3.21 \(\int x^3 (2+3 x^2) (5+x^4)^{3/2} \, dx\)

Optimal. Leaf size=67 \[ \frac{1}{20} \left (5 x^2+4\right ) \left (x^4+5\right )^{5/2}-\frac{5}{16} x^2 \left (x^4+5\right )^{3/2}-\frac{75}{32} x^2 \sqrt{x^4+5}-\frac{375}{32} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

[Out]

(-75*x^2*Sqrt[5 + x^4])/32 - (5*x^2*(5 + x^4)^(3/2))/16 + ((4 + 5*x^2)*(5 + x^4)^(5/2))/20 - (375*ArcSinh[x^2/
Sqrt[5]])/32

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Rubi [A]  time = 0.039551, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1252, 780, 195, 215} \[ \frac{1}{20} \left (5 x^2+4\right ) \left (x^4+5\right )^{5/2}-\frac{5}{16} x^2 \left (x^4+5\right )^{3/2}-\frac{75}{32} x^2 \sqrt{x^4+5}-\frac{375}{32} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*(2 + 3*x^2)*(5 + x^4)^(3/2),x]

[Out]

(-75*x^2*Sqrt[5 + x^4])/32 - (5*x^2*(5 + x^4)^(3/2))/16 + ((4 + 5*x^2)*(5 + x^4)^(5/2))/20 - (375*ArcSinh[x^2/
Sqrt[5]])/32

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 x^3 \left (2+3 x^2\right ) \left (5+x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (2+3 x) \left (5+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{1}{20} \left (4+5 x^2\right ) \left (5+x^4\right )^{5/2}-\frac{5}{4} \operatorname{Subst}\left (\int \left (5+x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac{5}{16} x^2 \left (5+x^4\right )^{3/2}+\frac{1}{20} \left (4+5 x^2\right ) \left (5+x^4\right )^{5/2}-\frac{75}{16} \operatorname{Subst}\left (\int \sqrt{5+x^2} \, dx,x,x^2\right )\\ &=-\frac{75}{32} x^2 \sqrt{5+x^4}-\frac{5}{16} x^2 \left (5+x^4\right )^{3/2}+\frac{1}{20} \left (4+5 x^2\right ) \left (5+x^4\right )^{5/2}-\frac{375}{32} \operatorname{Subst}\left (\int \frac{1}{\sqrt{5+x^2}} \, dx,x,x^2\right )\\ &=-\frac{75}{32} x^2 \sqrt{5+x^4}-\frac{5}{16} x^2 \left (5+x^4\right )^{3/2}+\frac{1}{20} \left (4+5 x^2\right ) \left (5+x^4\right )^{5/2}-\frac{375}{32} \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0414293, size = 54, normalized size = 0.81 \[ \frac{1}{160} \left (\sqrt{x^4+5} \left (40 x^{10}+32 x^8+350 x^6+320 x^4+375 x^2+800\right )-1875 \sinh ^{-1}\left (\frac{x^2}{\sqrt{5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(2 + 3*x^2)*(5 + x^4)^(3/2),x]

[Out]

(Sqrt[5 + x^4]*(800 + 375*x^2 + 320*x^4 + 350*x^6 + 32*x^8 + 40*x^10) - 1875*ArcSinh[x^2/Sqrt[5]])/160

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Maple [A]  time = 0.006, size = 58, normalized size = 0.9 \begin{align*}{\frac{{x}^{10}}{4}\sqrt{{x}^{4}+5}}+{\frac{35\,{x}^{6}}{16}\sqrt{{x}^{4}+5}}+{\frac{75\,{x}^{2}}{32}\sqrt{{x}^{4}+5}}-{\frac{375}{32}{\it Arcsinh} \left ({\frac{{x}^{2}\sqrt{5}}{5}} \right ) }+{\frac{1}{5} \left ({x}^{4}+5 \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(3*x^2+2)*(x^4+5)^(3/2),x)

[Out]

1/4*x^10*(x^4+5)^(1/2)+35/16*x^6*(x^4+5)^(1/2)+75/32*x^2*(x^4+5)^(1/2)-375/32*arcsinh(1/5*x^2*5^(1/2))+1/5*(x^
4+5)^(5/2)

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Maxima [B]  time = 1.41751, size = 159, normalized size = 2.37 \begin{align*} \frac{1}{5} \,{\left (x^{4} + 5\right )}^{\frac{5}{2}} - \frac{125 \,{\left (\frac{3 \, \sqrt{x^{4} + 5}}{x^{2}} - \frac{8 \,{\left (x^{4} + 5\right )}^{\frac{3}{2}}}{x^{6}} - \frac{3 \,{\left (x^{4} + 5\right )}^{\frac{5}{2}}}{x^{10}}\right )}}{32 \,{\left (\frac{3 \,{\left (x^{4} + 5\right )}}{x^{4}} - \frac{3 \,{\left (x^{4} + 5\right )}^{2}}{x^{8}} + \frac{{\left (x^{4} + 5\right )}^{3}}{x^{12}} - 1\right )}} - \frac{375}{64} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} + 1\right ) + \frac{375}{64} \, \log \left (\frac{\sqrt{x^{4} + 5}}{x^{2}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(3*x^2+2)*(x^4+5)^(3/2),x, algorithm="maxima")

[Out]

1/5*(x^4 + 5)^(5/2) - 125/32*(3*sqrt(x^4 + 5)/x^2 - 8*(x^4 + 5)^(3/2)/x^6 - 3*(x^4 + 5)^(5/2)/x^10)/(3*(x^4 +
5)/x^4 - 3*(x^4 + 5)^2/x^8 + (x^4 + 5)^3/x^12 - 1) - 375/64*log(sqrt(x^4 + 5)/x^2 + 1) + 375/64*log(sqrt(x^4 +
5)/x^2 - 1)

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Fricas [A]  time = 1.5947, size = 150, normalized size = 2.24 \begin{align*} \frac{1}{160} \,{\left (40 \, x^{10} + 32 \, x^{8} + 350 \, x^{6} + 320 \, x^{4} + 375 \, x^{2} + 800\right )} \sqrt{x^{4} + 5} + \frac{375}{32} \, \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(3*x^2+2)*(x^4+5)^(3/2),x, algorithm="fricas")

[Out]

1/160*(40*x^10 + 32*x^8 + 350*x^6 + 320*x^4 + 375*x^2 + 800)*sqrt(x^4 + 5) + 375/32*log(-x^2 + sqrt(x^4 + 5))

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Sympy [B]  time = 10.5095, size = 124, normalized size = 1.85 \begin{align*} \frac{x^{14}}{4 \sqrt{x^{4} + 5}} + \frac{55 x^{10}}{16 \sqrt{x^{4} + 5}} + \frac{x^{8} \sqrt{x^{4} + 5}}{5} + \frac{425 x^{6}}{32 \sqrt{x^{4} + 5}} + \frac{x^{4} \sqrt{x^{4} + 5}}{3} + \frac{375 x^{2}}{32 \sqrt{x^{4} + 5}} + \frac{5 \left (x^{4} + 5\right )^{\frac{3}{2}}}{3} - \frac{10 \sqrt{x^{4} + 5}}{3} - \frac{375 \operatorname{asinh}{\left (\frac{\sqrt{5} x^{2}}{5} \right )}}{32} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(3*x**2+2)*(x**4+5)**(3/2),x)

[Out]

x**14/(4*sqrt(x**4 + 5)) + 55*x**10/(16*sqrt(x**4 + 5)) + x**8*sqrt(x**4 + 5)/5 + 425*x**6/(32*sqrt(x**4 + 5))
+ x**4*sqrt(x**4 + 5)/3 + 375*x**2/(32*sqrt(x**4 + 5)) + 5*(x**4 + 5)**(3/2)/3 - 10*sqrt(x**4 + 5)/3 - 375*as
inh(sqrt(5)*x**2/5)/32

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Giac [A]  time = 1.15825, size = 80, normalized size = 1.19 \begin{align*} \frac{1}{160} \, \sqrt{x^{4} + 5}{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \, x^{2} + 4\right )} x^{2} + 175\right )} x^{2} + 160\right )} x^{2} + 375\right )} x^{2} + 800\right )} + \frac{375}{32} \, \log \left (-x^{2} + \sqrt{x^{4} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(3*x^2+2)*(x^4+5)^(3/2),x, algorithm="giac")

[Out]

1/160*sqrt(x^4 + 5)*((2*((4*(5*x^2 + 4)*x^2 + 175)*x^2 + 160)*x^2 + 375)*x^2 + 800) + 375/32*log(-x^2 + sqrt(x
^4 + 5))