3.80 \(\int \frac{x^5 (a+b \sinh ^{-1}(c x))}{\sqrt{\pi +c^2 \pi x^2}} \, dx\)

Optimal. Leaf size=149 \[ \frac{x^4 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi c^2}-\frac{4 x^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{15 \pi c^4}+\frac{8 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{15 \pi c^6}+\frac{4 b x^3}{45 \sqrt{\pi } c^3}-\frac{8 b x}{15 \sqrt{\pi } c^5}-\frac{b x^5}{25 \sqrt{\pi } c} \]

[Out]

(-8*b*x)/(15*c^5*Sqrt[Pi]) + (4*b*x^3)/(45*c^3*Sqrt[Pi]) - (b*x^5)/(25*c*Sqrt[Pi]) + (8*Sqrt[Pi + c^2*Pi*x^2]*
(a + b*ArcSinh[c*x]))/(15*c^6*Pi) - (4*x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(15*c^4*Pi) + (x^4*Sqrt
[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(5*c^2*Pi)

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Rubi [A]  time = 0.257035, antiderivative size = 215, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5758, 5717, 8, 30} \[ \frac{x^4 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi c^2}-\frac{4 x^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{15 \pi c^4}+\frac{8 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{15 \pi c^6}-\frac{b x^5 \sqrt{c^2 x^2+1}}{25 c \sqrt{\pi c^2 x^2+\pi }}+\frac{4 b x^3 \sqrt{c^2 x^2+1}}{45 c^3 \sqrt{\pi c^2 x^2+\pi }}-\frac{8 b x \sqrt{c^2 x^2+1}}{15 c^5 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

(-8*b*x*Sqrt[1 + c^2*x^2])/(15*c^5*Sqrt[Pi + c^2*Pi*x^2]) + (4*b*x^3*Sqrt[1 + c^2*x^2])/(45*c^3*Sqrt[Pi + c^2*
Pi*x^2]) - (b*x^5*Sqrt[1 + c^2*x^2])/(25*c*Sqrt[Pi + c^2*Pi*x^2]) + (8*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*
x]))/(15*c^6*Pi) - (4*x^2*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(15*c^4*Pi) + (x^4*Sqrt[Pi + c^2*Pi*x^2]
*(a + b*ArcSinh[c*x]))/(5*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^5 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{\pi +c^2 \pi x^2}} \, dx &=\frac{x^4 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }-\frac{4 \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{\pi +c^2 \pi x^2}} \, dx}{5 c^2}-\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int x^4 \, dx}{5 c \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{b x^5 \sqrt{1+c^2 x^2}}{25 c \sqrt{\pi +c^2 \pi x^2}}-\frac{4 x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 \pi }+\frac{x^4 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }+\frac{8 \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{\pi +c^2 \pi x^2}} \, dx}{15 c^4}+\frac{\left (4 b \sqrt{1+c^2 x^2}\right ) \int x^2 \, dx}{15 c^3 \sqrt{\pi +c^2 \pi x^2}}\\ &=\frac{4 b x^3 \sqrt{1+c^2 x^2}}{45 c^3 \sqrt{\pi +c^2 \pi x^2}}-\frac{b x^5 \sqrt{1+c^2 x^2}}{25 c \sqrt{\pi +c^2 \pi x^2}}+\frac{8 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 \pi }-\frac{4 x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 \pi }+\frac{x^4 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }-\frac{\left (8 b \sqrt{1+c^2 x^2}\right ) \int 1 \, dx}{15 c^5 \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{8 b x \sqrt{1+c^2 x^2}}{15 c^5 \sqrt{\pi +c^2 \pi x^2}}+\frac{4 b x^3 \sqrt{1+c^2 x^2}}{45 c^3 \sqrt{\pi +c^2 \pi x^2}}-\frac{b x^5 \sqrt{1+c^2 x^2}}{25 c \sqrt{\pi +c^2 \pi x^2}}+\frac{8 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^6 \pi }-\frac{4 x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{15 c^4 \pi }+\frac{x^4 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }\\ \end{align*}

Mathematica [A]  time = 0.184538, size = 108, normalized size = 0.72 \[ \frac{15 a \sqrt{c^2 x^2+1} \left (3 c^4 x^4-4 c^2 x^2+8\right )+b \left (-9 c^5 x^5+20 c^3 x^3-120 c x\right )+15 b \sqrt{c^2 x^2+1} \left (3 c^4 x^4-4 c^2 x^2+8\right ) \sinh ^{-1}(c x)}{225 \sqrt{\pi } c^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

(15*a*Sqrt[1 + c^2*x^2]*(8 - 4*c^2*x^2 + 3*c^4*x^4) + b*(-120*c*x + 20*c^3*x^3 - 9*c^5*x^5) + 15*b*Sqrt[1 + c^
2*x^2]*(8 - 4*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x])/(225*c^6*Sqrt[Pi])

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Maple [A]  time = 0.092, size = 193, normalized size = 1.3 \begin{align*} a \left ({\frac{{x}^{4}}{5\,\pi \,{c}^{2}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{4}{5\,{c}^{2}} \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 ) } \right ) +{\frac{b}{225\,{c}^{6}\sqrt{\pi }} \left ( 45\,{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}-15\,{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}-9\,{c}^{5}{x}^{5}\sqrt{{c}^{2}{x}^{2}+1}+60\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}+20\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+120\,{\it Arcsinh} \left ( cx \right ) -120\,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^5*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

a*(1/5*x^4/Pi/c^2*(Pi*c^2*x^2+Pi)^(1/2)-4/5/c^2*(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+P
i)^(1/2)))+1/225*b/c^6/Pi^(1/2)/(c^2*x^2+1)^(1/2)*(45*arcsinh(c*x)*c^6*x^6-15*arcsinh(c*x)*c^4*x^4-9*c^5*x^5*(
c^2*x^2+1)^(1/2)+60*arcsinh(c*x)*c^2*x^2+20*c^3*x^3*(c^2*x^2+1)^(1/2)+120*arcsinh(c*x)-120*c*x*(c^2*x^2+1)^(1/
2))

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Maxima [A]  time = 1.1725, size = 235, normalized size = 1.58 \begin{align*} \frac{1}{15} \,{\left (\frac{3 \, \sqrt{\pi + \pi c^{2} x^{2}} x^{4}}{\pi c^{2}} - \frac{4 \, \sqrt{\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{4}} + \frac{8 \, \sqrt{\pi + \pi c^{2} x^{2}}}{\pi c^{6}}\right )} b \operatorname{arsinh}\left (c x\right ) + \frac{1}{15} \,{\left (\frac{3 \, \sqrt{\pi + \pi c^{2} x^{2}} x^{4}}{\pi c^{2}} - \frac{4 \, \sqrt{\pi + \pi c^{2} x^{2}} x^{2}}{\pi c^{4}} + \frac{8 \, \sqrt{\pi + \pi c^{2} x^{2}}}{\pi c^{6}}\right )} a - \frac{{\left (9 \, c^{4} x^{5} - 20 \, c^{2} x^{3} + 120 \, x\right )} b}{225 \, \sqrt{\pi } c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*sqrt(pi + pi*c^2*x^2)*x^4/(pi*c^2) - 4*sqrt(pi + pi*c^2*x^2)*x^2/(pi*c^4) + 8*sqrt(pi + pi*c^2*x^2)/(p
i*c^6))*b*arcsinh(c*x) + 1/15*(3*sqrt(pi + pi*c^2*x^2)*x^4/(pi*c^2) - 4*sqrt(pi + pi*c^2*x^2)*x^2/(pi*c^4) + 8
*sqrt(pi + pi*c^2*x^2)/(pi*c^6))*a - 1/225*(9*c^4*x^5 - 20*c^2*x^3 + 120*x)*b/(sqrt(pi)*c^5)

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Fricas [A]  time = 2.60597, size = 363, normalized size = 2.44 \begin{align*} \frac{15 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (3 \, b c^{6} x^{6} - b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 8 \, b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + \sqrt{\pi + \pi c^{2} x^{2}}{\left (45 \, a c^{6} x^{6} - 15 \, a c^{4} x^{4} + 60 \, a c^{2} x^{2} -{\left (9 \, b c^{5} x^{5} - 20 \, b c^{3} x^{3} + 120 \, b c x\right )} \sqrt{c^{2} x^{2} + 1} + 120 \, a\right )}}{225 \,{\left (\pi c^{8} x^{2} + \pi c^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

1/225*(15*sqrt(pi + pi*c^2*x^2)*(3*b*c^6*x^6 - b*c^4*x^4 + 4*b*c^2*x^2 + 8*b)*log(c*x + sqrt(c^2*x^2 + 1)) + s
qrt(pi + pi*c^2*x^2)*(45*a*c^6*x^6 - 15*a*c^4*x^4 + 60*a*c^2*x^2 - (9*b*c^5*x^5 - 20*b*c^3*x^3 + 120*b*c*x)*sq
rt(c^2*x^2 + 1) + 120*a))/(pi*c^8*x^2 + pi*c^6)

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Sympy [A]  time = 26.581, size = 182, normalized size = 1.22 \begin{align*} \frac{a \left (\begin{cases} \frac{x^{4} \sqrt{c^{2} x^{2} + 1}}{5 c^{2}} - \frac{4 x^{2} \sqrt{c^{2} x^{2} + 1}}{15 c^{4}} + \frac{8 \sqrt{c^{2} x^{2} + 1}}{15 c^{6}} & \text{for}\: c \neq 0 \\\frac{x^{6}}{6} & \text{otherwise} \end{cases}\right )}{\sqrt{\pi }} + \frac{b \left (\begin{cases} - \frac{x^{5}}{25 c} + \frac{x^{4} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{5 c^{2}} + \frac{4 x^{3}}{45 c^{3}} - \frac{4 x^{2} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{15 c^{4}} - \frac{8 x}{15 c^{5}} + \frac{8 \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{15 c^{6}} & \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**5*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

a*Piecewise((x**4*sqrt(c**2*x**2 + 1)/(5*c**2) - 4*x**2*sqrt(c**2*x**2 + 1)/(15*c**4) + 8*sqrt(c**2*x**2 + 1)/
(15*c**6), Ne(c, 0)), (x**6/6, True))/sqrt(pi) + b*Piecewise((-x**5/(25*c) + x**4*sqrt(c**2*x**2 + 1)*asinh(c*
x)/(5*c**2) + 4*x**3/(45*c**3) - 4*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/(15*c**4) - 8*x/(15*c**5) + 8*sqrt(c**2
*x**2 + 1)*asinh(c*x)/(15*c**6), 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^{5}}{\sqrt{\pi + \pi c^{2} x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^5/sqrt(pi + pi*c^2*x^2), x)