3.1.34 \(\int \frac {x^3 (2+3 x^2)}{\sqrt {5+x^4}} \, dx\) [34]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {1266, 794, 221} \begin {gather*} \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 ) \end {gather*}

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 221

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 794

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 1266

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} \text {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} \text {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*}

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Mathematica [A]
time = 0.07, size = 39, normalized size = 1.11 \begin {gather*} \frac {1}{4} \left (4+3 x^2\right ) \sqrt {5+x^4}-\frac {15}{4} \tanh ^{-1}\left (\frac {x^2}{\sqrt {5+x^4}}\right ) \end {gather*}

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

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Maple [A]
time = 0.13, size = 32, normalized size = 0.91

method	result	size
risch	\(-\frac {15 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\frac {\left (3 x^{2}+4\right ) \sqrt {x^{4}+5}}{4}\)	\(29\)
trager	\(\left (\frac {3 x^{2}}{4}+1\right ) \sqrt {x^{4}+5}-\frac {15 \ln \left (x^{2}+\sqrt {x^{4}+5}\right )}{4}\)	\(31\)
default	\(\frac {3 x^{2} \sqrt {x^{4}+5}}{4}-\frac {15 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\sqrt {x^{4}+5}\)	\(32\)
elliptic	\(\frac {3 x^{2} \sqrt {x^{4}+5}}{4}-\frac {15 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\sqrt {x^{4}+5}\)	\(32\)
meijerg	\(\frac {\frac {3 \sqrt {\pi }\, x^{2} \sqrt {5}\, \sqrt {1+\frac {x^{4}}{5}}}{4}-\frac {15 \sqrt {\pi }\, \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}}{\sqrt {\pi }}+\frac {\sqrt {5}\, \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {1+\frac {x^{4}}{5}}\right )}{2 \sqrt {\pi }}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (28) = 56\).
time = 0.49, size = 65, normalized size = 1.86 \begin {gather*} \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 ) \end {gather*}

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)

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Fricas [A]
time = 0.36, size = 33, normalized size = 0.94 \begin {gather*} \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 ) \end {gather*}

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))

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Sympy [A]
time = 2.00, size = 53, normalized size = 1.51 \begin {gather*} \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} \end {gather*}

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

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Giac [A]
time = 4.22, size = 33, normalized size = 0.94 \begin {gather*} \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 ) \end {gather*}

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))

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Mupad [B]
time = 0.49, size = 27, normalized size = 0.77 \begin {gather*} \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} \end {gather*}

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

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