3.5.36 \(\int \sqrt {x^4 (a+b x^3)} \, dx\) [436]

Optimal. Leaf size=25 \[ \frac {2 \left (a x^4+b x^7\right )^{3/2}}{9 b x^6} \]

[Out]

2/9*(b*x^7+a*x^4)^(3/2)/b/x^6

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2004, 2025} \begin {gather*} \frac {2 \left (a x^4+b x^7\right )^{3/2}}{9 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^4*(a + b*x^3)],x]

[Out]

(2*(a*x^4 + b*x^7)^(3/2))/(9*b*x^6)

Rule 2004

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rubi steps

\begin {align*} \int \sqrt {x^4 \left (a+b x^3\right )} \, dx &=\int \sqrt {a x^4+b x^7} \, dx\\ &=\frac {2 \left (a x^4+b x^7\right )^{3/2}}{9 b x^6}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 25, normalized size = 1.00 \begin {gather*} \frac {2 \left (x^4 \left (a+b x^3\right )\right )^{3/2}}{9 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x^4*(a + b*x^3)],x]

[Out]

(2*(x^4*(a + b*x^3))^(3/2))/(9*b*x^6)

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Maple [A]
time = 0.38, size = 29, normalized size = 1.16

method	result	size
gosper	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {x^{4} \left (b \,x^{3}+a \right )}}{9 b \,x^{2}}\)	\(29\)
default	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {x^{4} \left (b \,x^{3}+a \right )}}{9 b \,x^{2}}\)	\(29\)
trager	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {b \,x^{7}+a \,x^{4}}}{9 b \,x^{2}}\)	\(29\)
risch	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {x^{4} \left (b \,x^{3}+a \right )}}{9 b \,x^{2}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(b*x^3+a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/9*(b*x^3+a)*(x^4*(b*x^3+a))^(1/2)/b/x^2

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Maxima [A]
time = 0.30, size = 14, normalized size = 0.56 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4*(b*x^3+a))^(1/2),x, algorithm="maxima")

[Out]

2/9*(b*x^3 + a)^(3/2)/b

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Fricas [A]
time = 1.52, size = 28, normalized size = 1.12 \begin {gather*} \frac {2 \, \sqrt {b x^{7} + a x^{4}} {\left (b x^{3} + a\right )}}{9 \, b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4*(b*x^3+a))^(1/2),x, algorithm="fricas")

[Out]

2/9*sqrt(b*x^7 + a*x^4)*(b*x^3 + a)/(b*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {x^{4} \left (a + b x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4*(b*x**3+a))**(1/2),x)

[Out]

Integral(sqrt(x**4*(a + b*x**3)), x)

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Giac [A]
time = 2.26, size = 14, normalized size = 0.56 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4*(b*x^3+a))^(1/2),x, algorithm="giac")

[Out]

2/9*(b*x^3 + a)^(3/2)/b

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Mupad [B]
time = 5.21, size = 22, normalized size = 0.88 \begin {gather*} \frac {2\,{\left (b\,x^3+a\right )}^{3/2}\,\sqrt {x^4}}{9\,b\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x^3))^(1/2),x)

[Out]

(2*(a + b*x^3)^(3/2)*(x^4)^(1/2))/(9*b*x^2)

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