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

Optimal. Leaf size=122 \[ \frac{1}{2} x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{1}{2} \sqrt{\pi } b c x^2 \left (a+b \sinh ^{-1}(c x)\right )+\frac{\sqrt{\pi } \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c}+\frac{1}{4} \sqrt{\pi } b^2 x \sqrt{c^2 x^2+1}-\frac{\sqrt{\pi } b^2 \sinh ^{-1}(c x)}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.112915, antiderivative size = 184, normalized size of antiderivative = 1.51, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5682, 5675, 5661, 321, 215} \[ \frac{\sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{c^2 x^2+1}}+\frac{1}{2} x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{b c x^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{c^2 x^2+1}}+\frac{1}{4} b^2 x \sqrt{\pi c^2 x^2+\pi }-\frac{b^2 \sqrt{\pi c^2 x^2+\pi } \sinh ^{-1}(c x)}{4 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{2} x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{\pi +c^2 \pi x^2} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}-\frac{\left (b c \sqrt{\pi +c^2 \pi x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=-\frac{b c x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}+\frac{\left (b^2 c^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{4} b^2 x \sqrt{\pi +c^2 \pi x^2}-\frac{b c x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}-\frac{\left (b^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{4} b^2 x \sqrt{\pi +c^2 \pi x^2}-\frac{b^2 \sqrt{\pi +c^2 \pi x^2} \sinh ^{-1}(c x)}{4 c \sqrt{1+c^2 x^2}}-\frac{b c x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.344711, size = 124, normalized size = 1.02 \[ \frac{\sqrt{\pi } \left (3 \left (4 a^2 c x \sqrt{c^2 x^2+1}-2 a b \cosh \left (2 \sinh ^{-1}(c x)\right )+b^2 \sinh \left (2 \sinh ^{-1}(c x)\right )\right )+6 \sinh ^{-1}(c x) \left (2 a \left (a+b \sinh \left (2 \sinh ^{-1}(c x)\right )\right )-b^2 \cosh \left (2 \sinh ^{-1}(c x)\right )\right )+6 b \sinh ^{-1}(c x)^2 \left (2 a+b \sinh \left (2 \sinh ^{-1}(c x)\right )\right )+4 b^2 \sinh ^{-1}(c x)^3\right )}{24 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi]*(4*b^2*ArcSinh[c*x]^3 + 6*b*ArcSinh[c*x]^2*(2*a + b*Sinh[2*ArcSinh[c*x]]) + 3*(4*a^2*c*x*Sqrt[1 + c^
2*x^2] - 2*a*b*Cosh[2*ArcSinh[c*x]] + b^2*Sinh[2*ArcSinh[c*x]]) + 6*ArcSinh[c*x]*(-(b^2*Cosh[2*ArcSinh[c*x]])
+ 2*a*(a + b*Sinh[2*ArcSinh[c*x]]))))/(24*c)

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Maple [B]  time = 0.071, size = 213, normalized size = 1.8 \begin{align*}{\frac{{a}^{2}x}{2}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{{a}^{2}\pi }{2}\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}^{2}\sqrt{\pi } \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}x}{2}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{b}^{2}\sqrt{\pi }c{\it Arcsinh} \left ( cx \right ){x}^{2}}{2}}+{\frac{{b}^{2}\sqrt{\pi } \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{3}}{6\,c}}+{\frac{{b}^{2}x\sqrt{\pi }}{4}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) \sqrt{\pi }}{4\,c}}+ab\sqrt{\pi }{\it Arcsinh} \left ( cx \right ) \sqrt{{c}^{2}{x}^{2}+1}x-{\frac{ab\sqrt{\pi }c{x}^{2}}{2}}+{\frac{ab\sqrt{\pi } \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,c}}-{\frac{ab\sqrt{\pi }}{2\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((Pi*c^2*x^2+Pi)^(1/2)*(a+b*arcsinh(c*x))^2,x)

[Out]

1/2*a^2*x*(Pi*c^2*x^2+Pi)^(1/2)+1/2*a^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/
2*b^2*Pi^(1/2)*arcsinh(c*x)^2*(c^2*x^2+1)^(1/2)*x-1/2*b^2*Pi^(1/2)*c*arcsinh(c*x)*x^2+1/6*b^2*Pi^(1/2)/c*arcsi
nh(c*x)^3+1/4*b^2*x*Pi^(1/2)*(c^2*x^2+1)^(1/2)-1/4*b^2*arcsinh(c*x)*Pi^(1/2)/c+a*b*Pi^(1/2)*arcsinh(c*x)*(c^2*
x^2+1)^(1/2)*x-1/2*a*b*Pi^(1/2)*c*x^2+1/2*a*b*Pi^(1/2)/c*arcsinh(c*x)^2-1/2*a*b*Pi^(1/2)/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)^(1/2)*(a+b*arcsinh(c*x))^2,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 (b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c**2*x**2+pi)**(1/2)*(a+b*asinh(c*x))**2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c^2*x^2+pi)^(1/2)*(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

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