d*x + c) - (sqrt(2)*a*d^5*cos(d*x + c)^2 + sqrt(2)*a*d^5*sin(d 
*x + c)^2 + 2*sqrt(2)*a*d^5*cos(d*x + c) + sqrt(2)*a*d^5)*integrate((x^2*c 
os(2*d*x + 2*c)*cos(1/2*d*x + 1/2*c) + 2*x^2*cos(d*x + c)*cos(1/2*d*x + 1/ 
2*c) + x^2*sin(2*d*x + 2*c)*sin(1/2*d*x + 1/2*c) + 2*x^2*sin(d*x + c)*sin( 
1/2*d*x + 1/2*c) + x^2*cos(1/2*d*x + 1/2*c))/(a*d^2*cos(2*d*x + 2*c)^2 + 4 
*a*d^2*cos(d*x + c)^2 + a*d^2*sin(2*d*x + 2*c)^2 + 4*a*d^2*sin(2*d*x + 2*c 
)*sin(d*x + c) + 4*a*d^2*sin(d*x + c)^2 + 4*a*d^2*cos(d*x + c) + a*d^2 + 2 
*(2*a*d^2*cos(d*x + c) + a*d^2)*cos(2*d*x + 2*c)), x) + 4*(sqrt(2)*a*d^4*c 
os(d*x + c)^2 + sqrt(2)*a*d^4*sin(d*x + c)^2 + 2*sqrt(2)*a*d^4*cos(d*x + c 
) + sqrt(2)*a*d^4)*integrate((x*cos(1/2*d*x + 1/2*c)*sin(2*d*x + 2*c) + 2* 
x*cos(1/2*d*x + 1/2*c)*sin(d*x + c) - x*cos(2*d*x + 2*c)*sin(1/2*d*x + 1/2 
*c) - 2*x*cos(d*x + c)*sin(1/2*d*x + 1/2*c) - x*sin(1/2*d*x + 1/2*c))/(a*d 
^2*cos(2*d*x + 2*c)^2 + 4*a*d^2*cos(d*x + c)^2 + a*d^2*sin(2*d*x + 2*c)^2 
+ 4*a*d^2*sin(2*d*x + 2*c)*sin(d*x + c) + 4*a*d^2*sin(d*x + c)^2 + 4*a*d^2 
*cos(d*x + c) + a*d^2 + 2*(2*a*d^2*cos(d*x + c) + a*d^2)*cos(2*d*x + 2*c)) 
, x) + 2*(sqrt(2)*cos(d*x + c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*sqrt(2)*cos( 
d*x + c) + sqrt(2))*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 
2*sin(1/2*d*x + 1/2*c) + 1) - 2*(sqrt(2)*cos(d*x + c)^2 + sqrt(2)*sin(d...
 

Giac [F]

\[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {x^{2}}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:

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

Output:

integrate(x^2/sqrt(a*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {x^2}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^2}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, x^{2}}{\cos \left (d x +c \right )+1}d x \right )}{a} \] Input:

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

Output:

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