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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 420, normalized size of antiderivative = 1.40, number of steps used = 16, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, 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 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]])

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 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 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 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 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 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 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]

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 = 0.93, size = 284, normalized size = 0.95 \[ \frac {\pi ^{5/2} \left (12 \sinh ^{-1}(c x) \left (360 a^2+540 a b \sinh \left (2 \sinh ^{-1}(c x)\right )+108 a b \sinh \left (4 \sinh ^{-1}(c x)\right )+12 a b \sinh \left (6 \sinh ^{-1}(c x)\right )-270 b^2 \cosh \left (2 \sinh ^{-1}(c x)\right )-27 b^2 \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b^2 \cosh \left (6 \sinh ^{-1}(c x)\right )\right )+9504 a^2 c x \sqrt {c^2 x^2+1}+2304 a^2 c^5 x^5 \sqrt {c^2 x^2+1}+7488 a^2 c^3 x^3 \sqrt {c^2 x^2+1}+72 b \sinh ^{-1}(c x)^2 \left (60 a+45 b \sinh \left (2 \sinh ^{-1}(c x)\right )+9 b \sinh \left (4 \sinh ^{-1}(c x)\right )+b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )-3240 a b \cosh \left (2 \sinh ^{-1}(c x)\right )-324 a b \cosh \left (4 \sinh ^{-1}(c x)\right )-24 a b \cosh \left (6 \sinh ^{-1}(c x)\right )+1440 b^2 \sinh ^{-1}(c x)^3+1620 b^2 \sinh \left (2 \sinh ^{-1}(c x)\right )+81 b^2 \sinh \left (4 \sinh ^{-1}(c x)\right )+4 b^2 \sinh \left (6 \sinh ^{-1}(c x)\right )\right )}{13824 c} \]

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)

________________________________________________________________________________________

fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\pi + \pi c^{2} x^{2}} {\left (\pi ^{2} a^{2} c^{4} x^{4} + 2 \, \pi ^{2} a^{2} c^{2} x^{2} + \pi ^{2} a^{2} + {\left (\pi ^{2} b^{2} c^{4} x^{4} + 2 \, \pi ^{2} b^{2} c^{2} x^{2} + \pi ^{2} b^{2}\right )} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, {\left (\pi ^{2} a b c^{4} x^{4} + 2 \, \pi ^{2} a b c^{2} x^{2} + \pi ^{2} a b\right )} \operatorname {arsinh}\left (c x\right )\right )}, x\right ) \]

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
)

________________________________________________________________________________________

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((pi*c^2*x^2+pi)^(5/2)*(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

________________________________________________________________________________________

maple [A]  time = 0.13, size = 486, normalized size = 1.62 \[ \frac {a^{2} x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{6}+\frac {5 a^{2} \pi x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{24}+\frac {5 a^{2} \pi ^{2} x \sqrt {\pi \,c^{2} x^{2}+\pi }}{16}+\frac {5 a^{2} \pi ^{3} \ln \left (\frac {\pi x \,c^{2}}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{16 \sqrt {\pi \,c^{2}}}+\frac {b^{2} \pi ^{\frac {5}{2}} c^{4} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right )^{2} x^{5}}{6}-\frac {b^{2} \pi ^{\frac {5}{2}} c^{5} \arcsinh \left (c x \right ) x^{6}}{18}+\frac {b^{2} \pi ^{\frac {5}{2}} c^{4} x^{5} \sqrt {c^{2} x^{2}+1}}{108}+\frac {13 b^{2} \pi ^{\frac {5}{2}} c^{2} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right )^{2} x^{3}}{24}-\frac {13 b^{2} \pi ^{\frac {5}{2}} c^{3} \arcsinh \left (c x \right ) x^{4}}{48}+\frac {97 b^{2} \pi ^{\frac {5}{2}} c^{2} x^{3} \sqrt {c^{2} x^{2}+1}}{1728}+\frac {11 b^{2} \pi ^{\frac {5}{2}} \arcsinh \left (c x \right )^{2} \sqrt {c^{2} x^{2}+1}\, x}{16}-\frac {11 b^{2} \pi ^{\frac {5}{2}} c \arcsinh \left (c x \right ) x^{2}}{16}+\frac {299 b^{2} \pi ^{\frac {5}{2}} x \sqrt {c^{2} x^{2}+1}}{1152}+\frac {5 b^{2} \pi ^{\frac {5}{2}} \arcsinh \left (c x \right )^{3}}{48 c}-\frac {299 b^{2} \pi ^{\frac {5}{2}} \arcsinh \left (c x \right )}{1152 c}+\frac {a b \,\pi ^{\frac {5}{2}} c^{4} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) x^{5}}{3}-\frac {a b \,\pi ^{\frac {5}{2}} c^{5} x^{6}}{18}+\frac {13 a b \,\pi ^{\frac {5}{2}} c^{2} \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) x^{3}}{12}-\frac {13 a b \,\pi ^{\frac {5}{2}} c^{3} x^{4}}{48}+\frac {11 a b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x}{8}-\frac {11 a b \,\pi ^{\frac {5}{2}} c \,x^{2}}{16}+\frac {5 a b \,\pi ^{\frac {5}{2}} \arcsinh \left (c x \right )^{2}}{16 c}-\frac {17 a b \,\pi ^{\frac {5}{2}}}{36 c} \]

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*(c^2*x^2+1)^(1
/2)*arcsinh(c*x)^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*(c^2*x^2+1)^(1/2)*arcsinh(c*x)^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+299/1152*b^2*Pi^(5/2)*x*(c^2*x^2+1)^(1/2)+5/48*b^2*Pi^(5/2)/c*arcsinh(c*x)^3-299/1152*b^2*Pi^(
5/2)*arcsinh(c*x)/c+1/3*a*b*Pi^(5/2)*c^4*(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^5-1/18*a*b*Pi^(5/2)*c^5*x^6+13/12*a*
b*Pi^(5/2)*c^2*(c^2*x^2+1)^(1/2)*arcsinh(c*x)*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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________