\(\int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx\) [272]

Optimal result

Integrand size = 30, antiderivative size = 15 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {\text {arcsinh}(a+b x)^4}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {5860, 5783} \[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {\text {arcsinh}(a+b x)^4}{4 b} \]

[In]

Int[ArcSinh[a + b*x]^3/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

ArcSinh[a + b*x]^4/(4*b)

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 5860

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[1/d, Subst[Int[(C/d^2 + (C/d^2)*x^2)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\text {arcsinh}(x)^3}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b} \\ & = \frac {\text {arcsinh}(a+b x)^4}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {\text {arcsinh}(a+b x)^4}{4 b} \]

[In]

Integrate[ArcSinh[a + b*x]^3/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

ArcSinh[a + b*x]^4/(4*b)

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\operatorname {arcsinh}\left (b x +a \right )^{4}}{4 b}\)	\(14\)

[In]

int(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (13) = 26\).

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.13 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {\log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )^{4}}{4 \, b} \]

[In]

integrate(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.73 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\begin {cases} \frac {\operatorname {asinh}^{4}{\left (a + b x \right )}}{4 b} & \text {for}\: b \neq 0 \\\frac {x \operatorname {asinh}^{3}{\left (a \right )}}{\sqrt {a^{2} + 1}} & \text {otherwise} \end {cases} \]

[In]

integrate(asinh(b*x+a)**3/(b**2*x**2+2*a*b*x+a**2+1)**(1/2),x)

[Out]

Piecewise((asinh(a + b*x)**4/(4*b), Ne(b, 0)), (x*asinh(a)**3/sqrt(a**2 + 1), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (13) = 26\).

Time = 0.30 (sec) , antiderivative size = 179, normalized size of antiderivative = 11.93 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {\operatorname {arsinh}\left (b x + a\right )^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{b} - \frac {3 \, \operatorname {arsinh}\left (b x + a\right )^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )^{2}}{2 \, b} + \frac {\operatorname {arsinh}\left (b x + a\right ) \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )^{3}}{b} - \frac {\operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )^{4}}{4 \, b} \]

[In]

integrate(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="maxima")

[Out]

arcsinh(b*x + a)^3*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b - 3/2*arcsinh(b*x + a)^2*arcs
inh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))^2/b + arcsinh(b*x + a)*arcsinh(2*(b^2*x + a*b)/sqrt(-4
*a^2*b^2 + 4*(a^2 + 1)*b^2))^3/b - 1/4*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))^4/b

Giac [F]

\[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}} \,d x } \]

[In]

integrate(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(arcsinh(b*x + a)^3/sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1), x)

Mupad [B] (verification not implemented)

Time = 2.64 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {{\mathrm {asinh}\left (a+b\,x\right )}^4}{4\,b} \]

[In]

int(asinh(a + b*x)^3/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(1/2),x)

[Out]

asinh(a + b*x)^4/(4*b)