3.3.15 \(\int \frac {x^2 (A+B x^3)}{\sqrt {a+b x^3}} \, dx\) [215]

Optimal. Leaf size=46 \[ \frac {2 (A b-a B) \sqrt {a+b x^3}}{3 b^2}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^2} \]

[Out]

2/9*B*(b*x^3+a)^(3/2)/b^2+2/3*(A*b-B*a)*(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 \sqrt {a+b x^3} (A b-a B)}{3 b^2}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*Sqrt[a + b*x^3])/(3*b^2) + (2*B*(a + b*x^3)^(3/2))/(9*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 )}{\sqrt {a+b x^3}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {A+B x}{\sqrt {a+b x}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (\frac {A b-a B}{b \sqrt {a+b x}}+\frac {B \sqrt {a+b x}}{b}\right ) \, dx,x,x^3\right )\\ &=\frac {2 (A b-a B) \sqrt {a+b x^3}}{3 b^2}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 33, normalized size = 0.72 \begin {gather*} \frac {2 \sqrt {a+b x^3} \left (3 A b-2 a B+b B x^3\right )}{9 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x^3]*(3*A*b - 2*a*B + b*B*x^3))/(9*b^2)

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Maple [A]
time = 0.34, size = 52, normalized size = 1.13

method	result	size
gosper	\(\frac {2 \sqrt {b \,x^{3}+a}\, \left (b B \,x^{3}+3 A b -2 B a \right )}{9 b^{2}}\)	\(30\)
trager	\(\frac {2 \sqrt {b \,x^{3}+a}\, \left (b B \,x^{3}+3 A b -2 B a \right )}{9 b^{2}}\)	\(30\)
risch	\(\frac {2 \sqrt {b \,x^{3}+a}\, \left (b B \,x^{3}+3 A b -2 B a \right )}{9 b^{2}}\)	\(30\)
elliptic	\(\frac {2 B \,x^{3} \sqrt {b \,x^{3}+a}}{9 b}+\frac {2 \left (A -\frac {2 a B}{3 b}\right ) \sqrt {b \,x^{3}+a}}{3 b}\)	\(43\)
default	\(B \left (\frac {2 x^{3} \sqrt {b \,x^{3}+a}}{9 b}-\frac {4 a \sqrt {b \,x^{3}+a}}{9 b^{2}}\right )+\frac {2 A \sqrt {b \,x^{3}+a}}{3 b}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(2/9*x^3*(b*x^3+a)^(1/2)/b-4/9*a*(b*x^3+a)^(1/2)/b^2)+2/3*A*(b*x^3+a)^(1/2)/b

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Maxima [A]
time = 0.30, size = 48, normalized size = 1.04 \begin {gather*} \frac {2}{9} \, B {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}}}{b^{2}} - \frac {3 \, \sqrt {b x^{3} + a} a}{b^{2}}\right )} + \frac {2 \, \sqrt {b x^{3} + a} A}{3 \, 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)^(1/2),x, algorithm="maxima")

[Out]

2/9*B*((b*x^3 + a)^(3/2)/b^2 - 3*sqrt(b*x^3 + a)*a/b^2) + 2/3*sqrt(b*x^3 + a)*A/b

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Fricas [A]
time = 1.87, size = 29, normalized size = 0.63 \begin {gather*} \frac {2 \, {\left (B b x^{3} - 2 \, B a + 3 \, A b\right )} \sqrt {b x^{3} + a}}{9 \, 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)^(1/2),x, algorithm="fricas")

[Out]

2/9*(B*b*x^3 - 2*B*a + 3*A*b)*sqrt(b*x^3 + a)/b^2

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Sympy [A]
time = 0.27, size = 75, normalized size = 1.63 \begin {gather*} \begin {cases} \frac {2 A \sqrt {a + b x^{3}}}{3 b} - \frac {4 B a \sqrt {a + b x^{3}}}{9 b^{2}} + \frac {2 B x^{3} \sqrt {a + b x^{3}}}{9 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{6}}{6}}{\sqrt {a}} & \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)**(1/2),x)

[Out]

Piecewise((2*A*sqrt(a + b*x**3)/(3*b) - 4*B*a*sqrt(a + b*x**3)/(9*b**2) + 2*B*x**3*sqrt(a + b*x**3)/(9*b), Ne(
b, 0)), ((A*x**3/3 + B*x**6/6)/sqrt(a), True))

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Giac [A]
time = 0.88, size = 38, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B}{9 \, b^{2}} - \frac {2 \, \sqrt {b x^{3} + a} {\left (B a - A b\right )}}{3 \, 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)^(1/2),x, algorithm="giac")

[Out]

2/9*(b*x^3 + a)^(3/2)*B/b^2 - 2/3*sqrt(b*x^3 + a)*(B*a - A*b)/b^2

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Mupad [B]
time = 2.60, size = 29, normalized size = 0.63 \begin {gather*} \frac {2\,\sqrt {b\,x^3+a}\,\left (B\,b\,x^3+3\,A\,b-2\,B\,a\right )}{9\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a + b*x^3)^(1/2)*(3*A*b - 2*B*a + B*b*x^3))/(9*b^2)

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