3.278 \(\int \frac {(d+c^2 d x^2)^{5/2} (a+b \sinh ^{-1}(c x))^2}{x} \, dx\)

Optimal. Leaf size=635 \[ -\frac {2 b d^2 \sqrt {c^2 d x^2+d} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {2 b d^2 \sqrt {c^2 d x^2+d} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}-\frac {2 a b c d^2 x \sqrt {c^2 d x^2+d}}{\sqrt {c^2 x^2+1}}-\frac {16 b c d^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {c^2 x^2+1}}+d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 d^2 \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {c^2 x^2+1}}+\frac {1}{5} \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 b c^5 d^2 x^5 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {c^2 x^2+1}}-\frac {22 b c^3 d^2 x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {c^2 x^2+1}}+\frac {2 b^2 d^2 \sqrt {c^2 d x^2+d} \text {Li}_3\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}-\frac {2 b^2 d^2 \sqrt {c^2 d x^2+d} \text {Li}_3\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}+\frac {598}{225} b^2 d^2 \sqrt {c^2 d x^2+d}+\frac {2}{125} b^2 d^2 \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d}+\frac {74}{675} b^2 d^2 \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}-\frac {2 b^2 c d^2 x \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}} \]

[Out]

1/3*d*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2+1/5*(c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2+598/225*b^2*d^2*(c
^2*d*x^2+d)^(1/2)+74/675*b^2*d^2*(c^2*x^2+1)*(c^2*d*x^2+d)^(1/2)+2/125*b^2*d^2*(c^2*x^2+1)^2*(c^2*d*x^2+d)^(1/
2)+d^2*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)-2*a*b*c*d^2*x*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2*b^2*c*d^
2*x*arcsinh(c*x)*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-16/15*b*c*d^2*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/
(c^2*x^2+1)^(1/2)-22/45*b*c^3*d^2*x^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2/25*b*c^5*d^2*
x^5*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2*d^2*(a+b*arcsinh(c*x))^2*arctanh(c*x+(c^2*x^2+1
)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2*b*d^2*(a+b*arcsinh(c*x))*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*(c
^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+2*b*d^2*(a+b*arcsinh(c*x))*polylog(2,c*x+(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^
(1/2)/(c^2*x^2+1)^(1/2)+2*b^2*d^2*polylog(3,-c*x-(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-2*b^
2*d^2*polylog(3,c*x+(c^2*x^2+1)^(1/2))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.91, antiderivative size = 635, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 16, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5744, 5742, 5760, 4182, 2531, 2282, 6589, 5653, 261, 5679, 444, 43, 194, 12, 1247, 698} \[ -\frac {2 b d^2 \sqrt {c^2 d x^2+d} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {2 b d^2 \sqrt {c^2 d x^2+d} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 x^2+1}}+\frac {2 b^2 d^2 \sqrt {c^2 d x^2+d} \text {PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}-\frac {2 b^2 d^2 \sqrt {c^2 d x^2+d} \text {PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 x^2+1}}-\frac {2 a b c d^2 x \sqrt {c^2 d x^2+d}}{\sqrt {c^2 x^2+1}}-\frac {2 b c^5 d^2 x^5 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {c^2 x^2+1}}-\frac {22 b c^3 d^2 x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {c^2 x^2+1}}-\frac {16 b c d^2 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {c^2 x^2+1}}+d^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 d^2 \sqrt {c^2 d x^2+d} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {c^2 x^2+1}}+\frac {1}{5} \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {598}{225} b^2 d^2 \sqrt {c^2 d x^2+d}+\frac {2}{125} b^2 d^2 \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d}+\frac {74}{675} b^2 d^2 \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}-\frac {2 b^2 c d^2 x \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/x,x]

[Out]

(598*b^2*d^2*Sqrt[d + c^2*d*x^2])/225 - (2*a*b*c*d^2*x*Sqrt[d + c^2*d*x^2])/Sqrt[1 + c^2*x^2] + (74*b^2*d^2*(1
+ c^2*x^2)*Sqrt[d + c^2*d*x^2])/675 + (2*b^2*d^2*(1 + c^2*x^2)^2*Sqrt[d + c^2*d*x^2])/125 - (2*b^2*c*d^2*x*Sq
rt[d + c^2*d*x^2]*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - (16*b*c*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(1
5*Sqrt[1 + c^2*x^2]) - (22*b*c^3*d^2*x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(45*Sqrt[1 + c^2*x^2]) - (2
*b*c^5*d^2*x^5*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(25*Sqrt[1 + c^2*x^2]) + d^2*Sqrt[d + c^2*d*x^2]*(a +
b*ArcSinh[c*x])^2 + (d*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/3 + ((d + c^2*d*x^2)^(5/2)*(a + b*ArcSin
h[c*x])^2)/5 - (2*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2*ArcTanh[E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] -
(2*b*d^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])*PolyLog[2, -E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] + (2*b*d^2*Sq
rt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])*PolyLog[2, E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] + (2*b^2*d^2*Sqrt[d + c^2
*d*x^2]*PolyLog[3, -E^ArcSinh[c*x]])/Sqrt[1 + c^2*x^2] - (2*b^2*d^2*Sqrt[d + c^2*d*x^2]*PolyLog[3, E^ArcSinh[c
*x]])/Sqrt[1 + c^2*x^2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 194

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

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 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 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 5679

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /;
FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

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 5744

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*d^IntPart[p]*(d + e*x^2)^FracPart[p]
)/(f*(m + 2*p + 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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 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 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 (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{x} \, dx &=\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+d \int \frac {\left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{x} \, dx-\frac {\left (2 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{5 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b c d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 \sqrt {1+c^2 x^2}}-\frac {4 b c^3 d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+\frac {1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+d^2 \int \frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{x} \, dx-\frac {\left (2 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{3 \sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x \left (15+10 c^2 x^2+3 c^4 x^4\right )}{15 \sqrt {1+c^2 x^2}} \, dx}{5 \sqrt {1+c^2 x^2}}\\ &=-\frac {16 b c d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {22 b c^3 d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \sqrt {1+c^2 x^2}} \, dx}{\sqrt {1+c^2 x^2}}-\frac {\left (2 b c d^2 \sqrt {d+c^2 d x^2}\right ) \int \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x \left (15+10 c^2 x^2+3 c^4 x^4\right )}{\sqrt {1+c^2 x^2}} \, dx}{75 \sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x \left (1+\frac {c^2 x^2}{3}\right )}{\sqrt {1+c^2 x^2}} \, dx}{3 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 a b c d^2 x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {16 b c d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {22 b c^3 d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x)^2 \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (2 b^2 c d^2 \sqrt {d+c^2 d x^2}\right ) \int \sinh ^{-1}(c x) \, dx}{\sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {15+10 c^2 x+3 c^4 x^2}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{75 \sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {1+\frac {c^2 x}{3}}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{3 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 a b c d^2 x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}-\frac {2 b^2 c d^2 x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {16 b c d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {22 b c^3 d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (2 b d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (2 b d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {8}{\sqrt {1+c^2 x}}+4 \sqrt {1+c^2 x}+3 \left (1+c^2 x\right )^{3/2}\right ) \, dx,x,x^2\right )}{75 \sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {2}{3 \sqrt {1+c^2 x}}+\frac {1}{3} \sqrt {1+c^2 x}\right ) \, dx,x,x^2\right )}{3 \sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 c^2 d^2 \sqrt {d+c^2 d x^2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {598}{225} b^2 d^2 \sqrt {d+c^2 d x^2}-\frac {2 a b c d^2 x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}+\frac {74}{675} b^2 d^2 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}+\frac {2}{125} b^2 d^2 \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}-\frac {2 b^2 c d^2 x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {16 b c d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {22 b c^3 d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {2 b d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {2 b d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (2 b^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}\\ &=\frac {598}{225} b^2 d^2 \sqrt {d+c^2 d x^2}-\frac {2 a b c d^2 x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}+\frac {74}{675} b^2 d^2 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}+\frac {2}{125} b^2 d^2 \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}-\frac {2 b^2 c d^2 x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {16 b c d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {22 b c^3 d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {2 b d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {2 b d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {\left (2 b^2 d^2 \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}\\ &=\frac {598}{225} b^2 d^2 \sqrt {d+c^2 d x^2}-\frac {2 a b c d^2 x \sqrt {d+c^2 d x^2}}{\sqrt {1+c^2 x^2}}+\frac {74}{675} b^2 d^2 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}+\frac {2}{125} b^2 d^2 \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}-\frac {2 b^2 c d^2 x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-\frac {16 b c d^2 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 \sqrt {1+c^2 x^2}}-\frac {22 b c^3 d^2 x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{45 \sqrt {1+c^2 x^2}}-\frac {2 b c^5 d^2 x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{25 \sqrt {1+c^2 x^2}}+d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{3} d \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{5} \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {2 b d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {2 b d^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}+\frac {2 b^2 d^2 \sqrt {d+c^2 d x^2} \text {Li}_3\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}-\frac {2 b^2 d^2 \sqrt {d+c^2 d x^2} \text {Li}_3\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 4.45, size = 710, normalized size = 1.12 \[ \frac {d^2 \left (54000 a^2 \sqrt {d} \sqrt {c^2 x^2+1} \log (c x)-54000 a^2 \sqrt {d} \sqrt {c^2 x^2+1} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+d\right )+3600 a^2 \sqrt {c^2 x^2+1} \left (3 c^4 x^4+11 c^2 x^2+23\right ) \sqrt {c^2 d x^2+d}-108000 a b \sqrt {c^2 d x^2+d} \left (-\sqrt {c^2 x^2+1} \sinh ^{-1}(c x)-\text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )+\text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+c x-\sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )-480 a b \sqrt {c^2 d x^2+d} \left (c x \left (9 c^4 x^4+5 c^2 x^2-30\right )-15 \sqrt {c^2 x^2+1} \left (3 c^4 x^4+c^2 x^2-2\right ) \sinh ^{-1}(c x)\right )-24000 a b \sqrt {c^2 d x^2+d} \left (c^3 x^3-3 \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)+3 c x\right )+54000 b^2 \sqrt {c^2 d x^2+d} \left (2 \sqrt {c^2 x^2+1}+\sqrt {c^2 x^2+1} \sinh ^{-1}(c x)^2+2 \sinh ^{-1}(c x) \left (\text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )\right )+2 \left (\text {Li}_3\left (-e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_3\left (e^{-\sinh ^{-1}(c x)}\right )\right )-2 c x \sinh ^{-1}(c x)+\sinh ^{-1}(c x)^2 \left (\log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )\right )+1000 b^2 \sqrt {c^2 d x^2+d} \left (27 \sqrt {c^2 x^2+1} \left (\sinh ^{-1}(c x)^2+2\right )-6 \sinh ^{-1}(c x) \left (9 c x+\sinh \left (3 \sinh ^{-1}(c x)\right )\right )+\left (9 \sinh ^{-1}(c x)^2+2\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )\right )-b^2 \sqrt {c^2 d x^2+d} \left (6750 \sqrt {c^2 x^2+1} \left (\sinh ^{-1}(c x)^2+2\right )+480 c x \left (9 c^4 x^4+5 c^2 x^2-30\right ) \sinh ^{-1}(c x)+125 \left (9 \sinh ^{-1}(c x)^2+2\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )-27 \left (25 \sinh ^{-1}(c x)^2+2\right ) \cosh \left (5 \sinh ^{-1}(c x)\right )\right )\right )}{54000 \sqrt {c^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/x,x]

[Out]

(d^2*(3600*a^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]*(23 + 11*c^2*x^2 + 3*c^4*x^4) - 24000*a*b*Sqrt[d + c^2*d*
x^2]*(3*c*x + c^3*x^3 - 3*(1 + c^2*x^2)^(3/2)*ArcSinh[c*x]) - 480*a*b*Sqrt[d + c^2*d*x^2]*(c*x*(-30 + 5*c^2*x^
2 + 9*c^4*x^4) - 15*Sqrt[1 + c^2*x^2]*(-2 + c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x]) - b^2*Sqrt[d + c^2*d*x^2]*(480*
c*x*(-30 + 5*c^2*x^2 + 9*c^4*x^4)*ArcSinh[c*x] + 6750*Sqrt[1 + c^2*x^2]*(2 + ArcSinh[c*x]^2) + 125*(2 + 9*ArcS
inh[c*x]^2)*Cosh[3*ArcSinh[c*x]] - 27*(2 + 25*ArcSinh[c*x]^2)*Cosh[5*ArcSinh[c*x]]) + 54000*a^2*Sqrt[d]*Sqrt[1
+ c^2*x^2]*Log[c*x] - 54000*a^2*Sqrt[d]*Sqrt[1 + c^2*x^2]*Log[d + Sqrt[d]*Sqrt[d + c^2*d*x^2]] - 108000*a*b*S
qrt[d + c^2*d*x^2]*(c*x - Sqrt[1 + c^2*x^2]*ArcSinh[c*x] - ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + ArcSinh[c
*x]*Log[1 + E^(-ArcSinh[c*x])] - PolyLog[2, -E^(-ArcSinh[c*x])] + PolyLog[2, E^(-ArcSinh[c*x])]) + 54000*b^2*S
qrt[d + c^2*d*x^2]*(2*Sqrt[1 + c^2*x^2] - 2*c*x*ArcSinh[c*x] + Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 + ArcSinh[c*x]
^2*(Log[1 - E^(-ArcSinh[c*x])] - Log[1 + E^(-ArcSinh[c*x])]) + 2*ArcSinh[c*x]*(PolyLog[2, -E^(-ArcSinh[c*x])]
- PolyLog[2, E^(-ArcSinh[c*x])]) + 2*(PolyLog[3, -E^(-ArcSinh[c*x])] - PolyLog[3, E^(-ArcSinh[c*x])])) + 1000*
b^2*Sqrt[d + c^2*d*x^2]*(27*Sqrt[1 + c^2*x^2]*(2 + ArcSinh[c*x]^2) + (2 + 9*ArcSinh[c*x]^2)*Cosh[3*ArcSinh[c*x
]] - 6*ArcSinh[c*x]*(9*c*x + Sinh[3*ArcSinh[c*x]]))))/(54000*Sqrt[1 + c^2*x^2])

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c^{4} d^{2} x^{4} + 2 \, a^{2} c^{2} d^{2} x^{2} + a^{2} d^{2} + {\left (b^{2} c^{4} d^{2} x^{4} + 2 \, b^{2} c^{2} d^{2} x^{2} + b^{2} d^{2}\right )} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, {\left (a b c^{4} d^{2} x^{4} + 2 \, a b c^{2} d^{2} x^{2} + a b d^{2}\right )} \operatorname {arsinh}\left (c x\right )\right )} \sqrt {c^{2} d x^{2} + d}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2/x,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))*sqrt(c^2*d*x^2 + d)/x, 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)^(5/2)*(a+b*arcsinh(c*x))^2/x,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 [B]  time = 0.46, size = 1321, normalized size = 2.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2/x,x)

[Out]

-2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*polylog(3,c*x+(c^2*x^2+1)^(1/2))*d^2+9394/3375*b^2*(d*(c^2*x^2+
1))^(1/2)*d^2/(c^2*x^2+1)+2/5*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^6*c^6+28/15*a*b*(d*(c^2
*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)*x^4*c^4+68/15*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*
x)*x^2*c^2-a^2*d^(5/2)*ln((2*d+2*d^(1/2)*(c^2*d*x^2+d)^(1/2))/x)+a^2*(c^2*d*x^2+d)^(1/2)*d^2+1/3*a^2*d*(c^2*d*
x^2+d)^(3/2)+1/5*(c^2*d*x^2+d)^(5/2)*a^2+2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*polylog(2,
c*x+(c^2*x^2+1)^(1/2))*d^2-b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/
2))*d^2+2/125*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*c^6*x^6+532/3375*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^
2+1)*c^4*x^4+9872/3375*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*c^2*x^2+b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)
^(1/2)*arcsinh(c*x)^2*ln(1-c*x-(c^2*x^2+1)^(1/2))*d^2-2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*
x)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*d^2+46/15*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)-2*a*b*(d
*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*d^2+2*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x
^2+1)^(1/2)*polylog(2,c*x+(c^2*x^2+1)^(1/2))*d^2+2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)*polylog(3,-c*x-
(c^2*x^2+1)^(1/2))*d^2+23/15*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)^2-2*a*b*(d*(c^2*x^2+1))^(1
/2)/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))*d^2+2*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/
2)*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))*d^2-46/15*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(1/2)*arcsinh(
c*x)*x*c-2/25*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(1/2)*c^5*x^5-22/45*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2
*x^2+1)^(1/2)*c^3*x^3-2/25*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^5*c^5-22/45*b^2*(d*(
c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^3*c^3-46/15*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)^(
1/2)*c*x+1/5*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)^2*x^6*c^6+14/15*b^2*(d*(c^2*x^2+1))^(1/2)*
d^2/(c^2*x^2+1)*arcsinh(c*x)^2*x^4*c^4+34/15*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)^2*x^2*c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{15} \, {\left (15 \, d^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - 3 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} - 5 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d - 15 \, \sqrt {c^{2} d x^{2} + d} d^{2}\right )} a^{2} + \int \frac {{\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{x} + \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*d*x^2+d)^(5/2)*(a+b*arcsinh(c*x))^2/x,x, algorithm="maxima")

[Out]

-1/15*(15*d^(5/2)*arcsinh(1/(c*abs(x))) - 3*(c^2*d*x^2 + d)^(5/2) - 5*(c^2*d*x^2 + d)^(3/2)*d - 15*sqrt(c^2*d*
x^2 + d)*d^2)*a^2 + integrate((c^2*d*x^2 + d)^(5/2)*b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/x + 2*(c^2*d*x^2 + d)^(
5/2)*a*b*log(c*x + sqrt(c^2*x^2 + 1))/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{5/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(5/2))/x,x)

[Out]

int(((a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(5/2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*d*x**2+d)**(5/2)*(a+b*asinh(c*x))**2/x,x)

[Out]

Integral((d*(c**2*x**2 + 1))**(5/2)*(a + b*asinh(c*x))**2/x, x)

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