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

Optimal. Leaf size=28 \[ -\frac {15}{2 x^{10/3}}-\frac {15}{2 x^{4/3}}+\frac {3 x^{2/3}}{2} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {276} \begin {gather*} \frac {3 x^{2/3}}{2}-\frac {15}{2 x^{4/3}}-\frac {15}{2 x^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\left (5+x^2\right )^2}{x^{13/3}} \, dx &=\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*}

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Mathematica [A]
time = 0.01, size = 19, normalized size = 0.68 \begin {gather*} \frac {3 \left (-5-5 x^2+x^4\right )}{2 x^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + x^2)^2/x^(13/3),x]

[Out]

(3*(-5 - 5*x^2 + x^4))/(2*x^(10/3))

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Maple [A]
time = 0.06, size = 17, normalized size = 0.61

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 \(-\frac {15}{2 x^{\frac {10}{3}}}-\frac {15}{2 x^{\frac {4}{3}}}+\frac {3 x^{\frac {2}{3}}}{2}\) \(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 1.52, size = 16, normalized size = 0.57 \begin {gather*} \frac {3}{2} \, x^{\frac {2}{3}} - \frac {15 \, {\left (x^{2} + 1\right )}}{2 \, x^{\frac {10}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.89, size = 15, normalized size = 0.54 \begin {gather*} \frac {3 \, {\left (x^{4} - 5 \, x^{2} - 5\right )}}{2 \, x^{\frac {10}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(x^4 - 5*x^2 - 5)/x^(10/3)

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Sympy [A]
time = 1.09, size = 24, normalized size = 0.86 \begin {gather*} \frac {3 x^{\frac {2}{3}}}{2} - \frac {15}{2 x^{\frac {4}{3}}} - \frac {15}{2 x^{\frac {10}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]
time = 0.67, size = 16, normalized size = 0.57 \begin {gather*} \frac {3}{2} \, x^{\frac {2}{3}} - \frac {15 \, {\left (x^{2} + 1\right )}}{2 \, x^{\frac {10}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 0.27, size = 17, normalized size = 0.61 \begin {gather*} -\frac {-3\,x^4+15\,x^2+15}{2\,x^{10/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 5)^2/x^(13/3),x)

[Out]

-(15*x^2 - 3*x^4 + 15)/(2*x^(10/3))

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