\(\int x^3 \text {arcsinh}(a+b x^4) \, dx\) [366]

Optimal result

Integrand size = 12, antiderivative size = 45 \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=-\frac {\sqrt {1+\left (a+b x^4\right )^2}}{4 b}+\frac {\left (a+b x^4\right ) \text {arcsinh}\left (a+b x^4\right )}{4 b} \]

[Out]

1/4*(b*x^4+a)*arcsinh(b*x^4+a)/b-1/4*(1+(b*x^4+a)^2)^(1/2)/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6847, 5858, 5772, 267} \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=\frac {\left (a+b x^4\right ) \text {arcsinh}\left (a+b x^4\right )}{4 b}-\frac {\sqrt {\left (a+b x^4\right )^2+1}}{4 b} \]

[In]

Int[x^3*ArcSinh[a + b*x^4],x]

[Out]

-1/4*Sqrt[1 + (a + b*x^4)^2]/b + ((a + b*x^4)*ArcSinh[a + b*x^4])/(4*b)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[x*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5858

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

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \text {arcsinh}(a+b x) \, dx,x,x^4\right ) \\ & = \frac {\text {Subst}\left (\int \text {arcsinh}(x) \, dx,x,a+b x^4\right )}{4 b} \\ & = \frac {\left (a+b x^4\right ) \text {arcsinh}\left (a+b x^4\right )}{4 b}-\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1+x^2}} \, dx,x,a+b x^4\right )}{4 b} \\ & = -\frac {\sqrt {1+\left (a+b x^4\right )^2}}{4 b}+\frac {\left (a+b x^4\right ) \text {arcsinh}\left (a+b x^4\right )}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=\frac {-\sqrt {1+\left (a+b x^4\right )^2}+\left (a+b x^4\right ) \text {arcsinh}\left (a+b x^4\right )}{4 b} \]

[In]

Integrate[x^3*ArcSinh[a + b*x^4],x]

[Out]

(-Sqrt[1 + (a + b*x^4)^2] + (a + b*x^4)*ArcSinh[a + b*x^4])/(4*b)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\left (b \,x^{4}+a \right ) \operatorname {arcsinh}\left (b \,x^{4}+a \right )-\sqrt {1+\left (b \,x^{4}+a \right )^{2}}}{4 b}\)	\(38\)
default	\(\frac {\left (b \,x^{4}+a \right ) \operatorname {arcsinh}\left (b \,x^{4}+a \right )-\sqrt {1+\left (b \,x^{4}+a \right )^{2}}}{4 b}\)	\(38\)

[In]

int(x^3*arcsinh(b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

1/4/b*((b*x^4+a)*arcsinh(b*x^4+a)-(1+(b*x^4+a)^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=\frac {{\left (b x^{4} + a\right )} \log \left (b x^{4} + a + \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}\right ) - \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}}{4 \, b} \]

[In]

integrate(x^3*arcsinh(b*x^4+a),x, algorithm="fricas")

[Out]

1/4*((b*x^4 + a)*log(b*x^4 + a + sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 + 1)) - sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 + 1))/b

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.36 \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=\begin {cases} \frac {a \operatorname {asinh}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {asinh}{\left (a + b x^{4} \right )}}{4} - \frac {\sqrt {a^{2} + 2 a b x^{4} + b^{2} x^{8} + 1}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {asinh}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*asinh(b*x**4+a),x)

[Out]

Piecewise((a*asinh(a + b*x**4)/(4*b) + x**4*asinh(a + b*x**4)/4 - sqrt(a**2 + 2*a*b*x**4 + b**2*x**8 + 1)/(4*b
), Ne(b, 0)), (x**4*asinh(a)/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=\frac {{\left (b x^{4} + a\right )} \operatorname {arsinh}\left (b x^{4} + a\right ) - \sqrt {{\left (b x^{4} + a\right )}^{2} + 1}}{4 \, b} \]

[In]

integrate(x^3*arcsinh(b*x^4+a),x, algorithm="maxima")

[Out]

1/4*((b*x^4 + a)*arcsinh(b*x^4 + a) - sqrt((b*x^4 + a)^2 + 1))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (39) = 78\).

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.33 \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (b x^{4} + a + \sqrt {{\left (b x^{4} + a\right )}^{2} + 1}\right ) - \frac {1}{4} \, b {\left (\frac {a \log \left (-a b - {\left (x^{4} {\left | b \right |} - \sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}\right )} {\left | b \right |}\right )}{b {\left | b \right |}} + \frac {\sqrt {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}}{b^{2}}\right )} \]

[In]

integrate(x^3*arcsinh(b*x^4+a),x, algorithm="giac")

[Out]

1/4*x^4*log(b*x^4 + a + sqrt((b*x^4 + a)^2 + 1)) - 1/4*b*(a*log(-a*b - (x^4*abs(b) - sqrt(b^2*x^8 + 2*a*b*x^4
+ a^2 + 1))*abs(b))/(b*abs(b)) + sqrt(b^2*x^8 + 2*a*b*x^4 + a^2 + 1)/b^2)

Mupad [B] (verification not implemented)

Time = 2.92 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.96 \[ \int x^3 \text {arcsinh}\left (a+b x^4\right ) \, dx=\frac {x^4\,\mathrm {asinh}\left (b\,x^4+a\right )}{4}-\frac {\sqrt {a^2+2\,a\,b\,x^4+b^2\,x^8+1}}{4\,b}+\frac {a\,\ln \left (\sqrt {a^2+2\,a\,b\,x^4+b^2\,x^8+1}+\frac {b^2\,x^4+a\,b}{\sqrt {b^2}}\right )}{4\,\sqrt {b^2}} \]

[In]

int(x^3*asinh(a + b*x^4),x)

[Out]

(x^4*asinh(a + b*x^4))/4 - (a^2 + b^2*x^8 + 2*a*b*x^4 + 1)^(1/2)/(4*b) + (a*log((a^2 + b^2*x^8 + 2*a*b*x^4 + 1
)^(1/2) + (a*b + b^2*x^4)/(b^2)^(1/2)))/(4*(b^2)^(1/2))