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

Optimal. Leaf size=59 \[ \frac {2 \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c d}+\frac {b \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )}{c d}-\frac {b \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )}{c d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5694, 4182, 2279, 2391} \[ \frac {b \text {PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{c d}-\frac {b \text {PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{c d}+\frac {2 \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcCosh[c*x])*ArcTanh[E^ArcCosh[c*x]])/(c*d) + (b*PolyLog[2, -E^ArcCosh[c*x]])/(c*d) - (b*PolyLog[2,
E^ArcCosh[c*x]])/(c*d)

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 4182

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

Rule 5694

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(c*d)^(-1), Subst[Int[
(a + b*x)^n*Csch[x], x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 64, normalized size = 1.08 \[ \frac {-\left (\left (\log \left (1-e^{\cosh ^{-1}(c x)}\right )-\log \left (e^{\cosh ^{-1}(c x)}+1\right )\right ) \left (a+b \cosh ^{-1}(c x)\right )\right )+b \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )-b \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )}{c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \operatorname {arcosh}\left (c x\right ) + a}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.61, size = 326, normalized size = 5.53 \[ \frac {a \arctanh \left (c x \right )}{c d}+\frac {b \arctanh \left (c x \right ) \mathrm {arccosh}\left (c x \right )}{c d}+\frac {2 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {c x}{2}}\, \sqrt {-\frac {1}{2}+\frac {c x}{2}}\, \arctanh \left (c x \right ) \ln \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d \left (c^{2} x^{2}-1\right )}-\frac {2 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {c x}{2}}\, \sqrt {-\frac {1}{2}+\frac {c x}{2}}\, \arctanh \left (c x \right ) \ln \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d \left (c^{2} x^{2}-1\right )}+\frac {2 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {c x}{2}}\, \sqrt {-\frac {1}{2}+\frac {c x}{2}}\, \dilog \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d \left (c^{2} x^{2}-1\right )}-\frac {2 i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {c x}{2}}\, \sqrt {-\frac {1}{2}+\frac {c x}{2}}\, \dilog \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )}{c d \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, b {\left (\frac {4 \, {\left (\log \left (c x + 1\right ) - \log \left (c x - 1\right )\right )} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right ) - \log \left (c x + 1\right )^{2} - 2 \, \log \left (c x + 1\right ) \log \left (c x - 1\right )}{c d} + 8 \, \int \frac {{\left (3 \, c x - 1\right )} \log \left (c x - 1\right )}{4 \, {\left (c^{2} d x^{2} - d\right )}}\,{d x} + 8 \, \int \frac {\log \left (c x + 1\right ) - \log \left (c x - 1\right )}{2 \, {\left (c^{3} d x^{3} - c d x + {\left (c^{2} d x^{2} - d\right )} \sqrt {c x + 1} \sqrt {c x - 1}\right )}}\,{d x}\right )} + \frac {1}{2} \, a {\left (\frac {\log \left (c x + 1\right )}{c d} - \frac {\log \left (c x - 1\right )}{c d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b*((4*(log(c*x + 1) - log(c*x - 1))*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) - log(c*x + 1)^2 - 2*log(c*x +
1)*log(c*x - 1))/(c*d) + 8*integrate(1/4*(3*c*x - 1)*log(c*x - 1)/(c^2*d*x^2 - d), x) + 8*integrate(1/2*(log(c
*x + 1) - log(c*x - 1))/(c^3*d*x^3 - c*d*x + (c^2*d*x^2 - d)*sqrt(c*x + 1)*sqrt(c*x - 1)), x)) + 1/2*a*(log(c*
x + 1)/(c*d) - log(c*x - 1)/(c*d))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {acosh}\left (c\,x\right )}{d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a}{c^{2} x^{2} - 1}\, dx + \int \frac {b \operatorname {acosh}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________