3.3.13 \(\int \frac {(d+c^2 d x^2)^2 (a+b \sinh ^{-1}(c x))^2}{x^2} \, dx\) [213]

Optimal. Leaf size=229 \[ \frac {32}{9} b^2 c^2 d^2 x+\frac {2}{27} b^2 c^4 d^2 x^3-\frac {10}{3} b c d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {8}{3} c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{3} c^2 d^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-2 b^2 c d^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \]

[Out]

32/9*b^2*c^2*d^2*x+2/27*b^2*c^4*d^2*x^3-2/9*b*c*d^2*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))+8/3*c^2*d^2*x*(a+b*ar
csinh(c*x))^2+4/3*c^2*d^2*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2-d^2*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))^2/x-4*b*c*d^
2*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))-2*b^2*c*d^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+2*b^2*c*d^2*
polylog(2,c*x+(c^2*x^2+1)^(1/2))-10/3*b*c*d^2*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.33, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {5807, 5786, 5772, 5798, 8, 5808, 5806, 5816, 4267, 2317, 2438} \begin {gather*} \frac {4}{3} c^2 d^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2}{9} b c d^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {10}{3} b c d^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\frac {8}{3} c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )^2-4 b c d^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {2}{27} b^2 c^4 d^2 x^3+\frac {32}{9} b^2 c^2 d^2 x-2 b^2 c d^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d^2 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^2*c^2*d^2*x)/9 + (2*b^2*c^4*d^2*x^3)/27 - (10*b*c*d^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/3 - (2*b*c
*d^2*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/9 + (8*c^2*d^2*x*(a + b*ArcSinh[c*x])^2)/3 + (4*c^2*d^2*x*(1 +
c^2*x^2)*(a + b*ArcSinh[c*x])^2)/3 - (d^2*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x])^2)/x - 4*b*c*d^2*(a + b*ArcSinh
[c*x])*ArcTanh[E^ArcSinh[c*x]] - 2*b^2*c*d^2*PolyLog[2, -E^ArcSinh[c*x]] + 2*b^2*c*d^2*PolyLog[2, E^ArcSinh[c*
x]]

Rule 8

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

Rule 2317

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 2438

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 4267

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 5772

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 5786

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

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5806

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[(1/(m + 2))*Simp[Sqrt[d + e*x^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/(f*(m + 2)))
*Simp[Sqrt[d + e*x^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] && IGtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])

Rule 5807

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/(f*(m + 1)))*Simp[(d + e*x^2)^p/(1
+ c^2*x^2)^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 5808

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 + 2*p + 1))), x] + (Dist[2*d*(p/(m + 2*p + 1)), Int
[(f*x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 2*p + 1)))*Simp[(d + e*x^2)^
p/(1 + c^2*x^2)^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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]

Rule 5816

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

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x^2} \, dx &=-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (4 c^2 d\right ) \int \left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx+\left (2 b c d^2\right ) \int \frac {\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx\\ &=\frac {2}{3} b c d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {4}{3} c^2 d^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (2 b c d^2\right ) \int \frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx+\frac {1}{3} \left (8 c^2 d^2\right ) \int \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {1}{3} \left (2 b^2 c^2 d^2\right ) \int \left (1+c^2 x^2\right ) \, dx-\frac {1}{3} \left (8 b c^3 d^2\right ) \int x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=-\frac {2}{3} b^2 c^2 d^2 x-\frac {2}{9} b^2 c^4 d^2 x^3+2 b c d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {8}{3} c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{3} c^2 d^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (2 b c d^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx+\frac {1}{9} \left (8 b^2 c^2 d^2\right ) \int \left (1+c^2 x^2\right ) \, dx-\left (2 b^2 c^2 d^2\right ) \int 1 \, dx-\frac {1}{3} \left (16 b c^3 d^2\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {16}{9} b^2 c^2 d^2 x+\frac {2}{27} b^2 c^4 d^2 x^3-\frac {10}{3} b c d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {8}{3} c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{3} c^2 d^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (2 b c d^2\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )+\frac {1}{3} \left (16 b^2 c^2 d^2\right ) \int 1 \, dx\\ &=\frac {32}{9} b^2 c^2 d^2 x+\frac {2}{27} b^2 c^4 d^2 x^3-\frac {10}{3} b c d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {8}{3} c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{3} c^2 d^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\left (2 b^2 c d^2\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )+\left (2 b^2 c d^2\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=\frac {32}{9} b^2 c^2 d^2 x+\frac {2}{27} b^2 c^4 d^2 x^3-\frac {10}{3} b c d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {8}{3} c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{3} c^2 d^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\left (2 b^2 c d^2\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )+\left (2 b^2 c d^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )\\ &=\frac {32}{9} b^2 c^2 d^2 x+\frac {2}{27} b^2 c^4 d^2 x^3-\frac {10}{3} b c d^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{9} b c d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {8}{3} c^2 d^2 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {4}{3} c^2 d^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-2 b^2 c d^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d^2 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.69, size = 306, normalized size = 1.34 \begin {gather*} \frac {1}{54} d^2 \left (-\frac {54 a^2}{x}+108 a^2 c^2 x+18 a^2 c^4 x^3-12 a b c \left (-2+c^2 x^2\right ) \sqrt {1+c^2 x^2}+36 a b c^4 x^3 \sinh ^{-1}(c x)-189 b^2 c \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+216 a b c \left (-\sqrt {1+c^2 x^2}+c x \sinh ^{-1}(c x)\right )+108 b^2 c^2 x \left (2+\sinh ^{-1}(c x)^2\right )+2 b^2 c^2 x \left (-12+2 c^2 x^2+9 c^2 x^2 \sinh ^{-1}(c x)^2\right )-\frac {108 a b \left (\sinh ^{-1}(c x)+c x \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )\right )}{x}-3 b^2 c \sinh ^{-1}(c x) \cosh \left (3 \sinh ^{-1}(c x)\right )-\frac {54 b^2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+2 c x \left (-\log \left (1-e^{-\sinh ^{-1}(c x)}\right )+\log \left (1+e^{-\sinh ^{-1}(c x)}\right )\right )\right )}{x}+108 b^2 c \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-108 b^2 c \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*((-54*a^2)/x + 108*a^2*c^2*x + 18*a^2*c^4*x^3 - 12*a*b*c*(-2 + c^2*x^2)*Sqrt[1 + c^2*x^2] + 36*a*b*c^4*x^
3*ArcSinh[c*x] - 189*b^2*c*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 216*a*b*c*(-Sqrt[1 + c^2*x^2] + c*x*ArcSinh[c*x])
+ 108*b^2*c^2*x*(2 + ArcSinh[c*x]^2) + 2*b^2*c^2*x*(-12 + 2*c^2*x^2 + 9*c^2*x^2*ArcSinh[c*x]^2) - (108*a*b*(Ar
cSinh[c*x] + c*x*ArcTanh[Sqrt[1 + c^2*x^2]]))/x - 3*b^2*c*ArcSinh[c*x]*Cosh[3*ArcSinh[c*x]] - (54*b^2*ArcSinh[
c*x]*(ArcSinh[c*x] + 2*c*x*(-Log[1 - E^(-ArcSinh[c*x])] + Log[1 + E^(-ArcSinh[c*x])])))/x + 108*b^2*c*PolyLog[
2, -E^(-ArcSinh[c*x])] - 108*b^2*c*PolyLog[2, E^(-ArcSinh[c*x])]))/54

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Maple [A]
time = 4.28, size = 364, normalized size = 1.59

method	result	size
derivativedivides	\(c \left (a^{2} d^{2} \left (\frac {c^{3} x^{3}}{3}+2 c x -\frac {1}{c x}\right )+\frac {b^{2} d^{2} \arcsinh \left (c x \right )^{2} c^{3} x^{3}}{3}+2 b^{2} d^{2} \arcsinh \left (c x \right )^{2} c x -\frac {32 b^{2} d^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{9}+2 b^{2} d^{2} \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+\frac {32 b^{2} d^{2} c x}{9}+\frac {2 b^{2} d^{2} c^{3} x^{3}}{27}+2 b^{2} d^{2} \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )-2 b^{2} d^{2} \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {2 b^{2} d^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}}{9}-2 b^{2} d^{2} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )-\frac {b^{2} d^{2} \arcsinh \left (c x \right )^{2}}{c x}+2 a \,d^{2} b \left (\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}+2 \arcsinh \left (c x \right ) c x -\frac {\arcsinh \left (c x \right )}{c x}-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {c^{2} x^{2}+1}}{9}-\arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\)	\(364\)
default	\(c \left (a^{2} d^{2} \left (\frac {c^{3} x^{3}}{3}+2 c x -\frac {1}{c x}\right )+\frac {b^{2} d^{2} \arcsinh \left (c x \right )^{2} c^{3} x^{3}}{3}+2 b^{2} d^{2} \arcsinh \left (c x \right )^{2} c x -\frac {32 b^{2} d^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{9}+2 b^{2} d^{2} \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+\frac {32 b^{2} d^{2} c x}{9}+\frac {2 b^{2} d^{2} c^{3} x^{3}}{27}+2 b^{2} d^{2} \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )-2 b^{2} d^{2} \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {2 b^{2} d^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}}{9}-2 b^{2} d^{2} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )-\frac {b^{2} d^{2} \arcsinh \left (c x \right )^{2}}{c x}+2 a \,d^{2} b \left (\frac {\arcsinh \left (c x \right ) c^{3} x^{3}}{3}+2 \arcsinh \left (c x \right ) c x -\frac {\arcsinh \left (c x \right )}{c x}-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {c^{2} x^{2}+1}}{9}-\arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\)	\(364\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(a^2*d^2*(1/3*c^3*x^3+2*c*x-1/c/x)+1/3*b^2*d^2*arcsinh(c*x)^2*c^3*x^3+2*b^2*d^2*arcsinh(c*x)^2*c*x-32/9*b^2*
d^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+2*b^2*d^2*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+32/9*b^2*d^2*c*x+2/27*b^
2*d^2*c^3*x^3+2*b^2*d^2*polylog(2,c*x+(c^2*x^2+1)^(1/2))-2*b^2*d^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-2/9*b^2*d
^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^2*x^2-2*b^2*d^2*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-b^2*d^2*arcsinh(c
*x)^2/c/x+2*a*d^2*b*(1/3*arcsinh(c*x)*c^3*x^3+2*arcsinh(c*x)*c*x-arcsinh(c*x)/c/x-1/9*c^2*x^2*(c^2*x^2+1)^(1/2
)-16/9*(c^2*x^2+1)^(1/2)-arctanh(1/(c^2*x^2+1)^(1/2))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*c^4*d^2*x^3 + 2/9*(3*x^3*arcsinh(c*x) - c*(sqrt(c^2*x^2 + 1)*x^2/c^2 - 2*sqrt(c^2*x^2 + 1)/c^4))*a*b*c
^4*d^2 + 2*b^2*c^2*d^2*x*arcsinh(c*x)^2 + 4*b^2*c^2*d^2*(x - sqrt(c^2*x^2 + 1)*arcsinh(c*x)/c) + 2*a^2*c^2*d^2
*x + 4*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*a*b*c*d^2 - 2*(c*arcsinh(1/(c*abs(x))) + arcsinh(c*x)/x)*a*b*d^2
- a^2*d^2/x + 1/3*(b^2*c^4*d^2*x^4 - 3*b^2*d^2)*log(c*x + sqrt(c^2*x^2 + 1))^2/x - integrate(2/3*(b^2*c^7*d^2
*x^6 + b^2*c^5*d^2*x^4 - 3*b^2*c^3*d^2*x^2 - 3*b^2*c*d^2 + (b^2*c^6*d^2*x^5 - 3*b^2*c^2*d^2*x)*sqrt(c^2*x^2 +
1))*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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*c^4*d^2*x^4 + 2*a^2*c^2*d^2*x^2 + a^2*d^2 + (b^2*c^4*d^2*x^4 + 2*b^2*c^2*d^2*x^2 + b^2*d^2)*arcs
inh(c*x)^2 + 2*(a*b*c^4*d^2*x^4 + 2*a*b*c^2*d^2*x^2 + a*b*d^2)*arcsinh(c*x))/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{2} \left (\int 2 a^{2} c^{2}\, dx + \int \frac {a^{2}}{x^{2}}\, dx + \int a^{2} c^{4} x^{2}\, dx + \int 2 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 4 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 + \int b^{2} c^{4} x^{2} \operatorname {asinh}^{2}{\left (c x \right )}\, dx + \int 2 a b c^{4} x^{2} \operatorname {asinh}{\left (c x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(2*a**2*c**2, x) + Integral(a**2/x**2, x) + Integral(a**2*c**4*x**2, x) + Integral(2*b**2*c**2*a
sinh(c*x)**2, x) + Integral(b**2*asinh(c*x)**2/x**2, x) + Integral(4*a*b*c**2*asinh(c*x), x) + Integral(2*a*b*
asinh(c*x)/x**2, x) + Integral(b**2*c**4*x**2*asinh(c*x)**2, x) + Integral(2*a*b*c**4*x**2*asinh(c*x), x))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^2*(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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^2}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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