3.3.63 \(\int \frac {x^3}{(a x^2+b x^3)^{3/2}} \, dx\) [263]

Optimal. Leaf size=21 \[ -\frac {2 x}{b \sqrt {a x^2+b x^3}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 21, 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 x}{b \sqrt {a x^2+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a*x^2 + b*x^3)^(3/2),x]

[Out]

(-2*x)/(b*Sqrt[a*x^2 + b*x^3])

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}{\left (a x^2+b x^3\right )^{3/2}} \, dx &=-\frac {2 x}{b \sqrt {a x^2+b x^3}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 19, normalized size = 0.90 \begin {gather*} -\frac {2 x}{b \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x)/(b*Sqrt[x^2*(a + b*x)])

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Maple [A]
time = 0.36, size = 27, normalized size = 1.29

method	result	size
gosper	\(-\frac {2 \left (b x +a \right ) x^{3}}{b \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(27\)
default	\(-\frac {2 \left (b x +a \right ) x^{3}}{b \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}\)	\(27\)
trager	\(-\frac {2 \sqrt {b \,x^{3}+a \,x^{2}}}{\left (b x +a \right ) b x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 12, normalized size = 0.57 \begin {gather*} -\frac {2}{\sqrt {b x + a} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(sqrt(b*x + a)*b)

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Fricas [A]
time = 1.67, size = 29, normalized size = 1.38 \begin {gather*} -\frac {2 \, \sqrt {b x^{3} + a x^{2}}}{b^{2} x^{2} + a b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.32, size = 27, normalized size = 1.29 \begin {gather*} \frac {2 \, \mathrm {sgn}\left (x\right )}{\sqrt {a} b} - \frac {2}{\sqrt {b x + a} b \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.07, size = 28, normalized size = 1.33 \begin {gather*} -\frac {2\,\sqrt {b\,x^3+a\,x^2}}{b\,x\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x^2 + b*x^3)^(3/2),x)

[Out]

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

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