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

Optimal. Leaf size=157 \[ -\frac {4 i \text {Li}_2\left (-i e^{\frac {c}{2}+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^2 \sqrt {a \cosh (c+d x)+a}}+\frac {4 i \text {Li}_2\left (i e^{\frac {c}{2}+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d^2 \sqrt {a \cosh (c+d x)+a}}+\frac {4 x \tan ^{-1}\left (e^{\frac {c}{2}+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a \cosh (c+d x)+a}} \]

[Out]

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)

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Rubi [A]  time = 0.08, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3319, 4180, 2279, 2391} \[ -\frac {4 i \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {PolyLog}\left (2,-i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a \cosh (c+d x)+a}}+\frac {4 i \cosh \left (\frac {c}{2}+\frac {d x}{2}\right ) \text {PolyLog}\left (2,i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a \cosh (c+d x)+a}}+\frac {4 x \tan ^{-1}\left (e^{\frac {c}{2}+\frac {d x}{2}}\right ) \cosh \left (\frac {c}{2}+\frac {d x}{2}\right )}{d \sqrt {a \cosh (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + a*Cosh[c + d*x]],x]

[Out]

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391

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 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((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 4180

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] + (-Dist[(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] + Dist[(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]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {a+a \cosh (c+d x)}} \, dx &=\frac {\sin \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right ) \int x \csc \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right ) \, dx}{\sqrt {a+a \cosh (c+d x)}}\\ &=\frac {4 x \tan ^{-1}\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 {\left (2 i \sin \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \int \log \left (1-i e^{\frac {c}{2}+\frac {d x}{2}}\right ) \, dx}{d \sqrt {a+a \cosh (c+d x)}}+\frac {\left (2 i \sin \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \int \log \left (1+i e^{\frac {c}{2}+\frac {d x}{2}}\right ) \, dx}{d \sqrt {a+a \cosh (c+d x)}}\\ &=\frac {4 x \tan ^{-1}\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 {\left (4 i \sin \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a+a \cosh (c+d x)}}+\frac {\left (4 i \sin \left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {\pi }{4}+\frac {i d x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a+a \cosh (c+d x)}}\\ &=\frac {4 x \tan ^{-1}\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 ) \text {Li}_2\left (-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 ) \text {Li}_2\left (i e^{\frac {c}{2}+\frac {d x}{2}}\right )}{d^2 \sqrt {a+a \cosh (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 117, normalized size = 0.75 \[ \frac {4 \cosh \left (\frac {1}{2} (c+d x)\right ) \left (-i \text {Li}_2\left (-i \left (\cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )\right )+i \text {Li}_2\left (i \left (\cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )\right )+d x \tan ^{-1}\left (\sinh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right )\right )\right )}{d^2 \sqrt {a (\cosh (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{\sqrt {a \cosh \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a \cosh \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a +a \cosh \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, \sqrt {2} d \int \frac {x e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{\sqrt {a} d e^{\left (2 \, d x + 2 \, c\right )} + 2 \, \sqrt {a} d e^{\left (d x + c\right )} + \sqrt {a} d}\,{d x} + 4 \, \sqrt {2} {\left (\frac {e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{{\left (\sqrt {a} d e^{\left (d x + c\right )} + \sqrt {a} d\right )} d} + \frac {\arctan \left (e^{\left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}\right )}{\sqrt {a} d^{2}}\right )} - \frac {2 \, {\left (\sqrt {2} \sqrt {a} d x e^{\left (\frac {1}{2} \, c\right )} + 2 \, \sqrt {2} \sqrt {a} e^{\left (\frac {1}{2} \, c\right )}\right )} e^{\left (\frac {1}{2} \, d x\right )}}{a d^{2} e^{\left (d x + c\right )} + a d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))/(sq
rt(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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{\sqrt {a+a\,\mathrm {cosh}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + a*cosh(c + d*x))^(1/2),x)

[Out]

int(x/(a + a*cosh(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a \left (\cosh {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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