3.81 \(\int \frac{x^4 (a+b \sinh ^{-1}(c x))}{\sqrt{\pi +c^2 \pi x^2}} \, dx\)

Optimal. Leaf size=126 \[ \frac{x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{4 \pi c^2}-\frac{3 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{8 \pi c^4}+\frac{3 \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \sqrt{\pi } b c^5}+\frac{3 b x^2}{16 \sqrt{\pi } c^3}-\frac{b x^4}{16 \sqrt{\pi } c} \]

[Out]

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

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Rubi [A]  time = 0.226495, antiderivative size = 170, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {5758, 5675, 30} \[ \frac{x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{4 \pi c^2}-\frac{3 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{8 \pi c^4}+\frac{3 \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \sqrt{\pi } b c^5}-\frac{b x^4 \sqrt{c^2 x^2+1}}{16 c \sqrt{\pi c^2 x^2+\pi }}+\frac{3 b x^2 \sqrt{c^2 x^2+1}}{16 c^3 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.237461, size = 111, normalized size = 0.88 \[ \frac{4 \sinh ^{-1}(c x) \left (12 a-8 b \sinh \left (2 \sinh ^{-1}(c x)\right )+b \sinh \left (4 \sinh ^{-1}(c x)\right )\right )+32 a c^3 x^3 \sqrt{c^2 x^2+1}-48 a c x \sqrt{c^2 x^2+1}+24 b \sinh ^{-1}(c x)^2+16 b \cosh \left (2 \sinh ^{-1}(c x)\right )-b \cosh \left (4 \sinh ^{-1}(c x)\right )}{128 \sqrt{\pi } c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

(-48*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*ArcSinh[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
^5*Sqrt[Pi])

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Maple [A]  time = 0.085, size = 188, normalized size = 1.5 \begin{align*}{\frac{a{x}^{3}}{4\,\pi \,{c}^{2}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{3\,ax}{8\,{c}^{4}\pi }\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{3\,a}{8\,{c}^{4}}\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{\it Arcsinh} \left ( cx \right ){x}^{3}}{4\,{c}^{2}\sqrt{\pi }}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{x}^{4}}{16\,c\sqrt{\pi }}}-{\frac{3\,b{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{4}\sqrt{\pi }}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{3\,b{x}^{2}}{16\,{c}^{3}\sqrt{\pi }}}+{\frac{3\,b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{16\,{c}^{5}\sqrt{\pi }}}+{\frac{b}{4\,{c}^{5}\sqrt{\pi }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

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

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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(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),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 (\frac{b x^{4} \operatorname{arsinh}\left (c x\right ) + a x^{4}}{\sqrt{\pi + \pi c^{2} x^{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 14.4618, size = 185, normalized size = 1.47 \begin{align*} \frac{a x^{5}}{4 \sqrt{\pi } \sqrt{c^{2} x^{2} + 1}} - \frac{a x^{3}}{8 \sqrt{\pi } c^{2} \sqrt{c^{2} x^{2} + 1}} - \frac{3 a x}{8 \sqrt{\pi } c^{4} \sqrt{c^{2} x^{2} + 1}} + \frac{3 a \operatorname{asinh}{\left (c x \right )}}{8 \sqrt{\pi } c^{5}} + \frac{b \left (\begin{cases} - \frac{x^{4}}{16 c} + \frac{x^{3} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{4 c^{2}} + \frac{3 x^{2}}{16 c^{3}} - \frac{3 x \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{8 c^{4}} + \frac{3 \operatorname{asinh}^{2}{\left (c x \right )}}{16 c^{5}} & \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**4*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

a*x**5/(4*sqrt(pi)*sqrt(c**2*x**2 + 1)) - a*x**3/(8*sqrt(pi)*c**2*sqrt(c**2*x**2 + 1)) - 3*a*x/(8*sqrt(pi)*c**
4*sqrt(c**2*x**2 + 1)) + 3*a*asinh(c*x)/(8*sqrt(pi)*c**5) + b*Piecewise((-x**4/(16*c) + x**3*sqrt(c**2*x**2 +
1)*asinh(c*x)/(4*c**2) + 3*x**2/(16*c**3) - 3*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(8*c**4) + 3*asinh(c*x)**2/(16*
c**5), 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^{4}}{\sqrt{\pi + \pi c^{2} x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="giac")

[Out]

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