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

Optimal. Leaf size=131 \[ -2 b c d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )-\frac {d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+2 c^2 d x \left (a+b \sinh ^{-1}(c x)\right )^2-4 b c d \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+2 b^2 c^2 d x-2 b^2 c d \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right ) \]

[Out]

2*b^2*c^2*d*x+2*c^2*d*x*(a+b*arcsinh(c*x))^2-d*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/x-4*b*c*d*(a+b*arcsinh(c*x))*a
rctanh(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*b^2*c*d*polylog(2,c*x+(c^2*x^2+1)^
(1/2))-2*b*c*d*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [A]  time = 0.43, size = 192, normalized size = 1.47 \[ \frac {d \left (a^2 c^2 x^2-a^2+2 a b c x \left (c x \sinh ^{-1}(c x)-\sqrt {c^2 x^2+1}\right )-2 a b \left (c x \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )+\sinh ^{-1}(c x)\right )+b^2 c x \left (-2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+2 c x+c x \sinh ^{-1}(c x)^2\right )-b^2 \left (-2 c x \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )+2 c x \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+2 c x \left (\log \left (e^{-\sinh ^{-1}(c x)}+1\right )-\log \left (1-e^{-\sinh ^{-1}(c x)}\right )\right )\right )\right )\right )}{x} \]

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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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c^{2} d x^{2} + a^{2} d + {\left (b^{2} c^{2} d x^{2} + b^{2} d\right )} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, {\left (a b c^{2} d x^{2} + a b d\right )} \operatorname {arsinh}\left (c x\right )}{x^{2}}, x\right ) \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.18, size = 252, normalized size = 1.92 \[ d \,a^{2} c^{2} x -\frac {d \,a^{2}}{x}+d \,b^{2} \arcsinh \left (c x \right )^{2} c^{2} x -2 c d \,b^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}+2 b^{2} c^{2} d x -\frac {d \,b^{2} \arcsinh \left (c x \right )^{2}}{x}-2 c d \,b^{2} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )-2 b^{2} c d \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+2 c d \,b^{2} \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+2 b^{2} c d \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )+2 d a b \arcsinh \left (c x \right ) c^{2} x -\frac {2 d a b \arcsinh \left (c x \right )}{x}-2 c d a b \sqrt {c^{2} x^{2}+1}-2 c d a b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right ) \]

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

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/(c*abs(x))) + arcsinh(c*x)/x)*a*b*d - b^2*d*(log(c*x + sqr
t(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\left (d\,c^2\,x^2+d\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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