\(\int \frac {x^9}{a+c x^4} \, dx\) [644]

Optimal result

Integrand size = 13, antiderivative size = 51 \[ \int \frac {x^9}{a+c x^4} \, dx=-\frac {a x^2}{2 c^2}+\frac {x^6}{6 c}+\frac {a^{3/2} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 c^{5/2}} \]

[Out]

-1/2*a*x^2/c^2+1/6*x^6/c+1/2*a^(3/2)*arctan(x^2*c^(1/2)/a^(1/2))/c^(5/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 308, 211} \[ \int \frac {x^9}{a+c x^4} \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 c^{5/2}}-\frac {a x^2}{2 c^2}+\frac {x^6}{6 c} \]

[In]

Int[x^9/(a + c*x^4),x]

[Out]

-1/2*(a*x^2)/c^2 + x^6/(6*c) + (a^(3/2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*c^(5/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^4}{a+c x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a}{c^2}+\frac {x^2}{c}+\frac {a^2}{c^2 \left (a+c x^2\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a x^2}{2 c^2}+\frac {x^6}{6 c}+\frac {a^2 \text {Subst}\left (\int \frac {1}{a+c x^2} \, dx,x,x^2\right )}{2 c^2} \\ & = -\frac {a x^2}{2 c^2}+\frac {x^6}{6 c}+\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int \frac {x^9}{a+c x^4} \, dx=\frac {1}{6} \left (\frac {-3 a x^2+c x^6}{c^2}+\frac {3 a^{3/2} \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{c^{5/2}}\right ) \]

[In]

Integrate[x^9/(a + c*x^4),x]

[Out]

((-3*a*x^2 + c*x^6)/c^2 + (3*a^(3/2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/c^(5/2))/6

Maple [A] (verified)

Time = 3.94 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {-\frac {1}{3} c \,x^{6}+a \,x^{2}}{2 c^{2}}+\frac {a^{2} \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 c^{2} \sqrt {a c}}\)	\(43\)
risch	\(\frac {x^{6}}{6 c}-\frac {a \,x^{2}}{2 c^{2}}+\frac {\sqrt {-a c}\, a \ln \left (c \,x^{2}+\sqrt {-a c}\right )}{4 c^{3}}-\frac {\sqrt {-a c}\, a \ln \left (c \,x^{2}-\sqrt {-a c}\right )}{4 c^{3}}\)	\(71\)

[In]

int(x^9/(c*x^4+a),x,method=_RETURNVERBOSE)

[Out]

-1/2/c^2*(-1/3*c*x^6+a*x^2)+1/2*a^2/c^2/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.10 \[ \int \frac {x^9}{a+c x^4} \, dx=\left [\frac {2 \, c x^{6} - 6 \, a x^{2} + 3 \, a \sqrt {-\frac {a}{c}} \log \left (\frac {c x^{4} + 2 \, c x^{2} \sqrt {-\frac {a}{c}} - a}{c x^{4} + a}\right )}{12 \, c^{2}}, \frac {c x^{6} - 3 \, a x^{2} + 3 \, a \sqrt {\frac {a}{c}} \arctan \left (\frac {c x^{2} \sqrt {\frac {a}{c}}}{a}\right )}{6 \, c^{2}}\right ] \]

[In]

integrate(x^9/(c*x^4+a),x, algorithm="fricas")

[Out]

[1/12*(2*c*x^6 - 6*a*x^2 + 3*a*sqrt(-a/c)*log((c*x^4 + 2*c*x^2*sqrt(-a/c) - a)/(c*x^4 + a)))/c^2, 1/6*(c*x^6 -
3*a*x^2 + 3*a*sqrt(a/c)*arctan(c*x^2*sqrt(a/c)/a))/c^2]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (42) = 84\).

Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.71 \[ \int \frac {x^9}{a+c x^4} \, dx=- \frac {a x^{2}}{2 c^{2}} - \frac {\sqrt {- \frac {a^{3}}{c^{5}}} \log {\left (x^{2} - \frac {c^{2} \sqrt {- \frac {a^{3}}{c^{5}}}}{a} \right )}}{4} + \frac {\sqrt {- \frac {a^{3}}{c^{5}}} \log {\left (x^{2} + \frac {c^{2} \sqrt {- \frac {a^{3}}{c^{5}}}}{a} \right )}}{4} + \frac {x^{6}}{6 c} \]

[In]

integrate(x**9/(c*x**4+a),x)

[Out]

-a*x**2/(2*c**2) - sqrt(-a**3/c**5)*log(x**2 - c**2*sqrt(-a**3/c**5)/a)/4 + sqrt(-a**3/c**5)*log(x**2 + c**2*s
qrt(-a**3/c**5)/a)/4 + x**6/(6*c)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {x^9}{a+c x^4} \, dx=\frac {a^{2} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} c^{2}} + \frac {c x^{6} - 3 \, a x^{2}}{6 \, c^{2}} \]

[In]

integrate(x^9/(c*x^4+a),x, algorithm="maxima")

[Out]

1/2*a^2*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/6*(c*x^6 - 3*a*x^2)/c^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.88 \[ \int \frac {x^9}{a+c x^4} \, dx=\frac {a^{2} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} c^{2}} + \frac {c^{2} x^{6} - 3 \, a c x^{2}}{6 \, c^{3}} \]

[In]

integrate(x^9/(c*x^4+a),x, algorithm="giac")

[Out]

1/2*a^2*arctan(c*x^2/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/6*(c^2*x^6 - 3*a*c*x^2)/c^3

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.73 \[ \int \frac {x^9}{a+c x^4} \, dx=\frac {x^6}{6\,c}-\frac {a\,x^2}{2\,c^2}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x^2}{\sqrt {a}}\right )}{2\,c^{5/2}} \]

[In]

int(x^9/(a + c*x^4),x)

[Out]

x^6/(6*c) - (a*x^2)/(2*c^2) + (a^(3/2)*atan((c^(1/2)*x^2)/a^(1/2)))/(2*c^(5/2))