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

Optimal. Leaf size=420 \[ \frac{5 d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \sqrt{c^2 x^2+1}}+\frac{5}{16} d^2 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}-\frac{5 b d^2 \left (c^2 x^2+1\right )^{3/2} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{5 b c d^2 x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{c^2 x^2+1}}+\frac{1}{6} x \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} d x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{108} b^2 d^2 x \left (c^2 x^2+1\right )^2 \sqrt{c^2 d x^2+d}+\frac{245 b^2 d^2 x \sqrt{c^2 d x^2+d}}{1152}+\frac{65 b^2 d^2 x \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d}}{1728}-\frac{115 b^2 d^2 \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)}{1152 c \sqrt{c^2 x^2+1}} \]

[Out]

(245*b^2*d^2*x*Sqrt[d + c^2*d*x^2])/1152 + (65*b^2*d^2*x*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/1728 + (b^2*d^2*x*
(1 + c^2*x^2)^2*Sqrt[d + c^2*d*x^2])/108 - (115*b^2*d^2*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x])/(1152*c*Sqrt[1 + c^2
*x^2]) - (5*b*c*d^2*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(16*Sqrt[1 + c^2*x^2]) - (5*b*d^2*(1 + c^2*x
^2)^(3/2)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(48*c) - (b*d^2*(1 + c^2*x^2)^(5/2)*Sqrt[d + c^2*d*x^2]*(a
+ b*ArcSinh[c*x]))/(18*c) + (5*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/16 + (5*d*x*(d + c^2*d*x^2)^
(3/2)*(a + b*ArcSinh[c*x])^2)/24 + (x*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/6 + (5*d^2*Sqrt[d + c^2*d*
x^2]*(a + b*ArcSinh[c*x])^3)/(48*b*c*Sqrt[1 + c^2*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.403579, antiderivative size = 420, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {5684, 5682, 5675, 5661, 321, 215, 5717, 195} \[ \frac{5 d^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \sqrt{c^2 x^2+1}}+\frac{5}{16} d^2 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}-\frac{5 b d^2 \left (c^2 x^2+1\right )^{3/2} \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{5 b c d^2 x^2 \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{c^2 x^2+1}}+\frac{1}{6} x \left (c^2 d x^2+d\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} d x \left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{108} b^2 d^2 x \left (c^2 x^2+1\right )^2 \sqrt{c^2 d x^2+d}+\frac{245 b^2 d^2 x \sqrt{c^2 d x^2+d}}{1152}+\frac{65 b^2 d^2 x \left (c^2 x^2+1\right ) \sqrt{c^2 d x^2+d}}{1728}-\frac{115 b^2 d^2 \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)}{1152 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(245*b^2*d^2*x*Sqrt[d + c^2*d*x^2])/1152 + (65*b^2*d^2*x*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/1728 + (b^2*d^2*x*
(1 + c^2*x^2)^2*Sqrt[d + c^2*d*x^2])/108 - (115*b^2*d^2*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x])/(1152*c*Sqrt[1 + c^2
*x^2]) - (5*b*c*d^2*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(16*Sqrt[1 + c^2*x^2]) - (5*b*d^2*(1 + c^2*x
^2)^(3/2)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(48*c) - (b*d^2*(1 + c^2*x^2)^(5/2)*Sqrt[d + c^2*d*x^2]*(a
+ b*ArcSinh[c*x]))/(18*c) + (5*d^2*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/16 + (5*d*x*(d + c^2*d*x^2)^
(3/2)*(a + b*ArcSinh[c*x])^2)/24 + (x*(d + c^2*d*x^2)^(5/2)*(a + b*ArcSinh[c*x])^2)/6 + (5*d^2*Sqrt[d + c^2*d*
x^2]*(a + b*ArcSinh[c*x])^3)/(48*b*c*Sqrt[1 + c^2*x^2])

Rule 5684

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1
+ c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && Gt
Q[n, 0] && GtQ[p, 0]

Rule 5682

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

Rule 5675

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rubi steps

\begin{align*} \int \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} (5 d) \int \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac{\left (b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{3 \sqrt{1+c^2 x^2}}\\ &=-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{8} \left (5 d^2\right ) \int \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx+\frac{\left (b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^{5/2} \, dx}{18 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{12 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{108} b^2 d^2 x \left (1+c^2 x^2\right )^2 \sqrt{d+c^2 d x^2}-\frac{5 b d^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\left (5 d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{1+c^2 x^2}} \, dx}{16 \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{108 \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{48 \sqrt{1+c^2 x^2}}-\frac{\left (5 b c d^2 \sqrt{d+c^2 d x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{8 \sqrt{1+c^2 x^2}}\\ &=\frac{65 b^2 d^2 x \left (1+c^2 x^2\right ) \sqrt{d+c^2 d x^2}}{1728}+\frac{1}{108} b^2 d^2 x \left (1+c^2 x^2\right )^2 \sqrt{d+c^2 d x^2}-\frac{5 b c d^2 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{1+c^2 x^2}}-\frac{5 b d^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5 d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \sqrt{1+c^2 x^2} \, dx}{144 \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \sqrt{1+c^2 x^2} \, dx}{64 \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 c^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{16 \sqrt{1+c^2 x^2}}\\ &=\frac{245 b^2 d^2 x \sqrt{d+c^2 d x^2}}{1152}+\frac{65 b^2 d^2 x \left (1+c^2 x^2\right ) \sqrt{d+c^2 d x^2}}{1728}+\frac{1}{108} b^2 d^2 x \left (1+c^2 x^2\right )^2 \sqrt{d+c^2 d x^2}-\frac{5 b c d^2 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{1+c^2 x^2}}-\frac{5 b d^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5 d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{288 \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{128 \sqrt{1+c^2 x^2}}-\frac{\left (5 b^2 d^2 \sqrt{d+c^2 d x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{32 \sqrt{1+c^2 x^2}}\\ &=\frac{245 b^2 d^2 x \sqrt{d+c^2 d x^2}}{1152}+\frac{65 b^2 d^2 x \left (1+c^2 x^2\right ) \sqrt{d+c^2 d x^2}}{1728}+\frac{1}{108} b^2 d^2 x \left (1+c^2 x^2\right )^2 \sqrt{d+c^2 d x^2}-\frac{115 b^2 d^2 \sqrt{d+c^2 d x^2} \sinh ^{-1}(c x)}{1152 c \sqrt{1+c^2 x^2}}-\frac{5 b c d^2 x^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{1+c^2 x^2}}-\frac{5 b d^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} d^2 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} d x \left (d+c^2 d x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (d+c^2 d x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5 d^2 \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.53273, size = 499, normalized size = 1.19 \[ \frac{d^2 \left (2304 a^2 c^5 x^5 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}+7488 a^2 c^3 x^3 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}+9504 a^2 c x \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}+4320 a^2 \sqrt{d} \sqrt{c^2 x^2+1} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+72 b \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)^2 \left (60 a+45 b \sinh \left (2 \sinh ^{-1}(c x)\right )+9 b \sinh \left (4 \sinh ^{-1}(c x)\right )+b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )-3240 a b \sqrt{c^2 d x^2+d} \cosh \left (2 \sinh ^{-1}(c x)\right )-324 a b \sqrt{c^2 d x^2+d} \cosh \left (4 \sinh ^{-1}(c x)\right )-24 a b \sqrt{c^2 d x^2+d} \cosh \left (6 \sinh ^{-1}(c x)\right )-12 b \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x) \left (-540 a \sinh \left (2 \sinh ^{-1}(c x)\right )-108 a \sinh \left (4 \sinh ^{-1}(c x)\right )-12 a \sinh \left (6 \sinh ^{-1}(c x)\right )+270 b \cosh \left (2 \sinh ^{-1}(c x)\right )+27 b \cosh \left (4 \sinh ^{-1}(c x)\right )+2 b \cosh \left (6 \sinh ^{-1}(c x)\right )\right )+1440 b^2 \sqrt{c^2 d x^2+d} \sinh ^{-1}(c x)^3+1620 b^2 \sqrt{c^2 d x^2+d} \sinh \left (2 \sinh ^{-1}(c x)\right )+81 b^2 \sqrt{c^2 d x^2+d} \sinh \left (4 \sinh ^{-1}(c x)\right )+4 b^2 \sqrt{c^2 d x^2+d} \sinh \left (6 \sinh ^{-1}(c x)\right )\right )}{13824 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(9504*a^2*c*x*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 7488*a^2*c^3*x^3*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x
^2] + 2304*a^2*c^5*x^5*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2] + 1440*b^2*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]^3 - 3
240*a*b*Sqrt[d + c^2*d*x^2]*Cosh[2*ArcSinh[c*x]] - 324*a*b*Sqrt[d + c^2*d*x^2]*Cosh[4*ArcSinh[c*x]] - 24*a*b*S
qrt[d + c^2*d*x^2]*Cosh[6*ArcSinh[c*x]] + 4320*a^2*Sqrt[d]*Sqrt[1 + c^2*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*
d*x^2]] + 1620*b^2*Sqrt[d + c^2*d*x^2]*Sinh[2*ArcSinh[c*x]] + 81*b^2*Sqrt[d + c^2*d*x^2]*Sinh[4*ArcSinh[c*x]]
+ 4*b^2*Sqrt[d + c^2*d*x^2]*Sinh[6*ArcSinh[c*x]] - 12*b*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]*(270*b*Cosh[2*ArcSinh
[c*x]] + 27*b*Cosh[4*ArcSinh[c*x]] + 2*b*Cosh[6*ArcSinh[c*x]] - 540*a*Sinh[2*ArcSinh[c*x]] - 108*a*Sinh[4*ArcS
inh[c*x]] - 12*a*Sinh[6*ArcSinh[c*x]]) + 72*b*Sqrt[d + c^2*d*x^2]*ArcSinh[c*x]^2*(60*a + 45*b*Sinh[2*ArcSinh[c
*x]] + 9*b*Sinh[4*ArcSinh[c*x]] + b*Sinh[6*ArcSinh[c*x]])))/(13824*c*Sqrt[1 + c^2*x^2])

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Maple [B]  time = 0.286, size = 966, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

5/16*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^2*d^2+11/8*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*
x^2+1)*arcsinh(c*x)*x-11/16*a*b*(d*(c^2*x^2+1))^(1/2)*d^2*c/(c^2*x^2+1)^(1/2)*x^2-1/18*a*b*(d*(c^2*x^2+1))^(1/
2)*d^2*c^5/(c^2*x^2+1)^(1/2)*x^6-13/48*a*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^3/(c^2*x^2+1)^(1/2)*x^4-1/18*b^2*(d*(c^
2*x^2+1))^(1/2)*d^2*c^5/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^6-13/48*b^2*(d*(c^2*x^2+1))^(1/2)*d^2*c^3/(c^2*x^2+1)
^(1/2)*arcsinh(c*x)*x^4-11/16*b^2*(d*(c^2*x^2+1))^(1/2)*d^2*c/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^2+1/6*b^2*(d*(c
^2*x^2+1))^(1/2)*d^2*c^6/(c^2*x^2+1)*arcsinh(c*x)^2*x^7+17/24*b^2*(d*(c^2*x^2+1))^(1/2)*d^2*c^4/(c^2*x^2+1)*ar
csinh(c*x)^2*x^5+59/48*b^2*(d*(c^2*x^2+1))^(1/2)*d^2*c^2/(c^2*x^2+1)*arcsinh(c*x)^2*x^3+5/24*a^2*d*x*(c^2*d*x^
2+d)^(3/2)+5/16*a^2*d^2*x*(c^2*d*x^2+d)^(1/2)+5/16*a^2*d^3*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*
d)^(1/2)+59/24*a*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^2/(c^2*x^2+1)*arcsinh(c*x)*x^3+1/3*a*b*(d*(c^2*x^2+1))^(1/2)*d^
2*c^6/(c^2*x^2+1)*arcsinh(c*x)*x^7+17/12*a*b*(d*(c^2*x^2+1))^(1/2)*d^2*c^4/(c^2*x^2+1)*arcsinh(c*x)*x^5+1/6*x*
(c^2*d*x^2+d)^(5/2)*a^2+299/1152*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*x+1091/3456*b^2*(d*(c^2*x^2+1))^(1/
2)*d^2*c^2/(c^2*x^2+1)*x^3+5/48*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(c*x)^3*d^2+11/16*b^2*(d*
(c^2*x^2+1))^(1/2)*d^2/(c^2*x^2+1)*arcsinh(c*x)^2*x-299/1152*b^2*(d*(c^2*x^2+1))^(1/2)*d^2/c/(c^2*x^2+1)^(1/2)
*arcsinh(c*x)-299/1152*a*b*(d*(c^2*x^2+1))^(1/2)*d^2/c/(c^2*x^2+1)^(1/2)+1/108*b^2*(d*(c^2*x^2+1))^(1/2)*d^2*c
^6/(c^2*x^2+1)*x^7+113/1728*b^2*(d*(c^2*x^2+1))^(1/2)*d^2*c^4/(c^2*x^2+1)*x^5

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\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\right ) \end{align*}

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, 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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}

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, algorithm="giac")

[Out]

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