3.30 \(\int x^7 (a+b x^2)^3 \, dx\)

Optimal. Leaf size=43 \[ \frac{3}{10} a^2 b x^{10}+\frac{a^3 x^8}{8}+\frac{1}{4} a b^2 x^{12}+\frac{b^3 x^{14}}{14} \]

[Out]

(a^3*x^8)/8 + (3*a^2*b*x^10)/10 + (a*b^2*x^12)/4 + (b^3*x^14)/14

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Rubi [A]  time = 0.0268245, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac{3}{10} a^2 b x^{10}+\frac{a^3 x^8}{8}+\frac{1}{4} a b^2 x^{12}+\frac{b^3 x^{14}}{14} \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*x^2)^3,x]

[Out]

(a^3*x^8)/8 + (3*a^2*b*x^10)/10 + (a*b^2*x^12)/4 + (b^3*x^14)/14

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^7 \left (a+b x^2\right )^3 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^3 (a+b x)^3 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^3 x^3+3 a^2 b x^4+3 a b^2 x^5+b^3 x^6\right ) \, dx,x,x^2\right )\\ &=\frac{a^3 x^8}{8}+\frac{3}{10} a^2 b x^{10}+\frac{1}{4} a b^2 x^{12}+\frac{b^3 x^{14}}{14}\\ \end{align*}

Mathematica [A]  time = 0.0015993, size = 43, normalized size = 1. \[ \frac{3}{10} a^2 b x^{10}+\frac{a^3 x^8}{8}+\frac{1}{4} a b^2 x^{12}+\frac{b^3 x^{14}}{14} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a + b*x^2)^3,x]

[Out]

(a^3*x^8)/8 + (3*a^2*b*x^10)/10 + (a*b^2*x^12)/4 + (b^3*x^14)/14

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Maple [A]  time = 0.001, size = 36, normalized size = 0.8 \begin{align*}{\frac{{a}^{3}{x}^{8}}{8}}+{\frac{3\,{a}^{2}b{x}^{10}}{10}}+{\frac{a{b}^{2}{x}^{12}}{4}}+{\frac{{b}^{3}{x}^{14}}{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(b*x^2+a)^3,x)

[Out]

1/8*a^3*x^8+3/10*a^2*b*x^10+1/4*a*b^2*x^12+1/14*b^3*x^14

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Maxima [A]  time = 1.19351, size = 47, normalized size = 1.09 \begin{align*} \frac{1}{14} \, b^{3} x^{14} + \frac{1}{4} \, a b^{2} x^{12} + \frac{3}{10} \, a^{2} b x^{10} + \frac{1}{8} \, a^{3} x^{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*b^3*x^14 + 1/4*a*b^2*x^12 + 3/10*a^2*b*x^10 + 1/8*a^3*x^8

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Fricas [A]  time = 1.31896, size = 86, normalized size = 2. \begin{align*} \frac{1}{14} x^{14} b^{3} + \frac{1}{4} x^{12} b^{2} a + \frac{3}{10} x^{10} b a^{2} + \frac{1}{8} x^{8} a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^2+a)^3,x, algorithm="fricas")

[Out]

1/14*x^14*b^3 + 1/4*x^12*b^2*a + 3/10*x^10*b*a^2 + 1/8*x^8*a^3

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Sympy [A]  time = 0.064422, size = 37, normalized size = 0.86 \begin{align*} \frac{a^{3} x^{8}}{8} + \frac{3 a^{2} b x^{10}}{10} + \frac{a b^{2} x^{12}}{4} + \frac{b^{3} x^{14}}{14} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(b*x**2+a)**3,x)

[Out]

a**3*x**8/8 + 3*a**2*b*x**10/10 + a*b**2*x**12/4 + b**3*x**14/14

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Giac [A]  time = 2.01414, size = 47, normalized size = 1.09 \begin{align*} \frac{1}{14} \, b^{3} x^{14} + \frac{1}{4} \, a b^{2} x^{12} + \frac{3}{10} \, a^{2} b x^{10} + \frac{1}{8} \, a^{3} x^{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*b^3*x^14 + 1/4*a*b^2*x^12 + 3/10*a^2*b*x^10 + 1/8*a^3*x^8