3.3.30 \(\int \frac {x^2 (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [230]

Optimal. Leaf size=46 \[ -\frac {2 (A b-a B)}{3 b^2 \sqrt {a+b x^3}}+\frac {2 B \sqrt {a+b x^3}}{3 b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {455, 45} \begin {gather*} \frac {2 B \sqrt {a+b x^3}}{3 b^2}-\frac {2 (A b-a B)}{3 b^2 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(-2*(A*b - a*B))/(3*b^2*Sqrt[a + b*x^3]) + (2*B*Sqrt[a + b*x^3])/(3*b^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 33, normalized size = 0.72 \begin {gather*} \frac {2 \left (-A b+2 a B+b B x^3\right )}{3 b^2 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(A*b) + 2*a*B + b*B*x^3))/(3*b^2*Sqrt[a + b*x^3])

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Maple [A]
time = 0.34, size = 53, normalized size = 1.15

method	result	size
gosper	\(-\frac {2 \left (-b B \,x^{3}+A b -2 B a \right )}{3 \sqrt {b \,x^{3}+a}\, b^{2}}\)	\(30\)
trager	\(-\frac {2 \left (-b B \,x^{3}+A b -2 B a \right )}{3 \sqrt {b \,x^{3}+a}\, b^{2}}\)	\(30\)
risch	\(-\frac {2 \left (A b -B a \right )}{3 b^{2} \sqrt {b \,x^{3}+a}}+\frac {2 B \sqrt {b \,x^{3}+a}}{3 b^{2}}\)	\(39\)
elliptic	\(-\frac {2 \left (A b -B a \right )}{3 b^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 B \sqrt {b \,x^{3}+a}}{3 b^{2}}\)	\(43\)
default	\(B \left (\frac {2 a}{3 b^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 \sqrt {b \,x^{3}+a}}{3 b^{2}}\right )-\frac {2 A}{3 b \sqrt {b \,x^{3}+a}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 47, normalized size = 1.02 \begin {gather*} \frac {2}{3} \, B {\left (\frac {\sqrt {b x^{3} + a}}{b^{2}} + \frac {a}{\sqrt {b x^{3} + a} b^{2}}\right )} - \frac {2 \, A}{3 \, \sqrt {b x^{3} + a} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.83, size = 41, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (B b x^{3} + 2 \, B a - A b\right )} \sqrt {b x^{3} + a}}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.33, size = 75, normalized size = 1.63 \begin {gather*} \begin {cases} - \frac {2 A}{3 b \sqrt {a + b x^{3}}} + \frac {4 B a}{3 b^{2} \sqrt {a + b x^{3}}} + \frac {2 B x^{3}}{3 b \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{6}}{6}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*A/(3*b*sqrt(a + b*x**3)) + 4*B*a/(3*b**2*sqrt(a + b*x**3)) + 2*B*x**3/(3*b*sqrt(a + b*x**3)), Ne
(b, 0)), ((A*x**3/3 + B*x**6/6)/a**(3/2), True))

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Giac [A]
time = 1.28, size = 38, normalized size = 0.83 \begin {gather*} \frac {2 \, \sqrt {b x^{3} + a} B}{3 \, b^{2}} + \frac {2 \, {\left (B a - A b\right )}}{3 \, \sqrt {b x^{3} + a} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.61, size = 33, normalized size = 0.72 \begin {gather*} \frac {2\,B\,a-2\,A\,b+2\,B\,\left (b\,x^3+a\right )}{3\,b^2\,\sqrt {b\,x^3+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(A + B*x^3))/(a + b*x^3)^(3/2),x)

[Out]

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

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