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

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

[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 = 0.20, antiderivative size = 181, normalized size of antiderivative = 1.52, 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 = {5742, 5758, 5675, 30} \[ \frac {1}{4} x^3 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac {x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac {\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {c^2 x^2+1}}-\frac {b c x^4 \sqrt {\pi c^2 x^2+\pi }}{16 \sqrt {c^2 x^2+1}}-\frac {b x^2 \sqrt {\pi c^2 x^2+\pi }}{16 c \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5742

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[Sqrt[d + e*x^2]/((m + 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*Sqrt[d + e*x^2])/(f*
(m + 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] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*Sqrt[d + e*x^2]
), Int[(f*x)^(m - 1)*(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] && GtQ[m, 1] && IntegerQ[m]

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 = 0.17, size = 79, normalized size = 0.66 \[ \frac {\sqrt {\pi } \left (\sinh ^{-1}(c x) \left (4 b \sinh \left (4 \sinh ^{-1}(c x)\right )-16 a\right )+16 a c x \sqrt {c^2 x^2+1} \left (2 c^2 x^2+1\right )-8 b \sinh ^{-1}(c x)^2-b \cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{128 c^3} \]

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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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\pi + \pi c^{2} x^{2}} {\left (b x^{2} \operatorname {arsinh}\left (c x\right ) + a x^{2}\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2}\,{d x} \]

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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maple [A]  time = 0.07, size = 170, normalized size = 1.43 \[ \frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{4 \pi \,c^{2}}-\frac {a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{2}}-\frac {a \pi \ln \left (\frac {\pi x \,c^{2}}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{2} \sqrt {\pi \,c^{2}}}+\frac {b \sqrt {\pi }\, \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3}}{4}-\frac {b c \,x^{4} \sqrt {\pi }}{16}+\frac {b \sqrt {\pi }\, \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x}{8 c^{2}}-\frac {b \,x^{2} \sqrt {\pi }}{16 c}-\frac {b \sqrt {\pi }\, \arcsinh \left (c x \right )^{2}}{16 c^{3}} \]

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)

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {\Pi \,c^2\,x^2+\Pi } \,d x \]

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

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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