3.3.29 \(\int \frac {x^5 (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [229]

Optimal. Leaf size=73 \[ \frac {2 a (A b-a B)}{3 b^3 \sqrt {a+b x^3}}+\frac {2 (A b-2 a B) \sqrt {a+b x^3}}{3 b^3}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^3} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 73, 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 = {457, 78} \begin {gather*} \frac {2 \sqrt {a+b x^3} (A b-2 a B)}{3 b^3}+\frac {2 a (A b-a B)}{3 b^3 \sqrt {a+b x^3}}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

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 {x^5 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x (A+B x)}{(a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (\frac {a (-A b+a B)}{b^2 (a+b x)^{3/2}}+\frac {A b-2 a B}{b^2 \sqrt {a+b x}}+\frac {B \sqrt {a+b x}}{b^2}\right ) \, dx,x,x^3\right )\\ &=\frac {2 a (A b-a B)}{3 b^3 \sqrt {a+b x^3}}+\frac {2 (A b-2 a B) \sqrt {a+b x^3}}{3 b^3}+\frac {2 B \left (a+b x^3\right )^{3/2}}{9 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 55, normalized size = 0.75 \begin {gather*} \frac {2 \left (6 a A b-8 a^2 B+3 A b^2 x^3-4 a b B x^3+b^2 B x^6\right )}{9 b^3 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(2*(6*a*A*b - 8*a^2*B + 3*A*b^2*x^3 - 4*a*b*B*x^3 + b^2*B*x^6))/(9*b^3*Sqrt[a + b*x^3])

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Maple [A]
time = 0.34, size = 94, normalized size = 1.29

method	result	size
gosper	\(\frac {\frac {2}{9} b^{2} B \,x^{6}+\frac {2}{3} A \,b^{2} x^{3}-\frac {8}{9} B a b \,x^{3}+\frac {4}{3} a b A -\frac {16}{9} a^{2} B}{\sqrt {b \,x^{3}+a}\, b^{3}}\)	\(52\)
trager	\(\frac {\frac {2}{9} b^{2} B \,x^{6}+\frac {2}{3} A \,b^{2} x^{3}-\frac {8}{9} B a b \,x^{3}+\frac {4}{3} a b A -\frac {16}{9} a^{2} B}{\sqrt {b \,x^{3}+a}\, b^{3}}\)	\(52\)
risch	\(\frac {2 \left (b B \,x^{3}+3 A b -5 B a \right ) \sqrt {b \,x^{3}+a}}{9 b^{3}}+\frac {2 a \left (A b -B a \right )}{3 b^{3} \sqrt {b \,x^{3}+a}}\)	\(54\)
elliptic	\(\frac {2 a \left (A b -B a \right )}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 B \,x^{3} \sqrt {b \,x^{3}+a}}{9 b^{2}}+\frac {2 \left (\frac {A b -B a}{b^{2}}-\frac {2 a B}{3 b^{2}}\right ) \sqrt {b \,x^{3}+a}}{3 b}\)	\(81\)
default	\(B \left (-\frac {2 a^{2}}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{3} \sqrt {b \,x^{3}+a}}{9 b^{2}}-\frac {10 a \sqrt {b \,x^{3}+a}}{9 b^{3}}\right )+A \left (\frac {2 a}{3 b^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 \sqrt {b \,x^{3}+a}}{3 b^{2}}\right )\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(-2/3/b^3*a^2/((x^3+a/b)*b)^(1/2)+2/9/b^2*x^3*(b*x^3+a)^(1/2)-10/9*a*(b*x^3+a)^(1/2)/b^3)+A*(2/3/b^2*a/((x^3
+a/b)*b)^(1/2)+2/3*(b*x^3+a)^(1/2)/b^2)

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Maxima [A]
time = 0.29, size = 81, normalized size = 1.11 \begin {gather*} \frac {2}{9} \, B {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}}}{b^{3}} - \frac {6 \, \sqrt {b x^{3} + a} a}{b^{3}} - \frac {3 \, a^{2}}{\sqrt {b x^{3} + a} b^{3}}\right )} + \frac {2}{3} \, A {\left (\frac {\sqrt {b x^{3} + a}}{b^{2}} + \frac {a}{\sqrt {b x^{3} + a} b^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*B*((b*x^3 + a)^(3/2)/b^3 - 6*sqrt(b*x^3 + a)*a/b^3 - 3*a^2/(sqrt(b*x^3 + a)*b^3)) + 2/3*A*(sqrt(b*x^3 + a)
/b^2 + a/(sqrt(b*x^3 + a)*b^2))

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Fricas [A]
time = 2.86, size = 63, normalized size = 0.86 \begin {gather*} \frac {2 \, {\left (B b^{2} x^{6} - {\left (4 \, B a b - 3 \, A b^{2}\right )} x^{3} - 8 \, B a^{2} + 6 \, A a b\right )} \sqrt {b x^{3} + a}}{9 \, {\left (b^{4} x^{3} + a b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

2/9*(B*b^2*x^6 - (4*B*a*b - 3*A*b^2)*x^3 - 8*B*a^2 + 6*A*a*b)*sqrt(b*x^3 + a)/(b^4*x^3 + a*b^3)

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Sympy [A]
time = 0.44, size = 124, normalized size = 1.70 \begin {gather*} \begin {cases} \frac {4 A a}{3 b^{2} \sqrt {a + b x^{3}}} + \frac {2 A x^{3}}{3 b \sqrt {a + b x^{3}}} - \frac {16 B a^{2}}{9 b^{3} \sqrt {a + b x^{3}}} - \frac {8 B a x^{3}}{9 b^{2} \sqrt {a + b x^{3}}} + \frac {2 B x^{6}}{9 b \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{6} + \frac {B x^{9}}{9}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**3+A)/(b*x**3+a)**(3/2),x)

[Out]

Piecewise((4*A*a/(3*b**2*sqrt(a + b*x**3)) + 2*A*x**3/(3*b*sqrt(a + b*x**3)) - 16*B*a**2/(9*b**3*sqrt(a + b*x*
*3)) - 8*B*a*x**3/(9*b**2*sqrt(a + b*x**3)) + 2*B*x**6/(9*b*sqrt(a + b*x**3)), Ne(b, 0)), ((A*x**6/6 + B*x**9/
9)/a**(3/2), True))

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Giac [A]
time = 1.05, size = 77, normalized size = 1.05 \begin {gather*} -\frac {2 \, {\left (B a^{2} - A a b\right )}}{3 \, \sqrt {b x^{3} + a} b^{3}} + \frac {2 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} B b^{6} - 6 \, \sqrt {b x^{3} + a} B a b^{6} + 3 \, \sqrt {b x^{3} + a} A b^{7}\right )}}{9 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

-2/3*(B*a^2 - A*a*b)/(sqrt(b*x^3 + a)*b^3) + 2/9*((b*x^3 + a)^(3/2)*B*b^6 - 6*sqrt(b*x^3 + a)*B*a*b^6 + 3*sqrt
(b*x^3 + a)*A*b^7)/b^9

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Mupad [B]
time = 2.68, size = 60, normalized size = 0.82 \begin {gather*} \frac {2\,B\,{\left (b\,x^3+a\right )}^2-6\,B\,a^2+6\,A\,b\,\left (b\,x^3+a\right )-12\,B\,a\,\left (b\,x^3+a\right )+6\,A\,a\,b}{9\,b^3\,\sqrt {b\,x^3+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(A + B*x^3))/(a + b*x^3)^(3/2),x)

[Out]

(2*B*(a + b*x^3)^2 - 6*B*a^2 + 6*A*b*(a + b*x^3) - 12*B*a*(a + b*x^3) + 6*A*a*b)/(9*b^3*(a + b*x^3)^(1/2))

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