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

Optimal. Leaf size=116 \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d} \]

[Out]

(-2*(a + b*ArcSinh[c*x])^2*ArcTanh[E^(2*ArcSinh[c*x])])/d - (b*(a + b*ArcSinh[c*x])*PolyLog[2, -E^(2*ArcSinh[c
*x])])/d + (b*(a + b*ArcSinh[c*x])*PolyLog[2, E^(2*ArcSinh[c*x])])/d + (b^2*PolyLog[3, -E^(2*ArcSinh[c*x])])/(
2*d) - (b^2*PolyLog[3, E^(2*ArcSinh[c*x])])/(2*d)

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Rubi [A]  time = 0.207527, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5720, 5461, 4182, 2531, 2282, 6589} \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSinh[c*x])^2*ArcTanh[E^(2*ArcSinh[c*x])])/d - (b*(a + b*ArcSinh[c*x])*PolyLog[2, -E^(2*ArcSinh[c
*x])])/d + (b*(a + b*ArcSinh[c*x])*PolyLog[2, E^(2*ArcSinh[c*x])])/d + (b^2*PolyLog[3, -E^(2*ArcSinh[c*x])])/(
2*d) - (b^2*PolyLog[3, E^(2*ArcSinh[c*x])])/(2*d)

Rule 5720

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

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

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 2531

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] + Dist[(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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )} \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x)^2 \text{csch}(x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int (a+b x)^2 \text{csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac{b^2 \text{Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{b^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.366319, size = 400, normalized size = 3.45 \[ -\frac{12 b^2 \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right ) \left (a+b \sinh ^{-1}(c x)\right )+12 b^2 \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right ) \left (a+b \sinh ^{-1}(c x)\right )-6 a b^2 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )-12 b^3 \text{PolyLog}\left (3,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )-12 b^3 \text{PolyLog}\left (3,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-6 b^3 \sinh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+3 b^3 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )+3 a^2 b \log \left (c^2 x^2+1\right )+6 a^2 b \sinh ^{-1}(c x)-6 a^2 b \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+2 a^3+12 a b^2 \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+12 a b^2 \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-12 a b^2 \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+6 b^3 \sinh ^{-1}(c x)^2 \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+6 b^3 \sinh ^{-1}(c x)^2 \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )-6 b^3 \sinh ^{-1}(c x)^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{6 b d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(2*a^3 + 6*a^2*b*ArcSinh[c*x] + 12*a*b^2*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 6*b^3*ArcSinh[
c*x]^2*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 12*a*b^2*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] +
6*b^3*ArcSinh[c*x]^2*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] - 6*a^2*b*Log[1 - E^(2*ArcSinh[c*x])] - 12*a*b^2*
ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] - 6*b^3*ArcSinh[c*x]^2*Log[1 - E^(2*ArcSinh[c*x])] + 3*a^2*b*Log[1 +
c^2*x^2] + 12*b^2*(a + b*ArcSinh[c*x])*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 12*b^2*(a + b*ArcSinh[c*x])
*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c] - 6*a*b^2*PolyLog[2, E^(2*ArcSinh[c*x])] - 6*b^3*ArcSinh[c*x]*PolyL
og[2, E^(2*ArcSinh[c*x])] - 12*b^3*PolyLog[3, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 12*b^3*PolyLog[3, (Sqrt[-c^2]*E
^ArcSinh[c*x])/c] + 3*b^3*PolyLog[3, E^(2*ArcSinh[c*x])])/(6*b*d)

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Maple [B]  time = 0.074, size = 354, normalized size = 3.1 \begin{align*}{\frac{{a}^{2}\ln \left ( cx \right ) }{d}}-{\frac{{a}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,d}}+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }+2\,{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ){\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }{d}}-2\,{\frac{{b}^{2}{\it polylog} \left ( 3,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }{d}}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) }{d}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{{b}^{2}}{2\,d}{\it polylog} \left ( 3,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+2\,{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ){\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }{d}}-2\,{\frac{{b}^{2}{\it polylog} \left ( 3,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }{d}}+2\,{\frac{ab{\it dilog} \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{-2} \right ) }{d}}-{\frac{ab}{2\,d}{\it dilog} \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{-4} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2/d*ln(c*x)-1/2*a^2/d*ln(c^2*x^2+1)+b^2/d*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/2))+2*b^2/d*arcsinh(c*x)*po
lylog(2,-c*x-(c^2*x^2+1)^(1/2))-2*b^2/d*polylog(3,-c*x-(c^2*x^2+1)^(1/2))-b^2/d*arcsinh(c*x)^2*ln(1+(c*x+(c^2*
x^2+1)^(1/2))^2)-b^2/d*arcsinh(c*x)*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)+1/2*b^2*polylog(3,-(c*x+(c^2*x^2+1)^
(1/2))^2)/d+b^2/d*arcsinh(c*x)^2*ln(1-c*x-(c^2*x^2+1)^(1/2))+2*b^2/d*arcsinh(c*x)*polylog(2,c*x+(c^2*x^2+1)^(1
/2))-2*b^2/d*polylog(3,c*x+(c^2*x^2+1)^(1/2))+2*a*b/d*dilog(1/(c*x+(c^2*x^2+1)^(1/2))^2)-1/2*a*b/d*dilog(1/(c*
x+(c^2*x^2+1)^(1/2))^4)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a^{2}{\left (\frac{\log \left (c^{2} x^{2} + 1\right )}{d} - \frac{2 \, \log \left (x\right )}{d}\right )} + \int \frac{b^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{c^{2} d x^{3} + d x} + \frac{2 \, a b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{2} d x^{3} + d x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}{c^{2} d x^{3} + d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^2*d*x^3 + d*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{2} x^{3} + x}\, dx + \int \frac{b^{2} \operatorname{asinh}^{2}{\left (c x \right )}}{c^{2} x^{3} + x}\, dx + \int \frac{2 a b \operatorname{asinh}{\left (c x \right )}}{c^{2} x^{3} + x}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**2*x**3 + x), x) + Integral(b**2*asinh(c*x)**2/(c**2*x**3 + x), x) + Integral(2*a*b*asinh(c*
x)/(c**2*x**3 + x), x))/d

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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