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

Optimal. Leaf size=77 \[ \frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi c^2}-\frac{1}{25} \pi ^{3/2} b c^3 x^5-\frac{2}{15} \pi ^{3/2} b c x^3-\frac{\pi ^{3/2} b x}{5 c} \]

[Out]

-(b*Pi^(3/2)*x)/(5*c) - (2*b*c*Pi^(3/2)*x^3)/15 - (b*c^3*Pi^(3/2)*x^5)/25 + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*Ar
cSinh[c*x]))/(5*c^2*Pi)

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Rubi [A]  time = 0.087301, antiderivative size = 146, normalized size of antiderivative = 1.9, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {5717, 194} \[ \frac{\left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 \pi c^2}-\frac{\pi b c^3 x^5 \sqrt{\pi c^2 x^2+\pi }}{25 \sqrt{c^2 x^2+1}}-\frac{2 \pi b c x^3 \sqrt{\pi c^2 x^2+\pi }}{15 \sqrt{c^2 x^2+1}}-\frac{\pi b x \sqrt{\pi c^2 x^2+\pi }}{5 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Pi*x*Sqrt[Pi + c^2*Pi*x^2])/(5*c*Sqrt[1 + c^2*x^2]) - (2*b*c*Pi*x^3*Sqrt[Pi + c^2*Pi*x^2])/(15*Sqrt[1 + c^
2*x^2]) - (b*c^3*Pi*x^5*Sqrt[Pi + c^2*Pi*x^2])/(25*Sqrt[1 + c^2*x^2]) + ((Pi + c^2*Pi*x^2)^(5/2)*(a + b*ArcSin
h[c*x]))/(5*c^2*Pi)

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 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int x \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }-\frac{\left (b \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right )^2 \, dx}{5 c \sqrt{1+c^2 x^2}}\\ &=\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }-\frac{\left (b \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx}{5 c \sqrt{1+c^2 x^2}}\\ &=-\frac{b \pi x \sqrt{\pi +c^2 \pi x^2}}{5 c \sqrt{1+c^2 x^2}}-\frac{2 b c \pi x^3 \sqrt{\pi +c^2 \pi x^2}}{15 \sqrt{1+c^2 x^2}}-\frac{b c^3 \pi x^5 \sqrt{\pi +c^2 \pi x^2}}{25 \sqrt{1+c^2 x^2}}+\frac{\left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c^2 \pi }\\ \end{align*}

Mathematica [A]  time = 0.124941, size = 72, normalized size = 0.94 \[ \frac{\pi ^{3/2} \left (15 a \left (c^2 x^2+1\right )^{5/2}-b c x \left (3 c^4 x^4+10 c^2 x^2+15\right )+15 b \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)\right )}{75 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(3/2)*(15*a*(1 + c^2*x^2)^(5/2) - b*c*x*(15 + 10*c^2*x^2 + 3*c^4*x^4) + 15*b*(1 + c^2*x^2)^(5/2)*ArcSinh[c
*x]))/(75*c^2)

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Maple [B]  time = 0.057, size = 139, normalized size = 1.8 \begin{align*}{\frac{a}{5\,\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}}+{\frac{b{\pi }^{{\frac{3}{2}}}}{75\,{c}^{2}} \left ( 15\,{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}+45\,{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}-3\,{c}^{5}{x}^{5}\sqrt{{c}^{2}{x}^{2}+1}+45\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}-10\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+15\,{\it Arcsinh} \left ( cx \right ) -15\,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*(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x)),x)

[Out]

1/5*a/Pi/c^2*(Pi*c^2*x^2+Pi)^(5/2)+1/75*b/c^2*Pi^(3/2)/(c^2*x^2+1)^(1/2)*(15*arcsinh(c*x)*c^6*x^6+45*arcsinh(c
*x)*c^4*x^4-3*c^5*x^5*(c^2*x^2+1)^(1/2)+45*arcsinh(c*x)*c^2*x^2-10*c^3*x^3*(c^2*x^2+1)^(1/2)+15*arcsinh(c*x)-1
5*c*x*(c^2*x^2+1)^(1/2))

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Maxima [A]  time = 1.16566, size = 115, normalized size = 1.49 \begin{align*} \frac{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{2}} b \operatorname{arsinh}\left (c x\right )}{5 \, \pi c^{2}} + \frac{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{2}} a}{5 \, \pi c^{2}} - \frac{{\left (3 \, \pi ^{\frac{5}{2}} c^{4} x^{5} + 10 \, \pi ^{\frac{5}{2}} c^{2} x^{3} + 15 \, \pi ^{\frac{5}{2}} x\right )} b}{75 \, \pi c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(pi + pi*c^2*x^2)^(5/2)*b*arcsinh(c*x)/(pi*c^2) + 1/5*(pi + pi*c^2*x^2)^(5/2)*a/(pi*c^2) - 1/75*(3*pi^(5/2
)*c^4*x^5 + 10*pi^(5/2)*c^2*x^3 + 15*pi^(5/2)*x)*b/(pi*c)

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Fricas [B]  time = 2.37377, size = 393, normalized size = 5.1 \begin{align*} \frac{15 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (\pi b c^{6} x^{6} + 3 \, \pi b c^{4} x^{4} + 3 \, \pi b c^{2} x^{2} + \pi b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + \sqrt{\pi + \pi c^{2} x^{2}}{\left (15 \, \pi a c^{6} x^{6} + 45 \, \pi a c^{4} x^{4} + 45 \, \pi a c^{2} x^{2} + 15 \, \pi a -{\left (3 \, \pi b c^{5} x^{5} + 10 \, \pi b c^{3} x^{3} + 15 \, \pi b c x\right )} \sqrt{c^{2} x^{2} + 1}\right )}}{75 \,{\left (c^{4} x^{2} + c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/75*(15*sqrt(pi + pi*c^2*x^2)*(pi*b*c^6*x^6 + 3*pi*b*c^4*x^4 + 3*pi*b*c^2*x^2 + pi*b)*log(c*x + sqrt(c^2*x^2
+ 1)) + sqrt(pi + pi*c^2*x^2)*(15*pi*a*c^6*x^6 + 45*pi*a*c^4*x^4 + 45*pi*a*c^2*x^2 + 15*pi*a - (3*pi*b*c^5*x^5
+ 10*pi*b*c^3*x^3 + 15*pi*b*c*x)*sqrt(c^2*x^2 + 1)))/(c^4*x^2 + c^2)

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Sympy [A]  time = 126.584, size = 221, normalized size = 2.87 \begin{align*} \begin{cases} \frac{\pi ^{\frac{3}{2}} a c^{2} x^{4} \sqrt{c^{2} x^{2} + 1}}{5} + \frac{2 \pi ^{\frac{3}{2}} a x^{2} \sqrt{c^{2} x^{2} + 1}}{5} + \frac{\pi ^{\frac{3}{2}} a \sqrt{c^{2} x^{2} + 1}}{5 c^{2}} - \frac{\pi ^{\frac{3}{2}} b c^{3} x^{5}}{25} + \frac{\pi ^{\frac{3}{2}} b c^{2} x^{4} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{2 \pi ^{\frac{3}{2}} b c x^{3}}{15} + \frac{2 \pi ^{\frac{3}{2}} b x^{2} \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{5} - \frac{\pi ^{\frac{3}{2}} b x}{5 c} + \frac{\pi ^{\frac{3}{2}} b \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{5 c^{2}} & \text{for}\: c \neq 0 \\\frac{\pi ^{\frac{3}{2}} a x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((pi**(3/2)*a*c**2*x**4*sqrt(c**2*x**2 + 1)/5 + 2*pi**(3/2)*a*x**2*sqrt(c**2*x**2 + 1)/5 + pi**(3/2)*
a*sqrt(c**2*x**2 + 1)/(5*c**2) - pi**(3/2)*b*c**3*x**5/25 + pi**(3/2)*b*c**2*x**4*sqrt(c**2*x**2 + 1)*asinh(c*
x)/5 - 2*pi**(3/2)*b*c*x**3/15 + 2*pi**(3/2)*b*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/5 - pi**(3/2)*b*x/(5*c) + p
i**(3/2)*b*sqrt(c**2*x**2 + 1)*asinh(c*x)/(5*c**2), Ne(c, 0)), (pi**(3/2)*a*x**2/2, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError