3.34 \(\int \frac {x^3 (2+3 x^2)}{\sqrt {5+x^4}} \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{4} \left (3 x^2+4\right ) \sqrt {x^4+5}-\frac {15}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

[Out]

-15/4*arcsinh(1/5*x^2*5^(1/2))+1/4*(3*x^2+4)*(x^4+5)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1252, 780, 215} \[ \frac {1}{4} \left (3 x^2+4\right ) \sqrt {x^4+5}-\frac {15}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4 + 3*x^2)*Sqrt[5 + x^4])/4 - (15*ArcSinh[x^2/Sqrt[5]])/4

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 \frac {x^3 \left (2+3 x^2\right )}{\sqrt {5+x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x (2+3 x)}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}-\frac {15}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}-\frac {15}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 34, normalized size = 0.97 \[ \frac {1}{4} \left (\left (3 x^2+4\right ) \sqrt {x^4+5}-15 \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4 + 3*x^2)*Sqrt[5 + x^4] - 15*ArcSinh[x^2/Sqrt[5]])/4

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 33, normalized size = 0.94 \[ \frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \frac {15}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(3*x^2+2)/(x^4+5)^(1/2),x, algorithm="fricas")

[Out]

1/4*sqrt(x^4 + 5)*(3*x^2 + 4) + 15/4*log(-x^2 + sqrt(x^4 + 5))

________________________________________________________________________________________

giac [A]  time = 0.22, size = 33, normalized size = 0.94 \[ \frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \frac {15}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(3*x^2+2)/(x^4+5)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(x^4 + 5)*(3*x^2 + 4) + 15/4*log(-x^2 + sqrt(x^4 + 5))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 32, normalized size = 0.91 \[ \frac {3 \sqrt {x^{4}+5}\, x^{2}}{4}-\frac {15 \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )}{4}+\sqrt {x^{4}+5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(3*x^2+2)/(x^4+5)^(1/2),x)

[Out]

3/4*(x^4+5)^(1/2)*x^2-15/4*arcsinh(1/5*5^(1/2)*x^2)+(x^4+5)^(1/2)

________________________________________________________________________________________

maxima [B]  time = 1.16, size = 65, normalized size = 1.86 \[ \sqrt {x^{4} + 5} + \frac {15 \, \sqrt {x^{4} + 5}}{4 \, x^{2} {\left (\frac {x^{4} + 5}{x^{4}} - 1\right )}} - \frac {15}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) + \frac {15}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(3*x^2+2)/(x^4+5)^(1/2),x, algorithm="maxima")

[Out]

sqrt(x^4 + 5) + 15/4*sqrt(x^4 + 5)/(x^2*((x^4 + 5)/x^4 - 1)) - 15/8*log(sqrt(x^4 + 5)/x^2 + 1) + 15/8*log(sqrt
(x^4 + 5)/x^2 - 1)

________________________________________________________________________________________

mupad [B]  time = 0.49, size = 27, normalized size = 0.77 \[ \sqrt {x^4+5}\,\left (\frac {3\,x^2}{4}+1\right )-\frac {15\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(3*x^2 + 2))/(x^4 + 5)^(1/2),x)

[Out]

(x^4 + 5)^(1/2)*((3*x^2)/4 + 1) - (15*asinh((5^(1/2)*x^2)/5))/4

________________________________________________________________________________________

sympy [A]  time = 4.04, size = 53, normalized size = 1.51 \[ \frac {3 x^{6}}{4 \sqrt {x^{4} + 5}} + \frac {15 x^{2}}{4 \sqrt {x^{4} + 5}} + \sqrt {x^{4} + 5} - \frac {15 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(3*x**2+2)/(x**4+5)**(1/2),x)

[Out]

3*x**6/(4*sqrt(x**4 + 5)) + 15*x**2/(4*sqrt(x**4 + 5)) + sqrt(x**4 + 5) - 15*asinh(sqrt(5)*x**2/5)/4

________________________________________________________________________________________