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

Optimal. Leaf size=246 \[ \frac{9\ 3^{3/4} \sqrt{2+\sqrt{3}} a b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{10 \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{\left (a+b x^3\right )^{3/2}}{2 x^2}+\frac{9}{10} b x \sqrt{a+b x^3} \]

[Out]

(9*b*x*Sqrt[a + b*x^3])/10 - (a + b*x^3)^(3/2)/(2*x^2) + (9*3^(3/4)*Sqrt[2 + Sqrt[3]]*a*b^(2/3)*(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[ArcS
in[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(10*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])

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Rubi [A]  time = 0.0627232, antiderivative size = 246, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {277, 195, 218} \[ \frac{9\ 3^{3/4} \sqrt{2+\sqrt{3}} a b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{10 \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{\left (a+b x^3\right )^{3/2}}{2 x^2}+\frac{9}{10} b x \sqrt{a+b x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(3/2)/x^3,x]

[Out]

(9*b*x*Sqrt[a + b*x^3])/10 - (a + b*x^3)^(3/2)/(2*x^2) + (9*3^(3/4)*Sqrt[2 + Sqrt[3]]*a*b^(2/3)*(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[ArcS
in[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(10*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 277

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

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 \frac{\left (a+b x^3\right )^{3/2}}{x^3} \, dx &=-\frac{\left (a+b x^3\right )^{3/2}}{2 x^2}+\frac{1}{4} (9 b) \int \sqrt{a+b x^3} \, dx\\ &=\frac{9}{10} b x \sqrt{a+b x^3}-\frac{\left (a+b x^3\right )^{3/2}}{2 x^2}+\frac{1}{20} (27 a b) \int \frac{1}{\sqrt{a+b x^3}} \, dx\\ &=\frac{9}{10} b x \sqrt{a+b x^3}-\frac{\left (a+b x^3\right )^{3/2}}{2 x^2}+\frac{9\ 3^{3/4} \sqrt{2+\sqrt{3}} a b^{2/3} \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 )}{10 \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 = 0.0084703, size = 52, normalized size = 0.21 \[ -\frac{a \sqrt{a+b x^3} \, _2F_1\left (-\frac{3}{2},-\frac{2}{3};\frac{1}{3};-\frac{b x^3}{a}\right )}{2 x^2 \sqrt{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(3/2)/x^3,x]

[Out]

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

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Maple [A]  time = 0.016, size = 310, normalized size = 1.3 \begin{align*} -{\frac{a}{2\,{x}^{2}}\sqrt{b{x}^{3}+a}}+{\frac{2\,bx}{5}\sqrt{b{x}^{3}+a}}-{{\frac{9\,i}{10}}a\sqrt{3}\sqrt [3]{-{b}^{2}a}\sqrt{{i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}-{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}\sqrt{{ \left ( x-{\frac{1}{b}\sqrt [3]{-{b}^{2}a}} \right ) \left ( -{\frac{3}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ) ^{-1}}}\sqrt{{-i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}{\it EllipticF} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}-{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}},\sqrt{{\frac{i\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a} \left ( -{\frac{3}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{b{x}^{3}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(b*x^3+a)^(1/2)/x^2+2/5*b*x*(b*x^3+a)^(1/2)-9/10*I*a*3^(1/2)*(-b^2*a)^(1/3)*(I*(x+1/2/b*(-b^2*a)^(1/3)-
1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2)*((x-1/b*(-b^2*a)^(1/3))/(-3/2/b*(-b^2*a)^(1/3)
+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*3^(1/2)*b/
(-b^2*a)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a
)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2),(I*3^(1/2)/b*(-b^2*a)^(1/3)/(-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-
b^2*a)^(1/3)))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2)/x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3 + a)^(3/2)/x^3, x)

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Sympy [A]  time = 1.00607, size = 42, normalized size = 0.17 \begin{align*} \frac{a^{\frac{3}{2}} \Gamma \left (- \frac{2}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, - \frac{2}{3} \\ \frac{1}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac{1}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{3}{2}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(3/2)/x^3, x)