\(\int \frac {x}{\sqrt {a+a \cos (c+d x)}} \, dx\) [172]

Optimal result

Integrand size = 16, antiderivative size = 156 \[ \int \frac {x}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {4 i x \arctan \left (e^{\frac {1}{2} i (c+d x)}\right ) \cos \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cos (c+d x)}}+\frac {4 i \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}}-\frac {4 i \cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )}{d^2 \sqrt {a+a \cos (c+d x)}} \] Output:

-4*I*x*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)+4*I*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)-4*I*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)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.57 \[ \int \frac {x}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {4 i \cos \left (\frac {1}{2} (c+d x)\right ) \left (d x \arctan \left (e^{\frac {1}{2} i (c+d x)}\right )-\operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} i (c+d x)}\right )+\operatorname {PolyLog}\left (2,i e^{\frac {1}{2} i (c+d x)}\right )\right )}{d^2 \sqrt {a (1+\cos (c+d x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3800, 3042, 4669, 2715, 2838}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

2*(sqrt(2)*d*x*cos(1/2*d*x + 1/2*c)*sin(d*x + c) - sqrt(2)*d*x*cos(d*x + c 
)*sin(1/2*d*x + 1/2*c) - sqrt(2)*d*x*sin(1/2*d*x + 1/2*c) - (sqrt(2)*cos(d 
*x + c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*arc 
tan2(cos(1/2*d*x + 1/2*c), sin(1/2*d*x + 1/2*c) + 1) - (sqrt(2)*cos(d*x + 
c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*arctan2( 
cos(1/2*d*x + 1/2*c), -sin(1/2*d*x + 1/2*c) + 1) + (sqrt(2)*a*d^3*cos(d*x 
+ c)^2 + sqrt(2)*a*d^3*sin(d*x + c)^2 + 2*sqrt(2)*a*d^3*cos(d*x + c) + sqr 
t(2)*a*d^3)*integrate((x*cos(2*d*x + 2*c)*cos(1/2*d*x + 1/2*c) + 2*x*cos(d 
*x + c)*cos(1/2*d*x + 1/2*c) + x*sin(2*d*x + 2*c)*sin(1/2*d*x + 1/2*c) + 2 
*x*sin(d*x + c)*sin(1/2*d*x + 1/2*c) + x*cos(1/2*d*x + 1/2*c))/(a*d*cos(2* 
d*x + 2*c)^2 + 4*a*d*cos(d*x + c)^2 + a*d*sin(2*d*x + 2*c)^2 + 4*a*d*sin(2 
*d*x + 2*c)*sin(d*x + c) + 4*a*d*sin(d*x + c)^2 + 4*a*d*cos(d*x + c) + a*d 
 + 2*(2*a*d*cos(d*x + c) + a*d)*cos(2*d*x + 2*c)), x))/((d^2*cos(d*x + c)^ 
2 + d^2*sin(d*x + c)^2 + 2*d^2*cos(d*x + c) + d^2)*sqrt(a))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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