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

Optimal. Leaf size=300 \[ -\frac{\pi ^{5/2} b \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{18 c}-\frac{5 \pi ^{5/2} b \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{48 c}+\frac{1}{6} x \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} \pi x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{16} \pi ^2 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{5}{16} \pi ^{5/2} b c x^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 \pi ^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c}+\frac{1}{108} \pi ^{5/2} b^2 x \left (c^2 x^2+1\right )^{5/2}+\frac{65 \pi ^{5/2} b^2 x \left (c^2 x^2+1\right )^{3/2}}{1728}+\frac{245 \pi ^{5/2} b^2 x \sqrt{c^2 x^2+1}}{1152}-\frac{115 \pi ^{5/2} b^2 \sinh ^{-1}(c x)}{1152 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.381804, antiderivative size = 420, normalized size of antiderivative = 1.4, 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} \[ \frac{5 \pi ^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \sqrt{c^2 x^2+1}}+\frac{1}{6} x \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} \pi x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{16} \pi ^2 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{\pi ^2 b \left (c^2 x^2+1\right )^{5/2} \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{18 c}-\frac{5 \pi ^2 b \left (c^2 x^2+1\right )^{3/2} \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{5 \pi ^2 b c x^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{c^2 x^2+1}}+\frac{1}{108} \pi ^2 b^2 x \left (c^2 x^2+1\right )^2 \sqrt{\pi c^2 x^2+\pi }+\frac{245 \pi ^2 b^2 x \sqrt{\pi c^2 x^2+\pi }}{1152}+\frac{65 \pi ^2 b^2 x \left (c^2 x^2+1\right ) \sqrt{\pi c^2 x^2+\pi }}{1728}-\frac{115 \pi ^2 b^2 \sqrt{\pi c^2 x^2+\pi } \sinh ^{-1}(c x)}{1152 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(245*b^2*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2])/1152 + (65*b^2*Pi^2*x*(1 + c^2*x^2)*Sqrt[Pi + c^2*Pi*x^2])/1728 + (b^2*
Pi^2*x*(1 + c^2*x^2)^2*Sqrt[Pi + c^2*Pi*x^2])/108 - (115*b^2*Pi^2*Sqrt[Pi + c^2*Pi*x^2]*ArcSinh[c*x])/(1152*c*
Sqrt[1 + c^2*x^2]) - (5*b*c*Pi^2*x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(16*Sqrt[1 + c^2*x^2]) - (5*b
*Pi^2*(1 + c^2*x^2)^(3/2)*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(48*c) - (b*Pi^2*(1 + c^2*x^2)^(5/2)*Sqr
t[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(18*c) + (5*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/16 +
(5*Pi*x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/24 + (x*(Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSinh[c*x])^
2)/6 + (5*Pi^2*Sqrt[Pi + c^2*Pi*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 (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} (5 \pi ) \int \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac{\left (b c \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{8} \left (5 \pi ^2\right ) \int \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx+\frac{\left (b^2 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 x \left (1+c^2 x^2\right )^2 \sqrt{\pi +c^2 \pi x^2}-\frac{5 b \pi ^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\left (5 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 x \left (1+c^2 x^2\right ) \sqrt{\pi +c^2 \pi x^2}}{1728}+\frac{1}{108} b^2 \pi ^2 x \left (1+c^2 x^2\right )^2 \sqrt{\pi +c^2 \pi x^2}-\frac{5 b c \pi ^2 x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{1+c^2 x^2}}-\frac{5 b \pi ^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \sqrt{1+c^2 x^2} \, dx}{144 \sqrt{1+c^2 x^2}}+\frac{\left (5 b^2 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{1152}+\frac{65 b^2 \pi ^2 x \left (1+c^2 x^2\right ) \sqrt{\pi +c^2 \pi x^2}}{1728}+\frac{1}{108} b^2 \pi ^2 x \left (1+c^2 x^2\right )^2 \sqrt{\pi +c^2 \pi x^2}-\frac{5 b c \pi ^2 x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{1+c^2 x^2}}-\frac{5 b \pi ^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi 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 \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{32 \sqrt{1+c^2 x^2}}\\ &=\frac{245 b^2 \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{1152}+\frac{65 b^2 \pi ^2 x \left (1+c^2 x^2\right ) \sqrt{\pi +c^2 \pi x^2}}{1728}+\frac{1}{108} b^2 \pi ^2 x \left (1+c^2 x^2\right )^2 \sqrt{\pi +c^2 \pi x^2}-\frac{115 b^2 \pi ^2 \sqrt{\pi +c^2 \pi x^2} \sinh ^{-1}(c x)}{1152 c \sqrt{1+c^2 x^2}}-\frac{5 b c \pi ^2 x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \sqrt{1+c^2 x^2}}-\frac{5 b \pi ^2 \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c}-\frac{b \pi ^2 \left (1+c^2 x^2\right )^{5/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac{5}{16} \pi ^2 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5}{24} \pi x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{6} x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{5 \pi ^2 \sqrt{\pi +c^2 \pi 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 = 0.95247, size = 284, normalized size = 0.95 \[ \frac{\pi ^{5/2} \left (12 \sinh ^{-1}(c x) \left (360 a^2+540 a b \sinh \left (2 \sinh ^{-1}(c x)\right )+108 a b \sinh \left (4 \sinh ^{-1}(c x)\right )+12 a b \sinh \left (6 \sinh ^{-1}(c x)\right )-270 b^2 \cosh \left (2 \sinh ^{-1}(c x)\right )-27 b^2 \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b^2 \cosh \left (6 \sinh ^{-1}(c x)\right )\right )+2304 a^2 c^5 x^5 \sqrt{c^2 x^2+1}+7488 a^2 c^3 x^3 \sqrt{c^2 x^2+1}+9504 a^2 c x \sqrt{c^2 x^2+1}+72 b \sinh ^{-1}(c x)^2 \left (60 a+45 b \sinh \left (2 \sinh ^{-1}(c x)\right )+9 b \sinh \left (4 \sinh ^{-1}(c x)\right )+b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )-3240 a b \cosh \left (2 \sinh ^{-1}(c x)\right )-324 a b \cosh \left (4 \sinh ^{-1}(c x)\right )-24 a b \cosh \left (6 \sinh ^{-1}(c x)\right )+1440 b^2 \sinh ^{-1}(c x)^3+1620 b^2 \sinh \left (2 \sinh ^{-1}(c x)\right )+81 b^2 \sinh \left (4 \sinh ^{-1}(c x)\right )+4 b^2 \sinh \left (6 \sinh ^{-1}(c x)\right )\right )}{13824 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(5/2)*(9504*a^2*c*x*Sqrt[1 + c^2*x^2] + 7488*a^2*c^3*x^3*Sqrt[1 + c^2*x^2] + 2304*a^2*c^5*x^5*Sqrt[1 + c^2
*x^2] + 1440*b^2*ArcSinh[c*x]^3 - 3240*a*b*Cosh[2*ArcSinh[c*x]] - 324*a*b*Cosh[4*ArcSinh[c*x]] - 24*a*b*Cosh[6
*ArcSinh[c*x]] + 1620*b^2*Sinh[2*ArcSinh[c*x]] + 81*b^2*Sinh[4*ArcSinh[c*x]] + 4*b^2*Sinh[6*ArcSinh[c*x]] + 72
*b*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]]) + 12*
ArcSinh[c*x]*(360*a^2 - 270*b^2*Cosh[2*ArcSinh[c*x]] - 27*b^2*Cosh[4*ArcSinh[c*x]] - 2*b^2*Cosh[6*ArcSinh[c*x]
] + 540*a*b*Sinh[2*ArcSinh[c*x]] + 108*a*b*Sinh[4*ArcSinh[c*x]] + 12*a*b*Sinh[6*ArcSinh[c*x]])))/(13824*c)

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Maple [A]  time = 0.103, size = 486, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}x}{6} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{a}^{2}\pi \,x}{24} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}{\pi }^{2}x}{16}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{5\,{a}^{2}{\pi }^{3}}{16}\ln \left ({\pi \,{c}^{2}x{\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+\sqrt{\pi \,{c}^{2}{x}^{2}+\pi } \right ){\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+{\frac{{b}^{2}{\pi }^{{\frac{5}{2}}}{c}^{4} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{x}^{5}}{6}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{b}^{2}{\pi }^{{\frac{5}{2}}}{c}^{5}{\it Arcsinh} \left ( cx \right ){x}^{6}}{18}}+{\frac{{b}^{2}{\pi }^{{\frac{5}{2}}}{c}^{4}{x}^{5}}{108}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{13\,{b}^{2}{\pi }^{5/2}{c}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{x}^{3}}{24}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{13\,{b}^{2}{\pi }^{5/2}{c}^{3}{\it Arcsinh} \left ( cx \right ){x}^{4}}{48}}+{\frac{97\,{b}^{2}{\pi }^{5/2}{c}^{2}{x}^{3}}{1728}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{11\,{b}^{2}{\pi }^{5/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}x}{16}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{11\,{b}^{2}{\pi }^{5/2}c{\it Arcsinh} \left ( cx \right ){x}^{2}}{16}}+{\frac{5\,{b}^{2}{\pi }^{5/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{3}}{48\,c}}+{\frac{299\,{b}^{2}{\pi }^{5/2}x}{1152}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{299\,{b}^{2}{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ) }{1152\,c}}+{\frac{ab{\pi }^{{\frac{5}{2}}}{c}^{4}{\it Arcsinh} \left ( cx \right ){x}^{5}}{3}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{ab{\pi }^{{\frac{5}{2}}}{c}^{5}{x}^{6}}{18}}+{\frac{13\,ab{\pi }^{5/2}{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{12}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{13\,ab{\pi }^{5/2}{c}^{3}{x}^{4}}{48}}+{\frac{11\,ab{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ) x}{8}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{11\,ab{\pi }^{5/2}c{x}^{2}}{16}}+{\frac{5\,ab{\pi }^{5/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{16\,c}}-{\frac{17\,ab{\pi }^{5/2}}{36\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^2*x*(Pi*c^2*x^2+Pi)^(5/2)+5/24*a^2*Pi*x*(Pi*c^2*x^2+Pi)^(3/2)+5/16*a^2*Pi^2*x*(Pi*c^2*x^2+Pi)^(1/2)+5/16
*a^2*Pi^3*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/6*b^2*Pi^(5/2)*c^4*arcsinh(c*x)^2
*(c^2*x^2+1)^(1/2)*x^5-1/18*b^2*Pi^(5/2)*c^5*arcsinh(c*x)*x^6+1/108*b^2*Pi^(5/2)*c^4*x^5*(c^2*x^2+1)^(1/2)+13/
24*b^2*Pi^(5/2)*c^2*arcsinh(c*x)^2*(c^2*x^2+1)^(1/2)*x^3-13/48*b^2*Pi^(5/2)*c^3*arcsinh(c*x)*x^4+97/1728*b^2*P
i^(5/2)*c^2*x^3*(c^2*x^2+1)^(1/2)+11/16*b^2*Pi^(5/2)*arcsinh(c*x)^2*(c^2*x^2+1)^(1/2)*x-11/16*b^2*Pi^(5/2)*c*a
rcsinh(c*x)*x^2+5/48*b^2*Pi^(5/2)/c*arcsinh(c*x)^3+299/1152*b^2*Pi^(5/2)*x*(c^2*x^2+1)^(1/2)-299/1152*b^2*Pi^(
5/2)*arcsinh(c*x)/c+1/3*a*b*Pi^(5/2)*c^4*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^5-1/18*a*b*Pi^(5/2)*c^5*x^6+13/12*a*
b*Pi^(5/2)*c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3-13/48*a*b*Pi^(5/2)*c^3*x^4+11/8*a*b*Pi^(5/2)*arcsinh(c*x)*(c
^2*x^2+1)^(1/2)*x-11/16*a*b*Pi^(5/2)*c*x^2+5/16*a*b*Pi^(5/2)/c*arcsinh(c*x)^2-17/36*a*b*Pi^(5/2)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{\pi + \pi c^{2} x^{2}}{\left (\pi ^{2} a^{2} c^{4} x^{4} + 2 \, \pi ^{2} a^{2} c^{2} x^{2} + \pi ^{2} a^{2} +{\left (\pi ^{2} b^{2} c^{4} x^{4} + 2 \, \pi ^{2} b^{2} c^{2} x^{2} + \pi ^{2} b^{2}\right )} \operatorname{arsinh}\left (c x\right )^{2} + 2 \,{\left (\pi ^{2} a b c^{4} x^{4} + 2 \, \pi ^{2} a b c^{2} x^{2} + \pi ^{2} a b\right )} \operatorname{arsinh}\left (c x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi^2*a^2*c^4*x^4 + 2*pi^2*a^2*c^2*x^2 + pi^2*a^2 + (pi^2*b^2*c^4*x^4 + 2*pi^2*
b^2*c^2*x^2 + pi^2*b^2)*arcsinh(c*x)^2 + 2*(pi^2*a*b*c^4*x^4 + 2*pi^2*a*b*c^2*x^2 + pi^2*a*b)*arcsinh(c*x)), 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((pi*c**2*x**2+pi)**(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 (\pi + \pi c^{2} x^{2}\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((pi*c^2*x^2+pi)^(5/2)*(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

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