3.3.86 \(\int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx\) [286]

Optimal. Leaf size=25 \[ \frac {2 \sqrt {a x^2+b x^5}}{3 b x} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1602} \begin {gather*} \frac {2 \sqrt {a x^2+b x^5}}{3 b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*Sqrt[a*x^2 + b*x^5])/(3*b*x)

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx &=\frac {2 \sqrt {a x^2+b x^5}}{3 b x}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*Sqrt[x^2*(a + b*x^3)])/(3*b*x)

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Maple [A]
time = 0.39, size = 27, normalized size = 1.08

method	result	size
trager	\(\frac {2 \sqrt {b \,x^{5}+a \,x^{2}}}{3 b x}\)	\(22\)
gosper	\(\frac {2 x \left (b \,x^{3}+a \right )}{3 b \sqrt {b \,x^{5}+a \,x^{2}}}\)	\(27\)
default	\(\frac {2 x \left (b \,x^{3}+a \right )}{3 b \sqrt {b \,x^{5}+a \,x^{2}}}\)	\(27\)
risch	\(\frac {2 x \left (b \,x^{3}+a \right )}{3 \sqrt {x^{2} \left (b \,x^{3}+a \right )}\, b}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b*x^5+a*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 14, normalized size = 0.56 \begin {gather*} \frac {2 \, \sqrt {b x^{3} + a}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

2/3*sqrt(b*x^3 + a)/b

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Fricas [A]
time = 1.52, size = 21, normalized size = 0.84 \begin {gather*} \frac {2 \, \sqrt {b x^{5} + a x^{2}}}{3 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

2/3*sqrt(b*x^5 + a*x^2)/(b*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b*x**5+a*x**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 1.77, size = 29, normalized size = 1.16 \begin {gather*} -\frac {2 \, \sqrt {a} \mathrm {sgn}\left (x\right )}{3 \, b} + \frac {2 \, \sqrt {b x^{3} + a}}{3 \, b \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 5.19, size = 21, normalized size = 0.84 \begin {gather*} \frac {2\,\sqrt {b\,x^5+a\,x^2}}{3\,b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x^2 + b*x^5)^(1/2),x)

[Out]

(2*(a*x^2 + b*x^5)^(1/2))/(3*b*x)

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