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

Optimal. Leaf size=348 \[ -\frac {51 a c x^2 \sqrt {a^2 c x^2+c}}{128 \sqrt {a^2 x^2+1}}-\frac {9 a c x^2 \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{16 \sqrt {a^2 x^2+1}}+\frac {1}{4} x \left (a^2 c x^2+c\right )^{3/2} \sinh ^{-1}(a x)^3+\frac {3}{8} c x \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^3+\frac {45}{64} c x \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)+\frac {3}{32} c x \left (a^2 x^2+1\right ) \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)+\frac {3 c \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^4}{32 a \sqrt {a^2 x^2+1}}-\frac {3 c \left (a^2 x^2+1\right )^{3/2} \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{16 a}-\frac {27 c \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{128 a \sqrt {a^2 x^2+1}}-\frac {3 a^3 c x^4 \sqrt {a^2 c x^2+c}}{128 \sqrt {a^2 x^2+1}} \]

[Out]

1/4*x*(a^2*c*x^2+c)^(3/2)*arcsinh(a*x)^3+45/64*c*x*arcsinh(a*x)*(a^2*c*x^2+c)^(1/2)+3/32*c*x*(a^2*x^2+1)*arcsi
nh(a*x)*(a^2*c*x^2+c)^(1/2)-3/16*c*(a^2*x^2+1)^(3/2)*arcsinh(a*x)^2*(a^2*c*x^2+c)^(1/2)/a+3/8*c*x*arcsinh(a*x)
^3*(a^2*c*x^2+c)^(1/2)-51/128*a*c*x^2*(a^2*c*x^2+c)^(1/2)/(a^2*x^2+1)^(1/2)-3/128*a^3*c*x^4*(a^2*c*x^2+c)^(1/2
)/(a^2*x^2+1)^(1/2)-27/128*c*arcsinh(a*x)^2*(a^2*c*x^2+c)^(1/2)/a/(a^2*x^2+1)^(1/2)-9/16*a*c*x^2*arcsinh(a*x)^
2*(a^2*c*x^2+c)^(1/2)/(a^2*x^2+1)^(1/2)+3/32*c*arcsinh(a*x)^4*(a^2*c*x^2+c)^(1/2)/a/(a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {5684, 5682, 5675, 5661, 5758, 30, 5717, 14} \[ -\frac {3 a^3 c x^4 \sqrt {a^2 c x^2+c}}{128 \sqrt {a^2 x^2+1}}-\frac {51 a c x^2 \sqrt {a^2 c x^2+c}}{128 \sqrt {a^2 x^2+1}}-\frac {9 a c x^2 \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{16 \sqrt {a^2 x^2+1}}+\frac {1}{4} x \left (a^2 c x^2+c\right )^{3/2} \sinh ^{-1}(a x)^3+\frac {3}{8} c x \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^3+\frac {45}{64} c x \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)+\frac {3}{32} c x \left (a^2 x^2+1\right ) \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)+\frac {3 c \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^4}{32 a \sqrt {a^2 x^2+1}}-\frac {3 c \left (a^2 x^2+1\right )^{3/2} \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{16 a}-\frac {27 c \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{128 a \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-51*a*c*x^2*Sqrt[c + a^2*c*x^2])/(128*Sqrt[1 + a^2*x^2]) - (3*a^3*c*x^4*Sqrt[c + a^2*c*x^2])/(128*Sqrt[1 + a^
2*x^2]) + (45*c*x*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x])/64 + (3*c*x*(1 + a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]
)/32 - (27*c*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^2)/(128*a*Sqrt[1 + a^2*x^2]) - (9*a*c*x^2*Sqrt[c + a^2*c*x^2]*Ar
cSinh[a*x]^2)/(16*Sqrt[1 + a^2*x^2]) - (3*c*(1 + a^2*x^2)^(3/2)*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^2)/(16*a) + (
3*c*x*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^3)/8 + (x*(c + a^2*c*x^2)^(3/2)*ArcSinh[a*x]^3)/4 + (3*c*Sqrt[c + a^2*c
*x^2]*ArcSinh[a*x]^4)/(32*a*Sqrt[1 + a^2*x^2])

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 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]

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 \left (c+a^2 c x^2\right )^{3/2} \sinh ^{-1}(a x)^3 \, dx &=\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \sinh ^{-1}(a x)^3+\frac {1}{4} (3 c) \int \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3 \, dx-\frac {\left (3 a c \sqrt {c+a^2 c x^2}\right ) \int x \left (1+a^2 x^2\right ) \sinh ^{-1}(a x)^2 \, dx}{4 \sqrt {1+a^2 x^2}}\\ &=-\frac {3 c \left (1+a^2 x^2\right )^{3/2} \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{16 a}+\frac {3}{8} c x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \sinh ^{-1}(a x)^3+\frac {\left (3 c \sqrt {c+a^2 c x^2}\right ) \int \left (1+a^2 x^2\right )^{3/2} \sinh ^{-1}(a x) \, dx}{8 \sqrt {1+a^2 x^2}}+\frac {\left (3 c \sqrt {c+a^2 c x^2}\right ) \int \frac {\sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx}{8 \sqrt {1+a^2 x^2}}-\frac {\left (9 a c \sqrt {c+a^2 c x^2}\right ) \int x \sinh ^{-1}(a x)^2 \, dx}{8 \sqrt {1+a^2 x^2}}\\ &=\frac {3}{32} c x \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)-\frac {9 a c x^2 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{16 \sqrt {1+a^2 x^2}}-\frac {3 c \left (1+a^2 x^2\right )^{3/2} \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{16 a}+\frac {3}{8} c x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \sinh ^{-1}(a x)^3+\frac {3 c \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^4}{32 a \sqrt {1+a^2 x^2}}+\frac {\left (9 c \sqrt {c+a^2 c x^2}\right ) \int \sqrt {1+a^2 x^2} \sinh ^{-1}(a x) \, dx}{32 \sqrt {1+a^2 x^2}}-\frac {\left (3 a c \sqrt {c+a^2 c x^2}\right ) \int x \left (1+a^2 x^2\right ) \, dx}{32 \sqrt {1+a^2 x^2}}+\frac {\left (9 a^2 c \sqrt {c+a^2 c x^2}\right ) \int \frac {x^2 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{8 \sqrt {1+a^2 x^2}}\\ &=\frac {45}{64} c x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)+\frac {3}{32} c x \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)-\frac {9 a c x^2 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{16 \sqrt {1+a^2 x^2}}-\frac {3 c \left (1+a^2 x^2\right )^{3/2} \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{16 a}+\frac {3}{8} c x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \sinh ^{-1}(a x)^3+\frac {3 c \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^4}{32 a \sqrt {1+a^2 x^2}}+\frac {\left (9 c \sqrt {c+a^2 c x^2}\right ) \int \frac {\sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{64 \sqrt {1+a^2 x^2}}-\frac {\left (9 c \sqrt {c+a^2 c x^2}\right ) \int \frac {\sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{16 \sqrt {1+a^2 x^2}}-\frac {\left (3 a c \sqrt {c+a^2 c x^2}\right ) \int \left (x+a^2 x^3\right ) \, dx}{32 \sqrt {1+a^2 x^2}}-\frac {\left (9 a c \sqrt {c+a^2 c x^2}\right ) \int x \, dx}{64 \sqrt {1+a^2 x^2}}-\frac {\left (9 a c \sqrt {c+a^2 c x^2}\right ) \int x \, dx}{16 \sqrt {1+a^2 x^2}}\\ &=-\frac {51 a c x^2 \sqrt {c+a^2 c x^2}}{128 \sqrt {1+a^2 x^2}}-\frac {3 a^3 c x^4 \sqrt {c+a^2 c x^2}}{128 \sqrt {1+a^2 x^2}}+\frac {45}{64} c x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)+\frac {3}{32} c x \left (1+a^2 x^2\right ) \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)-\frac {27 c \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{128 a \sqrt {1+a^2 x^2}}-\frac {9 a c x^2 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{16 \sqrt {1+a^2 x^2}}-\frac {3 c \left (1+a^2 x^2\right )^{3/2} \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{16 a}+\frac {3}{8} c x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \sinh ^{-1}(a x)^3+\frac {3 c \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^4}{32 a \sqrt {1+a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 136, normalized size = 0.39 \[ \frac {c \sqrt {a^2 c x^2+c} \left (96 \sinh ^{-1}(a x)^4+32 \left (8 \sinh \left (2 \sinh ^{-1}(a x)\right )+\sinh \left (4 \sinh ^{-1}(a x)\right )\right ) \sinh ^{-1}(a x)^3+12 \left (32 \sinh \left (2 \sinh ^{-1}(a x)\right )+\sinh \left (4 \sinh ^{-1}(a x)\right )\right ) \sinh ^{-1}(a x)-24 \sinh ^{-1}(a x)^2 \left (16 \cosh \left (2 \sinh ^{-1}(a x)\right )+\cosh \left (4 \sinh ^{-1}(a x)\right )\right )-3 \left (64 \cosh \left (2 \sinh ^{-1}(a x)\right )+\cosh \left (4 \sinh ^{-1}(a x)\right )\right )\right )}{1024 a \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c + a^2*c*x^2]*(96*ArcSinh[a*x]^4 - 24*ArcSinh[a*x]^2*(16*Cosh[2*ArcSinh[a*x]] + Cosh[4*ArcSinh[a*x]])
- 3*(64*Cosh[2*ArcSinh[a*x]] + Cosh[4*ArcSinh[a*x]]) + 32*ArcSinh[a*x]^3*(8*Sinh[2*ArcSinh[a*x]] + Sinh[4*Arc
Sinh[a*x]]) + 12*ArcSinh[a*x]*(32*Sinh[2*ArcSinh[a*x]] + Sinh[4*ArcSinh[a*x]])))/(1024*a*Sqrt[1 + a^2*x^2])

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (a x\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*c*x^2 + c)^(3/2)*arcsinh(a*x)^3, x)

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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((a^2*c*x^2+c)^(3/2)*arcsinh(a*x)^3,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 [A]  time = 0.20, size = 484, normalized size = 1.39 \[ \frac {3 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right )^{4} c}{32 \sqrt {a^{2} x^{2}+1}\, a}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (8 x^{5} a^{5}+8 \sqrt {a^{2} x^{2}+1}\, x^{4} a^{4}+12 x^{3} a^{3}+8 \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+4 a x +\sqrt {a^{2} x^{2}+1}\right ) \left (32 \arcsinh \left (a x \right )^{3}-24 \arcsinh \left (a x \right )^{2}+12 \arcsinh \left (a x \right )-3\right ) c}{2048 a \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (2 x^{3} a^{3}+2 \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+2 a x +\sqrt {a^{2} x^{2}+1}\right ) \left (4 \arcsinh \left (a x \right )^{3}-6 \arcsinh \left (a x \right )^{2}+6 \arcsinh \left (a x \right )-3\right ) c}{32 a \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (2 x^{3} a^{3}-2 \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+2 a x -\sqrt {a^{2} x^{2}+1}\right ) \left (4 \arcsinh \left (a x \right )^{3}+6 \arcsinh \left (a x \right )^{2}+6 \arcsinh \left (a x \right )+3\right ) c}{32 a \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (8 x^{5} a^{5}-8 \sqrt {a^{2} x^{2}+1}\, x^{4} a^{4}+12 x^{3} a^{3}-8 \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+4 a x -\sqrt {a^{2} x^{2}+1}\right ) \left (32 \arcsinh \left (a x \right )^{3}+24 \arcsinh \left (a x \right )^{2}+12 \arcsinh \left (a x \right )+3\right ) c}{2048 a \left (a^{2} x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a*arcsinh(a*x)^4*c+1/2048*(c*(a^2*x^2+1))^(1/2)*(8*x^5*a^5+8*(a^2
*x^2+1)^(1/2)*x^4*a^4+12*x^3*a^3+8*(a^2*x^2+1)^(1/2)*x^2*a^2+4*a*x+(a^2*x^2+1)^(1/2))*(32*arcsinh(a*x)^3-24*ar
csinh(a*x)^2+12*arcsinh(a*x)-3)*c/a/(a^2*x^2+1)+1/32*(c*(a^2*x^2+1))^(1/2)*(2*x^3*a^3+2*(a^2*x^2+1)^(1/2)*x^2*
a^2+2*a*x+(a^2*x^2+1)^(1/2))*(4*arcsinh(a*x)^3-6*arcsinh(a*x)^2+6*arcsinh(a*x)-3)*c/a/(a^2*x^2+1)+1/32*(c*(a^2
*x^2+1))^(1/2)*(2*x^3*a^3-2*(a^2*x^2+1)^(1/2)*x^2*a^2+2*a*x-(a^2*x^2+1)^(1/2))*(4*arcsinh(a*x)^3+6*arcsinh(a*x
)^2+6*arcsinh(a*x)+3)*c/a/(a^2*x^2+1)+1/2048*(c*(a^2*x^2+1))^(1/2)*(8*x^5*a^5-8*(a^2*x^2+1)^(1/2)*x^4*a^4+12*x
^3*a^3-8*(a^2*x^2+1)^(1/2)*x^2*a^2+4*a*x-(a^2*x^2+1)^(1/2))*(32*arcsinh(a*x)^3+24*arcsinh(a*x)^2+12*arcsinh(a*
x)+3)*c/a/(a^2*x^2+1)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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