3.305 \(\int \frac{(b x^2+c x^4)^2}{x^{3/2}} \, dx\)

Optimal. Leaf size=36 \[ \frac{2}{7} b^2 x^{7/2}+\frac{4}{11} b c x^{11/2}+\frac{2}{15} c^2 x^{15/2} \]

[Out]

(2*b^2*x^(7/2))/7 + (4*b*c*x^(11/2))/11 + (2*c^2*x^(15/2))/15

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Rubi [A]  time = 0.0157798, 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} \[ \frac{2}{7} b^2 x^{7/2}+\frac{4}{11} b c x^{11/2}+\frac{2}{15} c^2 x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^2/x^(3/2),x]

[Out]

(2*b^2*x^(7/2))/7 + (4*b*c*x^(11/2))/11 + (2*c^2*x^(15/2))/15

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 \frac{\left (b x^2+c x^4\right )^2}{x^{3/2}} \, dx &=\int x^{5/2} \left (b+c x^2\right )^2 \, dx\\ &=\int \left (b^2 x^{5/2}+2 b c x^{9/2}+c^2 x^{13/2}\right ) \, dx\\ &=\frac{2}{7} b^2 x^{7/2}+\frac{4}{11} b c x^{11/2}+\frac{2}{15} c^2 x^{15/2}\\ \end{align*}

Mathematica [A]  time = 0.0080964, size = 30, normalized size = 0.83 \[ \frac{2 x^{7/2} \left (165 b^2+210 b c x^2+77 c^2 x^4\right )}{1155} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^2/x^(3/2),x]

[Out]

(2*x^(7/2)*(165*b^2 + 210*b*c*x^2 + 77*c^2*x^4))/1155

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Maple [A]  time = 0.046, size = 27, normalized size = 0.8 \begin{align*}{\frac{154\,{c}^{2}{x}^{4}+420\,bc{x}^{2}+330\,{b}^{2}}{1155}{x}^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^2/x^(3/2),x)

[Out]

2/1155*x^(7/2)*(77*c^2*x^4+210*b*c*x^2+165*b^2)

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Maxima [A]  time = 0.985145, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{15} \, c^{2} x^{\frac{15}{2}} + \frac{4}{11} \, b c x^{\frac{11}{2}} + \frac{2}{7} \, b^{2} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^2/x^(3/2),x, algorithm="maxima")

[Out]

2/15*c^2*x^(15/2) + 4/11*b*c*x^(11/2) + 2/7*b^2*x^(7/2)

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Fricas [A]  time = 1.22613, size = 77, normalized size = 2.14 \begin{align*} \frac{2}{1155} \,{\left (77 \, c^{2} x^{7} + 210 \, b c x^{5} + 165 \, b^{2} x^{3}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^2/x^(3/2),x, algorithm="fricas")

[Out]

2/1155*(77*c^2*x^7 + 210*b*c*x^5 + 165*b^2*x^3)*sqrt(x)

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Sympy [A]  time = 6.09704, size = 34, normalized size = 0.94 \begin{align*} \frac{2 b^{2} x^{\frac{7}{2}}}{7} + \frac{4 b c x^{\frac{11}{2}}}{11} + \frac{2 c^{2} x^{\frac{15}{2}}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**2/x**(3/2),x)

[Out]

2*b**2*x**(7/2)/7 + 4*b*c*x**(11/2)/11 + 2*c**2*x**(15/2)/15

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Giac [A]  time = 1.18949, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{15} \, c^{2} x^{\frac{15}{2}} + \frac{4}{11} \, b c x^{\frac{11}{2}} + \frac{2}{7} \, b^{2} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^2/x^(3/2),x, algorithm="giac")

[Out]

2/15*c^2*x^(15/2) + 4/11*b*c*x^(11/2) + 2/7*b^2*x^(7/2)