3.4.36 \(\int \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3 \, dx\) [336]

Optimal. Leaf size=205 \[ -\frac {3 a x^2 \sqrt {c+a^2 c x^2}}{8 \sqrt {1+a^2 x^2}}+\frac {3}{4} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)-\frac {3 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{8 a \sqrt {1+a^2 x^2}}-\frac {3 a x^2 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{4 \sqrt {1+a^2 x^2}}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {\sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^4}{8 a \sqrt {1+a^2 x^2}} \]

[Out]

3/4*x*arcsinh(a*x)*(a^2*c*x^2+c)^(1/2)+1/2*x*arcsinh(a*x)^3*(a^2*c*x^2+c)^(1/2)-3/8*a*x^2*(a^2*c*x^2+c)^(1/2)/
(a^2*x^2+1)^(1/2)-3/8*arcsinh(a*x)^2*(a^2*c*x^2+c)^(1/2)/a/(a^2*x^2+1)^(1/2)-3/4*a*x^2*arcsinh(a*x)^2*(a^2*c*x
^2+c)^(1/2)/(a^2*x^2+1)^(1/2)+1/8*arcsinh(a*x)^4*(a^2*c*x^2+c)^(1/2)/a/(a^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5785, 5783, 5776, 5812, 30} \begin {gather*} -\frac {3 a x^2 \sqrt {a^2 c x^2+c}}{8 \sqrt {a^2 x^2+1}}+\frac {\sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^4}{8 a \sqrt {a^2 x^2+1}}+\frac {1}{2} x \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^3-\frac {3 a x^2 \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{4 \sqrt {a^2 x^2+1}}-\frac {3 \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x)^2}{8 a \sqrt {a^2 x^2+1}}+\frac {3}{4} x \sqrt {a^2 c x^2+c} \sinh ^{-1}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^3,x]

[Out]

(-3*a*x^2*Sqrt[c + a^2*c*x^2])/(8*Sqrt[1 + a^2*x^2]) + (3*x*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x])/4 - (3*Sqrt[c +
a^2*c*x^2]*ArcSinh[a*x]^2)/(8*a*Sqrt[1 + a^2*x^2]) - (3*a*x^2*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^2)/(4*Sqrt[1 +
a^2*x^2]) + (x*Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^3)/2 + (Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^4)/(8*a*Sqrt[1 + a^2*
x^2])

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 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 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3 \, dx &=\frac {1}{2} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {\sqrt {c+a^2 c x^2} \int \frac {\sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {1+a^2 x^2}}-\frac {\left (3 a \sqrt {c+a^2 c x^2}\right ) \int x \sinh ^{-1}(a x)^2 \, dx}{2 \sqrt {1+a^2 x^2}}\\ &=-\frac {3 a x^2 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{4 \sqrt {1+a^2 x^2}}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {\sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^4}{8 a \sqrt {1+a^2 x^2}}+\frac {\left (3 a^2 \sqrt {c+a^2 c x^2}\right ) \int \frac {x^2 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {1+a^2 x^2}}\\ &=\frac {3}{4} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)-\frac {3 a x^2 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{4 \sqrt {1+a^2 x^2}}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {\sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^4}{8 a \sqrt {1+a^2 x^2}}-\frac {\left (3 \sqrt {c+a^2 c x^2}\right ) \int \frac {\sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{4 \sqrt {1+a^2 x^2}}-\frac {\left (3 a \sqrt {c+a^2 c x^2}\right ) \int x \, dx}{4 \sqrt {1+a^2 x^2}}\\ &=-\frac {3 a x^2 \sqrt {c+a^2 c x^2}}{8 \sqrt {1+a^2 x^2}}+\frac {3}{4} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)-\frac {3 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{8 a \sqrt {1+a^2 x^2}}-\frac {3 a x^2 \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^2}{4 \sqrt {1+a^2 x^2}}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^3+\frac {\sqrt {c+a^2 c x^2} \sinh ^{-1}(a x)^4}{8 a \sqrt {1+a^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 86, normalized size = 0.42 \begin {gather*} \frac {\sqrt {c \left (1+a^2 x^2\right )} \left (-3 \left (1+2 \sinh ^{-1}(a x)^2\right ) \cosh \left (2 \sinh ^{-1}(a x)\right )+2 \sinh ^{-1}(a x) \left (\sinh ^{-1}(a x)^3+\left (3+2 \sinh ^{-1}(a x)^2\right ) \sinh \left (2 \sinh ^{-1}(a x)\right )\right )\right )}{16 a \sqrt {1+a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + a^2*c*x^2]*ArcSinh[a*x]^3,x]

[Out]

(Sqrt[c*(1 + a^2*x^2)]*(-3*(1 + 2*ArcSinh[a*x]^2)*Cosh[2*ArcSinh[a*x]] + 2*ArcSinh[a*x]*(ArcSinh[a*x]^3 + (3 +
2*ArcSinh[a*x]^2)*Sinh[2*ArcSinh[a*x]])))/(16*a*Sqrt[1 + a^2*x^2])

________________________________________________________________________________________

Maple [A]
time = 2.46, size = 231, normalized size = 1.13

method	result	size
default	\(\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right )^{4}}{8 \sqrt {a^{2} x^{2}+1}\, a}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (2 a^{3} x^{3}+2 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+2 a x +\sqrt {a^{2} x^{2}+1}\right ) \left (4 \arcsinh \left (a x \right )^{3}-6 \arcsinh \left (a x \right )^{2}+6 \arcsinh \left (a x \right )-3\right )}{32 a \left (a^{2} x^{2}+1\right )}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (2 a^{3} x^{3}-2 \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+2 a x -\sqrt {a^{2} x^{2}+1}\right ) \left (4 \arcsinh \left (a x \right )^{3}+6 \arcsinh \left (a x \right )^{2}+6 \arcsinh \left (a x \right )+3\right )}{32 a \left (a^{2} x^{2}+1\right )}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a*arcsinh(a*x)^4+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)/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)/a/(a^2*x^2+1)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(arcsinh(a*x)^3*(a^2*c*x^2+c)^(1/2),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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________