3.1.56 \(\int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \sinh ^{-1}(c x)) \, dx\) [56]

Optimal. Leaf size=119 \[ -\frac {b \sqrt {\pi } x^2}{16 c}-\frac {1}{16} b c \sqrt {\pi } x^4+\frac {\sqrt {\pi } x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\sqrt {\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3} \]

[Out]

-1/16*b*x^2*Pi^(1/2)/c-1/16*b*c*x^4*Pi^(1/2)-1/16*(a+b*arcsinh(c*x))^2*Pi^(1/2)/b/c^3+1/8*x*(a+b*arcsinh(c*x))
*Pi^(1/2)*(c^2*x^2+1)^(1/2)/c^2+1/4*x^3*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5806, 5812, 5783, 30} \begin {gather*} -\frac {\sqrt {\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3}+\frac {\sqrt {\pi } x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{16} \sqrt {\pi } b c x^4-\frac {\sqrt {\pi } b x^2}{16 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(b*Sqrt[Pi]*x^2)/c - (b*c*Sqrt[Pi]*x^4)/16 + (Sqrt[Pi]*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(8*c^2)
+ (x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/4 - (Sqrt[Pi]*(a + b*ArcSinh[c*x])^2)/(16*b*c^3)

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 5806

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(
f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcSinh[c*x])^n/(f*(m + 2))), x] + (Dist[(1/(m + 2))*Simp[Sqrt[d + e*x^2]
/Sqrt[1 + c^2*x^2]], Int[(f*x)^m*((a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2]), x], x] - Dist[b*c*(n/(f*(m + 2)))
*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[n, 0] && (IGtQ[m, -2] || EqQ[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 x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} x^3 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\sqrt {\pi +c^2 \pi x^2} \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{4 \sqrt {1+c^2 x^2}}-\frac {\left (b c \sqrt {\pi +c^2 \pi x^2}\right ) \int x^3 \, dx}{4 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c x^4 \sqrt {\pi +c^2 \pi x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\sqrt {\pi +c^2 \pi x^2} \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{8 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (b \sqrt {\pi +c^2 \pi x^2}\right ) \int x \, dx}{8 c \sqrt {1+c^2 x^2}}\\ &=-\frac {b x^2 \sqrt {\pi +c^2 \pi x^2}}{16 c \sqrt {1+c^2 x^2}}-\frac {b c x^4 \sqrt {\pi +c^2 \pi x^2}}{16 \sqrt {1+c^2 x^2}}+\frac {x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 79, normalized size = 0.66 \begin {gather*} \frac {\sqrt {\pi } \left (16 a c x \sqrt {1+c^2 x^2} \left (1+2 c^2 x^2\right )-8 b \sinh ^{-1}(c x)^2-b \cosh \left (4 \sinh ^{-1}(c x)\right )+\sinh ^{-1}(c x) \left (-16 a+4 b \sinh \left (4 \sinh ^{-1}(c x)\right )\right )\right )}{128 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi]*(16*a*c*x*Sqrt[1 + c^2*x^2]*(1 + 2*c^2*x^2) - 8*b*ArcSinh[c*x]^2 - b*Cosh[4*ArcSinh[c*x]] + ArcSinh[
c*x]*(-16*a + 4*b*Sinh[4*ArcSinh[c*x]])))/(128*c^3)

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Maple [A]
time = 0.80, size = 156, normalized size = 1.31

method	result	size
default	\(\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4 \pi \,c^{2}}-\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{2}}-\frac {a \pi \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \sqrt {\pi }\, \left (-4 \sqrt {c^{2} x^{2}+1}\, \arcsinh \left (c x \right ) x^{3} c^{3}+c^{4} x^{4}-2 \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x c +c^{2} x^{2}+\arcsinh \left (c x \right )^{2}\right )}{16 c^{3}}\)	\(156\)

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]

1/4*a*x*(Pi*c^2*x^2+Pi)^(3/2)/Pi/c^2-1/8*a/c^2*x*(Pi*c^2*x^2+Pi)^(1/2)-1/8*a/c^2*Pi*ln(Pi*c^2*x/(Pi*c^2)^(1/2)
+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)-1/16*b*Pi^(1/2)*(-4*(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^3*c^3+c^4*x^4-2*ar
csinh(c*x)*(c^2*x^2+1)^(1/2)*x*c+c^2*x^2+arcsinh(c*x)^2)/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(sqrt(pi + pi*c^2*x^2)*(b*x^2*arcsinh(c*x) + a*x^2), x)

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

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

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int 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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