3.3.22 \(\int \frac {(d+c^2 d x^2)^3 (a+b \sinh ^{-1}(c x))^2}{x^2} \, dx\) [222]

Optimal. Leaf size=307 \[ \frac {122}{25} b^2 c^2 d^3 x+\frac {14}{75} b^2 c^4 d^3 x^3+\frac {2}{125} b^2 c^6 d^3 x^5-\frac {22}{5} b c d^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {16}{5} c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {8}{5} c^2 d^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-2 b^2 c d^3 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d^3 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \]

[Out]

122/25*b^2*c^2*d^3*x+14/75*b^2*c^4*d^3*x^3+2/125*b^2*c^6*d^3*x^5-2/5*b*c*d^3*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*
x))-2/25*b*c*d^3*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))+16/5*c^2*d^3*x*(a+b*arcsinh(c*x))^2+8/5*c^2*d^3*x*(c^2*x
^2+1)*(a+b*arcsinh(c*x))^2+6/5*c^2*d^3*x*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))^2-d^3*(c^2*x^2+1)^3*(a+b*arcsinh(c*x
))^2/x-4*b*c*d^3*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))-2*b^2*c*d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2
))+2*b^2*c*d^3*polylog(2,c*x+(c^2*x^2+1)^(1/2))-22/5*b*c*d^3*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.47, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 12, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5807, 5786, 5772, 5798, 8, 200, 5808, 5806, 5816, 4267, 2317, 2438} \begin {gather*} \frac {6}{5} c^2 d^3 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {8}{5} c^2 d^3 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2}{25} b c d^3 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{5} b c d^3 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {22}{5} b c d^3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )-\frac {d^3 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\frac {16}{5} c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )^2-4 b c d^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )+\frac {2}{125} b^2 c^6 d^3 x^5+\frac {14}{75} b^2 c^4 d^3 x^3+\frac {122}{25} b^2 c^2 d^3 x-2 b^2 c d^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)^3*(a + b*ArcSinh[c*x])^2)/x^2,x]

[Out]

(122*b^2*c^2*d^3*x)/25 + (14*b^2*c^4*d^3*x^3)/75 + (2*b^2*c^6*d^3*x^5)/125 - (22*b*c*d^3*Sqrt[1 + c^2*x^2]*(a
+ b*ArcSinh[c*x]))/5 - (2*b*c*d^3*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/5 - (2*b*c*d^3*(1 + c^2*x^2)^(5/2)
*(a + b*ArcSinh[c*x]))/25 + (16*c^2*d^3*x*(a + b*ArcSinh[c*x])^2)/5 + (8*c^2*d^3*x*(1 + c^2*x^2)*(a + b*ArcSin
h[c*x])^2)/5 + (6*c^2*d^3*x*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x])^2)/5 - (d^3*(1 + c^2*x^2)^3*(a + b*ArcSinh[c*
x])^2)/x - 4*b*c*d^3*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]] - 2*b^2*c*d^3*PolyLog[2, -E^ArcSinh[c*x]] +
2*b^2*c*d^3*PolyLog[2, E^ArcSinh[c*x]]

Rule 8

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

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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 )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x^2} \, dx &=-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (6 c^2 d\right ) \int \left (d+c^2 d x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx+\left (2 b c d^3\right ) \int \frac {\left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx\\ &=\frac {2}{5} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\frac {1}{5} \left (24 c^2 d^2\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^3\right ) \int \frac {\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx-\frac {1}{5} \left (2 b^2 c^2 d^3\right ) \int \left (1+c^2 x^2\right )^2 \, dx-\frac {1}{5} \left (12 b c^3 d^3\right ) \int x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=\frac {2}{3} b c d^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {8}{5} c^2 d^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (2 b c d^3\right ) \int \frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x} \, dx+\frac {1}{5} \left (16 c^2 d^3\right ) \int \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {1}{5} \left (2 b^2 c^2 d^3\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx+\frac {1}{25} \left (12 b^2 c^2 d^3\right ) \int \left (1+c^2 x^2\right )^2 \, dx-\frac {1}{3} \left (2 b^2 c^2 d^3\right ) \int \left (1+c^2 x^2\right ) \, dx-\frac {1}{5} \left (16 b c^3 d^3\right ) \int x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx\\ &=-\frac {16}{15} b^2 c^2 d^3 x-\frac {22}{45} b^2 c^4 d^3 x^3-\frac {2}{25} b^2 c^6 d^3 x^5+2 b c d^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {16}{5} c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {8}{5} c^2 d^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (2 b c d^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx+\frac {1}{25} \left (12 b^2 c^2 d^3\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx+\frac {1}{15} \left (16 b^2 c^2 d^3\right ) \int \left (1+c^2 x^2\right ) \, dx-\left (2 b^2 c^2 d^3\right ) \int 1 \, dx-\frac {1}{5} \left (32 b c^3 d^3\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {38}{25} b^2 c^2 d^3 x+\frac {14}{75} b^2 c^4 d^3 x^3+\frac {2}{125} b^2 c^6 d^3 x^5-\frac {22}{5} b c d^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {16}{5} c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {8}{5} c^2 d^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}+\left (2 b c d^3\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )+\frac {1}{5} \left (32 b^2 c^2 d^3\right ) \int 1 \, dx\\ &=\frac {122}{25} b^2 c^2 d^3 x+\frac {14}{75} b^2 c^4 d^3 x^3+\frac {2}{125} b^2 c^6 d^3 x^5-\frac {22}{5} b c d^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {16}{5} c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {8}{5} c^2 d^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\left (2 b^2 c d^3\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )+\left (2 b^2 c d^3\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=\frac {122}{25} b^2 c^2 d^3 x+\frac {14}{75} b^2 c^4 d^3 x^3+\frac {2}{125} b^2 c^6 d^3 x^5-\frac {22}{5} b c d^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {16}{5} c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {8}{5} c^2 d^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\left (2 b^2 c d^3\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )+\left (2 b^2 c d^3\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )\\ &=\frac {122}{25} b^2 c^2 d^3 x+\frac {14}{75} b^2 c^4 d^3 x^3+\frac {2}{125} b^2 c^6 d^3 x^5-\frac {22}{5} b c d^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{5} b c d^3 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {2}{25} b c d^3 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {16}{5} c^2 d^3 x \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {8}{5} c^2 d^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {6}{5} c^2 d^3 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {d^3 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{x}-4 b c d^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-2 b^2 c d^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+2 b^2 c d^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.86, size = 466, normalized size = 1.52 \begin {gather*} \frac {1}{720} d^3 \left (-\frac {720 a^2}{x}+2160 a^2 c^2 x+3460 b^2 c^2 x+720 a^2 c^4 x^3+144 a^2 c^6 x^5-\frac {17568}{5} a b c \sqrt {1+c^2 x^2}-\frac {2016}{5} a b c^3 x^2 \sqrt {1+c^2 x^2}-\frac {288}{5} a b c^5 x^4 \sqrt {1+c^2 x^2}-\frac {1440 a b \sinh ^{-1}(c x)}{x}+4320 a b c^2 x \sinh ^{-1}(c x)+1440 a b c^4 x^3 \sinh ^{-1}(c x)+288 a b c^6 x^5 \sinh ^{-1}(c x)-3420 b^2 c \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)-\frac {720 b^2 \sinh ^{-1}(c x)^2}{x}+1890 b^2 c^2 x \sinh ^{-1}(c x)^2-1440 a b c \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )+80 b^2 c^2 x \cosh \left (2 \sinh ^{-1}(c x)\right )+360 b^2 c^2 x \sinh ^{-1}(c x)^2 \cosh \left (2 \sinh ^{-1}(c x)\right )-90 b^2 c \sinh ^{-1}(c x) \cosh \left (3 \sinh ^{-1}(c x)\right )-\frac {18}{5} b^2 c \sinh ^{-1}(c x) \cosh \left (5 \sinh ^{-1}(c x)\right )+1440 b^2 c \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-1440 b^2 c \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )+1440 b^2 c \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-1440 b^2 c \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-10 b^2 c \sinh \left (3 \sinh ^{-1}(c x)\right )-45 b^2 c \sinh ^{-1}(c x)^2 \sinh \left (3 \sinh ^{-1}(c x)\right )+\frac {18}{25} b^2 c \sinh \left (5 \sinh ^{-1}(c x)\right )+9 b^2 c \sinh ^{-1}(c x)^2 \sinh \left (5 \sinh ^{-1}(c x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + c^2*d*x^2)^3*(a + b*ArcSinh[c*x])^2)/x^2,x]

[Out]

(d^3*((-720*a^2)/x + 2160*a^2*c^2*x + 3460*b^2*c^2*x + 720*a^2*c^4*x^3 + 144*a^2*c^6*x^5 - (17568*a*b*c*Sqrt[1
+ c^2*x^2])/5 - (2016*a*b*c^3*x^2*Sqrt[1 + c^2*x^2])/5 - (288*a*b*c^5*x^4*Sqrt[1 + c^2*x^2])/5 - (1440*a*b*Ar
cSinh[c*x])/x + 4320*a*b*c^2*x*ArcSinh[c*x] + 1440*a*b*c^4*x^3*ArcSinh[c*x] + 288*a*b*c^6*x^5*ArcSinh[c*x] - 3
420*b^2*c*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] - (720*b^2*ArcSinh[c*x]^2)/x + 1890*b^2*c^2*x*ArcSinh[c*x]^2 - 1440*a
*b*c*ArcTanh[Sqrt[1 + c^2*x^2]] + 80*b^2*c^2*x*Cosh[2*ArcSinh[c*x]] + 360*b^2*c^2*x*ArcSinh[c*x]^2*Cosh[2*ArcS
inh[c*x]] - 90*b^2*c*ArcSinh[c*x]*Cosh[3*ArcSinh[c*x]] - (18*b^2*c*ArcSinh[c*x]*Cosh[5*ArcSinh[c*x]])/5 + 1440
*b^2*c*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - 1440*b^2*c*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 1440*b^2
*c*PolyLog[2, -E^(-ArcSinh[c*x])] - 1440*b^2*c*PolyLog[2, E^(-ArcSinh[c*x])] - 10*b^2*c*Sinh[3*ArcSinh[c*x]] -
45*b^2*c*ArcSinh[c*x]^2*Sinh[3*ArcSinh[c*x]] + (18*b^2*c*Sinh[5*ArcSinh[c*x]])/25 + 9*b^2*c*ArcSinh[c*x]^2*Si
nh[5*ArcSinh[c*x]]))/720

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Maple [A]
time = 4.66, size = 463, normalized size = 1.51

method	result	size
derivativedivides	\(c \left (d^{3} a^{2} \left (\frac {c^{5} x^{5}}{5}+c^{3} x^{3}+3 c x -\frac {1}{c x}\right )-2 b^{2} d^{3} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+\frac {b^{2} d^{3} \arcsinh \left (c x \right )^{2} c^{5} x^{5}}{5}+b^{2} d^{3} \arcsinh \left (c x \right )^{2} c^{3} x^{3}+3 b^{2} d^{3} \arcsinh \left (c x \right )^{2} c x -\frac {122 b^{2} d^{3} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{25}-2 b^{2} d^{3} \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+\frac {122 b^{2} d^{3} c x}{25}+\frac {14 b^{2} d^{3} c^{3} x^{3}}{75}+\frac {2 b^{2} d^{3} c^{5} x^{5}}{125}+2 b^{2} d^{3} \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )-\frac {2 b^{2} d^{3} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}}{25}-\frac {14 b^{2} d^{3} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}}{25}+2 b^{2} d^{3} \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {b^{2} d^{3} \arcsinh \left (c x \right )^{2}}{c x}+2 d^{3} a b \left (\frac {\arcsinh \left (c x \right ) c^{5} x^{5}}{5}+\arcsinh \left (c x \right ) c^{3} x^{3}+3 \arcsinh \left (c x \right ) c x -\frac {\arcsinh \left (c x \right )}{c x}-\frac {c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{25}-\frac {61 \sqrt {c^{2} x^{2}+1}}{25}-\arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\)	\(463\)
default	\(c \left (d^{3} a^{2} \left (\frac {c^{5} x^{5}}{5}+c^{3} x^{3}+3 c x -\frac {1}{c x}\right )-2 b^{2} d^{3} \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+\frac {b^{2} d^{3} \arcsinh \left (c x \right )^{2} c^{5} x^{5}}{5}+b^{2} d^{3} \arcsinh \left (c x \right )^{2} c^{3} x^{3}+3 b^{2} d^{3} \arcsinh \left (c x \right )^{2} c x -\frac {122 b^{2} d^{3} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{25}-2 b^{2} d^{3} \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+\frac {122 b^{2} d^{3} c x}{25}+\frac {14 b^{2} d^{3} c^{3} x^{3}}{75}+\frac {2 b^{2} d^{3} c^{5} x^{5}}{125}+2 b^{2} d^{3} \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )-\frac {2 b^{2} d^{3} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}}{25}-\frac {14 b^{2} d^{3} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}}{25}+2 b^{2} d^{3} \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )-\frac {b^{2} d^{3} \arcsinh \left (c x \right )^{2}}{c x}+2 d^{3} a b \left (\frac {\arcsinh \left (c x \right ) c^{5} x^{5}}{5}+\arcsinh \left (c x \right ) c^{3} x^{3}+3 \arcsinh \left (c x \right ) c x -\frac {\arcsinh \left (c x \right )}{c x}-\frac {c^{4} x^{4} \sqrt {c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{25}-\frac {61 \sqrt {c^{2} x^{2}+1}}{25}-\arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )\right )\right )\)	\(463\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^3*(a+b*arcsinh(c*x))^2/x^2,x,method=_RETURNVERBOSE)

[Out]

c*(d^3*a^2*(1/5*c^5*x^5+c^3*x^3+3*c*x-1/c/x)-2*b^2*d^3*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))+1/5*b^2*d^3*ar
csinh(c*x)^2*c^5*x^5+b^2*d^3*arcsinh(c*x)^2*c^3*x^3+3*b^2*d^3*arcsinh(c*x)^2*c*x-122/25*b^2*d^3*arcsinh(c*x)*(
c^2*x^2+1)^(1/2)-2*b^2*d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+122/25*b^2*d^3*c*x+14/75*b^2*d^3*c^3*x^3+2/125*b^
2*d^3*c^5*x^5+2*b^2*d^3*polylog(2,c*x+(c^2*x^2+1)^(1/2))-2/25*b^2*d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^4*x^4-1
4/25*b^2*d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^2*x^2+2*b^2*d^3*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))-b^2*d^3
*arcsinh(c*x)^2/c/x+2*d^3*a*b*(1/5*arcsinh(c*x)*c^5*x^5+arcsinh(c*x)*c^3*x^3+3*arcsinh(c*x)*c*x-arcsinh(c*x)/c
/x-1/25*c^4*x^4*(c^2*x^2+1)^(1/2)-7/25*c^2*x^2*(c^2*x^2+1)^(1/2)-61/25*(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)^3*(a+b*arcsinh(c*x))^2/x^2,x, algorithm="maxima")

[Out]

1/5*a^2*c^6*d^3*x^5 + 2/75*(15*x^5*arcsinh(c*x) - (3*sqrt(c^2*x^2 + 1)*x^4/c^2 - 4*sqrt(c^2*x^2 + 1)*x^2/c^4 +
8*sqrt(c^2*x^2 + 1)/c^6)*c)*a*b*c^6*d^3 + a^2*c^4*d^3*x^3 + 2/3*(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^3 + 3*b^2*c^2*d^3*x*arcsinh(c*x)^2 + 6*b^2*c^2*d^3*(x - sqrt(c^2*x
^2 + 1)*arcsinh(c*x)/c) + 3*a^2*c^2*d^3*x + 6*(c*x*arcsinh(c*x) - sqrt(c^2*x^2 + 1))*a*b*c*d^3 - 2*(c*arcsinh(
1/(c*abs(x))) + arcsinh(c*x)/x)*a*b*d^3 - a^2*d^3/x + 1/5*(b^2*c^6*d^3*x^6 + 5*b^2*c^4*d^3*x^4 - 5*b^2*d^3)*lo
g(c*x + sqrt(c^2*x^2 + 1))^2/x - integrate(2/5*(b^2*c^9*d^3*x^8 + 6*b^2*c^7*d^3*x^6 + 5*b^2*c^5*d^3*x^4 - 5*b^
2*c^3*d^3*x^2 - 5*b^2*c*d^3 + (b^2*c^8*d^3*x^7 + 5*b^2*c^6*d^3*x^5 - 5*b^2*c^2*d^3*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)^3*(a+b*arcsinh(c*x))^2/x^2,x, algorithm="fricas")

[Out]

integral((a^2*c^6*d^3*x^6 + 3*a^2*c^4*d^3*x^4 + 3*a^2*c^2*d^3*x^2 + a^2*d^3 + (b^2*c^6*d^3*x^6 + 3*b^2*c^4*d^3
*x^4 + 3*b^2*c^2*d^3*x^2 + b^2*d^3)*arcsinh(c*x)^2 + 2*(a*b*c^6*d^3*x^6 + 3*a*b*c^4*d^3*x^4 + 3*a*b*c^2*d^3*x^
2 + a*b*d^3)*arcsinh(c*x))/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{3} \left (\int 3 a^{2} c^{2}\, dx + \int \frac {a^{2}}{x^{2}}\, dx + \int 3 a^{2} c^{4} x^{2}\, dx + \int a^{2} c^{6} x^{4}\, dx + \int 3 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 6 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 3 b^{2} c^{4} x^{2} \operatorname {asinh}^{2}{\left (c x \right )}\, dx + \int b^{2} c^{6} x^{4} \operatorname {asinh}^{2}{\left (c x \right )}\, dx + \int 6 a b c^{4} x^{2} \operatorname {asinh}{\left (c x \right )}\, dx + \int 2 a b c^{6} x^{4} \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)**3*(a+b*asinh(c*x))**2/x**2,x)

[Out]

d**3*(Integral(3*a**2*c**2, x) + Integral(a**2/x**2, x) + Integral(3*a**2*c**4*x**2, x) + Integral(a**2*c**6*x
**4, x) + Integral(3*b**2*c**2*asinh(c*x)**2, x) + Integral(b**2*asinh(c*x)**2/x**2, x) + Integral(6*a*b*c**2*
asinh(c*x), x) + Integral(2*a*b*asinh(c*x)/x**2, x) + Integral(3*b**2*c**4*x**2*asinh(c*x)**2, x) + Integral(b
**2*c**6*x**4*asinh(c*x)**2, x) + Integral(6*a*b*c**4*x**2*asinh(c*x), x) + Integral(2*a*b*c**6*x**4*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)^3*(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 )}^3}{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)^3)/x^2,x)

[Out]

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

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