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

Optimal. Leaf size=272 \[ \frac {54 a^2 x \sqrt {a+b x^3}}{935 b}-\frac {36\ 3^{3/4} \sqrt {2+\sqrt {3}} a^3 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{935 b^{4/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {2}{17} x^4 \left (a+b x^3\right )^{3/2}+\frac {18}{187} a x^4 \sqrt {a+b x^3} \]

[Out]

2/17*x^4*(b*x^3+a)^(3/2)+54/935*a^2*x*(b*x^3+a)^(1/2)/b+18/187*a*x^4*(b*x^3+a)^(1/2)-36/935*3^(3/4)*a^3*(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)/b^(4/3)/(
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)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {279, 321, 218} \[ -\frac {36\ 3^{3/4} \sqrt {2+\sqrt {3}} a^3 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{935 b^{4/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {54 a^2 x \sqrt {a+b x^3}}{935 b}+\frac {2}{17} x^4 \left (a+b x^3\right )^{3/2}+\frac {18}{187} a x^4 \sqrt {a+b x^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^3)^(3/2),x]

[Out]

(54*a^2*x*Sqrt[a + b*x^3])/(935*b) + (18*a*x^4*Sqrt[a + b*x^3])/187 + (2*x^4*(a + b*x^3)^(3/2))/17 - (36*3^(3/
4)*Sqrt[2 + Sqrt[3]]*a^3*(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]])/(935*b^(4/3)*Sqrt[(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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 x^3 \left (a+b x^3\right )^{3/2} \, dx &=\frac {2}{17} x^4 \left (a+b x^3\right )^{3/2}+\frac {1}{17} (9 a) \int x^3 \sqrt {a+b x^3} \, dx\\ &=\frac {18}{187} a x^4 \sqrt {a+b x^3}+\frac {2}{17} x^4 \left (a+b x^3\right )^{3/2}+\frac {1}{187} \left (27 a^2\right ) \int \frac {x^3}{\sqrt {a+b x^3}} \, dx\\ &=\frac {54 a^2 x \sqrt {a+b x^3}}{935 b}+\frac {18}{187} a x^4 \sqrt {a+b x^3}+\frac {2}{17} x^4 \left (a+b x^3\right )^{3/2}-\frac {\left (54 a^3\right ) \int \frac {1}{\sqrt {a+b x^3}} \, dx}{935 b}\\ &=\frac {54 a^2 x \sqrt {a+b x^3}}{935 b}+\frac {18}{187} a x^4 \sqrt {a+b x^3}+\frac {2}{17} x^4 \left (a+b x^3\right )^{3/2}-\frac {36\ 3^{3/4} \sqrt {2+\sqrt {3}} a^3 \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 )}{935 b^{4/3} \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*}

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 67, normalized size = 0.25 \[ \frac {2 x \sqrt {a+b x^3} \left (\left (a+b x^3\right )^2-\frac {a^2 \, _2F_1\left (-\frac {3}{2},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\sqrt {\frac {b x^3}{a}+1}}\right )}{17 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^3)^(3/2),x]

[Out]

(2*x*Sqrt[a + b*x^3]*((a + b*x^3)^2 - (a^2*Hypergeometric2F1[-3/2, 1/3, 4/3, -((b*x^3)/a)])/Sqrt[1 + (b*x^3)/a
]))/(17*b)

________________________________________________________________________________________

fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{6} + a x^{3}\right )} \sqrt {b x^{3} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2)*x^3, x)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 335, normalized size = 1.23 \[ \frac {2 \sqrt {b \,x^{3}+a}\, b \,x^{7}}{17}+\frac {40 \sqrt {b \,x^{3}+a}\, a \,x^{4}}{187}+\frac {36 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, a^{3} \EllipticF \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) b}}\right )}{935 \sqrt {b \,x^{3}+a}\, b^{2}}+\frac {54 \sqrt {b \,x^{3}+a}\, a^{2} x}{935 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/17*b*x^7*(b*x^3+a)^(1/2)+40/187*a*x^4*(b*x^3+a)^(1/2)+54/935*a^2*x*(b*x^3+a)^(1/2)/b+36/935*I*a^3/b^2*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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2)*x^3, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^3)^(3/2),x)

[Out]

int(x^3*(a + b*x^3)^(3/2), x)

________________________________________________________________________________________

sympy [A]  time = 2.57, size = 39, normalized size = 0.14 \[ \frac {a^{\frac {3}{2}} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*x**4*gamma(4/3)*hyper((-3/2, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))

________________________________________________________________________________________