3.1.67 \(\int x^3 \sinh ^{-1}(a+b x)^2 \, dx\) [67]

Optimal. Leaf size=331 \[ \frac {4 a x}{3 b^3}-\frac {2 a^3 x}{b^3}-\frac {3 (a+b x)^2}{32 b^4}+\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac {3 \sinh ^{-1}(a+b x)^2}{32 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2 \]

[Out]

4/3*a*x/b^3-2*a^3*x/b^3-3/32*(b*x+a)^2/b^4+3/4*a^2*(b*x+a)^2/b^4-2/9*a*(b*x+a)^3/b^4+1/32*(b*x+a)^4/b^4-3/32*a
rcsinh(b*x+a)^2/b^4+3/4*a^2*arcsinh(b*x+a)^2/b^4-1/4*a^4*arcsinh(b*x+a)^2/b^4+1/4*x^4*arcsinh(b*x+a)^2-4/3*a*a
rcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^4+2*a^3*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^4+3/16*(b*x+a)*arcsinh(b*x+a)
*(1+(b*x+a)^2)^(1/2)/b^4-3/2*a^2*(b*x+a)*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^4+2/3*a*(b*x+a)^2*arcsinh(b*x+a)
*(1+(b*x+a)^2)^(1/2)/b^4-1/8*(b*x+a)^3*arcsinh(b*x+a)*(1+(b*x+a)^2)^(1/2)/b^4

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Rubi [A]
time = 0.38, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5859, 5828, 5838, 5783, 5798, 8, 5812, 30} \begin {gather*} -\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {2 a^3 \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{b^4}-\frac {2 a^3 x}{b^3}+\frac {3 a^2 (a+b x)^2}{4 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {3 a^2 (a+b x) \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{2 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {3 (a+b x)^2}{32 b^4}+\frac {2 a (a+b x)^2 \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {4 a \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {3 \sinh ^{-1}(a+b x)^2}{32 b^4}-\frac {(a+b x)^3 \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{8 b^4}+\frac {3 (a+b x) \sqrt {(a+b x)^2+1} \sinh ^{-1}(a+b x)}{16 b^4}+\frac {4 a x}{3 b^3}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSinh[a + b*x]^2,x]

[Out]

(4*a*x)/(3*b^3) - (2*a^3*x)/b^3 - (3*(a + b*x)^2)/(32*b^4) + (3*a^2*(a + b*x)^2)/(4*b^4) - (2*a*(a + b*x)^3)/(
9*b^4) + (a + b*x)^4/(32*b^4) - (4*a*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(3*b^4) + (2*a^3*Sqrt[1 + (a + b*
x)^2]*ArcSinh[a + b*x])/b^4 + (3*(a + b*x)*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(16*b^4) - (3*a^2*(a + b*x)
*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(2*b^4) + (2*a*(a + b*x)^2*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(3
*b^4) - ((a + b*x)^3*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/(8*b^4) - (3*ArcSinh[a + b*x]^2)/(32*b^4) + (3*a^
2*ArcSinh[a + b*x]^2)/(4*b^4) - (a^4*ArcSinh[a + b*x]^2)/(4*b^4) + (x^4*ArcSinh[a + b*x]^2)/4

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rule 5828

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

Rule 5838

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

Rule 5859

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

\begin {align*} \int x^3 \sinh ^{-1}(a+b x)^2 \, dx &=\frac {\text {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \sinh ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {1}{2} \text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {1}{2} \text {Subst}\left (\int \left (\frac {a^4 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}-\frac {4 a^3 x \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}+\frac {6 a^2 x^2 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}-\frac {4 a x^3 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}+\frac {x^4 \sinh ^{-1}(x)}{b^4 \sqrt {1+x^2}}\right ) \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {\text {Subst}\left (\int \frac {x^4 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^4}+\frac {(2 a) \text {Subst}\left (\int \frac {x^3 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^4}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {x^2 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^4}+\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {x \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b^4}-\frac {a^4 \text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^4}\\ &=\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2+\frac {\text {Subst}\left (\int x^3 \, dx,x,a+b x\right )}{8 b^4}+\frac {3 \text {Subst}\left (\int \frac {x^2 \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{8 b^4}-\frac {(2 a) \text {Subst}\left (\int x^2 \, dx,x,a+b x\right )}{3 b^4}-\frac {(4 a) \text {Subst}\left (\int \frac {x \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{3 b^4}+\frac {\left (3 a^2\right ) \text {Subst}(\int x \, dx,x,a+b x)}{2 b^4}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b^4}-\frac {\left (2 a^3\right ) \text {Subst}(\int 1 \, dx,x,a+b x)}{b^4}\\ &=-\frac {2 a^3 x}{b^3}+\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2-\frac {3 \text {Subst}(\int x \, dx,x,a+b x)}{16 b^4}-\frac {3 \text {Subst}\left (\int \frac {\sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{16 b^4}+\frac {(4 a) \text {Subst}(\int 1 \, dx,x,a+b x)}{3 b^4}\\ &=\frac {4 a x}{3 b^3}-\frac {2 a^3 x}{b^3}-\frac {3 (a+b x)^2}{32 b^4}+\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {2 a (a+b x)^3}{9 b^4}+\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}+\frac {2 a^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{16 b^4}-\frac {3 a^2 (a+b x) \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^2 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{3 b^4}-\frac {(a+b x)^3 \sqrt {1+(a+b x)^2} \sinh ^{-1}(a+b x)}{8 b^4}-\frac {3 \sinh ^{-1}(a+b x)^2}{32 b^4}+\frac {3 a^2 \sinh ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sinh ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sinh ^{-1}(a+b x)^2\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 145, normalized size = 0.44 \begin {gather*} \frac {b x \left (-300 a^3+78 a^2 b x+a \left (330-28 b^2 x^2\right )+9 b x \left (-3+b^2 x^2\right )\right )+6 \sqrt {1+a^2+2 a b x+b^2 x^2} \left (50 a^3+9 b x-26 a^2 b x-6 b^3 x^3+a \left (-55+14 b^2 x^2\right )\right ) \sinh ^{-1}(a+b x)-9 \left (3-24 a^2+8 a^4-8 b^4 x^4\right ) \sinh ^{-1}(a+b x)^2}{288 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSinh[a + b*x]^2,x]

[Out]

(b*x*(-300*a^3 + 78*a^2*b*x + a*(330 - 28*b^2*x^2) + 9*b*x*(-3 + b^2*x^2)) + 6*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^
2]*(50*a^3 + 9*b*x - 26*a^2*b*x - 6*b^3*x^3 + a*(-55 + 14*b^2*x^2))*ArcSinh[a + b*x] - 9*(3 - 24*a^2 + 8*a^4 -
8*b^4*x^4)*ArcSinh[a + b*x]^2)/(288*b^4)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{3} \arcsinh \left (b x +a \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(b*x+a)^2,x, algorithm="maxima")

[Out]

1/4*x^4*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - integrate(1/2*(b^3*x^6 + 2*a*b^2*x^5 + (a^2*b + b
)*x^4 + (b^2*x^5 + a*b*x^4)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
)/(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b + b)*x + (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + a), x)

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Fricas [A]
time = 0.37, size = 182, normalized size = 0.55 \begin {gather*} \frac {9 \, b^{4} x^{4} - 28 \, a b^{3} x^{3} + 3 \, {\left (26 \, a^{2} - 9\right )} b^{2} x^{2} - 30 \, {\left (10 \, a^{3} - 11 \, a\right )} b x + 9 \, {\left (8 \, b^{4} x^{4} - 8 \, a^{4} + 24 \, a^{2} - 3\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{2} - 6 \, {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} + {\left (26 \, a^{2} - 9\right )} b x + 55 \, a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{288 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(9*b^4*x^4 - 28*a*b^3*x^3 + 3*(26*a^2 - 9)*b^2*x^2 - 30*(10*a^3 - 11*a)*b*x + 9*(8*b^4*x^4 - 8*a^4 + 24*
a^2 - 3)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - 6*(6*b^3*x^3 - 14*a*b^2*x^2 - 50*a^3 + (26*a^2 -
9)*b*x + 55*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)))/b^4

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Sympy [A]
time = 0.46, size = 366, normalized size = 1.11 \begin {gather*} \begin {cases} - \frac {a^{4} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{4}} - \frac {25 a^{3} x}{24 b^{3}} + \frac {25 a^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{24 b^{4}} + \frac {13 a^{2} x^{2}}{48 b^{2}} - \frac {13 a^{2} x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{24 b^{3}} + \frac {3 a^{2} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4 b^{4}} - \frac {7 a x^{3}}{72 b} + \frac {7 a x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{24 b^{2}} + \frac {55 a x}{48 b^{3}} - \frac {55 a \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{48 b^{4}} + \frac {x^{4} \operatorname {asinh}^{2}{\left (a + b x \right )}}{4} + \frac {x^{4}}{32} - \frac {x^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{8 b} - \frac {3 x^{2}}{32 b^{2}} + \frac {3 x \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \operatorname {asinh}{\left (a + b x \right )}}{16 b^{3}} - \frac {3 \operatorname {asinh}^{2}{\left (a + b x \right )}}{32 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {asinh}^{2}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**4*asinh(a + b*x)**2/(4*b**4) - 25*a**3*x/(24*b**3) + 25*a**3*sqrt(a**2 + 2*a*b*x + b**2*x**2 +
1)*asinh(a + b*x)/(24*b**4) + 13*a**2*x**2/(48*b**2) - 13*a**2*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a
+ b*x)/(24*b**3) + 3*a**2*asinh(a + b*x)**2/(4*b**4) - 7*a*x**3/(72*b) + 7*a*x**2*sqrt(a**2 + 2*a*b*x + b**2*x
**2 + 1)*asinh(a + b*x)/(24*b**2) + 55*a*x/(48*b**3) - 55*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x
)/(48*b**4) + x**4*asinh(a + b*x)**2/4 + x**4/32 - x**3*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(8
*b) - 3*x**2/(32*b**2) + 3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*asinh(a + b*x)/(16*b**3) - 3*asinh(a + b*x)*
*2/(32*b**4), Ne(b, 0)), (x**4*asinh(a)**2/4, True))

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Giac [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(x^3*arcsinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^3*arcsinh(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*asinh(a + b*x)^2,x)

[Out]

int(x^3*asinh(a + b*x)^2, x)

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