3.2.58 \(\int \frac {x^2 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [158]

Optimal. Leaf size=130 \[ -\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {d+c^2 d x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {d+c^2 d x^2}}+\frac {b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{2 c^3 d \sqrt {d+c^2 d x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 130, 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 = {5810, 5783, 266} \begin {gather*} -\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {c^2 d x^2+d}}+\frac {\sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {c^2 d x^2+d}}+\frac {b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 c^3 d \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 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 5810

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/(2*e*(p + 1))), x] + (-Dist[f^2*((m - 1)/(2*e*(p +
1))), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(2*c*(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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.12, size = 146, normalized size = 1.12 \begin {gather*} -\frac {a x \sqrt {d \left (1+c^2 x^2\right )}}{c^2 d^2 \left (1+c^2 x^2\right )}+\frac {b \left (-2 c x \sinh ^{-1}(c x)+\sqrt {1+c^2 x^2} \left (\sinh ^{-1}(c x)^2+2 \log \left (\sqrt {1+c^2 x^2}\right )\right )\right )}{2 c^3 d \sqrt {d \left (1+c^2 x^2\right )}}+\frac {a \log \left (c d x+\sqrt {d} \sqrt {d \left (1+c^2 x^2\right )}\right )}{c^3 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 3.56, size = 232, normalized size = 1.78

method	result	size
default	\(-\frac {a x}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {a \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{c^{2} d \sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )^{2}}{2 \sqrt {c^{2} x^{2}+1}\, c^{3} d^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{3} d^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{\sqrt {c^{2} x^{2}+1}\, c^{3} d^{2}}\)	\(232\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*x^2*arcsinh(c*x) + a*x^2)/(c^4*d^2*x^4 + 2*c^2*d^2*x^2 + d^2), 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 )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, 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)**(3/2),x)

[Out]

Integral(x**2*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(3/2), 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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^2/(c^2*d*x^2 + d)^(3/2), 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 )}{{\left (d\,c^2\,x^2+d\right )}^{3/2}} \,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)^(3/2),x)

[Out]

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

________________________________________________________________________________________