3.1.64 \(\int x^2 (\pi +c^2 \pi x^2)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\) [64]

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

[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 = 0.22, antiderivative size = 165, normalized size of antiderivative = 1.00, 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 = {5808, 5806, 5812, 5783, 30, 14} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 5808

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/(f*(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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -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 \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 = 0.24, size = 154, normalized size = 0.93 \begin {gather*} \frac {\pi ^{3/2} \left (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}-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 )-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 )\right )}{2304 c^3} \end {gather*}

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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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx\]

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]

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

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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*(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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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*(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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Sympy [A]
time = 7.01, size = 262, normalized size = 1.59 \begin {gather*} \begin {cases} \frac {\pi ^{\frac {3}{2}} a c^{2} x^{5} \sqrt {c^{2} x^{2} + 1}}{6} + \frac {7 \pi ^{\frac {3}{2}} a x^{3} \sqrt {c^{2} x^{2} + 1}}{24} + \frac {\pi ^{\frac {3}{2}} a x \sqrt {c^{2} x^{2} + 1}}{16 c^{2}} - \frac {\pi ^{\frac {3}{2}} a \operatorname {asinh}{\left (c x \right )}}{16 c^{3}} - \frac {\pi ^{\frac {3}{2}} b c^{3} x^{6}}{36} + \frac {\pi ^{\frac {3}{2}} b c^{2} x^{5} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{6} - \frac {7 \pi ^{\frac {3}{2}} b c x^{4}}{96} + \frac {7 \pi ^{\frac {3}{2}} b x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{24} - \frac {\pi ^{\frac {3}{2}} b x^{2}}{32 c} + \frac {\pi ^{\frac {3}{2}} b x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{16 c^{2}} - \frac {\pi ^{\frac {3}{2}} b \operatorname {asinh}^{2}{\left (c x \right )}}{32 c^{3}} & \text {for}\: c \neq 0 \\\frac {\pi ^{\frac {3}{2}} a x^{3}}{3} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((pi**(3/2)*a*c**2*x**5*sqrt(c**2*x**2 + 1)/6 + 7*pi**(3/2)*a*x**3*sqrt(c**2*x**2 + 1)/24 + pi**(3/2)
*a*x*sqrt(c**2*x**2 + 1)/(16*c**2) - pi**(3/2)*a*asinh(c*x)/(16*c**3) - pi**(3/2)*b*c**3*x**6/36 + pi**(3/2)*b
*c**2*x**5*sqrt(c**2*x**2 + 1)*asinh(c*x)/6 - 7*pi**(3/2)*b*c*x**4/96 + 7*pi**(3/2)*b*x**3*sqrt(c**2*x**2 + 1)
*asinh(c*x)/24 - pi**(3/2)*b*x**2/(32*c) + pi**(3/2)*b*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(16*c**2) - pi**(3/2)*
b*asinh(c*x)**2/(32*c**3), Ne(c, 0)), (pi**(3/2)*a*x**3/3, True))

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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*(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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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 )\,{\left (\Pi \,c^2\,x^2+\Pi \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))*(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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