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

Optimal. Leaf size=45 \[ 3 \sqrt {x^4+5}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\left (3 x^2+2\right ) x^2}{2 \sqrt {x^4+5}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1252, 819, 641, 215} \[ -\frac {\left (3 x^2+2\right ) x^2}{2 \sqrt {x^4+5}}+3 \sqrt {x^4+5}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 641

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

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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^5 \left (2+3 x^2\right )}{\left (5+x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2 (2+3 x)}{\left (5+x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (2+3 x^2\right )}{2 \sqrt {5+x^4}}+\frac {1}{10} \operatorname {Subst}\left (\int \frac {10+30 x}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (2+3 x^2\right )}{2 \sqrt {5+x^4}}+3 \sqrt {5+x^4}+\operatorname {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (2+3 x^2\right )}{2 \sqrt {5+x^4}}+3 \sqrt {5+x^4}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 46, normalized size = 1.02 \[ \frac {3 x^4-2 x^2+2 \sqrt {x^4+5} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+30}{2 \sqrt {x^4+5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(30 - 2*x^2 + 3*x^4 + 2*Sqrt[5 + x^4]*ArcSinh[x^2/Sqrt[5]])/(2*Sqrt[5 + x^4])

________________________________________________________________________________________

fricas [A]  time = 0.67, size = 58, normalized size = 1.29 \[ -\frac {2 \, x^{4} + 2 \, {\left (x^{4} + 5\right )} \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) - {\left (3 \, x^{4} - 2 \, x^{2} + 30\right )} \sqrt {x^{4} + 5} + 10}{2 \, {\left (x^{4} + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x^4 + 2*(x^4 + 5)*log(-x^2 + sqrt(x^4 + 5)) - (3*x^4 - 2*x^2 + 30)*sqrt(x^4 + 5) + 10)/(x^4 + 5)

________________________________________________________________________________________

giac [A]  time = 0.23, size = 39, normalized size = 0.87 \[ \frac {{\left (3 \, x^{2} - 2\right )} x^{2} + 30}{2 \, \sqrt {x^{4} + 5}} - \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((3*x^2 - 2)*x^2 + 30)/sqrt(x^4 + 5) - log(-x^2 + sqrt(x^4 + 5))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 37, normalized size = 0.82 \[ -\frac {x^{2}}{\sqrt {x^{4}+5}}+\arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )+\frac {\frac {3 x^{4}}{2}+15}{\sqrt {x^{4}+5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2/(x^4+5)^(1/2)*(x^4+10)-1/(x^4+5)^(1/2)*x^2+arcsinh(1/5*5^(1/2)*x^2)

________________________________________________________________________________________

maxima [A]  time = 1.41, size = 63, normalized size = 1.40 \[ -\frac {x^{2}}{\sqrt {x^{4} + 5}} + \frac {3}{2} \, \sqrt {x^{4} + 5} + \frac {15}{2 \, \sqrt {x^{4} + 5}} + \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.89, size = 89, normalized size = 1.98 \[ \mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )+\frac {3\,\sqrt {x^4+5}}{2}-\frac {\sqrt {5}\,\left (-15+\sqrt {5}\,2{}\mathrm {i}\right )\,\sqrt {x^4+5}\,1{}\mathrm {i}}{20\,\left (-x^2+\sqrt {5}\,1{}\mathrm {i}\right )}+\frac {\sqrt {5}\,\left (15+\sqrt {5}\,2{}\mathrm {i}\right )\,\sqrt {x^4+5}\,1{}\mathrm {i}}{20\,\left (x^2+\sqrt {5}\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

asinh((5^(1/2)*x^2)/5) + (3*(x^4 + 5)^(1/2))/2 - (5^(1/2)*(5^(1/2)*2i - 15)*(x^4 + 5)^(1/2)*1i)/(20*(5^(1/2)*1
i - x^2)) + (5^(1/2)*(5^(1/2)*2i + 15)*(x^4 + 5)^(1/2)*1i)/(20*(5^(1/2)*1i + x^2))

________________________________________________________________________________________

sympy [A]  time = 12.34, size = 48, normalized size = 1.07 \[ \frac {3 x^{4}}{2 \sqrt {x^{4} + 5}} - \frac {x^{2}}{\sqrt {x^{4} + 5}} + \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )} + \frac {15}{\sqrt {x^{4} + 5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________