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

Optimal. Leaf size=246 \[ \frac {18\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {18}{55} a x \sqrt {a+b x^3}+\frac {2}{11} x \left (a+b x^3\right )^{3/2} \]

[Out]

2/11*x*(b*x^3+a)^(3/2)+18/55*a*x*(b*x^3+a)^(1/2)+18/55*3^(3/4)*a^2*(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^(1/3)/(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 = 0.06, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {195, 218} \[ \frac {18\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{55 \sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}+\frac {18}{55} a x \sqrt {a+b x^3}+\frac {2}{11} x \left (a+b x^3\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(3/2),x]

[Out]

(18*a*x*Sqrt[a + b*x^3])/55 + (2*x*(a + b*x^3)^(3/2))/11 + (18*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*(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]])/(55*b^(1/3)*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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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]

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^{3/2} \, dx &=\frac {2}{11} x \left (a+b x^3\right )^{3/2}+\frac {1}{11} (9 a) \int \sqrt {a+b x^3} \, dx\\ &=\frac {18}{55} a x \sqrt {a+b x^3}+\frac {2}{11} x \left (a+b x^3\right )^{3/2}+\frac {1}{55} \left (27 a^2\right ) \int \frac {1}{\sqrt {a+b x^3}} \, dx\\ &=\frac {18}{55} a x \sqrt {a+b x^3}+\frac {2}{11} x \left (a+b x^3\right )^{3/2}+\frac {18\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \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 )}{55 \sqrt [3]{b} \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 = 0.01, size = 47, normalized size = 0.19 \[ \frac {a x \sqrt {a+b x^3} \, _2F_1\left (-\frac {3}{2},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\sqrt {\frac {b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(3/2),x]

[Out]

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

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3 + a)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2), x)

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maple [A]  time = 0.02, size = 315, normalized size = 1.28 \[ \frac {2 \sqrt {b \,x^{3}+a}\, b \,x^{4}}{11}-\frac {18 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, a^{2} \EllipticF \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) b}}\right )}{55 \sqrt {b \,x^{3}+a}\, b}+\frac {28 \sqrt {b \,x^{3}+a}\, a x}{55} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*b*x^4*(b*x^3+a)^(1/2)+28/55*a*x*(b*x^3+a)^(1/2)-18/55*I*a^2*3^(1/2)*(-a*b^2)^(1/3)/b*(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 \[ \int {\left (b x^{3} + a\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2), x)

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mupad [B]  time = 1.02, size = 37, normalized size = 0.15 \[ \frac {x\,{\left (b\,x^3+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{3};\ \frac {4}{3};\ -\frac {b\,x^3}{a}\right )}{{\left (\frac {b\,x^3}{a}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(3/2),x)

[Out]

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

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sympy [A]  time = 1.33, size = 37, normalized size = 0.15 \[ \frac {a^{\frac {3}{2}} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*x*gamma(1/3)*hyper((-3/2, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3))

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