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

Optimal. Leaf size=131 \[ -2 b^2 c d \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )-2 b c d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-4 b c d \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+2 b^2 c^2 d x \]

[Out]

2*b^2*c^2*d*x - 2*b*c*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]) + 2*c^2*d*x*(a + b*ArcSinh[c*x])^2 - (d*(1 + c^
2*x^2)*(a + b*ArcSinh[c*x])^2)/x - 4*b*c*d*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]] - 2*b^2*c*d*PolyLog[2,
-E^ArcSinh[c*x]] + 2*b^2*c*d*PolyLog[2, E^ArcSinh[c*x]]

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Rubi [A]  time = 0.318425, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5739, 5653, 5717, 8, 5742, 5760, 4182, 2279, 2391} \[ -2 b^2 c d \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )-2 b c d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )-\frac{d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-4 b c d \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+2 b^2 c^2 d x \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*b^2*c^2*d*x - 2*b*c*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]) + 2*c^2*d*x*(a + b*ArcSinh[c*x])^2 - (d*(1 + c^
2*x^2)*(a + b*ArcSinh[c*x])^2)/x - 4*b*c*d*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]] - 2*b^2*c*d*PolyLog[2,
-E^ArcSinh[c*x]] + 2*b^2*c*d*PolyLog[2, E^ArcSinh[c*x]]

Rule 5739

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

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5742

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
(f*x)^(m + 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1
+ c^2*x^2]), Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(f*
(m + 2)*Sqrt[1 + c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f
, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 5760

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

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 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]

Rubi steps

\begin{align*} \int \frac{\left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x^2} \, dx &=-\frac{d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+(2 b c d) \int \frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx+\left (2 c^2 d\right ) \int \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx\\ &=2 b c d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+(2 b c d) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{1+c^2 x^2}} \, dx-\left (2 b^2 c^2 d\right ) \int 1 \, dx-\left (4 b c^3 d\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx\\ &=-2 b^2 c^2 d x-2 b c d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+(2 b c d) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )+\left (4 b^2 c^2 d\right ) \int 1 \, dx\\ &=2 b^2 c^2 d x-2 b c d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\left (2 b^2 c d\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )+\left (2 b^2 c d\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=2 b^2 c^2 d x-2 b c d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\left (2 b^2 c d\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )+\left (2 b^2 c d\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )\\ &=2 b^2 c^2 d x-2 b c d \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-2 b^2 c d \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.427355, size = 192, normalized size = 1.47 \[ \frac{d \left (-b^2 \left (-2 c x \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+2 c x \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+2 c x \left (\log \left (e^{-\sinh ^{-1}(c x)}+1\right )-\log \left (1-e^{-\sinh ^{-1}(c x)}\right )\right )\right )\right )+a^2 c^2 x^2-a^2+2 a b c x \left (c x \sinh ^{-1}(c x)-\sqrt{c^2 x^2+1}\right )-2 a b \left (c x \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )+\sinh ^{-1}(c x)\right )+b^2 c x \left (-2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+2 c x+c x \sinh ^{-1}(c x)^2\right )\right )}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.105, size = 252, normalized size = 1.9 \begin{align*} d{a}^{2}{c}^{2}x-{\frac{d{a}^{2}}{x}}+d{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{c}^{2}x-2\,cd{b}^{2}{\it Arcsinh} \left ( cx \right ) \sqrt{{c}^{2}{x}^{2}+1}+2\,{b}^{2}{c}^{2}dx-{\frac{d{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{x}}-2\,cd{b}^{2}{\it Arcsinh} \left ( cx \right ) \ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) -2\,{b}^{2}cd{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +2\,cd{b}^{2}{\it Arcsinh} \left ( cx \right ) \ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) +2\,{b}^{2}cd{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) +2\,dab{\it Arcsinh} \left ( cx \right ){c}^{2}x-2\,{\frac{dab{\it Arcsinh} \left ( cx \right ) }{x}}-2\,cdab\sqrt{{c}^{2}{x}^{2}+1}-2\,cdab{\it Artanh} \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*a^2*c^2*x-d*a^2/x+d*b^2*arcsinh(c*x)^2*c^2*x-2*c*d*b^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+2*b^2*c^2*d*x-d*b^2*ar
csinh(c*x)^2/x-2*c*d*b^2*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-2*b^2*c*d*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+
2*c*d*b^2*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+2*b^2*c*d*polylog(2,c*x+(c^2*x^2+1)^(1/2))+2*d*a*b*arcsinh(
c*x)*c^2*x-2*d*a*b*arcsinh(c*x)/x-2*c*d*a*b*(c^2*x^2+1)^(1/2)-2*c*d*a*b*arctanh(1/(c^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b^{2} c^{2} d x \operatorname{arsinh}\left (c x\right )^{2} + 2 \, b^{2} c^{2} d{\left (x - \frac{\sqrt{c^{2} x^{2} + 1} \operatorname{arsinh}\left (c x\right )}{c}\right )} + a^{2} c^{2} d x + 2 \,{\left (c x \operatorname{arsinh}\left (c x\right ) - \sqrt{c^{2} x^{2} + 1}\right )} a b c d - 2 \,{\left (c \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arsinh}\left (c x\right )}{x}\right )} a b d - b^{2} d{\left (\frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{x} - \int \frac{2 \,{\left (c^{3} x^{2} + \sqrt{c^{2} x^{2} + 1} c^{2} x + c\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{3} x^{4} + c x^{2} +{\left (c^{2} x^{3} + x\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x}\right )} - \frac{a^{2} d}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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