3.4.85 \(\int \frac {\sqrt {a x^4}}{\sqrt {1+x^3}} \, dx\) [385]

Optimal. Leaf size=25 \[ \frac {2 \sqrt {a x^4} \sqrt {1+x^3}}{3 x^2} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {15, 267} \begin {gather*} \frac {2 \sqrt {x^3+1} \sqrt {a x^4}}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^4]/Sqrt[1 + x^3],x]

[Out]

(2*Sqrt[a*x^4]*Sqrt[1 + x^3])/(3*x^2)

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 267

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a x^4}}{\sqrt {1+x^3}} \, dx &=\frac {\sqrt {a x^4} \int \frac {x^2}{\sqrt {1+x^3}} \, dx}{x^2}\\ &=\frac {2 \sqrt {a x^4} \sqrt {1+x^3}}{3 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 25, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {a x^4} \sqrt {1+x^3}}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^4]/Sqrt[1 + x^3],x]

[Out]

(2*Sqrt[a*x^4]*Sqrt[1 + x^3])/(3*x^2)

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Maple [A]
time = 0.20, size = 20, normalized size = 0.80

method	result	size
default	\(\frac {2 \sqrt {a \,x^{4}}\, \sqrt {x^{3}+1}}{3 x^{2}}\)	\(20\)
risch	\(\frac {2 \sqrt {a \,x^{4}}\, \sqrt {x^{3}+1}}{3 x^{2}}\)	\(20\)
gosper	\(\frac {2 \left (1+x \right ) \left (x^{2}-x +1\right ) \sqrt {a \,x^{4}}}{3 x^{2} \sqrt {x^{3}+1}}\)	\(31\)
meijerg	\(\frac {\sqrt {a \,x^{4}}\, \left (-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {x^{3}+1}\right )}{3 x^{2} \sqrt {\pi }}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 28, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (\sqrt {a} x^{3} + \sqrt {a}\right )}}{3 \, \sqrt {x^{2} - x + 1} \sqrt {x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(sqrt(a)*x^3 + sqrt(a))/(sqrt(x^2 - x + 1)*sqrt(x + 1))

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Fricas [A]
time = 0.34, size = 19, normalized size = 0.76 \begin {gather*} \frac {2 \, \sqrt {a x^{4}} \sqrt {x^{3} + 1}}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(a*x^4)*sqrt(x^3 + 1)/x^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a x^{4}}}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*x**4)/sqrt((x + 1)*(x**2 - x + 1)), x)

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Giac [A]
time = 3.27, size = 12, normalized size = 0.48 \begin {gather*} \frac {2}{3} \, \sqrt {x^{3} + 1} \sqrt {a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(x^3 + 1)*sqrt(a)

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Mupad [B]
time = 2.91, size = 20, normalized size = 0.80 \begin {gather*} \frac {2\,\sqrt {a}\,\sqrt {x^3+1}\,\sqrt {x^4}}{3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4)^(1/2)/(x^3 + 1)^(1/2),x)

[Out]

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

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