3.748 \(\int \frac {\sqrt [3]{a+b x^2}}{(c x)^{2/3}} \, dx\)

Optimal. Leaf size=381 \[ \frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\cos ^{-1}\left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} c^{5/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}+\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c} \]

[Out]

(c*x)^(1/3)*(b*x^2+a)^(1/3)/c+1/3*(c*x)^(1/3)*(b*x^2+a)^(1/3)*(c^(2/3)-b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))*((
c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1/3))^2/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^
(1/3))^2)^(1/2)/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1/3))*(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1
/2))/(b*x^2+a)^(1/3))*EllipticF((1-(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1-3^(1/2))/(b*x^2+a)^(1/3))^2/(c^(2/3)-b^(1/3
)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*((c^(4/3)+b^(2/3)*(c*x)^(4/3)/(b*
x^2+a)^(2/3)+b^(1/3)*c^(2/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(
1/3))^2)^(1/2)*3^(3/4)/c^(5/3)/(-b^(1/3)*(c*x)^(2/3)*(c^(2/3)-b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))/(b*x^2+a)^(
1/3)/(c^(2/3)-b^(1/3)*(c*x)^(2/3)*(1+3^(1/2))/(b*x^2+a)^(1/3))^2)^(1/2)

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Rubi [A]  time = 0.66, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {279, 329, 241, 225} \[ \frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}+c^{4/3}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{b x^2+a}}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} c^{5/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}+\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/3)/(c*x)^(2/3),x]

[Out]

((c*x)^(1/3)*(a + b*x^2)^(1/3))/c + ((c*x)^(1/3)*(a + b*x^2)^(1/3)*(c^(2/3) - (b^(1/3)*(c*x)^(2/3))/(a + b*x^2
)^(1/3))*Sqrt[(c^(4/3) + (b^(2/3)*(c*x)^(4/3))/(a + b*x^2)^(2/3) + (b^(1/3)*c^(2/3)*(c*x)^(2/3))/(a + b*x^2)^(
1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2]*EllipticF[ArcCos[(c^(2/3) - ((1 - S
qrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))/(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/
3))], (2 + Sqrt[3])/4])/(3^(1/4)*c^(5/3)*Sqrt[-((b^(1/3)*(c*x)^(2/3)*(c^(2/3) - (b^(1/3)*(c*x)^(2/3))/(a + b*x
^2)^(1/3)))/((a + b*x^2)^(1/3)*(c^(2/3) - ((1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(a + b*x^2)^(1/3))^2))])

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rule 241

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a/(a + b*x^n))^(p + 1/n)*(a + b*x^n)^(p + 1/n), Subst[In
t[1/(1 - b*x^n)^(p + 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)] && LtQ[Denominator[p + 1/n], Denominator[p]]

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{2/3}} \, dx &=\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}+\frac {1}{3} (2 a) \int \frac {1}{(c x)^{2/3} \left (a+b x^2\right )^{2/3}} \, dx\\ &=\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{\left (a+\frac {b x^6}{c^2}\right )^{2/3}} \, dx,x,\sqrt [3]{c x}\right )}{c}\\ &=\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {b x^6}{c^2}}} \, dx,x,\frac {\sqrt [3]{c x}}{\sqrt [6]{a+b x^2}}\right )}{c \sqrt {\frac {a}{a+b x^2}} \sqrt {a+b x^2}}\\ &=\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}+\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} c^{5/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.14 \[ \frac {3 x \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {1}{3},\frac {1}{6};\frac {7}{6};-\frac {b x^2}{a}\right )}{(c x)^{2/3} \sqrt [3]{\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/3)/(c*x)^(2/3),x]

[Out]

(3*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-1/3, 1/6, 7/6, -((b*x^2)/a)])/((c*x)^(2/3)*(1 + (b*x^2)/a)^(1/3))

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}}}{c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(2/3),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/3)*(c*x)^(1/3)/(c*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}}}{\left (c x\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(2/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(2/3), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{\left (c x \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/3)/(c*x)^(2/3),x)

[Out]

int((b*x^2+a)^(1/3)/(c*x)^(2/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}}}{\left (c x\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(2/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(2/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/3)/(c*x)^(2/3),x)

[Out]

int((a + b*x^2)^(1/3)/(c*x)^(2/3), x)

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sympy [C]  time = 1.31, size = 46, normalized size = 0.12 \[ \frac {\sqrt [3]{a} \sqrt [3]{x} \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{6} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {2}{3}} \Gamma \left (\frac {7}{6}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/3)/(c*x)**(2/3),x)

[Out]

a**(1/3)*x**(1/3)*gamma(1/6)*hyper((-1/3, 1/6), (7/6,), b*x**2*exp_polar(I*pi)/a)/(2*c**(2/3)*gamma(7/6))

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