3.1.25 \(\int x^7 (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\)

Optimal. Leaf size=167 \[ \frac {3 a b^2 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {3 a^2 b x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {b^3 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {a^3 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1355, 270} \begin {gather*} \frac {b^3 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {3 a b^2 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {3 a^2 b x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {a^3 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(a^3*x^8*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*(a + b*x^3)) + (3*a^2*b*x^11*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(11
*(a + b*x^3)) + (3*a*b^2*x^14*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(14*(a + b*x^3)) + (b^3*x^17*Sqrt[a^2 + 2*a*b*x
^3 + b^2*x^6])/(17*(a + b*x^3))

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]

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^7 \left (a b+b^2 x^3\right )^3 \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (a^3 b^3 x^7+3 a^2 b^4 x^{10}+3 a b^5 x^{13}+b^6 x^{16}\right ) \, dx}{b^2 \left (a b+b^2 x^3\right )}\\ &=\frac {a^3 x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {3 a^2 b x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {3 a b^2 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {b^3 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.37 \begin {gather*} \frac {x^8 \sqrt {\left (a+b x^3\right )^2} \left (1309 a^3+2856 a^2 b x^3+2244 a b^2 x^6+616 b^3 x^9\right )}{10472 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(x^8*Sqrt[(a + b*x^3)^2]*(1309*a^3 + 2856*a^2*b*x^3 + 2244*a*b^2*x^6 + 616*b^3*x^9))/(10472*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 9.62, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^3\right )^2} \left (1309 a^3 x^8+2856 a^2 b x^{11}+2244 a b^2 x^{14}+616 b^3 x^{17}\right )}{10472 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(Sqrt[(a + b*x^3)^2]*(1309*a^3*x^8 + 2856*a^2*b*x^11 + 2244*a*b^2*x^14 + 616*b^3*x^17))/(10472*(a + b*x^3))

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fricas [A]  time = 1.32, size = 35, normalized size = 0.21 \begin {gather*} \frac {1}{17} \, b^{3} x^{17} + \frac {3}{14} \, a b^{2} x^{14} + \frac {3}{11} \, a^{2} b x^{11} + \frac {1}{8} \, a^{3} x^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/17*b^3*x^17 + 3/14*a*b^2*x^14 + 3/11*a^2*b*x^11 + 1/8*a^3*x^8

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giac [A]  time = 0.29, size = 67, normalized size = 0.40 \begin {gather*} \frac {1}{17} \, b^{3} x^{17} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{14} \, a b^{2} x^{14} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{11} \, a^{2} b x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{8} \, a^{3} x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

1/17*b^3*x^17*sgn(b*x^3 + a) + 3/14*a*b^2*x^14*sgn(b*x^3 + a) + 3/11*a^2*b*x^11*sgn(b*x^3 + a) + 1/8*a^3*x^8*s
gn(b*x^3 + a)

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maple [A]  time = 0.01, size = 58, normalized size = 0.35 \begin {gather*} \frac {\left (616 b^{3} x^{9}+2244 a \,b^{2} x^{6}+2856 a^{2} b \,x^{3}+1309 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}} x^{8}}{10472 \left (b \,x^{3}+a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)

[Out]

1/10472*x^8*(616*b^3*x^9+2244*a*b^2*x^6+2856*a^2*b*x^3+1309*a^3)*((b*x^3+a)^2)^(3/2)/(b*x^3+a)^3

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maxima [A]  time = 0.45, size = 35, normalized size = 0.21 \begin {gather*} \frac {1}{17} \, b^{3} x^{17} + \frac {3}{14} \, a b^{2} x^{14} + \frac {3}{11} \, a^{2} b x^{11} + \frac {1}{8} \, a^{3} x^{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/17*b^3*x^17 + 3/14*a*b^2*x^14 + 3/11*a^2*b*x^11 + 1/8*a^3*x^8

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^7\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)

[Out]

int(x^7*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{7} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

Integral(x**7*((a + b*x**3)**2)**(3/2), x)

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