3.3.28 \(\int \frac {x^8 (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [228]

Optimal. Leaf size=103 \[ -\frac {2 a^2 (A b-a B)}{3 b^4 \sqrt {a+b x^3}}-\frac {2 a (2 A b-3 a B) \sqrt {a+b x^3}}{3 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{3/2}}{9 b^4}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b^4} \]

[Out]

2/9*(A*b-3*B*a)*(b*x^3+a)^(3/2)/b^4+2/15*B*(b*x^3+a)^(5/2)/b^4-2/3*a^2*(A*b-B*a)/b^4/(b*x^3+a)^(1/2)-2/3*a*(2*
A*b-3*B*a)*(b*x^3+a)^(1/2)/b^4

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Rubi [A]
time = 0.05, antiderivative size = 103, 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 a^2 (A b-a B)}{3 b^4 \sqrt {a+b x^3}}+\frac {2 \left (a+b x^3\right )^{3/2} (A b-3 a B)}{9 b^4}-\frac {2 a \sqrt {a+b x^3} (2 A b-3 a B)}{3 b^4}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(-2*a^2*(A*b - a*B))/(3*b^4*Sqrt[a + b*x^3]) - (2*a*(2*A*b - 3*a*B)*Sqrt[a + b*x^3])/(3*b^4) + (2*(A*b - 3*a*B
)*(a + b*x^3)^(3/2))/(9*b^4) + (2*B*(a + b*x^3)^(5/2))/(15*b^4)

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^8 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2 (A+B x)}{(a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {a^2 (-A b+a B)}{b^3 (a+b x)^{3/2}}+\frac {a (-2 A b+3 a B)}{b^3 \sqrt {a+b x}}+\frac {(A b-3 a B) \sqrt {a+b x}}{b^3}+\frac {B (a+b x)^{3/2}}{b^3}\right ) \, dx,x,x^3\right )\\ &=-\frac {2 a^2 (A b-a B)}{3 b^4 \sqrt {a+b x^3}}-\frac {2 a (2 A b-3 a B) \sqrt {a+b x^3}}{3 b^4}+\frac {2 (A b-3 a B) \left (a+b x^3\right )^{3/2}}{9 b^4}+\frac {2 B \left (a+b x^3\right )^{5/2}}{15 b^4}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 77, normalized size = 0.75 \begin {gather*} \frac {2 \left (48 a^3 B-8 a^2 b \left (5 A-3 B x^3\right )+b^3 x^6 \left (5 A+3 B x^3\right )-2 a b^2 x^3 \left (10 A+3 B x^3\right )\right )}{45 b^4 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(2*(48*a^3*B - 8*a^2*b*(5*A - 3*B*x^3) + b^3*x^6*(5*A + 3*B*x^3) - 2*a*b^2*x^3*(10*A + 3*B*x^3)))/(45*b^4*Sqrt
[a + b*x^3])

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Maple [A]
time = 0.35, size = 134, normalized size = 1.30

method	result	size
gosper	\(-\frac {2 \left (-3 B \,x^{9} b^{3}-5 A \,b^{3} x^{6}+6 B a \,b^{2} x^{6}+20 A a \,b^{2} x^{3}-24 B \,a^{2} b \,x^{3}+40 A \,a^{2} b -48 B \,a^{3}\right )}{45 \sqrt {b \,x^{3}+a}\, b^{4}}\)	\(77\)
trager	\(-\frac {2 \left (-3 B \,x^{9} b^{3}-5 A \,b^{3} x^{6}+6 B a \,b^{2} x^{6}+20 A a \,b^{2} x^{3}-24 B \,a^{2} b \,x^{3}+40 A \,a^{2} b -48 B \,a^{3}\right )}{45 \sqrt {b \,x^{3}+a}\, b^{4}}\)	\(77\)
risch	\(-\frac {2 \left (-3 b^{2} B \,x^{6}-5 A \,b^{2} x^{3}+9 B a b \,x^{3}+25 a b A -33 a^{2} B \right ) \sqrt {b \,x^{3}+a}}{45 b^{4}}-\frac {2 a^{2} \left (A b -B a \right )}{3 b^{4} \sqrt {b \,x^{3}+a}}\)	\(79\)
default	\(B \left (\frac {2 a^{3}}{3 b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{6} \sqrt {b \,x^{3}+a}}{15 b^{2}}-\frac {2 a \,x^{3} \sqrt {b \,x^{3}+a}}{5 b^{3}}+\frac {22 a^{2} \sqrt {b \,x^{3}+a}}{15 b^{4}}\right )+A \left (-\frac {2 a^{2}}{3 b^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x^{3} \sqrt {b \,x^{3}+a}}{9 b^{2}}-\frac {10 a \sqrt {b \,x^{3}+a}}{9 b^{3}}\right )\)	\(134\)
elliptic	\(-\frac {2 a^{2} \left (A b -B a \right )}{3 b^{4} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 B \,x^{6} \sqrt {b \,x^{3}+a}}{15 b^{2}}+\frac {2 \left (\frac {A b -B a}{b^{2}}-\frac {4 a B}{5 b^{2}}\right ) x^{3} \sqrt {b \,x^{3}+a}}{9 b}+\frac {2 \left (-\frac {a \left (A b -B a \right )}{b^{3}}-\frac {2 \left (\frac {A b -B a}{b^{2}}-\frac {4 a B}{5 b^{2}}\right ) a}{3 b}\right ) \sqrt {b \,x^{3}+a}}{3 b}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

B*(2/3/b^4*a^3/((x^3+a/b)*b)^(1/2)+2/15/b^2*x^6*(b*x^3+a)^(1/2)-2/5*a/b^3*x^3*(b*x^3+a)^(1/2)+22/15*a^2*(b*x^3
+a)^(1/2)/b^4)+A*(-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)

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Maxima [A]
time = 0.28, size = 116, normalized size = 1.13 \begin {gather*} \frac {2}{15} \, B {\left (\frac {{\left (b x^{3} + a\right )}^{\frac {5}{2}}}{b^{4}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a}{b^{4}} + \frac {15 \, \sqrt {b x^{3} + a} a^{2}}{b^{4}} + \frac {5 \, a^{3}}{\sqrt {b x^{3} + a} b^{4}}\right )} + \frac {2}{9} \, A {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

2/15*B*((b*x^3 + a)^(5/2)/b^4 - 5*(b*x^3 + a)^(3/2)*a/b^4 + 15*sqrt(b*x^3 + a)*a^2/b^4 + 5*a^3/(sqrt(b*x^3 + a
)*b^4)) + 2/9*A*((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))

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Fricas [A]
time = 1.91, size = 88, normalized size = 0.85 \begin {gather*} \frac {2 \, {\left (3 \, B b^{3} x^{9} - {\left (6 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 48 \, B a^{3} - 40 \, A a^{2} b + 4 \, {\left (6 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3}\right )} \sqrt {b x^{3} + a}}{45 \, {\left (b^{5} x^{3} + a b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

2/45*(3*B*b^3*x^9 - (6*B*a*b^2 - 5*A*b^3)*x^6 + 48*B*a^3 - 40*A*a^2*b + 4*(6*B*a^2*b - 5*A*a*b^2)*x^3)*sqrt(b*
x^3 + a)/(b^5*x^3 + a*b^4)

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Sympy [A]
time = 0.60, size = 175, normalized size = 1.70 \begin {gather*} \begin {cases} - \frac {16 A a^{2}}{9 b^{3} \sqrt {a + b x^{3}}} - \frac {8 A a x^{3}}{9 b^{2} \sqrt {a + b x^{3}}} + \frac {2 A x^{6}}{9 b \sqrt {a + b x^{3}}} + \frac {32 B a^{3}}{15 b^{4} \sqrt {a + b x^{3}}} + \frac {16 B a^{2} x^{3}}{15 b^{3} \sqrt {a + b x^{3}}} - \frac {4 B a x^{6}}{15 b^{2} \sqrt {a + b x^{3}}} + \frac {2 B x^{9}}{15 b \sqrt {a + b x^{3}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{9}}{9} + \frac {B x^{12}}{12}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**3+A)/(b*x**3+a)**(3/2),x)

[Out]

Piecewise((-16*A*a**2/(9*b**3*sqrt(a + b*x**3)) - 8*A*a*x**3/(9*b**2*sqrt(a + b*x**3)) + 2*A*x**6/(9*b*sqrt(a
+ b*x**3)) + 32*B*a**3/(15*b**4*sqrt(a + b*x**3)) + 16*B*a**2*x**3/(15*b**3*sqrt(a + b*x**3)) - 4*B*a*x**6/(15
*b**2*sqrt(a + b*x**3)) + 2*B*x**9/(15*b*sqrt(a + b*x**3)), Ne(b, 0)), ((A*x**9/9 + B*x**12/12)/a**(3/2), True
))

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Giac [A]
time = 0.78, size = 114, normalized size = 1.11 \begin {gather*} \frac {2 \, {\left (B a^{3} - A a^{2} b\right )}}{3 \, \sqrt {b x^{3} + a} b^{4}} + \frac {2 \, {\left (3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} B b^{16} - 15 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} B a b^{16} + 45 \, \sqrt {b x^{3} + a} B a^{2} b^{16} + 5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} A b^{17} - 30 \, \sqrt {b x^{3} + a} A a b^{17}\right )}}{45 \, b^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

2/3*(B*a^3 - A*a^2*b)/(sqrt(b*x^3 + a)*b^4) + 2/45*(3*(b*x^3 + a)^(5/2)*B*b^16 - 15*(b*x^3 + a)^(3/2)*B*a*b^16
+ 45*sqrt(b*x^3 + a)*B*a^2*b^16 + 5*(b*x^3 + a)^(3/2)*A*b^17 - 30*sqrt(b*x^3 + a)*A*a*b^17)/b^20

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Mupad [B]
time = 2.77, size = 152, normalized size = 1.48 \begin {gather*} \frac {\sqrt {b\,x^3+a}\,\left (\frac {2\,\left (B\,a^2-A\,a\,b\right )}{b^3}-\frac {2\,a\,\left (\frac {2\,\left (A\,b^2-B\,a\,b\right )}{b^3}-\frac {8\,B\,a}{5\,b^2}\right )}{3\,b}\right )}{3\,b}+\frac {x^3\,\sqrt {b\,x^3+a}\,\left (\frac {2\,\left (A\,b^2-B\,a\,b\right )}{b^3}-\frac {8\,B\,a}{5\,b^2}\right )}{9\,b}-\frac {a^2\,\left (\frac {2\,A}{3\,b}-\frac {2\,B\,a}{3\,b^2}\right )}{b^2\,\sqrt {b\,x^3+a}}+\frac {2\,B\,x^6\,\sqrt {b\,x^3+a}}{15\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8*(A + B*x^3))/(a + b*x^3)^(3/2),x)

[Out]

((a + b*x^3)^(1/2)*((2*(B*a^2 - A*a*b))/b^3 - (2*a*((2*(A*b^2 - B*a*b))/b^3 - (8*B*a)/(5*b^2)))/(3*b)))/(3*b)
+ (x^3*(a + b*x^3)^(1/2)*((2*(A*b^2 - B*a*b))/b^3 - (8*B*a)/(5*b^2)))/(9*b) - (a^2*((2*A)/(3*b) - (2*B*a)/(3*b
^2)))/(b^2*(a + b*x^3)^(1/2)) + (2*B*x^6*(a + b*x^3)^(1/2))/(15*b^2)

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