3.315 \(\int \frac{(b x^2+c x^4)^3}{x^{7/2}} \, dx\)

Optimal. Leaf size=51 \[ \frac{6}{11} b^2 c x^{11/2}+\frac{2}{7} b^3 x^{7/2}+\frac{2}{5} b c^2 x^{15/2}+\frac{2}{19} c^3 x^{19/2} \]

[Out]

(2*b^3*x^(7/2))/7 + (6*b^2*c*x^(11/2))/11 + (2*b*c^2*x^(15/2))/5 + (2*c^3*x^(19/2))/19

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Rubi [A]  time = 0.0222626, antiderivative size = 51, 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} \[ \frac{6}{11} b^2 c x^{11/2}+\frac{2}{7} b^3 x^{7/2}+\frac{2}{5} b c^2 x^{15/2}+\frac{2}{19} c^3 x^{19/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^(7/2),x]

[Out]

(2*b^3*x^(7/2))/7 + (6*b^2*c*x^(11/2))/11 + (2*b*c^2*x^(15/2))/5 + (2*c^3*x^(19/2))/19

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 )^3}{x^{7/2}} \, dx &=\int x^{5/2} \left (b+c x^2\right )^3 \, dx\\ &=\int \left (b^3 x^{5/2}+3 b^2 c x^{9/2}+3 b c^2 x^{13/2}+c^3 x^{17/2}\right ) \, dx\\ &=\frac{2}{7} b^3 x^{7/2}+\frac{6}{11} b^2 c x^{11/2}+\frac{2}{5} b c^2 x^{15/2}+\frac{2}{19} c^3 x^{19/2}\\ \end{align*}

Mathematica [A]  time = 0.0097158, size = 41, normalized size = 0.8 \[ \frac{2 x^{7/2} \left (1995 b^2 c x^2+1045 b^3+1463 b c^2 x^4+385 c^3 x^6\right )}{7315} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^3/x^(7/2),x]

[Out]

(2*x^(7/2)*(1045*b^3 + 1995*b^2*c*x^2 + 1463*b*c^2*x^4 + 385*c^3*x^6))/7315

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Maple [A]  time = 0.046, size = 38, normalized size = 0.8 \begin{align*}{\frac{770\,{c}^{3}{x}^{6}+2926\,b{c}^{2}{x}^{4}+3990\,{b}^{2}c{x}^{2}+2090\,{b}^{3}}{7315}{x}^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^3/x^(7/2),x)

[Out]

2/7315*x^(7/2)*(385*c^3*x^6+1463*b*c^2*x^4+1995*b^2*c*x^2+1045*b^3)

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Maxima [A]  time = 0.96881, size = 47, normalized size = 0.92 \begin{align*} \frac{2}{19} \, c^{3} x^{\frac{19}{2}} + \frac{2}{5} \, b c^{2} x^{\frac{15}{2}} + \frac{6}{11} \, b^{2} c x^{\frac{11}{2}} + \frac{2}{7} \, b^{3} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^(7/2),x, algorithm="maxima")

[Out]

2/19*c^3*x^(19/2) + 2/5*b*c^2*x^(15/2) + 6/11*b^2*c*x^(11/2) + 2/7*b^3*x^(7/2)

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Fricas [A]  time = 1.21656, size = 107, normalized size = 2.1 \begin{align*} \frac{2}{7315} \,{\left (385 \, c^{3} x^{9} + 1463 \, b c^{2} x^{7} + 1995 \, b^{2} c x^{5} + 1045 \, b^{3} 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)^3/x^(7/2),x, algorithm="fricas")

[Out]

2/7315*(385*c^3*x^9 + 1463*b*c^2*x^7 + 1995*b^2*c*x^5 + 1045*b^3*x^3)*sqrt(x)

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Sympy [A]  time = 28.3509, size = 49, normalized size = 0.96 \begin{align*} \frac{2 b^{3} x^{\frac{7}{2}}}{7} + \frac{6 b^{2} c x^{\frac{11}{2}}}{11} + \frac{2 b c^{2} x^{\frac{15}{2}}}{5} + \frac{2 c^{3} x^{\frac{19}{2}}}{19} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**3/x**(7/2),x)

[Out]

2*b**3*x**(7/2)/7 + 6*b**2*c*x**(11/2)/11 + 2*b*c**2*x**(15/2)/5 + 2*c**3*x**(19/2)/19

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Giac [A]  time = 1.115, size = 47, normalized size = 0.92 \begin{align*} \frac{2}{19} \, c^{3} x^{\frac{19}{2}} + \frac{2}{5} \, b c^{2} x^{\frac{15}{2}} + \frac{6}{11} \, b^{2} c x^{\frac{11}{2}} + \frac{2}{7} \, b^{3} x^{\frac{7}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^3/x^(7/2),x, algorithm="giac")

[Out]

2/19*c^3*x^(19/2) + 2/5*b*c^2*x^(15/2) + 6/11*b^2*c*x^(11/2) + 2/7*b^3*x^(7/2)