3.4.35 \(\int (c+a^2 c x^2)^{3/2} \sinh ^{-1}(a x)^3 \, dx\) [335]

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

[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 = 0.23, 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 = {5786, 5785, 5783, 5776, 5812, 30, 5798, 14} \begin {gather*} -\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}} \end {gather*}

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 5776

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 5783

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

Rule 5785

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[(1/2)*Simp[Sqrt[d + e*x^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/2)*Simp[Sqrt[d + e*x^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 5786

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/(2*p + 1))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*A
rcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0]

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5812

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

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 = 0.20, size = 136, normalized size = 0.39 \begin {gather*} \frac {c \sqrt {c+a^2 c x^2} \left (96 \sinh ^{-1}(a x)^4-24 \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 )+32 \sinh ^{-1}(a x)^3 \left (8 \sinh \left (2 \sinh ^{-1}(a x)\right )+\sinh \left (4 \sinh ^{-1}(a x)\right )\right )+12 \sinh ^{-1}(a x) \left (32 \sinh \left (2 \sinh ^{-1}(a x)\right )+\sinh \left (4 \sinh ^{-1}(a x)\right )\right )\right )}{1024 a \sqrt {1+a^2 x^2}} \end {gather*}

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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Maple [A]
time = 2.27, size = 484, normalized size = 1.39

method	result	size
default	\(\frac {3 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right )^{4} c}{32 \sqrt {a^{2} x^{2}+1}\, a}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (8 a^{5} x^{5}+8 a^{4} \sqrt {a^{2} x^{2}+1}\, x^{4}+12 a^{3} x^{3}+8 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+4 a x +\sqrt {a^{2} x^{2}+1}\right ) \left (32 \arcsinh \left (a x \right )^{3}-24 \arcsinh \left (a x \right )^{2}+12 \arcsinh \left (a x \right )-3\right ) c}{2048 a \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (2 a^{3} x^{3}+2 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+2 a x +\sqrt {a^{2} x^{2}+1}\right ) \left (4 \arcsinh \left (a x \right )^{3}-6 \arcsinh \left (a x \right )^{2}+6 \arcsinh \left (a x \right )-3\right ) c}{32 a \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (2 a^{3} x^{3}-2 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+2 a x -\sqrt {a^{2} x^{2}+1}\right ) \left (4 \arcsinh \left (a x \right )^{3}+6 \arcsinh \left (a x \right )^{2}+6 \arcsinh \left (a x \right )+3\right ) c}{32 a \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (8 a^{5} x^{5}-8 a^{4} \sqrt {a^{2} x^{2}+1}\, x^{4}+12 a^{3} x^{3}-8 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+4 a x -\sqrt {a^{2} x^{2}+1}\right ) \left (32 \arcsinh \left (a x \right )^{3}+24 \arcsinh \left (a x \right )^{2}+12 \arcsinh \left (a x \right )+3\right ) c}{2048 a \left (a^{2} x^{2}+1\right )}\)	\(484\)

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,method=_RETURNVERBOSE)

[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*a^5*x^5+8*a^4*
(a^2*x^2+1)^(1/2)*x^4+12*a^3*x^3+8*(a^2*x^2+1)^(1/2)*a^2*x^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*a^3*x^3+2*(a^2*x^2+1)^(1/2)*a^2*
x^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*a^3*x^3-2*(a^2*x^2+1)^(1/2)*a^2*x^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*a^5*x^5-8*a^4*(a^2*x^2+1)^(1/2)*x^4+12*a
^3*x^3-8*(a^2*x^2+1)^(1/2)*a^2*x^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 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

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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