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

Optimal. Leaf size=111 \[ \frac{1}{4} x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3}{8} \pi x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 \pi ^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c}-\frac{1}{16} \pi ^{3/2} b c^3 x^4-\frac{5}{16} \pi ^{3/2} b c x^2 \]

[Out]

(-5*b*c*Pi^(3/2)*x^2)/16 - (b*c^3*Pi^(3/2)*x^4)/16 + (3*Pi*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/8 + (
x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/4 + (3*Pi^(3/2)*(a + b*ArcSinh[c*x])^2)/(16*b*c)

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Rubi [A]  time = 0.112484, antiderivative size = 180, normalized size of antiderivative = 1.62, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {5684, 5682, 5675, 30, 14} \[ \frac{1}{4} x \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3}{8} \pi x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 \pi \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{c^2 x^2+1}}-\frac{\pi b c^3 x^4 \sqrt{\pi c^2 x^2+\pi }}{16 \sqrt{c^2 x^2+1}}-\frac{5 \pi b c x^2 \sqrt{\pi c^2 x^2+\pi }}{16 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*c*Pi*x^2*Sqrt[Pi + c^2*Pi*x^2])/(16*Sqrt[1 + c^2*x^2]) - (b*c^3*Pi*x^4*Sqrt[Pi + c^2*Pi*x^2])/(16*Sqrt[1
+ c^2*x^2]) + (3*Pi*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/8 + (x*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSi
nh[c*x]))/4 + (3*Pi*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/(16*b*c*Sqrt[1 + c^2*x^2])

Rule 5684

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

Rule 5682

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

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 30

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

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]]

Rubi steps

\begin{align*} \int \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} (3 \pi ) \int \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 \left (1+c^2 x^2\right ) \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{3}{8} \pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (3 \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\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 \left (x+c^2 x^3\right ) \, dx}{4 \sqrt{1+c^2 x^2}}-\frac{\left (3 b c \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int x \, dx}{8 \sqrt{1+c^2 x^2}}\\ &=-\frac{5 b c \pi x^2 \sqrt{\pi +c^2 \pi x^2}}{16 \sqrt{1+c^2 x^2}}-\frac{b c^3 \pi x^4 \sqrt{\pi +c^2 \pi x^2}}{16 \sqrt{1+c^2 x^2}}+\frac{3}{8} \pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{4} x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 \pi \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.224868, size = 111, normalized size = 1. \[ \frac{\pi ^{3/2} \left (4 \sinh ^{-1}(c x) \left (12 a+8 b \sinh \left (2 \sinh ^{-1}(c x)\right )+b \sinh \left (4 \sinh ^{-1}(c x)\right )\right )+32 a c^3 x^3 \sqrt{c^2 x^2+1}+80 a c x \sqrt{c^2 x^2+1}+24 b \sinh ^{-1}(c x)^2-16 b \cosh \left (2 \sinh ^{-1}(c x)\right )-b \cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{128 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(3/2)*(80*a*c*x*Sqrt[1 + c^2*x^2] + 32*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 24*b*ArcSinh[c*x]^2 - 16*b*Cosh[2*Arc
Sinh[c*x]] - b*Cosh[4*ArcSinh[c*x]] + 4*ArcSinh[c*x]*(12*a + 8*b*Sinh[2*ArcSinh[c*x]] + b*Sinh[4*ArcSinh[c*x]]
)))/(128*c)

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Maple [A]  time = 0.048, size = 170, normalized size = 1.5 \begin{align*}{\frac{ax}{4} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}+{\frac{3\,a\pi \,x}{8}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{3\,a{\pi }^{2}}{8}\ln \left ({\pi \,{c}^{2}x{\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+\sqrt{\pi \,{c}^{2}{x}^{2}+\pi } \right ){\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+{\frac{b{\pi }^{{\frac{3}{2}}}{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{4}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{c}^{3}{\pi }^{{\frac{3}{2}}}{x}^{4}}{16}}+{\frac{5\,b{\pi }^{3/2}{\it Arcsinh} \left ( cx \right ) x}{8}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{5\,bc{\pi }^{3/2}{x}^{2}}{16}}+{\frac{3\,b{\pi }^{3/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{16\,c}}-{\frac{b{\pi }^{{\frac{3}{2}}}}{4\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(Pi*c^2*x^2+Pi)^(3/2)*a+3/8*a*Pi*x*(Pi*c^2*x^2+Pi)^(1/2)+3/8*a*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)+1/4*b*Pi^(3/2)*c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3-1/16*b*c^3*Pi^(3/2)*x^4+5/8
*b*Pi^(3/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-5/16*b*c*Pi^(3/2)*x^2+3/16*b*Pi^(3/2)/c*arcsinh(c*x)^2-1/4*b*Pi^(
3/2)/c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{\pi + \pi c^{2} x^{2}}{\left (\pi a c^{2} x^{2} + \pi a +{\left (\pi b c^{2} x^{2} + \pi b\right )} \operatorname{arsinh}\left (c x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 52.3446, size = 185, normalized size = 1.67 \begin{align*} \begin{cases} \frac{\pi ^{\frac{3}{2}} a c^{2} x^{3} \sqrt{c^{2} x^{2} + 1}}{4} + \frac{5 \pi ^{\frac{3}{2}} a x \sqrt{c^{2} x^{2} + 1}}{8} + \frac{3 \pi ^{\frac{3}{2}} a \operatorname{asinh}{\left (c x \right )}}{8 c} - \frac{\pi ^{\frac{3}{2}} b c^{3} x^{4}}{16} + \frac{\pi ^{\frac{3}{2}} b c^{2} x^{3} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{4} - \frac{5 \pi ^{\frac{3}{2}} b c x^{2}}{16} + \frac{5 \pi ^{\frac{3}{2}} b x \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{8} + \frac{3 \pi ^{\frac{3}{2}} b \operatorname{asinh}^{2}{\left (c x \right )}}{16 c} & \text{for}\: c \neq 0 \\\pi ^{\frac{3}{2}} a x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((pi**(3/2)*a*c**2*x**3*sqrt(c**2*x**2 + 1)/4 + 5*pi**(3/2)*a*x*sqrt(c**2*x**2 + 1)/8 + 3*pi**(3/2)*a
*asinh(c*x)/(8*c) - pi**(3/2)*b*c**3*x**4/16 + pi**(3/2)*b*c**2*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)/4 - 5*pi**
(3/2)*b*c*x**2/16 + 5*pi**(3/2)*b*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/8 + 3*pi**(3/2)*b*asinh(c*x)**2/(16*c), Ne(
c, 0)), (pi**(3/2)*a*x, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}}{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)