\(\int \frac {x}{\sqrt {a+a \cosh (c+d x)}} \, dx\) [141]

Optimal result

Integrand size = 16, antiderivative size = 157 \[ \int \frac {x}{\sqrt {a+a \cosh (c+d x)}} \, dx=\frac {4 x \arctan \left (e^{\frac {c}{2}+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cosh (c+d x)}}-\frac {4 i \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a+a \cosh (c+d x)}}+\frac {4 i \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a+a \cosh (c+d x)}} \] Output:

4*x*arctan(exp(1/2*d*x+1/2*c))*cosh(1/2*d*x+1/2*c)/d/(a+a*cosh(d*x+c))^(1/ 
2)-4*I*cosh(1/2*d*x+1/2*c)*polylog(2,-I*exp(1/2*d*x+1/2*c))/d^2/(a+a*cosh( 
d*x+c))^(1/2)+4*I*cosh(1/2*d*x+1/2*c)*polylog(2,I*exp(1/2*d*x+1/2*c))/d^2/ 
(a+a*cosh(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\sqrt {a+a \cosh (c+d x)}} \, dx=\frac {4 i \cosh \left (\frac {1}{2} (c+d x)\right ) \left (i c \arctan \left (e^{\frac {1}{2} (c+d x)}\right )+\frac {1}{2} (c+d x) \log \left (1-i e^{\frac {1}{2} (c+d x)}\right )-\frac {1}{2} (c+d x) \log \left (1+i e^{\frac {1}{2} (c+d x)}\right )-\operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} (c+d x)}\right )+\operatorname {PolyLog}\left (2,i e^{\frac {1}{2} (c+d x)}\right )\right )}{d^2 \sqrt {a (1+\cosh (c+d x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 104, 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, 4668, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \int x \text {sech}\left (\frac {c}{2}+\frac {d x}{2}\right )dx}{\sqrt {a \cosh (c+d x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4668

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

(Cosh[c/2 + (d*x)/2]*((4*x*ArcTan[E^(c/2 + (d*x)/2)])/d - ((4*I)*PolyLog[2 
, (-I)*E^(c/2 + (d*x)/2)])/d^2 + ((4*I)*PolyLog[2, I*E^(c/2 + (d*x)/2)])/d 
^2))/Sqrt[a + a*Cosh[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 4668

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

2*sqrt(2)*d*integrate(x*e^(1/2*d*x + 1/2*c)/(sqrt(a)*d*e^(2*d*x + 2*c) + 2 
*sqrt(a)*d*e^(d*x + c) + sqrt(a)*d), x) + 4*sqrt(2)*(e^(1/2*d*x + 1/2*c)/( 
(sqrt(a)*d*e^(d*x + c) + sqrt(a)*d)*d) + arctan(e^(1/2*d*x + 1/2*c))/(sqrt 
(a)*d^2)) - 2*(sqrt(2)*sqrt(a)*d*x*e^(1/2*c) + 2*sqrt(2)*sqrt(a)*e^(1/2*c) 
)*e^(1/2*d*x)/(a*d^2*e^(d*x + c) + a*d^2)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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