\(\int \frac {(5+x^2)^2}{x^{13/3}} \, dx\) [471]

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=-\frac {15}{2 x^{10/3}}-\frac {15}{2 x^{4/3}}+\frac {3 x^{2/3}}{2} \]

[Out]

-15/2/x^(10/3)-15/2/x^(4/3)+3/2*x^(2/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {276} \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=\frac {3 x^{2/3}}{2}-\frac {15}{2 x^{4/3}}-\frac {15}{2 x^{10/3}} \]

[In]

Int[(5 + x^2)^2/x^(13/3),x]

[Out]

-15/(2*x^(10/3)) - 15/(2*x^(4/3)) + (3*x^(2/3))/2

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {25}{x^{13/3}}+\frac {10}{x^{7/3}}+\frac {1}{\sqrt [3]{x}}\right ) \, dx \\ & = -\frac {15}{2 x^{10/3}}-\frac {15}{2 x^{4/3}}+\frac {3 x^{2/3}}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=\frac {3 \left (-5-5 x^2+x^4\right )}{2 x^{10/3}} \]

[In]

Integrate[(5 + x^2)^2/x^(13/3),x]

[Out]

(3*(-5 - 5*x^2 + x^4))/(2*x^(10/3))

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.57

method	result	size
gosper	\(\frac {\frac {3}{2} x^{4}-\frac {15}{2}-\frac {15}{2} x^{2}}{x^{\frac {10}{3}}}\)	\(16\)
trager	\(\frac {\frac {3}{2} x^{4}-\frac {15}{2}-\frac {15}{2} x^{2}}{x^{\frac {10}{3}}}\)	\(16\)
risch	\(\frac {\frac {3}{2} x^{4}-\frac {15}{2}-\frac {15}{2} x^{2}}{x^{\frac {10}{3}}}\)	\(16\)
derivativedivides	\(-\frac {15}{2 x^{\frac {10}{3}}}-\frac {15}{2 x^{\frac {4}{3}}}+\frac {3 x^{\frac {2}{3}}}{2}\)	\(17\)
default	\(-\frac {15}{2 x^{\frac {10}{3}}}-\frac {15}{2 x^{\frac {4}{3}}}+\frac {3 x^{\frac {2}{3}}}{2}\)	\(17\)

[In]

int((x^2+5)^2/x^(13/3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=\frac {3 \, {\left (x^{4} - 5 \, x^{2} - 5\right )}}{2 \, x^{\frac {10}{3}}} \]

[In]

integrate((x^2+5)^2/x^(13/3),x, algorithm="fricas")

[Out]

3/2*(x^4 - 5*x^2 - 5)/x^(10/3)

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=\frac {3 x^{\frac {2}{3}}}{2} - \frac {15}{2 x^{\frac {4}{3}}} - \frac {15}{2 x^{\frac {10}{3}}} \]

[In]

integrate((x**2+5)**2/x**(13/3),x)

[Out]

3*x**(2/3)/2 - 15/(2*x**(4/3)) - 15/(2*x**(10/3))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.57 \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=\frac {3}{2} \, x^{\frac {2}{3}} - \frac {15 \, {\left (x^{2} + 1\right )}}{2 \, x^{\frac {10}{3}}} \]

[In]

integrate((x^2+5)^2/x^(13/3),x, algorithm="maxima")

[Out]

3/2*x^(2/3) - 15/2*(x^2 + 1)/x^(10/3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.57 \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=\frac {3}{2} \, x^{\frac {2}{3}} - \frac {15 \, {\left (x^{2} + 1\right )}}{2 \, x^{\frac {10}{3}}} \]

[In]

integrate((x^2+5)^2/x^(13/3),x, algorithm="giac")

[Out]

3/2*x^(2/3) - 15/2*(x^2 + 1)/x^(10/3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx=-\frac {-3\,x^4+15\,x^2+15}{2\,x^{10/3}} \]

[In]

int((x^2 + 5)^2/x^(13/3),x)

[Out]

-(15*x^2 - 3*x^4 + 15)/(2*x^(10/3))