3.1.28 \(\int x^4 (a^2+2 a b x^3+b^2 x^6)^{3/2} \, dx\) [28]

Optimal. Leaf size=167 \[ \frac {a^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )}+\frac {3 a^2 b x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {3 a b^2 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {b^3 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )} \]

[Out]

1/5*a^3*x^5*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+3/8*a^2*b*x^8*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+3/11*a*b^2*x^11*((b*x^3+
a)^2)^(1/2)/(b*x^3+a)+1/14*b^3*x^14*((b*x^3+a)^2)^(1/2)/(b*x^3+a)

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Rubi [A]
time = 0.03, 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 = {1369, 276} \begin {gather*} \frac {3 a b^2 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {3 a^2 b x^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 \left (a+b x^3\right )}+\frac {b^3 x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {a^3 x^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(a^3*x^5*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*(a + b*x^3)) + (3*a^2*b*x^8*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*(
a + b*x^3)) + (3*a*b^2*x^11*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(11*(a + b*x^3)) + (b^3*x^14*Sqrt[a^2 + 2*a*b*x^3
+ b^2*x^6])/(14*(a + b*x^3))

Rule 276

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 1369

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

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Mathematica [A]
time = 0.01, size = 61, normalized size = 0.37 \begin {gather*} \frac {x^5 \sqrt {\left (a+b x^3\right )^2} \left (616 a^3+1155 a^2 b x^3+840 a b^2 x^6+220 b^3 x^9\right )}{3080 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]

[Out]

(x^5*Sqrt[(a + b*x^3)^2]*(616*a^3 + 1155*a^2*b*x^3 + 840*a*b^2*x^6 + 220*b^3*x^9))/(3080*(a + b*x^3))

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Maple [A]
time = 0.06, size = 58, normalized size = 0.35

method	result	size
gosper	\(\frac {x^{5} \left (220 b^{3} x^{9}+840 a \,b^{2} x^{6}+1155 a^{2} b \,x^{3}+616 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}{3080 \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
default	\(\frac {x^{5} \left (220 b^{3} x^{9}+840 a \,b^{2} x^{6}+1155 a^{2} b \,x^{3}+616 a^{3}\right ) \left (\left (b \,x^{3}+a \right )^{2}\right )^{\frac {3}{2}}}{3080 \left (b \,x^{3}+a \right )^{3}}\)	\(58\)
risch	\(\frac {a^{3} x^{5} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{5 b \,x^{3}+5 a}+\frac {3 a^{2} b \,x^{8} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{8 \left (b \,x^{3}+a \right )}+\frac {3 a \,b^{2} x^{11} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{11 \left (b \,x^{3}+a \right )}+\frac {b^{3} x^{14} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{14 b \,x^{3}+14 a}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/3080*x^5*(220*b^3*x^9+840*a*b^2*x^6+1155*a^2*b*x^3+616*a^3)*((b*x^3+a)^2)^(3/2)/(b*x^3+a)^3

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Maxima [A]
time = 0.28, size = 35, normalized size = 0.21 \begin {gather*} \frac {1}{14} \, b^{3} x^{14} + \frac {3}{11} \, a b^{2} x^{11} + \frac {3}{8} \, a^{2} b x^{8} + \frac {1}{5} \, a^{3} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/14*b^3*x^14 + 3/11*a*b^2*x^11 + 3/8*a^2*b*x^8 + 1/5*a^3*x^5

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Fricas [A]
time = 0.38, size = 35, normalized size = 0.21 \begin {gather*} \frac {1}{14} \, b^{3} x^{14} + \frac {3}{11} \, a b^{2} x^{11} + \frac {3}{8} \, a^{2} b x^{8} + \frac {1}{5} \, a^{3} x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/14*b^3*x^14 + 3/11*a*b^2*x^11 + 3/8*a^2*b*x^8 + 1/5*a^3*x^5

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \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**4*(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

Integral(x**4*((a + b*x**3)**2)**(3/2), x)

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Giac [A]
time = 3.70, size = 67, normalized size = 0.40 \begin {gather*} \frac {1}{14} \, b^{3} x^{14} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{11} \, a b^{2} x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {3}{8} \, a^{2} b x^{8} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{5} \, a^{3} x^{5} \mathrm {sgn}\left (b x^{3} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

1/14*b^3*x^14*sgn(b*x^3 + a) + 3/11*a*b^2*x^11*sgn(b*x^3 + a) + 3/8*a^2*b*x^8*sgn(b*x^3 + a) + 1/5*a^3*x^5*sgn
(b*x^3 + a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,{\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^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)

[Out]

int(x^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2), x)

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