\(\int \frac {x^9}{(1+x^4)^{3/2}} \, dx\) [942]

Optimal result

Integrand size = 13, antiderivative size = 41 \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {x^6}{2 \sqrt {1+x^4}}+\frac {3}{4} x^2 \sqrt {1+x^4}-\frac {3 \text {arcsinh}\left (x^2\right )}{4} \]

[Out]

-3/4*arcsinh(x^2)-1/2*x^6/(x^4+1)^(1/2)+3/4*x^2*(x^4+1)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 294, 327, 221} \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {3 \text {arcsinh}\left (x^2\right )}{4}-\frac {x^6}{2 \sqrt {x^4+1}}+\frac {3}{4} \sqrt {x^4+1} x^2 \]

[In]

Int[x^9/(1 + x^4)^(3/2),x]

[Out]

-1/2*x^6/Sqrt[1 + x^4] + (3*x^2*Sqrt[1 + x^4])/4 - (3*ArcSinh[x^2])/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 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]

Rule 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^{3/2}} \, dx,x,x^2\right ) \\ & = -\frac {x^6}{2 \sqrt {1+x^4}}+\frac {3}{2} \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2}} \, dx,x,x^2\right ) \\ & = -\frac {x^6}{2 \sqrt {1+x^4}}+\frac {3}{4} x^2 \sqrt {1+x^4}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,x^2\right ) \\ & = -\frac {x^6}{2 \sqrt {1+x^4}}+\frac {3}{4} x^2 \sqrt {1+x^4}-\frac {3}{4} \sinh ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=\frac {3 x^2+x^6}{4 \sqrt {1+x^4}}-\frac {3}{4} \log \left (x^2+\sqrt {1+x^4}\right ) \]

[In]

Integrate[x^9/(1 + x^4)^(3/2),x]

[Out]

(3*x^2 + x^6)/(4*Sqrt[1 + x^4]) - (3*Log[x^2 + Sqrt[1 + x^4]])/4

Maple [A] (verified)

Time = 4.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {x^{2} \left (x^{4}+3\right )}{4 \sqrt {x^{4}+1}}-\frac {3 \,\operatorname {arcsinh}\left (x^{2}\right )}{4}\)	\(25\)
default	\(\frac {x^{6}}{4 \sqrt {x^{4}+1}}+\frac {3 x^{2}}{4 \sqrt {x^{4}+1}}-\frac {3 \,\operatorname {arcsinh}\left (x^{2}\right )}{4}\)	\(32\)
elliptic	\(\frac {x^{6}}{4 \sqrt {x^{4}+1}}+\frac {3 x^{2}}{4 \sqrt {x^{4}+1}}-\frac {3 \,\operatorname {arcsinh}\left (x^{2}\right )}{4}\)	\(32\)
pseudoelliptic	\(\frac {x^{6}-3 \,\operatorname {arcsinh}\left (x^{2}\right ) \sqrt {x^{4}+1}+3 x^{2}}{4 \sqrt {x^{4}+1}}\)	\(32\)
trager	\(\frac {x^{2} \left (x^{4}+3\right )}{4 \sqrt {x^{4}+1}}+\frac {3 \ln \left (x^{2}-\sqrt {x^{4}+1}\right )}{4}\)	\(35\)
meijerg	\(\frac {\frac {\sqrt {\pi }\, x^{2} \left (5 x^{4}+15\right )}{10 \sqrt {x^{4}+1}}-\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{2}\right )}{2}}{2 \sqrt {\pi }}\)	\(38\)

[In]

int(x^9/(x^4+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.32 \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=\frac {2 \, x^{4} + 3 \, {\left (x^{4} + 1\right )} \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) + {\left (x^{6} + 3 \, x^{2}\right )} \sqrt {x^{4} + 1} + 2}{4 \, {\left (x^{4} + 1\right )}} \]

[In]

integrate(x^9/(x^4+1)^(3/2),x, algorithm="fricas")

[Out]

1/4*(2*x^4 + 3*(x^4 + 1)*log(-x^2 + sqrt(x^4 + 1)) + (x^6 + 3*x^2)*sqrt(x^4 + 1) + 2)/(x^4 + 1)

Sympy [A] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=\frac {x^{6}}{4 \sqrt {x^{4} + 1}} + \frac {3 x^{2}}{4 \sqrt {x^{4} + 1}} - \frac {3 \operatorname {asinh}{\left (x^{2} \right )}}{4} \]

[In]

integrate(x**9/(x**4+1)**(3/2),x)

[Out]

x**6/(4*sqrt(x**4 + 1)) + 3*x**2/(4*sqrt(x**4 + 1)) - 3*asinh(x**2)/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (31) = 62\).

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.78 \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=-\frac {\frac {3 \, {\left (x^{4} + 1\right )}}{x^{4}} - 2}{4 \, {\left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - \frac {{\left (x^{4} + 1\right )}^{\frac {3}{2}}}{x^{6}}\right )}} - \frac {3}{8} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} + 1\right ) + \frac {3}{8} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - 1\right ) \]

[In]

integrate(x^9/(x^4+1)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(3*(x^4 + 1)/x^4 - 2)/(sqrt(x^4 + 1)/x^2 - (x^4 + 1)^(3/2)/x^6) - 3/8*log(sqrt(x^4 + 1)/x^2 + 1) + 3/8*lo
g(sqrt(x^4 + 1)/x^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=\frac {{\left (x^{4} + 3\right )} x^{2}}{4 \, \sqrt {x^{4} + 1}} + \frac {3}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) \]

[In]

integrate(x^9/(x^4+1)^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^9}{\left (1+x^4\right )^{3/2}} \, dx=\int \frac {x^9}{{\left (x^4+1\right )}^{3/2}} \,d x \]

[In]

int(x^9/(x^4 + 1)^(3/2),x)

[Out]

int(x^9/(x^4 + 1)^(3/2), x)