\(\int \frac {x^5}{(1+x^4)^{3/2}} \, dx\) [943]

Optimal result

Integrand size = 13, antiderivative size = 25 \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^2}{2 \sqrt {1+x^4}}+\frac {\text {arcsinh}\left (x^2\right )}{2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 294, 221} \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=\frac {\text {arcsinh}\left (x^2\right )}{2}-\frac {x^2}{2 \sqrt {x^4+1}} \]

[In]

Int[x^5/(1 + x^4)^(3/2),x]

[Out]

-1/2*x^2/Sqrt[1 + x^4] + ArcSinh[x^2]/2

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {x^2}{2 \sqrt {1+x^4}}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^2\right ) \\ & = -\frac {x^2}{2 \sqrt {1+x^4}}+\frac {1}{2} \sinh ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^2}{2 \sqrt {1+x^4}}+\frac {1}{2} \log \left (x^2+\sqrt {1+x^4}\right ) \]

[In]

Integrate[x^5/(1 + x^4)^(3/2),x]

[Out]

-1/2*x^2/Sqrt[1 + x^4] + Log[x^2 + Sqrt[1 + x^4]]/2

Maple [A] (verified)

Time = 4.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {x^{2}}{2 \sqrt {x^{4}+1}}\)	\(20\)
risch	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {x^{2}}{2 \sqrt {x^{4}+1}}\)	\(20\)
elliptic	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {x^{2}}{2 \sqrt {x^{4}+1}}\)	\(20\)
pseudoelliptic	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {x^{2}}{2 \sqrt {x^{4}+1}}\)	\(20\)
trager	\(-\frac {x^{2}}{2 \sqrt {x^{4}+1}}+\frac {\ln \left (x^{2}+\sqrt {x^{4}+1}\right )}{2}\)	\(28\)
meijerg	\(\frac {-\frac {\sqrt {\pi }\, x^{2}}{\sqrt {x^{4}+1}}+\sqrt {\pi }\, \operatorname {arcsinh}\left (x^{2}\right )}{2 \sqrt {\pi }}\)	\(30\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^{4} + \sqrt {x^{4} + 1} x^{2} + {\left (x^{4} + 1\right )} \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) + 1}{2 \, {\left (x^{4} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=- \frac {x^{2}}{2 \sqrt {x^{4} + 1}} + \frac {\operatorname {asinh}{\left (x^{2} \right )}}{2} \]

[In]

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

[Out]

-x**2/(2*sqrt(x**4 + 1)) + asinh(x**2)/2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (19) = 38\).

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^{2}}{2 \, \sqrt {x^{4} + 1}} + \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} + 1\right ) - \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^{2}}{2 \, \sqrt {x^{4} + 1}} - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5}{\left (1+x^4\right )^{3/2}} \, dx=\int \frac {x^5}{{\left (x^4+1\right )}^{3/2}} \,d x \]

[In]

int(x^5/(x^4 + 1)^(3/2),x)

[Out]

int(x^5/(x^4 + 1)^(3/2), x)