3.266 \(\int x^3 (d+c^2 d x^2)^{3/2} (a+b \sinh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=482 \[ \frac {d x^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^2}-\frac {16 b c d x^5 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{175 \sqrt {c^2 x^2+1}}+\frac {1}{7} x^4 \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {3}{35} d x^4 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 b d x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{105 c \sqrt {c^2 x^2+1}}-\frac {2 d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^4}+\frac {4 a b d x \sqrt {c^2 d x^2+d}}{35 c^3 \sqrt {c^2 x^2+1}}-\frac {2 b c^3 d x^7 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {c^2 x^2+1}}+\frac {2 b^2 d \left (c^2 x^2+1\right )^3 \sqrt {c^2 d x^2+d}}{343 c^4}-\frac {38 b^2 d \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d}}{6125 c^4}-\frac {304 b^2 d \sqrt {c^2 d x^2+d}}{3675 c^4}-\frac {152 b^2 d \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}}{11025 c^4}+\frac {4 b^2 d x \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{35 c^3 \sqrt {c^2 x^2+1}} \]

[Out]

1/7*x^4*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2-304/3675*b^2*d*(c^2*d*x^2+d)^(1/2)/c^4-152/11025*b^2*d*(c^2*x
^2+1)*(c^2*d*x^2+d)^(1/2)/c^4-38/6125*b^2*d*(c^2*x^2+1)^2*(c^2*d*x^2+d)^(1/2)/c^4+2/343*b^2*d*(c^2*x^2+1)^3*(c
^2*d*x^2+d)^(1/2)/c^4-2/35*d*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/c^4+1/35*d*x^2*(a+b*arcsinh(c*x))^2*(c^2
*d*x^2+d)^(1/2)/c^2+3/35*d*x^4*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)+4/35*a*b*d*x*(c^2*d*x^2+d)^(1/2)/c^3/(
c^2*x^2+1)^(1/2)+4/35*b^2*d*x*arcsinh(c*x)*(c^2*d*x^2+d)^(1/2)/c^3/(c^2*x^2+1)^(1/2)-2/105*b*d*x^3*(a+b*arcsin
h(c*x))*(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-16/175*b*c*d*x^5*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x
^2+1)^(1/2)-2/49*b*c^3*d*x^7*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.81, antiderivative size = 482, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 14, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5744, 5742, 5758, 5717, 5653, 261, 5661, 266, 43, 14, 5730, 12, 446, 77} \[ \frac {4 a b d x \sqrt {c^2 d x^2+d}}{35 c^3 \sqrt {c^2 x^2+1}}-\frac {2 b c^3 d x^7 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {c^2 x^2+1}}-\frac {16 b c d x^5 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{175 \sqrt {c^2 x^2+1}}+\frac {1}{7} x^4 \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {3}{35} d x^4 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {2 b d x^3 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{105 c \sqrt {c^2 x^2+1}}+\frac {d x^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^2}-\frac {2 d \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^4}+\frac {2 b^2 d \left (c^2 x^2+1\right )^3 \sqrt {c^2 d x^2+d}}{343 c^4}-\frac {38 b^2 d \left (c^2 x^2+1\right )^2 \sqrt {c^2 d x^2+d}}{6125 c^4}-\frac {304 b^2 d \sqrt {c^2 d x^2+d}}{3675 c^4}-\frac {152 b^2 d \left (c^2 x^2+1\right ) \sqrt {c^2 d x^2+d}}{11025 c^4}+\frac {4 b^2 d x \sqrt {c^2 d x^2+d} \sinh ^{-1}(c x)}{35 c^3 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-304*b^2*d*Sqrt[d + c^2*d*x^2])/(3675*c^4) + (4*a*b*d*x*Sqrt[d + c^2*d*x^2])/(35*c^3*Sqrt[1 + c^2*x^2]) - (15
2*b^2*d*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/(11025*c^4) - (38*b^2*d*(1 + c^2*x^2)^2*Sqrt[d + c^2*d*x^2])/(6125*
c^4) + (2*b^2*d*(1 + c^2*x^2)^3*Sqrt[d + c^2*d*x^2])/(343*c^4) + (4*b^2*d*x*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x])/
(35*c^3*Sqrt[1 + c^2*x^2]) - (2*b*d*x^3*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(105*c*Sqrt[1 + c^2*x^2]) -
(16*b*c*d*x^5*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(175*Sqrt[1 + c^2*x^2]) - (2*b*c^3*d*x^7*Sqrt[d + c^2*
d*x^2]*(a + b*ArcSinh[c*x]))/(49*Sqrt[1 + c^2*x^2]) - (2*d*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(35*c^4
) + (d*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(35*c^2) + (3*d*x^4*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[
c*x])^2)/35 + (x^4*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/7

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 266

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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 5730

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(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, f, m}, 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 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*Sqrt[d + e*x^2]
), Int[(f*x)^(m - 1)*(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[m, 1] && IntegerQ[m]

Rubi steps

\begin {align*} \int x^3 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{7} x^4 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} (3 d) \int x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac {\left (2 b c d \sqrt {d+c^2 d x^2}\right ) \int x^4 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{7 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b c d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{35 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^7 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {1+c^2 x^2}}+\frac {3}{35} d x^4 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} x^4 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (3 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{35 \sqrt {1+c^2 x^2}}-\frac {\left (6 b c d \sqrt {d+c^2 d x^2}\right ) \int x^4 \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{35 \sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^5 \left (7+5 c^2 x^2\right )}{35 \sqrt {1+c^2 x^2}} \, dx}{7 \sqrt {1+c^2 x^2}}\\ &=-\frac {16 b c d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^7 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {1+c^2 x^2}}+\frac {d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^2}+\frac {3}{35} d x^4 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} x^4 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac {\left (2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{35 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (2 b d \sqrt {d+c^2 d x^2}\right ) \int x^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{35 c \sqrt {1+c^2 x^2}}+\frac {\left (2 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^5 \left (7+5 c^2 x^2\right )}{\sqrt {1+c^2 x^2}} \, dx}{245 \sqrt {1+c^2 x^2}}+\frac {\left (6 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^5}{\sqrt {1+c^2 x^2}} \, dx}{175 \sqrt {1+c^2 x^2}}\\ &=-\frac {2 b d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c \sqrt {1+c^2 x^2}}-\frac {16 b c d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^7 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {1+c^2 x^2}}-\frac {2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^4}+\frac {d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^2}+\frac {3}{35} d x^4 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} x^4 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (2 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x^3}{\sqrt {1+c^2 x^2}} \, dx}{105 \sqrt {1+c^2 x^2}}+\frac {\left (4 b d \sqrt {d+c^2 d x^2}\right ) \int \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{35 c^3 \sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (7+5 c^2 x\right )}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{245 \sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{175 \sqrt {1+c^2 x^2}}\\ &=\frac {4 a b d x \sqrt {d+c^2 d x^2}}{35 c^3 \sqrt {1+c^2 x^2}}-\frac {2 b d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c \sqrt {1+c^2 x^2}}-\frac {16 b c d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^7 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {1+c^2 x^2}}-\frac {2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^4}+\frac {d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^2}+\frac {3}{35} d x^4 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} x^4 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (b^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{105 \sqrt {1+c^2 x^2}}+\frac {\left (4 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \sinh ^{-1}(c x) \, dx}{35 c^3 \sqrt {1+c^2 x^2}}+\frac {\left (b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {2}{c^4 \sqrt {1+c^2 x}}+\frac {\sqrt {1+c^2 x}}{c^4}-\frac {8 \left (1+c^2 x\right )^{3/2}}{c^4}+\frac {5 \left (1+c^2 x\right )^{5/2}}{c^4}\right ) \, dx,x,x^2\right )}{245 \sqrt {1+c^2 x^2}}+\frac {\left (3 b^2 c^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^4 \sqrt {1+c^2 x}}-\frac {2 \sqrt {1+c^2 x}}{c^4}+\frac {\left (1+c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )}{175 \sqrt {1+c^2 x^2}}\\ &=\frac {62 b^2 d \sqrt {d+c^2 d x^2}}{1225 c^4}+\frac {4 a b d x \sqrt {d+c^2 d x^2}}{35 c^3 \sqrt {1+c^2 x^2}}-\frac {74 b^2 d \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{3675 c^4}-\frac {38 b^2 d \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}}{6125 c^4}+\frac {2 b^2 d \left (1+c^2 x^2\right )^3 \sqrt {d+c^2 d x^2}}{343 c^4}+\frac {4 b^2 d x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{35 c^3 \sqrt {1+c^2 x^2}}-\frac {2 b d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c \sqrt {1+c^2 x^2}}-\frac {16 b c d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^7 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {1+c^2 x^2}}-\frac {2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^4}+\frac {d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^2}+\frac {3}{35} d x^4 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} x^4 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (b^2 d \sqrt {d+c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{c^2 \sqrt {1+c^2 x}}+\frac {\sqrt {1+c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{105 \sqrt {1+c^2 x^2}}-\frac {\left (4 b^2 d \sqrt {d+c^2 d x^2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{35 c^2 \sqrt {1+c^2 x^2}}\\ &=-\frac {304 b^2 d \sqrt {d+c^2 d x^2}}{3675 c^4}+\frac {4 a b d x \sqrt {d+c^2 d x^2}}{35 c^3 \sqrt {1+c^2 x^2}}-\frac {152 b^2 d \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{11025 c^4}-\frac {38 b^2 d \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2}}{6125 c^4}+\frac {2 b^2 d \left (1+c^2 x^2\right )^3 \sqrt {d+c^2 d x^2}}{343 c^4}+\frac {4 b^2 d x \sqrt {d+c^2 d x^2} \sinh ^{-1}(c x)}{35 c^3 \sqrt {1+c^2 x^2}}-\frac {2 b d x^3 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{105 c \sqrt {1+c^2 x^2}}-\frac {16 b c d x^5 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{175 \sqrt {1+c^2 x^2}}-\frac {2 b c^3 d x^7 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{49 \sqrt {1+c^2 x^2}}-\frac {2 d \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^4}+\frac {d x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{35 c^2}+\frac {3}{35} d x^4 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {1}{7} x^4 \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 251, normalized size = 0.52 \[ \frac {d \sqrt {c^2 d x^2+d} \left (11025 a^2 \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^3-210 a b c x \left (75 c^6 x^6+168 c^4 x^4+35 c^2 x^2-210\right ) \sqrt {c^2 x^2+1}-210 b \sinh ^{-1}(c x) \left (b c x \sqrt {c^2 x^2+1} \left (75 c^6 x^6+168 c^4 x^4+35 c^2 x^2-210\right )-105 a \left (c^2 x^2+1\right )^3 \left (5 c^2 x^2-2\right )\right )+11025 b^2 \left (5 c^2 x^2-2\right ) \left (c^2 x^2+1\right )^3 \sinh ^{-1}(c x)^2+2 b^2 \left (1125 c^8 x^8+3303 c^6 x^6+499 c^4 x^4-20371 c^2 x^2-18692\right )\right )}{385875 c^4 \left (c^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d + c^2*d*x^2]*(11025*a^2*(1 + c^2*x^2)^3*(-2 + 5*c^2*x^2) - 210*a*b*c*x*Sqrt[1 + c^2*x^2]*(-210 + 35*
c^2*x^2 + 168*c^4*x^4 + 75*c^6*x^6) + 2*b^2*(-18692 - 20371*c^2*x^2 + 499*c^4*x^4 + 3303*c^6*x^6 + 1125*c^8*x^
8) - 210*b*(-105*a*(1 + c^2*x^2)^3*(-2 + 5*c^2*x^2) + b*c*x*Sqrt[1 + c^2*x^2]*(-210 + 35*c^2*x^2 + 168*c^4*x^4
+ 75*c^6*x^6))*ArcSinh[c*x] + 11025*b^2*(1 + c^2*x^2)^3*(-2 + 5*c^2*x^2)*ArcSinh[c*x]^2))/(385875*c^4*(1 + c^
2*x^2))

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fricas [A]  time = 0.67, size = 402, normalized size = 0.83 \[ \frac {11025 \, {\left (5 \, b^{2} c^{8} d x^{8} + 13 \, b^{2} c^{6} d x^{6} + 9 \, b^{2} c^{4} d x^{4} - b^{2} c^{2} d x^{2} - 2 \, b^{2} d\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 210 \, {\left (525 \, a b c^{8} d x^{8} + 1365 \, a b c^{6} d x^{6} + 945 \, a b c^{4} d x^{4} - 105 \, a b c^{2} d x^{2} - 210 \, a b d - {\left (75 \, b^{2} c^{7} d x^{7} + 168 \, b^{2} c^{5} d x^{5} + 35 \, b^{2} c^{3} d x^{3} - 210 \, b^{2} c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (1125 \, {\left (49 \, a^{2} + 2 \, b^{2}\right )} c^{8} d x^{8} + 9 \, {\left (15925 \, a^{2} + 734 \, b^{2}\right )} c^{6} d x^{6} + {\left (99225 \, a^{2} + 998 \, b^{2}\right )} c^{4} d x^{4} - {\left (11025 \, a^{2} + 40742 \, b^{2}\right )} c^{2} d x^{2} - 2 \, {\left (11025 \, a^{2} + 18692 \, b^{2}\right )} d - 210 \, {\left (75 \, a b c^{7} d x^{7} + 168 \, a b c^{5} d x^{5} + 35 \, a b c^{3} d x^{3} - 210 \, a b c d x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {c^{2} d x^{2} + d}}{385875 \, {\left (c^{6} x^{2} + c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

1/385875*(11025*(5*b^2*c^8*d*x^8 + 13*b^2*c^6*d*x^6 + 9*b^2*c^4*d*x^4 - b^2*c^2*d*x^2 - 2*b^2*d)*sqrt(c^2*d*x^
2 + d)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 210*(525*a*b*c^8*d*x^8 + 1365*a*b*c^6*d*x^6 + 945*a*b*c^4*d*x^4 - 105*
a*b*c^2*d*x^2 - 210*a*b*d - (75*b^2*c^7*d*x^7 + 168*b^2*c^5*d*x^5 + 35*b^2*c^3*d*x^3 - 210*b^2*c*d*x)*sqrt(c^2
*x^2 + 1))*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + (1125*(49*a^2 + 2*b^2)*c^8*d*x^8 + 9*(15925*a^2
+ 734*b^2)*c^6*d*x^6 + (99225*a^2 + 998*b^2)*c^4*d*x^4 - (11025*a^2 + 40742*b^2)*c^2*d*x^2 - 2*(11025*a^2 + 18
692*b^2)*d - 210*(75*a*b*c^7*d*x^7 + 168*a*b*c^5*d*x^5 + 35*a*b*c^3*d*x^3 - 210*a*b*c*d*x)*sqrt(c^2*x^2 + 1))*
sqrt(c^2*d*x^2 + d))/(c^6*x^2 + c^4)

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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(x^3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*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 [B]  time = 0.54, size = 1766, normalized size = 3.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(1/7*x^2*(c^2*d*x^2+d)^(5/2)/c^2/d-2/35/d/c^4*(c^2*d*x^2+d)^(5/2))+b^2*(1/43904*(d*(c^2*x^2+1))^(1/2)*(64*
c^8*x^8+64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c^6*x^6+112*c^5*x^5*(c^2*x^2+1)^(1/2)+104*c^4*x^4+56*c^3*x^3*(c^2*x^2
+1)^(1/2)+25*c^2*x^2+7*c*x*(c^2*x^2+1)^(1/2)+1)*(49*arcsinh(c*x)^2-14*arcsinh(c*x)+2)*d/c^4/(c^2*x^2+1)+1/1600
0*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^
2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+1)*(25*arcsinh(c*x)^2-10*arcsinh(c*x)+2)*d/c^4/(c^2*x^2+1)-1/1152*(d*(c^2*x^2+1)
)^(1/2)*(4*c^4*x^4+4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1/2)+1)*(9*arcsinh(c*x)^2-6*arcsin
h(c*x)+2)*d/c^4/(c^2*x^2+1)-3/128*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(arcsinh(c*x)^2-2*ar
csinh(c*x)+2)*d/c^4/(c^2*x^2+1)-3/128*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(arcsinh(c*x)^2+
2*arcsinh(c*x)+2)*d/c^4/(c^2*x^2+1)-1/1152*(d*(c^2*x^2+1))^(1/2)*(4*c^4*x^4-4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*
x^2-3*c*x*(c^2*x^2+1)^(1/2)+1)*(9*arcsinh(c*x)^2+6*arcsinh(c*x)+2)*d/c^4/(c^2*x^2+1)+1/16000*(d*(c^2*x^2+1))^(
1/2)*(16*c^6*x^6-16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4-20*c^3*x^3*(c^2*x^2+1)^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^
2+1)^(1/2)+1)*(25*arcsinh(c*x)^2+10*arcsinh(c*x)+2)*d/c^4/(c^2*x^2+1)+1/43904*(d*(c^2*x^2+1))^(1/2)*(64*c^8*x^
8-64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c^6*x^6-112*c^5*x^5*(c^2*x^2+1)^(1/2)+104*c^4*x^4-56*c^3*x^3*(c^2*x^2+1)^(1
/2)+25*c^2*x^2-7*c*x*(c^2*x^2+1)^(1/2)+1)*(49*arcsinh(c*x)^2+14*arcsinh(c*x)+2)*d/c^4/(c^2*x^2+1))+2*a*b*(1/62
72*(d*(c^2*x^2+1))^(1/2)*(64*c^8*x^8+64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c^6*x^6+112*c^5*x^5*(c^2*x^2+1)^(1/2)+10
4*c^4*x^4+56*c^3*x^3*(c^2*x^2+1)^(1/2)+25*c^2*x^2+7*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+7*arcsinh(c*x))*d/c^4/(c^2*x^
2+1)+1/3200*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3*(c^2*x^2+1)^(
1/2)+13*c^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+5*arcsinh(c*x))*d/c^4/(c^2*x^2+1)-1/384*(d*(c^2*x^2+1))^(1/2)*(
4*c^4*x^4+4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1/2)+1)*(-1+3*arcsinh(c*x))*d/c^4/(c^2*x^2+
1)-3/128*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*(-1+arcsinh(c*x))*d/c^4/(c^2*x^2+1)-3/128*(d*
(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*(1+arcsinh(c*x))*d/c^4/(c^2*x^2+1)-1/384*(d*(c^2*x^2+1))^
(1/2)*(4*c^4*x^4-4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2-3*c*x*(c^2*x^2+1)^(1/2)+1)*(1+3*arcsinh(c*x))*d/c^4/(c^
2*x^2+1)+1/3200*(d*(c^2*x^2+1))^(1/2)*(16*c^6*x^6-16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4-20*c^3*x^3*(c^2*x^2+
1)^(1/2)+13*c^2*x^2-5*c*x*(c^2*x^2+1)^(1/2)+1)*(1+5*arcsinh(c*x))*d/c^4/(c^2*x^2+1)+1/6272*(d*(c^2*x^2+1))^(1/
2)*(64*c^8*x^8-64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c^6*x^6-112*c^5*x^5*(c^2*x^2+1)^(1/2)+104*c^4*x^4-56*c^3*x^3*(
c^2*x^2+1)^(1/2)+25*c^2*x^2-7*c*x*(c^2*x^2+1)^(1/2)+1)*(1+7*arcsinh(c*x))*d/c^4/(c^2*x^2+1))

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maxima [A]  time = 0.44, size = 346, normalized size = 0.72 \[ \frac {1}{35} \, {\left (\frac {5 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{2} d} - \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{4} d}\right )} b^{2} \operatorname {arsinh}\left (c x\right )^{2} + \frac {2}{35} \, {\left (\frac {5 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{2} d} - \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{4} d}\right )} a b \operatorname {arsinh}\left (c x\right ) + \frac {1}{35} \, {\left (\frac {5 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{2} d} - \frac {2 \, {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{4} d}\right )} a^{2} + \frac {2}{385875} \, b^{2} {\left (\frac {1125 \, \sqrt {c^{2} x^{2} + 1} c^{4} d^{\frac {3}{2}} x^{6} + 2178 \, \sqrt {c^{2} x^{2} + 1} c^{2} d^{\frac {3}{2}} x^{4} - 1679 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {3}{2}} x^{2} - \frac {18692 \, \sqrt {c^{2} x^{2} + 1} d^{\frac {3}{2}}}{c^{2}}}{c^{2}} - \frac {105 \, {\left (75 \, c^{6} d^{\frac {3}{2}} x^{7} + 168 \, c^{4} d^{\frac {3}{2}} x^{5} + 35 \, c^{2} d^{\frac {3}{2}} x^{3} - 210 \, d^{\frac {3}{2}} x\right )} \operatorname {arsinh}\left (c x\right )}{c^{3}}\right )} - \frac {2 \, {\left (75 \, c^{6} d^{\frac {3}{2}} x^{7} + 168 \, c^{4} d^{\frac {3}{2}} x^{5} + 35 \, c^{2} d^{\frac {3}{2}} x^{3} - 210 \, d^{\frac {3}{2}} x\right )} a b}{3675 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="maxima")

[Out]

1/35*(5*(c^2*d*x^2 + d)^(5/2)*x^2/(c^2*d) - 2*(c^2*d*x^2 + d)^(5/2)/(c^4*d))*b^2*arcsinh(c*x)^2 + 2/35*(5*(c^2
*d*x^2 + d)^(5/2)*x^2/(c^2*d) - 2*(c^2*d*x^2 + d)^(5/2)/(c^4*d))*a*b*arcsinh(c*x) + 1/35*(5*(c^2*d*x^2 + d)^(5
/2)*x^2/(c^2*d) - 2*(c^2*d*x^2 + d)^(5/2)/(c^4*d))*a^2 + 2/385875*b^2*((1125*sqrt(c^2*x^2 + 1)*c^4*d^(3/2)*x^6
+ 2178*sqrt(c^2*x^2 + 1)*c^2*d^(3/2)*x^4 - 1679*sqrt(c^2*x^2 + 1)*d^(3/2)*x^2 - 18692*sqrt(c^2*x^2 + 1)*d^(3/
2)/c^2)/c^2 - 105*(75*c^6*d^(3/2)*x^7 + 168*c^4*d^(3/2)*x^5 + 35*c^2*d^(3/2)*x^3 - 210*d^(3/2)*x)*arcsinh(c*x)
/c^3) - 2/3675*(75*c^6*d^(3/2)*x^7 + 168*c^4*d^(3/2)*x^5 + 35*c^2*d^(3/2)*x^3 - 210*d^(3/2)*x)*a*b/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(d*(c**2*x**2 + 1))**(3/2)*(a + b*asinh(c*x))**2, x)

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