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

Optimal. Leaf size=210 \[ \frac{1}{4} x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{3}{8} \pi x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{\pi ^{3/2} b \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{3}{8} \pi ^{3/2} b c x^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac{\pi ^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c}+\frac{1}{32} \pi ^{3/2} b^2 x \left (c^2 x^2+1\right )^{3/2}+\frac{15}{64} \pi ^{3/2} b^2 x \sqrt{c^2 x^2+1}-\frac{9 \pi ^{3/2} b^2 \sinh ^{-1}(c x)}{64 c} \]

[Out]

(15*b^2*Pi^(3/2)*x*Sqrt[1 + c^2*x^2])/64 + (b^2*Pi^(3/2)*x*(1 + c^2*x^2)^(3/2))/32 - (9*b^2*Pi^(3/2)*ArcSinh[c
*x])/(64*c) - (3*b*c*Pi^(3/2)*x^2*(a + b*ArcSinh[c*x]))/8 - (b*Pi^(3/2)*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/
(8*c) + (3*Pi*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/8 + (x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*
x])^2)/4 + (Pi^(3/2)*(a + b*ArcSinh[c*x])^3)/(8*b*c)

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Rubi [A]  time = 0.227046, antiderivative size = 294, normalized size of antiderivative = 1.4, number of steps used = 10, 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} \[ \frac{\pi \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt{c^2 x^2+1}}+\frac{1}{4} x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{3}{8} \pi x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{\pi b \left (c^2 x^2+1\right )^{3/2} \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{8 c}-\frac{3 \pi b c x^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt{c^2 x^2+1}}+\frac{15}{64} \pi b^2 x \sqrt{\pi c^2 x^2+\pi }+\frac{1}{32} \pi b^2 x \left (c^2 x^2+1\right ) \sqrt{\pi c^2 x^2+\pi }-\frac{9 \pi b^2 \sqrt{\pi c^2 x^2+\pi } \sinh ^{-1}(c x)}{64 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*b^2*Pi*x*Sqrt[Pi + c^2*Pi*x^2])/64 + (b^2*Pi*x*(1 + c^2*x^2)*Sqrt[Pi + c^2*Pi*x^2])/32 - (9*b^2*Pi*Sqrt[Pi
+ c^2*Pi*x^2]*ArcSinh[c*x])/(64*c*Sqrt[1 + c^2*x^2]) - (3*b*c*Pi*x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x
]))/(8*Sqrt[1 + c^2*x^2]) - (b*Pi*(1 + c^2*x^2)^(3/2)*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(8*c) + (3*P
i*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/8 + (x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x])^2)/4 + (
Pi*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^3)/(8*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 )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{4} (3 \pi ) \int \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx-\frac{\left (b c \pi \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}{2 \sqrt{1+c^2 x^2}}\\ &=-\frac{b \pi \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{3}{8} \pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\left (3 \pi \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}{8 \sqrt{1+c^2 x^2}}+\frac{\left (b^2 \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{8 \sqrt{1+c^2 x^2}}-\frac{\left (3 b c \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{32} b^2 \pi x \left (1+c^2 x^2\right ) \sqrt{\pi +c^2 \pi x^2}-\frac{3 b c \pi x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt{1+c^2 x^2}}-\frac{b \pi \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{3}{8} \pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\pi \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt{1+c^2 x^2}}+\frac{\left (3 b^2 \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \sqrt{1+c^2 x^2} \, dx}{32 \sqrt{1+c^2 x^2}}+\frac{\left (3 b^2 c^2 \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{8 \sqrt{1+c^2 x^2}}\\ &=\frac{15}{64} b^2 \pi x \sqrt{\pi +c^2 \pi x^2}+\frac{1}{32} b^2 \pi x \left (1+c^2 x^2\right ) \sqrt{\pi +c^2 \pi x^2}-\frac{3 b c \pi x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt{1+c^2 x^2}}-\frac{b \pi \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{3}{8} \pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\pi \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt{1+c^2 x^2}}+\frac{\left (3 b^2 \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{64 \sqrt{1+c^2 x^2}}-\frac{\left (3 b^2 \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{16 \sqrt{1+c^2 x^2}}\\ &=\frac{15}{64} b^2 \pi x \sqrt{\pi +c^2 \pi x^2}+\frac{1}{32} b^2 \pi x \left (1+c^2 x^2\right ) \sqrt{\pi +c^2 \pi x^2}-\frac{9 b^2 \pi \sqrt{\pi +c^2 \pi x^2} \sinh ^{-1}(c x)}{64 c \sqrt{1+c^2 x^2}}-\frac{3 b c \pi x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \sqrt{1+c^2 x^2}}-\frac{b \pi \left (1+c^2 x^2\right )^{3/2} \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c}+\frac{3}{8} \pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\pi \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{8 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.621436, size = 202, normalized size = 0.96 \[ \frac{\pi ^{3/2} \left (64 a^2 c^3 x^3 \sqrt{c^2 x^2+1}+160 a^2 c x \sqrt{c^2 x^2+1}+4 \sinh ^{-1}(c x) \left (4 a \left (6 a+8 b \sinh \left (2 \sinh ^{-1}(c x)\right )+b \sinh \left (4 \sinh ^{-1}(c x)\right )\right )-16 b^2 \cosh \left (2 \sinh ^{-1}(c x)\right )-b^2 \cosh \left (4 \sinh ^{-1}(c x)\right )\right )+8 b \sinh ^{-1}(c x)^2 \left (12 a+8 b \sinh \left (2 \sinh ^{-1}(c x)\right )+b \sinh \left (4 \sinh ^{-1}(c x)\right )\right )-64 a b \cosh \left (2 \sinh ^{-1}(c x)\right )-4 a b \cosh \left (4 \sinh ^{-1}(c x)\right )+32 b^2 \sinh ^{-1}(c x)^3+32 b^2 \sinh \left (2 \sinh ^{-1}(c x)\right )+b^2 \sinh \left (4 \sinh ^{-1}(c x)\right )\right )}{256 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(3/2)*(160*a^2*c*x*Sqrt[1 + c^2*x^2] + 64*a^2*c^3*x^3*Sqrt[1 + c^2*x^2] + 32*b^2*ArcSinh[c*x]^3 - 64*a*b*C
osh[2*ArcSinh[c*x]] - 4*a*b*Cosh[4*ArcSinh[c*x]] + 32*b^2*Sinh[2*ArcSinh[c*x]] + b^2*Sinh[4*ArcSinh[c*x]] + 8*
b*ArcSinh[c*x]^2*(12*a + 8*b*Sinh[2*ArcSinh[c*x]] + b*Sinh[4*ArcSinh[c*x]]) + 4*ArcSinh[c*x]*(-16*b^2*Cosh[2*A
rcSinh[c*x]] - b^2*Cosh[4*ArcSinh[c*x]] + 4*a*(6*a + 8*b*Sinh[2*ArcSinh[c*x]] + b*Sinh[4*ArcSinh[c*x]]))))/(25
6*c)

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Maple [B]  time = 0.081, size = 350, normalized size = 1.7 \begin{align*}{\frac{{a}^{2}x}{4} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{2}\pi \,x}{8}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{3\,{a}^{2}{\pi }^{2}}{8}\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{3}{2}}}{c}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}{x}^{3}}{4}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{b}^{2}{\pi }^{{\frac{3}{2}}}{c}^{3}{\it Arcsinh} \left ( cx \right ){x}^{4}}{8}}+{\frac{{b}^{2}{\pi }^{{\frac{3}{2}}}{c}^{2}{x}^{3}}{32}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{5\,{b}^{2}{\pi }^{3/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}x}{8}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{5\,{b}^{2}{\pi }^{3/2}c{\it Arcsinh} \left ( cx \right ){x}^{2}}{8}}+{\frac{{b}^{2}{\pi }^{{\frac{3}{2}}} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{3}}{8\,c}}+{\frac{17\,{b}^{2}{\pi }^{3/2}x}{64}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{17\,{b}^{2}{\pi }^{3/2}{\it Arcsinh} \left ( cx \right ) }{64\,c}}+{\frac{ab{\pi }^{{\frac{3}{2}}}{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{2}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{ab{\pi }^{{\frac{3}{2}}}{c}^{3}{x}^{4}}{8}}+{\frac{5\,ab{\pi }^{3/2}{\it Arcsinh} \left ( cx \right ) x}{4}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{5\,ab{\pi }^{3/2}c{x}^{2}}{8}}+{\frac{3\,ab{\pi }^{3/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{8\,c}}-{\frac{ab{\pi }^{{\frac{3}{2}}}}{2\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x*(Pi*c^2*x^2+Pi)^(3/2)+3/8*a^2*Pi*x*(Pi*c^2*x^2+Pi)^(1/2)+3/8*a^2*Pi^2*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi
*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/4*b^2*Pi^(3/2)*c^2*arcsinh(c*x)^2*(c^2*x^2+1)^(1/2)*x^3-1/8*b^2*Pi^(3/2)*
c^3*arcsinh(c*x)*x^4+1/32*b^2*Pi^(3/2)*c^2*x^3*(c^2*x^2+1)^(1/2)+5/8*b^2*Pi^(3/2)*arcsinh(c*x)^2*(c^2*x^2+1)^(
1/2)*x-5/8*b^2*Pi^(3/2)*c*arcsinh(c*x)*x^2+1/8*b^2*Pi^(3/2)/c*arcsinh(c*x)^3+17/64*b^2*Pi^(3/2)*x*(c^2*x^2+1)^
(1/2)-17/64*b^2*Pi^(3/2)*arcsinh(c*x)/c+1/2*a*b*Pi^(3/2)*c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3-1/8*a*b*Pi^(3/
2)*c^3*x^4+5/4*a*b*Pi^(3/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-5/8*a*b*Pi^(3/2)*c*x^2+3/8*a*b*Pi^(3/2)/c*arcsinh
(c*x)^2-1/2*a*b*Pi^(3/2)/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)^(3/2)*(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 a^{2} c^{2} x^{2} + \pi a^{2} +{\left (\pi b^{2} c^{2} x^{2} + \pi b^{2}\right )} \operatorname{arsinh}\left (c x\right )^{2} + 2 \,{\left (\pi a b c^{2} x^{2} + \pi 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)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi*a^2*c^2*x^2 + pi*a^2 + (pi*b^2*c^2*x^2 + pi*b^2)*arcsinh(c*x)^2 + 2*(pi*a*b
*c^2*x^2 + pi*a*b)*arcsinh(c*x)), x)

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Sympy [A]  time = 151.738, size = 405, normalized size = 1.93 \begin{align*} \begin{cases} \frac{\pi ^{\frac{3}{2}} a^{2} c^{2} x^{3} \sqrt{c^{2} x^{2} + 1}}{4} + \frac{5 \pi ^{\frac{3}{2}} a^{2} x \sqrt{c^{2} x^{2} + 1}}{8} + \frac{3 \pi ^{\frac{3}{2}} a^{2} \operatorname{asinh}{\left (c x \right )}}{8 c} - \frac{\pi ^{\frac{3}{2}} a b c^{3} x^{4}}{8} + \frac{\pi ^{\frac{3}{2}} a b c^{2} x^{3} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{2} - \frac{5 \pi ^{\frac{3}{2}} a b c x^{2}}{8} + \frac{5 \pi ^{\frac{3}{2}} a b x \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{4} + \frac{3 \pi ^{\frac{3}{2}} a b \operatorname{asinh}^{2}{\left (c x \right )}}{8 c} - \frac{\pi ^{\frac{3}{2}} b^{2} c^{3} x^{4} \operatorname{asinh}{\left (c x \right )}}{8} + \frac{\pi ^{\frac{3}{2}} b^{2} c^{2} x^{3} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (c x \right )}}{4} + \frac{\pi ^{\frac{3}{2}} b^{2} c^{2} x^{3} \sqrt{c^{2} x^{2} + 1}}{32} - \frac{5 \pi ^{\frac{3}{2}} b^{2} c x^{2} \operatorname{asinh}{\left (c x \right )}}{8} + \frac{5 \pi ^{\frac{3}{2}} b^{2} x \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (c x \right )}}{8} + \frac{17 \pi ^{\frac{3}{2}} b^{2} x \sqrt{c^{2} x^{2} + 1}}{64} + \frac{\pi ^{\frac{3}{2}} b^{2} \operatorname{asinh}^{3}{\left (c x \right )}}{8 c} - \frac{17 \pi ^{\frac{3}{2}} b^{2} \operatorname{asinh}{\left (c x \right )}}{64 c} & \text{for}\: c \neq 0 \\\pi ^{\frac{3}{2}} a^{2} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((pi**(3/2)*a**2*c**2*x**3*sqrt(c**2*x**2 + 1)/4 + 5*pi**(3/2)*a**2*x*sqrt(c**2*x**2 + 1)/8 + 3*pi**(
3/2)*a**2*asinh(c*x)/(8*c) - pi**(3/2)*a*b*c**3*x**4/8 + pi**(3/2)*a*b*c**2*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x
)/2 - 5*pi**(3/2)*a*b*c*x**2/8 + 5*pi**(3/2)*a*b*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/4 + 3*pi**(3/2)*a*b*asinh(c*
x)**2/(8*c) - pi**(3/2)*b**2*c**3*x**4*asinh(c*x)/8 + pi**(3/2)*b**2*c**2*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)*
*2/4 + pi**(3/2)*b**2*c**2*x**3*sqrt(c**2*x**2 + 1)/32 - 5*pi**(3/2)*b**2*c*x**2*asinh(c*x)/8 + 5*pi**(3/2)*b*
*2*x*sqrt(c**2*x**2 + 1)*asinh(c*x)**2/8 + 17*pi**(3/2)*b**2*x*sqrt(c**2*x**2 + 1)/64 + pi**(3/2)*b**2*asinh(c
*x)**3/(8*c) - 17*pi**(3/2)*b**2*asinh(c*x)/(64*c), Ne(c, 0)), (pi**(3/2)*a**2*x, True))

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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{3}{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)^(3/2)*(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

integrate((pi + pi*c^2*x^2)^(3/2)*(b*arcsinh(c*x) + a)^2, x)