3.428 \(\int \frac {x^6}{(a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=251 \[ -\frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {16 x \sqrt {a+b x^3}}{15 b^2}-\frac {2 x^4}{3 b \sqrt {a+b x^3}} \]

[Out]

-2/3*x^4/b/(b*x^3+a)^(1/2)+16/15*x*(b*x^3+a)^(1/2)/b^2-32/45*a*(a^(1/3)+b^(1/3)*x)*EllipticF((b^(1/3)*x+a^(1/3
)*(1-3^(1/2)))/(b^(1/3)*x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)-a^(1/3)*b^(1
/3)*x+b^(2/3)*x^2)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/b^(7/3)/(b*x^3+a)^(1/2)/(a^(1/3)*(a^(1/3)+
b^(1/3)*x)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {288, 321, 218} \[ -\frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {16 x \sqrt {a+b x^3}}{15 b^2}-\frac {2 x^4}{3 b \sqrt {a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^3)^(3/2),x]

[Out]

(-2*x^4)/(3*b*Sqrt[a + b*x^3]) + (16*x*Sqrt[a + b*x^3])/(15*b^2) - (32*Sqrt[2 + Sqrt[3]]*a*(a^(1/3) + b^(1/3)*
x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((
1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(15*3^(1/4)*b^(7/3)*S
qrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, 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]

Rubi steps

\begin {align*} \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx &=-\frac {2 x^4}{3 b \sqrt {a+b x^3}}+\frac {8 \int \frac {x^3}{\sqrt {a+b x^3}} \, dx}{3 b}\\ &=-\frac {2 x^4}{3 b \sqrt {a+b x^3}}+\frac {16 x \sqrt {a+b x^3}}{15 b^2}-\frac {(16 a) \int \frac {1}{\sqrt {a+b x^3}} \, dx}{15 b^2}\\ &=-\frac {2 x^4}{3 b \sqrt {a+b x^3}}+\frac {16 x \sqrt {a+b x^3}}{15 b^2}-\frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 65, normalized size = 0.26 \[ \frac {2 x \left (-8 a \sqrt {\frac {b x^3}{a}+1} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {b x^3}{a}\right )+8 a+3 b x^3\right )}{15 b^2 \sqrt {a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^3)^(3/2),x]

[Out]

(2*x*(8*a + 3*b*x^3 - 8*a*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x^3)/a)]))/(15*b^2*Sqrt[a
+ b*x^3])

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{3} + a} x^{6}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^3 + a)*x^6/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

integrate(x^6/(b*x^3 + a)^(3/2), x)

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maple [A]  time = 0.02, size = 320, normalized size = 1.27 \[ \frac {2 a x}{3 \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}\, b^{2}}+\frac {32 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, a \EllipticF \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) b}}\right )}{45 \sqrt {b \,x^{3}+a}\, b^{3}}+\frac {2 \sqrt {b \,x^{3}+a}\, x}{5 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^3+a)^(3/2),x)

[Out]

2/3/b^2*a*x/((x^3+a/b)*b)^(1/2)+2/5*x*(b*x^3+a)^(1/2)/b^2+32/45*I*a/b^3*3^(1/2)*(-a*b^2)^(1/3)*(I*(x+1/2*(-a*b
^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2)*((x-(-a*b^2)^(1/3)/b)/(-3/2*(-a*b^
2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b))^(1/2)*(-I*(x+1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*
3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)
*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2),(I*3^(1/2)*(-a*b^2)^(1/3)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1
/2)*(-a*b^2)^(1/3)/b)/b)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{6}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^6/(b*x^3 + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^6}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a + b*x^3)^(3/2),x)

[Out]

int(x^6/(a + b*x^3)^(3/2), x)

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sympy [A]  time = 2.74, size = 37, normalized size = 0.15 \[ \frac {x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**3+a)**(3/2),x)

[Out]

x**7*gamma(7/3)*hyper((3/2, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(10/3))

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