\(\int x \sqrt {a x^2+b x^5} \, dx\) [435]

Optimal result

Integrand size = 17, antiderivative size = 25 \[ \int x \sqrt {a x^2+b x^5} \, dx=\frac {2 \left (a x^2+b x^5\right )^{3/2}}{9 b x^3} \]

[Out]

2/9*(b*x^5+a*x^2)^(3/2)/b/x^3

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1602} \[ \int x \sqrt {a x^2+b x^5} \, dx=\frac {2 \left (a x^2+b x^5\right )^{3/2}}{9 b x^3} \]

[In]

Int[x*Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*(a*x^2 + b*x^5)^(3/2))/(9*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*} \text {integral}& = \frac {2 \left (a x^2+b x^5\right )^{3/2}}{9 b x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int x \sqrt {a x^2+b x^5} \, dx=\frac {2 \left (x^2 \left (a+b x^3\right )\right )^{3/2}}{9 b x^3} \]

[In]

Integrate[x*Sqrt[a*x^2 + b*x^5],x]

[Out]

(2*(x^2*(a + b*x^3))^(3/2))/(9*b*x^3)

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16

method	result	size
gosper	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {b \,x^{5}+a \,x^{2}}}{9 b x}\)	\(29\)
default	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {b \,x^{5}+a \,x^{2}}}{9 b x}\)	\(29\)
trager	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {b \,x^{5}+a \,x^{2}}}{9 b x}\)	\(29\)
risch	\(\frac {2 \left (b \,x^{3}+a \right ) \sqrt {x^{2} \left (b \,x^{3}+a \right )}}{9 b x}\)	\(29\)

[In]

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

[Out]

2/9*(b*x^3+a)/b/x*(b*x^5+a*x^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int x \sqrt {a x^2+b x^5} \, dx=\frac {2 \, \sqrt {b x^{5} + a x^{2}} {\left (b x^{3} + a\right )}}{9 \, b x} \]

[In]

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

[Out]

2/9*sqrt(b*x^5 + a*x^2)*(b*x^3 + a)/(b*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.56 \[ \int x \sqrt {a x^2+b x^5} \, dx=\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{9 \, b} \]

[In]

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

[Out]

2/9*(b*x^3 + a)^(3/2)/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int x \sqrt {a x^2+b x^5} \, dx=\frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} \mathrm {sgn}\left (x\right )}{9 \, b} - \frac {2 \, a^{\frac {3}{2}} \mathrm {sgn}\left (x\right )}{9 \, b} \]

[In]

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

[Out]

2/9*(b*x^3 + a)^(3/2)*sgn(x)/b - 2/9*a^(3/2)*sgn(x)/b

Mupad [B] (verification not implemented)

Time = 8.94 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int x \sqrt {a x^2+b x^5} \, dx=\frac {\left (\frac {2\,a}{9\,b}+\frac {2\,x^3}{9}\right )\,\sqrt {b\,x^5+a\,x^2}}{x} \]

[In]

int(x*(a*x^2 + b*x^5)^(1/2),x)

[Out]

(((2*a)/(9*b) + (2*x^3)/9)*(a*x^2 + b*x^5)^(1/2))/x