3.64 \(\int x^2 (\pi +c^2 \pi x^2)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=165 \[ -\frac {\pi ^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3}+\frac {\pi ^{3/2} x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}+\frac {1}{6} x^3 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} \pi x^3 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{36} \pi ^{3/2} b c^3 x^6-\frac {7}{96} \pi ^{3/2} b c x^4-\frac {\pi ^{3/2} b x^2}{32 c} \]

[Out]

-1/32*b*Pi^(3/2)*x^2/c-7/96*b*c*Pi^(3/2)*x^4-1/36*b*c^3*Pi^(3/2)*x^6+1/6*x^3*(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsin
h(c*x))-1/32*Pi^(3/2)*(a+b*arcsinh(c*x))^2/b/c^3+1/16*Pi^(3/2)*x*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^2+1/8*
Pi*x^3*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 254, normalized size of antiderivative = 1.54, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5744, 5742, 5758, 5675, 30, 14} \[ \frac {1}{6} x^3 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{8} \pi x^3 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac {\pi x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}-\frac {\pi \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3 \sqrt {c^2 x^2+1}}-\frac {\pi b c^3 x^6 \sqrt {\pi c^2 x^2+\pi }}{36 \sqrt {c^2 x^2+1}}-\frac {7 \pi b c x^4 \sqrt {\pi c^2 x^2+\pi }}{96 \sqrt {c^2 x^2+1}}-\frac {\pi b x^2 \sqrt {\pi c^2 x^2+\pi }}{32 c \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Pi*x^2*Sqrt[Pi + c^2*Pi*x^2])/(32*c*Sqrt[1 + c^2*x^2]) - (7*b*c*Pi*x^4*Sqrt[Pi + c^2*Pi*x^2])/(96*Sqrt[1 +
c^2*x^2]) - (b*c^3*Pi*x^6*Sqrt[Pi + c^2*Pi*x^2])/(36*Sqrt[1 + c^2*x^2]) + (Pi*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*
ArcSinh[c*x]))/(16*c^2) + (Pi*x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/8 + (x^3*(Pi + c^2*Pi*x^2)^(3/2)
*(a + b*ArcSinh[c*x]))/6 - (Pi*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/(32*b*c^3*Sqrt[1 + c^2*x^2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 5744

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n)/(f*(m + 2*p + 1)), x] + (Dist[(2*d*p)/(m + 2*p + 1), Int
[(f*x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p]
)/(f*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 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 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac {1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{2} \pi \int x^2 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac {\left (b c \pi \sqrt {\pi +c^2 \pi x^2}\right ) \int x^3 \left (1+c^2 x^2\right ) \, dx}{6 \sqrt {1+c^2 x^2}}\\ &=\frac {1}{8} \pi x^3 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\left (\pi \sqrt {\pi +c^2 \pi x^2}\right ) \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{8 \sqrt {1+c^2 x^2}}-\frac {\left (b c \pi \sqrt {\pi +c^2 \pi x^2}\right ) \int x^3 \, dx}{8 \sqrt {1+c^2 x^2}}-\frac {\left (b c \pi \sqrt {\pi +c^2 \pi x^2}\right ) \int \left (x^3+c^2 x^5\right ) \, dx}{6 \sqrt {1+c^2 x^2}}\\ &=-\frac {7 b c \pi x^4 \sqrt {\pi +c^2 \pi x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {b c^3 \pi x^6 \sqrt {\pi +c^2 \pi x^2}}{36 \sqrt {1+c^2 x^2}}+\frac {\pi x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}+\frac {1}{8} \pi x^3 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\left (\pi \sqrt {\pi +c^2 \pi x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{16 c^2 \sqrt {1+c^2 x^2}}-\frac {\left (b \pi \sqrt {\pi +c^2 \pi x^2}\right ) \int x \, dx}{16 c \sqrt {1+c^2 x^2}}\\ &=-\frac {b \pi x^2 \sqrt {\pi +c^2 \pi x^2}}{32 c \sqrt {1+c^2 x^2}}-\frac {7 b c \pi x^4 \sqrt {\pi +c^2 \pi x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {b c^3 \pi x^6 \sqrt {\pi +c^2 \pi x^2}}{36 \sqrt {1+c^2 x^2}}+\frac {\pi x \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}+\frac {1}{8} \pi x^3 \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {\pi \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3 \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 154, normalized size = 0.93 \[ \frac {\pi ^{3/2} \left (-12 \sinh ^{-1}(c x) \left (12 a+3 b \sinh \left (2 \sinh ^{-1}(c x)\right )-3 b \sinh \left (4 \sinh ^{-1}(c x)\right )-b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )+144 a c x \sqrt {c^2 x^2+1}+384 a c^5 x^5 \sqrt {c^2 x^2+1}+672 a c^3 x^3 \sqrt {c^2 x^2+1}-72 b \sinh ^{-1}(c x)^2+18 b \cosh \left (2 \sinh ^{-1}(c x)\right )-9 b \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b \cosh \left (6 \sinh ^{-1}(c x)\right )\right )}{2304 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(3/2)*(144*a*c*x*Sqrt[1 + c^2*x^2] + 672*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 384*a*c^5*x^5*Sqrt[1 + c^2*x^2] - 7
2*b*ArcSinh[c*x]^2 + 18*b*Cosh[2*ArcSinh[c*x]] - 9*b*Cosh[4*ArcSinh[c*x]] - 2*b*Cosh[6*ArcSinh[c*x]] - 12*ArcS
inh[c*x]*(12*a + 3*b*Sinh[2*ArcSinh[c*x]] - 3*b*Sinh[4*ArcSinh[c*x]] - b*Sinh[6*ArcSinh[c*x]])))/(2304*c^3)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\pi + \pi c^{2} x^{2}} {\left (\pi a c^{2} x^{4} + \pi a x^{2} + {\left (\pi b c^{2} x^{4} + \pi b x^{2}\right )} \operatorname {arsinh}\left (c x\right )\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(pi*c^2*x^2+pi)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi*a*c^2*x^4 + pi*a*x^2 + (pi*b*c^2*x^4 + pi*b*x^2)*arcsinh(c*x)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(pi*c^2*x^2+pi)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

integrate((pi + pi*c^2*x^2)^(3/2)*(b*arcsinh(c*x) + a)*x^2, x)

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maple [A]  time = 0.11, size = 240, normalized size = 1.45 \[ \frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{6 \pi \,c^{2}}-\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{24 c^{2}}-\frac {a \pi x \sqrt {\pi \,c^{2} x^{2}+\pi }}{16 c^{2}}-\frac {a \,\pi ^{2} \ln \left (\frac {\pi x \,c^{2}}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{16 c^{2} \sqrt {\pi \,c^{2}}}+\frac {b \,\pi ^{\frac {3}{2}} c^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{5}}{6}-\frac {b \,c^{3} \pi ^{\frac {3}{2}} x^{6}}{36}+\frac {7 b \,\pi ^{\frac {3}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3}}{24}-\frac {7 b c \,\pi ^{\frac {3}{2}} x^{4}}{96}+\frac {b \,\pi ^{\frac {3}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x}{16 c^{2}}-\frac {b \,\pi ^{\frac {3}{2}} x^{2}}{32 c}-\frac {b \,\pi ^{\frac {3}{2}} \arcsinh \left (c x \right )^{2}}{32 c^{3}}+\frac {b \,\pi ^{\frac {3}{2}}}{72 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x)),x)

[Out]

1/6*a*x*(Pi*c^2*x^2+Pi)^(5/2)/Pi/c^2-1/24*a/c^2*x*(Pi*c^2*x^2+Pi)^(3/2)-1/16*a/c^2*Pi*x*(Pi*c^2*x^2+Pi)^(1/2)-
1/16*a/c^2*Pi^2*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/6*b*Pi^(3/2)*c^2*arcsinh(c*
x)*(c^2*x^2+1)^(1/2)*x^5-1/36*b*c^3*Pi^(3/2)*x^6+7/24*b*Pi^(3/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3-7/96*b*c*P
i^(3/2)*x^4+1/16*b*Pi^(3/2)/c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-1/32*b*Pi^(3/2)*x^2/c-1/32*b*Pi^(3/2)/c^3*arc
sinh(c*x)^2+1/72*b*Pi^(3/2)/c^3

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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*(pi*c^2*x^2+pi)^(3/2)*(a+b*arcsinh(c*x)),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 \[ \int x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2} \,d x \]

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)^(3/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(pi*c**2*x**2+pi)**(3/2)*(a+b*asinh(c*x)),x)

[Out]

Timed out

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