3.110 \(\int \frac {a+b \cosh ^{-1}(c x)}{x \sqrt {d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=151 \[ \frac {2 \sqrt {c x-1} \sqrt {c x+1} \tan ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {c x-1} \sqrt {c x+1} \text {Li}_2\left (-i e^{\cosh ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {c x-1} \sqrt {c x+1} \text {Li}_2\left (i e^{\cosh ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}} \]

[Out]

2*(a+b*arccosh(c*x))*arctan(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)-
I*b*polylog(2,-I*(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)))*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+I*b*polyl
og(2,I*(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)))*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5798, 5761, 4180, 2279, 2391} \[ -\frac {i b \sqrt {c x-1} \sqrt {c x+1} \text {PolyLog}\left (2,-i e^{\cosh ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {c x-1} \sqrt {c x+1} \text {PolyLog}\left (2,i e^{\cosh ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {2 \sqrt {c x-1} \sqrt {c x+1} \tan ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c*x])/(x*Sqrt[d - c^2*d*x^2]),x]

[Out]

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

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 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]

Rule 5761

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]
), x_Symbol] :> Dist[1/(c^(m + 1)*Sqrt[-(d1*d2)]), Subst[Int[(a + b*x)^n*Cosh[x]^m, x], x, ArcCosh[c*x]], x] /
; FreeQ[{a, b, c, d1, e1, d2, e2}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IGtQ[n, 0] && GtQ[d1, 0] &&
LtQ[d2, 0] && IntegerQ[m]

Rule 5798

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist
[((-d)^IntPart[p]*(d + e*x^2)^FracPart[p])/((1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^m*(1 + c*
x)^p*(-1 + c*x)^p*(a + b*ArcCosh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[c^2*d + e, 0]
&&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 153, normalized size = 1.01 \[ -\frac {a \log \left (\sqrt {d} \sqrt {d-c^2 d x^2}+d\right )}{\sqrt {d}}+\frac {a \log (x)}{\sqrt {d}}-\frac {i b \sqrt {\frac {c x-1}{c x+1}} (c x+1) \left (\text {Li}_2\left (-i e^{-\cosh ^{-1}(c x)}\right )-\text {Li}_2\left (i e^{-\cosh ^{-1}(c x)}\right )+\cosh ^{-1}(c x) \left (\log \left (1-i e^{-\cosh ^{-1}(c x)}\right )-\log \left (1+i e^{-\cosh ^{-1}(c x)}\right )\right )\right )}{\sqrt {d-c^2 d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[c*x])/(x*Sqrt[d - c^2*d*x^2]),x]

[Out]

(a*Log[x])/Sqrt[d] - (a*Log[d + Sqrt[d]*Sqrt[d - c^2*d*x^2]])/Sqrt[d] - (I*b*Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c
*x)*(ArcCosh[c*x]*(Log[1 - I/E^ArcCosh[c*x]] - Log[1 + I/E^ArcCosh[c*x]]) + PolyLog[2, (-I)/E^ArcCosh[c*x]] -
PolyLog[2, I/E^ArcCosh[c*x]]))/Sqrt[d - c^2*d*x^2]

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}}{c^{2} d x^{3} - d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*arccosh(c*x) + a)/(c^2*d*x^3 - d*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcosh}\left (c x\right ) + a}{\sqrt {-c^{2} d x^{2} + d} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccosh(c*x) + a)/(sqrt(-c^2*d*x^2 + d)*x), x)

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maple [A]  time = 0.37, size = 327, normalized size = 2.17 \[ -\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{\sqrt {d}}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \mathrm {arccosh}\left (c x \right ) \ln \left (1+i \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )\right )}{d \left (c^{2} x^{2}-1\right )}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \mathrm {arccosh}\left (c x \right ) \ln \left (1-i \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )\right )}{d \left (c^{2} x^{2}-1\right )}+\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \dilog \left (1+i \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )\right )}{d \left (c^{2} x^{2}-1\right )}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \dilog \left (1-i \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )\right )}{d \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(c*x))/x/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-a/d^(1/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)+I*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/d
/(c^2*x^2-1)*arccosh(c*x)*ln(1+I*(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)))-I*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(
c*x+1)^(1/2)/d/(c^2*x^2-1)*arccosh(c*x)*ln(1-I*(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)))+I*b*(-d*(c^2*x^2-1))^(1/2)*(
c*x-1)^(1/2)*(c*x+1)^(1/2)/d/(c^2*x^2-1)*dilog(1+I*(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)))-I*b*(-d*(c^2*x^2-1))^(1/
2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/d/(c^2*x^2-1)*dilog(1-I*(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))/(sqrt(-c^2*d*x^2 + d)*x), x) - a*log(2*sqrt(-c^2*d*x^2 + d)
*sqrt(d)/abs(x) + 2*d/abs(x))/sqrt(d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acosh}\left (c\,x\right )}{x\,\sqrt {d-c^2\,d\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/(x*(d - c^2*d*x^2)^(1/2)),x)

[Out]

int((a + b*acosh(c*x))/(x*(d - c^2*d*x^2)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(c*x))/x/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*acosh(c*x))/(x*sqrt(-d*(c*x - 1)*(c*x + 1))), x)

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