3.2.48 \(\int \frac {x^2 (a+b \sinh ^{-1}(c x))}{\sqrt {d+c^2 d x^2}} \, dx\) [148]

Optimal. Leaf size=119 \[ -\frac {b x^2 \sqrt {1+c^2 x^2}}{4 c \sqrt {d+c^2 d x^2}}+\frac {x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}-\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {d+c^2 d x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {5812, 5783, 30} \begin {gather*} \frac {x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 d}-\frac {\sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {c^2 d x^2+d}}-\frac {b x^2 \sqrt {c^2 x^2+1}}{4 c \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcSinh[c*x]))/Sqrt[d + c^2*d*x^2],x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.50, size = 121, normalized size = 1.02 \begin {gather*} -\frac {-\frac {4 a c x \sqrt {d+c^2 d x^2}}{d}+\frac {4 a \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{\sqrt {d}}+\frac {b \sqrt {1+c^2 x^2} \left (\cosh \left (2 \sinh ^{-1}(c x)\right )+2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-\sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )}{\sqrt {d+c^2 d x^2}}}{8 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcSinh[c*x]))/Sqrt[d + c^2*d*x^2],x]

[Out]

-1/8*((-4*a*c*x*Sqrt[d + c^2*d*x^2])/d + (4*a*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/Sqrt[d] + (b*Sqrt[1 +
c^2*x^2]*(Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*(ArcSinh[c*x] - Sinh[2*ArcSinh[c*x]])))/Sqrt[d + c^2*d*x^2])/c
^3

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(272\) vs. \(2(103)=206\).
time = 3.65, size = 273, normalized size = 2.29

method	result	size
default	\(\frac {a x \sqrt {c^{2} d \,x^{2}+d}}{2 c^{2} d}-\frac {a \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 c^{2} \sqrt {c^{2} d}}+b \left (-\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2}}{4 \sqrt {c^{2} x^{2}+1}\, c^{3} d}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+2 \arcsinh \left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) \left (1+2 \arcsinh \left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}\right )\)	\(273\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x/c^2/d*(c^2*d*x^2+d)^(1/2)-1/2*a/c^2*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+b*(-1/
4*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^3/d*arcsinh(c*x)^2+1/16*(d*(c^2*x^2+1))^(1/2)*(2*c^3*x^3+2*c^2*x^2
*(c^2*x^2+1)^(1/2)+2*c*x+(c^2*x^2+1)^(1/2))*(-1+2*arcsinh(c*x))/c^3/d/(c^2*x^2+1)+1/16*(d*(c^2*x^2+1))^(1/2)*(
2*c^3*x^3-2*c^2*x^2*(c^2*x^2+1)^(1/2)+2*c*x-(c^2*x^2+1)^(1/2))*(1+2*arcsinh(c*x))/c^3/d/(c^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(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(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(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral((b*x^2*arcsinh(c*x) + a*x^2)/sqrt(c^2*d*x^2 + d), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*asinh(c*x))/sqrt(d*(c**2*x**2 + 1)), x)

________________________________________________________________________________________

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

[Out]

integrate((b*arcsinh(c*x) + a)*x^2/sqrt(c^2*d*x^2 + d), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {d\,c^2\,x^2+d}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________