3.1.83 \(\int \frac {x^2 (a+b \sinh ^{-1}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx\) [83]

Optimal. Leaf size=75 \[ -\frac {b x^2}{4 c \sqrt {\pi }}+\frac {x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^2 \pi }-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {\pi }} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 75, 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 {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 \sqrt {\pi } b c^3}+\frac {x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi c^2}-\frac {b x^2}{4 \sqrt {\pi } c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.11, size = 69, normalized size = 0.92 \begin {gather*} \frac {4 a c x \sqrt {1+c^2 x^2}-2 b \sinh ^{-1}(c x)^2-b \cosh \left (2 \sinh ^{-1}(c x)\right )+\sinh ^{-1}(c x) \left (-4 a+2 b \sinh \left (2 \sinh ^{-1}(c x)\right )\right )}{8 c^3 \sqrt {\pi }} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.21, size = 107, normalized size = 1.43

method result size
default \(\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{2 \pi \,c^{2}}-\frac {a \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \left (-2 \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x c +c^{2} x^{2}+\arcsinh \left (c x \right )^{2}+1\right )}{4 \sqrt {\pi }\, c^{3}}\) \(107\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

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

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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))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 2.66, size = 92, normalized size = 1.23 \begin {gather*} \frac {a x \sqrt {c^{2} x^{2} + 1}}{2 \sqrt {\pi } c^{2}} - \frac {a \operatorname {asinh}{\left (c x \right )}}{2 \sqrt {\pi } c^{3}} + \frac {b \left (\begin {cases} - \frac {x^{2}}{4 c} + \frac {x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{2 c^{2}} - \frac {\operatorname {asinh}^{2}{\left (c x \right )}}{4 c^{3}} & \text {for}\: c \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="giac")

[Out]

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

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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 {\Pi \,c^2\,x^2+\Pi }} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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