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

Optimal. Leaf size=36 \[ \frac{2}{17} b^2 x^{17/2}+\frac{4}{21} b c x^{21/2}+\frac{2}{25} c^2 x^{25/2} \]

[Out]

(2*b^2*x^(17/2))/17 + (4*b*c*x^(21/2))/21 + (2*c^2*x^(25/2))/25

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Rubi [A]  time = 0.0160419, 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} \[ \frac{2}{17} b^2 x^{17/2}+\frac{4}{21} b c x^{21/2}+\frac{2}{25} c^2 x^{25/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*(b*x^2 + c*x^4)^2,x]

[Out]

(2*b^2*x^(17/2))/17 + (4*b*c*x^(21/2))/21 + (2*c^2*x^(25/2))/25

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^{7/2} \left (b x^2+c x^4\right )^2 \, dx &=\int x^{15/2} \left (b+c x^2\right )^2 \, dx\\ &=\int \left (b^2 x^{15/2}+2 b c x^{19/2}+c^2 x^{23/2}\right ) \, dx\\ &=\frac{2}{17} b^2 x^{17/2}+\frac{4}{21} b c x^{21/2}+\frac{2}{25} c^2 x^{25/2}\\ \end{align*}

Mathematica [A]  time = 0.008169, size = 30, normalized size = 0.83 \[ \frac{2 x^{17/2} \left (525 b^2+850 b c x^2+357 c^2 x^4\right )}{8925} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(17/2)*(525*b^2 + 850*b*c*x^2 + 357*c^2*x^4))/8925

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Maple [A]  time = 0.045, size = 27, normalized size = 0.8 \begin{align*}{\frac{714\,{c}^{2}{x}^{4}+1700\,bc{x}^{2}+1050\,{b}^{2}}{8925}{x}^{{\frac{17}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/8925*x^(17/2)*(357*c^2*x^4+850*b*c*x^2+525*b^2)

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Maxima [A]  time = 0.963226, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{25} \, c^{2} x^{\frac{25}{2}} + \frac{4}{21} \, b c x^{\frac{21}{2}} + \frac{2}{17} \, b^{2} x^{\frac{17}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*c^2*x^(25/2) + 4/21*b*c*x^(21/2) + 2/17*b^2*x^(17/2)

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Fricas [A]  time = 1.2389, size = 81, normalized size = 2.25 \begin{align*} \frac{2}{8925} \,{\left (357 \, c^{2} x^{12} + 850 \, b c x^{10} + 525 \, b^{2} x^{8}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/8925*(357*c^2*x^12 + 850*b*c*x^10 + 525*b^2*x^8)*sqrt(x)

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Sympy [A]  time = 41.1167, size = 34, normalized size = 0.94 \begin{align*} \frac{2 b^{2} x^{\frac{17}{2}}}{17} + \frac{4 b c x^{\frac{21}{2}}}{21} + \frac{2 c^{2} x^{\frac{25}{2}}}{25} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b**2*x**(17/2)/17 + 4*b*c*x**(21/2)/21 + 2*c**2*x**(25/2)/25

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Giac [A]  time = 1.15506, size = 32, normalized size = 0.89 \begin{align*} \frac{2}{25} \, c^{2} x^{\frac{25}{2}} + \frac{4}{21} \, b c x^{\frac{21}{2}} + \frac{2}{17} \, b^{2} x^{\frac{17}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*c^2*x^(25/2) + 4/21*b*c*x^(21/2) + 2/17*b^2*x^(17/2)