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

Optimal result

Integrand size = 19, antiderivative size = 25 \[ \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 \sqrt {a x^2+b x^5}}{3 b x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1602} \[ \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 \sqrt {a x^2+b x^5}}{3 b x} \]

[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*} \text {integral}& = \frac {2 \sqrt {a x^2+b x^5}}{3 b x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 \sqrt {x^2 \left (a+b x^3\right )}}{3 b x} \]

[In]

Integrate[x^3/Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*Sqrt[x^2*(a + b*x^3)])/(3*b*x)

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

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	\(\frac {2 x \left (b \,x^{3}+a \right )}{3 \sqrt {x^{2} \left (b \,x^{3}+a \right )}\, b}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 \, \sqrt {b x^{5} + a x^{2}}}{3 \, b x} \]

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.56 \[ \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx=\frac {2 \, \sqrt {b x^{3} + a}}{3 \, b} \]

[In]

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

[Out]

2/3*sqrt(b*x^3 + a)/b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx=-\frac {2 \, \sqrt {a} \mathrm {sgn}\left (x\right )}{3 \, b} + \frac {2 \, \sqrt {b x^{3} + a}}{3 \, b \mathrm {sgn}\left (x\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {x^3}{\sqrt {a x^2+b x^5}} \, dx=\frac {2\,\sqrt {b\,x^5+a\,x^2}}{3\,b\,x} \]

[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)