3.2.72 \(\int \frac {\sqrt {-b+a x^3} (2 b+a x^3)}{x^4} \, dx\)

Optimal. Leaf size=20 \[ \frac {2 \left (a x^3-b\right )^{3/2}}{3 x^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {446, 74} \begin {gather*} \frac {2 \left (a x^3-b\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-b + a*x^3]*(2*b + a*x^3))/x^4,x]

[Out]

(2*(-b + a*x^3)^(3/2))/(3*x^3)

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
 NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\sqrt {-b+a x^3} \left (2 b+a x^3\right )}{x^4} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {-b+a x} (2 b+a x)}{x^2} \, dx,x,x^3\right )\\ &=\frac {2 \left (-b+a x^3\right )^{3/2}}{3 x^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \left (a x^3-b\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[-b + a*x^3]*(2*b + a*x^3))/x^4,x]

[Out]

(2*(-b + a*x^3)^(3/2))/(3*x^3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.04, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 \left (-b+a x^3\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[-b + a*x^3]*(2*b + a*x^3))/x^4,x]

[Out]

(2*(-b + a*x^3)^(3/2))/(3*x^3)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 16, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (a x^{3} - b\right )}^{\frac {3}{2}}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a*x^3 - b)^(3/2)/x^3

________________________________________________________________________________________

giac [B]  time = 0.61, size = 39, normalized size = 1.95 \begin {gather*} \frac {2 \, {\left (\sqrt {a x^{3} - b} a^{2} - \frac {\sqrt {a x^{3} - b} a b}{x^{3}}\right )}}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.10, size = 17, normalized size = 0.85

method result size
gosper \(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{3 x^{3}}\) \(17\)
trager \(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{3 x^{3}}\) \(17\)
risch \(\frac {2 \left (a \,x^{3}-b \right )^{\frac {3}{2}}}{3 x^{3}}\) \(17\)
elliptic \(-\frac {2 b \sqrt {a \,x^{3}-b}}{3 x^{3}}+\frac {2 a \sqrt {a \,x^{3}-b}}{3}\) \(33\)
default \(a \left (\frac {2 \sqrt {a \,x^{3}-b}}{3}+\frac {2 b \arctanh \left (\frac {\sqrt {a \,x^{3}-b}}{\sqrt {-b}}\right )}{3 \sqrt {-b}}\right )+2 b \left (-\frac {\sqrt {a \,x^{3}-b}}{3 x^{3}}-\frac {a \arctanh \left (\frac {\sqrt {a \,x^{3}-b}}{\sqrt {-b}}\right )}{3 \sqrt {-b}}\right )\) \(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.73, size = 79, normalized size = 3.95 \begin {gather*} -\frac {2}{3} \, {\left (\sqrt {b} \arctan \left (\frac {\sqrt {a x^{3} - b}}{\sqrt {b}}\right ) - \sqrt {a x^{3} - b}\right )} a + \frac {2}{3} \, {\left (\frac {a \arctan \left (\frac {\sqrt {a x^{3} - b}}{\sqrt {b}}\right )}{\sqrt {b}} - \frac {\sqrt {a x^{3} - b}}{x^{3}}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.24, size = 16, normalized size = 0.80 \begin {gather*} \frac {2\,{\left (a\,x^3-b\right )}^{3/2}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x^3 - b)^(1/2)*(2*b + a*x^3))/x^4,x)

[Out]

(2*(a*x^3 - b)^(3/2))/(3*x^3)

________________________________________________________________________________________

sympy [C]  time = 22.40, size = 309, normalized size = 15.45 \begin {gather*} a \left (\begin {cases} - \frac {2 i \sqrt {a} x^{\frac {3}{2}}}{3 \sqrt {-1 + \frac {b}{a x^{3}}}} - \frac {2 i \sqrt {b} \operatorname {acosh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{3} + \frac {2 i b}{3 \sqrt {a} x^{\frac {3}{2}} \sqrt {-1 + \frac {b}{a x^{3}}}} & \text {for}\: \left |{\frac {b}{a x^{3}}}\right | > 1 \\\frac {2 \sqrt {a} x^{\frac {3}{2}}}{3 \sqrt {1 - \frac {b}{a x^{3}}}} + \frac {2 \sqrt {b} \operatorname {asin}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{3} - \frac {2 b}{3 \sqrt {a} x^{\frac {3}{2}} \sqrt {1 - \frac {b}{a x^{3}}}} & \text {otherwise} \end {cases}\right ) + 2 b \left (\begin {cases} \frac {i \sqrt {a}}{3 x^{\frac {3}{2}} \sqrt {-1 + \frac {b}{a x^{3}}}} + \frac {i a \operatorname {acosh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{3 \sqrt {b}} - \frac {i b}{3 \sqrt {a} x^{\frac {9}{2}} \sqrt {-1 + \frac {b}{a x^{3}}}} & \text {for}\: \left |{\frac {b}{a x^{3}}}\right | > 1 \\- \frac {\sqrt {a} \sqrt {1 - \frac {b}{a x^{3}}}}{3 x^{\frac {3}{2}}} - \frac {a \operatorname {asin}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{3 \sqrt {b}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((-2*I*sqrt(a)*x**(3/2)/(3*sqrt(-1 + b/(a*x**3))) - 2*I*sqrt(b)*acosh(sqrt(b)/(sqrt(a)*x**(3/2)))/3
 + 2*I*b/(3*sqrt(a)*x**(3/2)*sqrt(-1 + b/(a*x**3))), Abs(b/(a*x**3)) > 1), (2*sqrt(a)*x**(3/2)/(3*sqrt(1 - b/(
a*x**3))) + 2*sqrt(b)*asin(sqrt(b)/(sqrt(a)*x**(3/2)))/3 - 2*b/(3*sqrt(a)*x**(3/2)*sqrt(1 - b/(a*x**3))), True
)) + 2*b*Piecewise((I*sqrt(a)/(3*x**(3/2)*sqrt(-1 + b/(a*x**3))) + I*a*acosh(sqrt(b)/(sqrt(a)*x**(3/2)))/(3*sq
rt(b)) - I*b/(3*sqrt(a)*x**(9/2)*sqrt(-1 + b/(a*x**3))), Abs(b/(a*x**3)) > 1), (-sqrt(a)*sqrt(1 - b/(a*x**3))/
(3*x**(3/2)) - a*asin(sqrt(b)/(sqrt(a)*x**(3/2)))/(3*sqrt(b)), True))

________________________________________________________________________________________