3.301 \(\int x^{5/2} (b x^2+c x^4)^2 \, dx\)

Optimal. Leaf size=36 \[ \frac{2}{15} b^2 x^{15/2}+\frac{4}{19} b c x^{19/2}+\frac{2}{23} c^2 x^{23/2} \]

[Out]

(2*b^2*x^(15/2))/15 + (4*b*c*x^(19/2))/19 + (2*c^2*x^(23/2))/23

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Rubi [A]  time = 0.0179989, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {1584, 270} \[ \frac{2}{15} b^2 x^{15/2}+\frac{4}{19} b c x^{19/2}+\frac{2}{23} c^2 x^{23/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(b*x^2 + c*x^4)^2,x]

[Out]

(2*b^2*x^(15/2))/15 + (4*b*c*x^(19/2))/19 + (2*c^2*x^(23/2))/23

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 270

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 x^{5/2} \left (b x^2+c x^4\right )^2 \, dx &=\int x^{13/2} \left (b+c x^2\right )^2 \, dx\\ &=\int \left (b^2 x^{13/2}+2 b c x^{17/2}+c^2 x^{21/2}\right ) \, dx\\ &=\frac{2}{15} b^2 x^{15/2}+\frac{4}{19} b c x^{19/2}+\frac{2}{23} c^2 x^{23/2}\\ \end{align*}

Mathematica [A]  time = 0.0077621, size = 30, normalized size = 0.83 \[ \frac{2 x^{15/2} \left (437 b^2+690 b c x^2+285 c^2 x^4\right )}{6555} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(b*x^2 + c*x^4)^2,x]

[Out]

(2*x^(15/2)*(437*b^2 + 690*b*c*x^2 + 285*c^2*x^4))/6555

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Maple [A]  time = 0.045, size = 27, normalized size = 0.8 \begin{align*}{\frac{570\,{c}^{2}{x}^{4}+1380\,bc{x}^{2}+874\,{b}^{2}}{6555}{x}^{{\frac{15}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(c*x^4+b*x^2)^2,x)

[Out]

2/6555*x^(15/2)*(285*c^2*x^4+690*b*c*x^2+437*b^2)

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Maxima [A]  time = 0.969372, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{23} \, c^{2} x^{\frac{23}{2}} + \frac{4}{19} \, b c x^{\frac{19}{2}} + \frac{2}{15} \, b^{2} x^{\frac{15}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

2/23*c^2*x^(23/2) + 4/19*b*c*x^(19/2) + 2/15*b^2*x^(15/2)

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Fricas [A]  time = 1.26604, size = 80, normalized size = 2.22 \begin{align*} \frac{2}{6555} \,{\left (285 \, c^{2} x^{11} + 690 \, b c x^{9} + 437 \, b^{2} x^{7}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

2/6555*(285*c^2*x^11 + 690*b*c*x^9 + 437*b^2*x^7)*sqrt(x)

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Sympy [A]  time = 24.1289, size = 34, normalized size = 0.94 \begin{align*} \frac{2 b^{2} x^{\frac{15}{2}}}{15} + \frac{4 b c x^{\frac{19}{2}}}{19} + \frac{2 c^{2} x^{\frac{23}{2}}}{23} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(c*x**4+b*x**2)**2,x)

[Out]

2*b**2*x**(15/2)/15 + 4*b*c*x**(19/2)/19 + 2*c**2*x**(23/2)/23

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Giac [A]  time = 1.12866, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{23} \, c^{2} x^{\frac{23}{2}} + \frac{4}{19} \, b c x^{\frac{19}{2}} + \frac{2}{15} \, b^{2} x^{\frac{15}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

2/23*c^2*x^(23/2) + 4/19*b*c*x^(19/2) + 2/15*b^2*x^(15/2)