3.567 \(\int \frac {x^7}{(a+b x^3)^{2/3}} \, dx\)

Optimal. Leaf size=123 \[ -\frac {5 a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{8/3}}-\frac {5 a^2 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3}}-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b} \]

[Out]

-5/18*a*x^2*(b*x^3+a)^(1/3)/b^2+1/6*x^5*(b*x^3+a)^(1/3)/b-5/18*a^2*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(8/3)-5/27*
a^2*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(8/3)*3^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 176, normalized size of antiderivative = 1.43, number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {321, 331, 292, 31, 634, 617, 204, 628} \[ -\frac {5 a^2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{8/3}}+\frac {5 a^2 \log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{54 b^{8/3}}-\frac {5 a^2 \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3}}-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(a + b*x^3)^(2/3),x]

[Out]

(-5*a*x^2*(a + b*x^3)^(1/3))/(18*b^2) + (x^5*(a + b*x^3)^(1/3))/(6*b) - (5*a^2*ArcTan[(1 + (2*b^(1/3)*x)/(a +
b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(8/3)) - (5*a^2*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(27*b^(8/3)) + (5
*a^2*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(54*b^(8/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {x^7}{\left (a+b x^3\right )^{2/3}} \, dx &=\frac {x^5 \sqrt [3]{a+b x^3}}{6 b}-\frac {(5 a) \int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx}{6 b}\\ &=-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b}+\frac {\left (5 a^2\right ) \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx}{9 b^2}\\ &=-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 b^2}\\ &=-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{7/3}}-\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1-\sqrt [3]{b} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{7/3}}\\ &=-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b}-\frac {5 a^2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{8/3}}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{b}+2 b^{2/3} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{54 b^{8/3}}-\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{18 b^{7/3}}\\ &=-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b}-\frac {5 a^2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{8/3}}+\frac {5 a^2 \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{54 b^{8/3}}+\frac {\left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{8/3}}\\ &=-\frac {5 a x^2 \sqrt [3]{a+b x^3}}{18 b^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 b}-\frac {5 a^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3}}-\frac {5 a^2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{27 b^{8/3}}+\frac {5 a^2 \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{54 b^{8/3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 69, normalized size = 0.56 \[ \frac {x^2 \left (5 a^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {b x^3}{b x^3+a}\right )-5 a^2-2 a b x^3+3 b^2 x^6\right )}{18 b^2 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a + b*x^3)^(2/3),x]

[Out]

(x^2*(-5*a^2 - 2*a*b*x^3 + 3*b^2*x^6 + 5*a^2*Hypergeometric2F1[2/3, 1, 5/3, (b*x^3)/(a + b*x^3)]))/(18*b^2*(a
+ b*x^3)^(2/3))

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fricas [B]  time = 0.66, size = 212, normalized size = 1.72 \[ \frac {10 \, \sqrt {3} a^{2} b \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) - 10 \, \left (-b^{2}\right )^{\frac {2}{3}} a^{2} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + 5 \, \left (-b^{2}\right )^{\frac {2}{3}} a^{2} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + 3 \, {\left (3 \, b^{3} x^{5} - 5 \, a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

1/54*(10*sqrt(3)*a^2*b*sqrt(-(-b^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-b^2)^(1/3)*b*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)
*(-b^2)^(2/3))*sqrt(-(-b^2)^(1/3))/(b^2*x)) - 10*(-b^2)^(2/3)*a^2*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/
x) + 5*(-b^2)^(2/3)*a^2*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x - (b*x^3 + a)^(2/3)*b)/x^2
) + 3*(3*b^3*x^5 - 5*a*b^2*x^2)*(b*x^3 + a)^(1/3))/b^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{7}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

integrate(x^7/(b*x^3 + a)^(2/3), x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {x^{7}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^3+a)^(2/3),x)

[Out]

int(x^7/(b*x^3+a)^(2/3),x)

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maxima [A]  time = 3.00, size = 184, normalized size = 1.50 \[ \frac {5 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{27 \, b^{\frac {8}{3}}} + \frac {5 \, a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{54 \, b^{\frac {8}{3}}} - \frac {5 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{27 \, b^{\frac {8}{3}}} + \frac {\frac {8 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} b}{x} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}}{x^{4}}}{18 \, {\left (b^{4} - \frac {2 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {{\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

5/27*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(8/3) + 5/54*a^2*log(b^(2/3)
+ (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(8/3) - 5/27*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/
b^(8/3) + 1/18*(8*(b*x^3 + a)^(1/3)*a^2*b/x - 5*(b*x^3 + a)^(4/3)*a^2/x^4)/(b^4 - 2*(b*x^3 + a)*b^3/x^3 + (b*x
^3 + a)^2*b^2/x^6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^7}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(a + b*x^3)^(2/3),x)

[Out]

int(x^7/(a + b*x^3)^(2/3), x)

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sympy [C]  time = 2.84, size = 37, normalized size = 0.30 \[ \frac {x^{8} \Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {11}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(b*x**3+a)**(2/3),x)

[Out]

x**8*gamma(8/3)*hyper((2/3, 8/3), (11/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(11/3))

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