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

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

[Out]

-5/16*x^2*(x^4+5)^(3/2)+1/20*(5*x^2+4)*(x^4+5)^(5/2)-375/32*arcsinh(1/5*x^2*5^(1/2))-75/32*x^2*(x^4+5)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.63, size = 53, normalized size = 0.79 \[ \frac {1}{160} \, {\left (40 \, x^{10} + 32 \, x^{8} + 350 \, x^{6} + 320 \, x^{4} + 375 \, x^{2} + 800\right )} \sqrt {x^{4} + 5} + \frac {375}{32} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

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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giac [A]  time = 0.20, size = 71, normalized size = 1.06 \[ \frac {1}{32} \, {\left (2 \, {\left (4 \, x^{4} + 5\right )} x^{4} - 75\right )} \sqrt {x^{4} + 5} x^{2} + \frac {15}{16} \, {\left (2 \, x^{4} + 5\right )} \sqrt {x^{4} + 5} x^{2} + \frac {1}{5} \, {\left (x^{4} + 5\right )}^{\frac {5}{2}} + \frac {375}{32} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

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]

1/32*(2*(4*x^4 + 5)*x^4 - 75)*sqrt(x^4 + 5)*x^2 + 15/16*(2*x^4 + 5)*sqrt(x^4 + 5)*x^2 + 1/5*(x^4 + 5)^(5/2) +
375/32*log(-x^2 + sqrt(x^4 + 5))

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maple [A]  time = 0.01, size = 58, normalized size = 0.87 \[ \frac {\sqrt {x^{4}+5}\, x^{10}}{4}+\frac {35 \sqrt {x^{4}+5}\, x^{6}}{16}+\frac {75 \sqrt {x^{4}+5}\, x^{2}}{32}-\frac {375 \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )}{32}+\frac {\left (x^{4}+5\right )^{\frac {5}{2}}}{5} \]

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

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maxima [B]  time = 1.13, size = 118, normalized size = 1.76 \[ \frac {1}{5} \, {\left (x^{4} + 5\right )}^{\frac {5}{2}} - \frac {125 \, {\left (\frac {3 \, \sqrt {x^{4} + 5}}{x^{2}} - \frac {8 \, {\left (x^{4} + 5\right )}^{\frac {3}{2}}}{x^{6}} - \frac {3 \, {\left (x^{4} + 5\right )}^{\frac {5}{2}}}{x^{10}}\right )}}{32 \, {\left (\frac {3 \, {\left (x^{4} + 5\right )}}{x^{4}} - \frac {3 \, {\left (x^{4} + 5\right )}^{2}}{x^{8}} + \frac {{\left (x^{4} + 5\right )}^{3}}{x^{12}} - 1\right )}} - \frac {375}{64} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) + \frac {375}{64} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

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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mupad [B]  time = 0.43, size = 47, normalized size = 0.70 \[ \sqrt {x^4+5}\,\left (\frac {x^{10}}{4}+\frac {x^8}{5}+\frac {35\,x^6}{16}+2\,x^4+\frac {75\,x^2}{32}+5\right )-\frac {375\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^4 + 5)^(1/2)*((75*x^2)/32 + 2*x^4 + (35*x^6)/16 + x^8/5 + x^10/4 + 5) - (375*asinh((5^(1/2)*x^2)/5))/32

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sympy [B]  time = 11.57, size = 124, normalized size = 1.85 \[ \frac {x^{14}}{4 \sqrt {x^{4} + 5}} + \frac {55 x^{10}}{16 \sqrt {x^{4} + 5}} + \frac {x^{8} \sqrt {x^{4} + 5}}{5} + \frac {425 x^{6}}{32 \sqrt {x^{4} + 5}} + \frac {x^{4} \sqrt {x^{4} + 5}}{3} + \frac {375 x^{2}}{32 \sqrt {x^{4} + 5}} + \frac {5 \left (x^{4} + 5\right )^{\frac {3}{2}}}{3} - \frac {10 \sqrt {x^{4} + 5}}{3} - \frac {375 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{32} \]

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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