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

Optimal. Leaf size=98 \[ \frac{x^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2}-\frac{2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^4}+\frac{2 b x}{3 \sqrt{\pi } c^3}-\frac{b x^3}{9 \sqrt{\pi } c} \]

[Out]

(2*b*x)/(3*c^3*Sqrt[Pi]) - (b*x^3)/(9*c*Sqrt[Pi]) - (2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(3*c^4*Pi)
+ (x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(3*c^2*Pi)

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Rubi [A]  time = 0.157292, antiderivative size = 142, normalized size of antiderivative = 1.45, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5758, 5717, 8, 30} \[ \frac{x^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2}-\frac{2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^4}-\frac{b x^3 \sqrt{c^2 x^2+1}}{9 c \sqrt{\pi c^2 x^2+\pi }}+\frac{2 b x \sqrt{c^2 x^2+1}}{3 c^3 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x*Sqrt[1 + c^2*x^2])/(3*c^3*Sqrt[Pi + c^2*Pi*x^2]) - (b*x^3*Sqrt[1 + c^2*x^2])/(9*c*Sqrt[Pi + c^2*Pi*x^2]
) - (2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(3*c^4*Pi) + (x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]
))/(3*c^2*Pi)

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]

Rule 5717

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{\pi +c^2 \pi x^2}} \, dx &=\frac{x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi }-\frac{2 \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{\pi +c^2 \pi x^2}} \, dx}{3 c^2}-\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int x^2 \, dx}{3 c \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{b x^3 \sqrt{1+c^2 x^2}}{9 c \sqrt{\pi +c^2 \pi x^2}}-\frac{2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi }+\frac{x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi }+\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \int 1 \, dx}{3 c^3 \sqrt{\pi +c^2 \pi x^2}}\\ &=\frac{2 b x \sqrt{1+c^2 x^2}}{3 c^3 \sqrt{\pi +c^2 \pi x^2}}-\frac{b x^3 \sqrt{1+c^2 x^2}}{9 c \sqrt{\pi +c^2 \pi x^2}}-\frac{2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi }+\frac{x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi }\\ \end{align*}

Mathematica [A]  time = 0.135346, size = 82, normalized size = 0.84 \[ \frac{3 a \sqrt{c^2 x^2+1} \left (c^2 x^2-2\right )+b \left (6 c x-c^3 x^3\right )+3 b \sqrt{c^2 x^2+1} \left (c^2 x^2-2\right ) \sinh ^{-1}(c x)}{9 \sqrt{\pi } c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(-2 + c^2*x^2)*Sqrt[1 + c^2*x^2] + b*(6*c*x - c^3*x^3) + 3*b*(-2 + c^2*x^2)*Sqrt[1 + c^2*x^2]*ArcSinh[c*x
])/(9*c^4*Sqrt[Pi])

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Maple [A]  time = 0.083, size = 133, normalized size = 1.4 \begin{align*} a \left ({\frac{{x}^{2}}{3\,\pi \,{c}^{2}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{2}{3\,\pi \,{c}^{4}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }} \right ) +{\frac{b}{9\,{c}^{4}\sqrt{\pi }} \left ( 3\,{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}-3\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}-{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}-6\,{\it Arcsinh} \left ( cx \right ) +6\,cx\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

a*(1/3*x^2/Pi/c^2*(Pi*c^2*x^2+Pi)^(1/2)-2/3/Pi/c^4*(Pi*c^2*x^2+Pi)^(1/2))+1/9*b/c^4/Pi^(1/2)/(c^2*x^2+1)^(1/2)
*(3*arcsinh(c*x)*c^4*x^4-3*arcsinh(c*x)*c^2*x^2-c^3*x^3*(c^2*x^2+1)^(1/2)-6*arcsinh(c*x)+6*c*x*(c^2*x^2+1)^(1/
2))

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Maxima [A]  time = 1.14162, size = 158, normalized size = 1.61 \begin{align*} \frac{1}{3} \, b{\left (\frac{\sqrt{\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{2}} - \frac{2 \, \sqrt{\pi + \pi c^{2} x^{2}}}{\pi c^{4}}\right )} \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \, a{\left (\frac{\sqrt{\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{2}} - \frac{2 \, \sqrt{\pi + \pi c^{2} x^{2}}}{\pi c^{4}}\right )} - \frac{{\left (c^{2} x^{3} - 6 \, x\right )} b}{9 \, \sqrt{\pi } c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

1/3*b*(sqrt(pi + pi*c^2*x^2)*x^2/(pi*c^2) - 2*sqrt(pi + pi*c^2*x^2)/(pi*c^4))*arcsinh(c*x) + 1/3*a*(sqrt(pi +
pi*c^2*x^2)*x^2/(pi*c^2) - 2*sqrt(pi + pi*c^2*x^2)/(pi*c^4)) - 1/9*(c^2*x^3 - 6*x)*b/(sqrt(pi)*c^3)

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Fricas [A]  time = 2.59313, size = 286, normalized size = 2.92 \begin{align*} \frac{3 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (b c^{4} x^{4} - b c^{2} x^{2} - 2 \, b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + \sqrt{\pi + \pi c^{2} x^{2}}{\left (3 \, a c^{4} x^{4} - 3 \, a c^{2} x^{2} -{\left (b c^{3} x^{3} - 6 \, b c x\right )} \sqrt{c^{2} x^{2} + 1} - 6 \, a\right )}}{9 \,{\left (\pi c^{6} x^{2} + \pi c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

1/9*(3*sqrt(pi + pi*c^2*x^2)*(b*c^4*x^4 - b*c^2*x^2 - 2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(pi + pi*c^2*x^2
)*(3*a*c^4*x^4 - 3*a*c^2*x^2 - (b*c^3*x^3 - 6*b*c*x)*sqrt(c^2*x^2 + 1) - 6*a))/(pi*c^6*x^2 + pi*c^4)

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Sympy [A]  time = 5.43048, size = 122, normalized size = 1.24 \begin{align*} \frac{a \left (\begin{cases} \frac{x^{2} \sqrt{c^{2} x^{2} + 1}}{3 c^{2}} - \frac{2 \sqrt{c^{2} x^{2} + 1}}{3 c^{4}} & \text{for}\: c \neq 0 \\\frac{x^{4}}{4} & \text{otherwise} \end{cases}\right )}{\sqrt{\pi }} + \frac{b \left (\begin{cases} - \frac{x^{3}}{9 c} + \frac{x^{2} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{3 c^{2}} + \frac{2 x}{3 c^{3}} - \frac{2 \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{3 c^{4}} & \text{for}\: c \neq 0 \\0 & \text{otherwise} \end{cases}\right )}{\sqrt{\pi }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

a*Piecewise((x**2*sqrt(c**2*x**2 + 1)/(3*c**2) - 2*sqrt(c**2*x**2 + 1)/(3*c**4), Ne(c, 0)), (x**4/4, True))/sq
rt(pi) + b*Piecewise((-x**3/(9*c) + x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(3*c**2) + 2*x/(3*c**3) - 2*sqrt(c**2*
x**2 + 1)*asinh(c*x)/(3*c**4), Ne(c, 0)), (0, True))/sqrt(pi)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="giac")

[Out]

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