\(\int \frac {x^3}{\sqrt {a+a \cosh (c+d x)}} \, dx\) [139]

Optimal result

Integrand size = 18, antiderivative size = 383 \[ \int \frac {x^3}{\sqrt {a+a \cosh (c+d x)}} \, dx=\frac {4 x^3 \arctan \left (e^{\frac {c}{2}+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a+a \cosh (c+d x)}}-\frac {12 i x^2 \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a+a \cosh (c+d x)}}+\frac {12 i x^2 \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a+a \cosh (c+d x)}}+\frac {48 i x \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^3 \sqrt {a+a \cosh (c+d x)}}-\frac {48 i x \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (3,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^3 \sqrt {a+a \cosh (c+d x)}}-\frac {96 i \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (4,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^4 \sqrt {a+a \cosh (c+d x)}}+\frac {96 i \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \operatorname {PolyLog}\left (4,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^4 \sqrt {a+a \cosh (c+d x)}} \] Output:

4*x^3*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)-12*I*x^2*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)+12*I*x^2*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)+48*I*x*cosh(1/2*d*x+1/2*c)*polylog(3,- 
I*exp(1/2*d*x+1/2*c))/d^3/(a+a*cosh(d*x+c))^(1/2)-48*I*x*cosh(1/2*d*x+1/2* 
c)*polylog(3,I*exp(1/2*d*x+1/2*c))/d^3/(a+a*cosh(d*x+c))^(1/2)-96*I*cosh(1 
/2*d*x+1/2*c)*polylog(4,-I*exp(1/2*d*x+1/2*c))/d^4/(a+a*cosh(d*x+c))^(1/2) 
+96*I*cosh(1/2*d*x+1/2*c)*polylog(4,I*exp(1/2*d*x+1/2*c))/d^4/(a+a*cosh(d* 
x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.56 \[ \int \frac {x^3}{\sqrt {a+a \cosh (c+d x)}} \, dx=\frac {2 i \cosh \left (\frac {1}{2} (c+d x)\right ) \left (d^3 x^3 \log \left (1-i e^{\frac {1}{2} (c+d x)}\right )-d^3 x^3 \log \left (1+i e^{\frac {1}{2} (c+d x)}\right )-6 d^2 x^2 \operatorname {PolyLog}\left (2,-i e^{\frac {1}{2} (c+d x)}\right )+6 d^2 x^2 \operatorname {PolyLog}\left (2,i e^{\frac {1}{2} (c+d x)}\right )+24 d x \operatorname {PolyLog}\left (3,-i e^{\frac {1}{2} (c+d x)}\right )-24 d x \operatorname {PolyLog}\left (3,i e^{\frac {1}{2} (c+d x)}\right )-48 \operatorname {PolyLog}\left (4,-i e^{\frac {1}{2} (c+d x)}\right )+48 \operatorname {PolyLog}\left (4,i e^{\frac {1}{2} (c+d x)}\right )\right )}{d^4 \sqrt {a (1+\cosh (c+d x))}} \] Input:

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

Output:

((2*I)*Cosh[(c + d*x)/2]*(d^3*x^3*Log[1 - I*E^((c + d*x)/2)] - d^3*x^3*Log 
[1 + I*E^((c + d*x)/2)] - 6*d^2*x^2*PolyLog[2, (-I)*E^((c + d*x)/2)] + 6*d 
^2*x^2*PolyLog[2, I*E^((c + d*x)/2)] + 24*d*x*PolyLog[3, (-I)*E^((c + d*x) 
/2)] - 24*d*x*PolyLog[3, I*E^((c + d*x)/2)] - 48*PolyLog[4, (-I)*E^((c + d 
*x)/2)] + 48*PolyLog[4, I*E^((c + d*x)/2)]))/(d^4*Sqrt[a*(1 + Cosh[c + d*x 
])])
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.62, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 3800, 3042, 4668, 3011, 7163, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4668

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {6 i \int x^2 \log \left (1-i e^{\frac {c}{2}+\frac {d x}{2}}\right )dx}{d}+\frac {6 i \int x^2 \log \left (1+i e^{\frac {c}{2}+\frac {d x}{2}}\right )dx}{d}+\frac {4 x^3 \arctan \left (e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{\sqrt {a \cosh (c+d x)+a}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {6 i \left (\frac {4 \int x \operatorname {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )dx}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}-\frac {6 i \left (\frac {4 \int x \operatorname {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )dx}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 x^3 \arctan \left (e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{\sqrt {a \cosh (c+d x)+a}}\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {6 i \left (\frac {4 \left (\frac {2 x \operatorname {PolyLog}\left (3,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}-\frac {2 \int \operatorname {PolyLog}\left (3,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )dx}{d}\right )}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}-\frac {6 i \left (\frac {4 \left (\frac {2 x \operatorname {PolyLog}\left (3,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}-\frac {2 \int \operatorname {PolyLog}\left (3,i e^{\frac {c}{2}+\frac {d x}{2}}\right )dx}{d}\right )}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 x^3 \arctan \left (e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{\sqrt {a \cosh (c+d x)+a}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {6 i \left (\frac {4 \left (\frac {2 x \operatorname {PolyLog}\left (3,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}-\frac {4 \int e^{-\frac {c}{2}-\frac {d x}{2}} \operatorname {PolyLog}\left (3,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )de^{\frac {c}{2}+\frac {d x}{2}}}{d^2}\right )}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}-\frac {6 i \left (\frac {4 \left (\frac {2 x \operatorname {PolyLog}\left (3,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}-\frac {4 \int e^{-\frac {c}{2}-\frac {d x}{2}} \operatorname {PolyLog}\left (3,i e^{\frac {c}{2}+\frac {d x}{2}}\right )de^{\frac {c}{2}+\frac {d x}{2}}}{d^2}\right )}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}+\frac {4 x^3 \arctan \left (e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{\sqrt {a \cosh (c+d x)+a}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {4 x^3 \arctan \left (e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}+\frac {6 i \left (\frac {4 \left (\frac {2 x \operatorname {PolyLog}\left (3,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}-\frac {4 \operatorname {PolyLog}\left (4,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2}\right )}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}-\frac {6 i \left (\frac {4 \left (\frac {2 x \operatorname {PolyLog}\left (3,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}-\frac {4 \operatorname {PolyLog}\left (4,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2}\right )}{d}-\frac {2 x^2 \operatorname {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d}\right )}{d}\right )}{\sqrt {a \cosh (c+d x)+a}}\)

Input:

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

Output:

(Cosh[c/2 + (d*x)/2]*((4*x^3*ArcTan[E^(c/2 + (d*x)/2)])/d + ((6*I)*((-2*x^ 
2*PolyLog[2, (-I)*E^(c/2 + (d*x)/2)])/d + (4*((2*x*PolyLog[3, (-I)*E^(c/2 
+ (d*x)/2)])/d - (4*PolyLog[4, (-I)*E^(c/2 + (d*x)/2)])/d^2))/d))/d - ((6* 
I)*((-2*x^2*PolyLog[2, I*E^(c/2 + (d*x)/2)])/d + (4*((2*x*PolyLog[3, I*E^( 
c/2 + (d*x)/2)])/d - (4*PolyLog[4, I*E^(c/2 + (d*x)/2)])/d^2))/d))/d))/Sqr 
t[a + a*Cosh[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 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]
 

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]
 

rule 7163

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

Maple [F]

\[\int \frac {x^{3}}{\sqrt {a +a \cosh \left (d x +c \right )}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^3/sqrt(a*cosh(d*x + c) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

2*sqrt(2)*d^3*integrate(x^3*e^(1/2*d*x + 1/2*c)/(sqrt(a)*d^3*e^(2*d*x + 2* 
c) + 2*sqrt(a)*d^3*e^(d*x + c) + sqrt(a)*d^3), x) + 12*sqrt(2)*d^2*integra 
te(x^2*e^(1/2*d*x + 1/2*c)/(sqrt(a)*d^3*e^(2*d*x + 2*c) + 2*sqrt(a)*d^3*e^ 
(d*x + c) + sqrt(a)*d^3), x) + 48*sqrt(2)*d*integrate(x*e^(1/2*d*x + 1/2*c 
)/(sqrt(a)*d^3*e^(2*d*x + 2*c) + 2*sqrt(a)*d^3*e^(d*x + c) + sqrt(a)*d^3), 
 x) + 96*sqrt(2)*(e^(1/2*d*x + 1/2*c)/((sqrt(a)*d^3*e^(d*x + c) + sqrt(a)* 
d^3)*d) + arctan(e^(1/2*d*x + 1/2*c))/(sqrt(a)*d^4)) - 2*(sqrt(2)*sqrt(a)* 
d^3*x^3*e^(1/2*c) + 6*sqrt(2)*sqrt(a)*d^2*x^2*e^(1/2*c) + 24*sqrt(2)*sqrt( 
a)*d*x*e^(1/2*c) + 48*sqrt(2)*sqrt(a)*e^(1/2*c))*e^(1/2*d*x)/(a*d^4*e^(d*x 
 + c) + a*d^4)
 

Giac [F]

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

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

Output:

integrate(x^3/sqrt(a*cosh(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {a+a \cosh (c+d x)}} \, dx=\int \frac {x^3}{\sqrt {a+a\,\mathrm {cosh}\left (c+d\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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