3.8.50 \(\int \frac {\sqrt [3]{a+b x^2}}{(c x)^{14/3}} \, dx\) [750]

Optimal. Leaf size=422 \[ -\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}-\frac {6 b \sqrt [3]{a+b x^2}}{55 a c^3 (c x)^{5/3}}-\frac {3\ 3^{3/4} b^2 \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} F\left (\cos ^{-1}\left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{55 a^2 c^{17/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}} \]

[Out]

-3/11*(b*x^2+a)^(1/3)/c/(c*x)^(11/3)-6/55*b*(b*x^2+a)^(1/3)/a/c^3/(c*x)^(5/3)-3/55*3^(3/4)*b^2*(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)/a^2/c^(17/3)/(-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)

________________________________________________________________________________________

Rubi [A]
time = 0.51, antiderivative size = 422, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {283, 331, 335, 247, 231} \begin {gather*} -\frac {3\ 3^{3/4} b^2 \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 (\text {ArcCos}\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 )}{55 a^2 c^{17/3} \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 {6 b \sqrt [3]{a+b x^2}}{55 a c^3 (c x)^{5/3}}-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/3)/(c*x)^(14/3),x]

[Out]

(-3*(a + b*x^2)^(1/3))/(11*c*(c*x)^(11/3)) - (6*b*(a + b*x^2)^(1/3))/(55*a*c^3*(c*x)^(5/3)) - (3*3^(3/4)*b^2*(
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 - Sqrt[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])/(55*a^2*c^(17/
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 231

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]/(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)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 247

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 283

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 331

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

Rule 335

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)^{14/3}} \, dx &=-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}+\frac {(2 b) \int \frac {1}{(c x)^{8/3} \left (a+b x^2\right )^{2/3}} \, dx}{11 c^2}\\ &=-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}-\frac {6 b \sqrt [3]{a+b x^2}}{55 a c^3 (c x)^{5/3}}-\frac {\left (6 b^2\right ) \int \frac {1}{(c x)^{2/3} \left (a+b x^2\right )^{2/3}} \, dx}{55 a c^4}\\ &=-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}-\frac {6 b \sqrt [3]{a+b x^2}}{55 a c^3 (c x)^{5/3}}-\frac {\left (18 b^2\right ) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^6}{c^2}\right )^{2/3}} \, dx,x,\sqrt [3]{c x}\right )}{55 a c^5}\\ &=-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}-\frac {6 b \sqrt [3]{a+b x^2}}{55 a c^3 (c x)^{5/3}}-\frac {\left (18 b^2\right ) \text {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 )}{55 a c^5 \sqrt {\frac {a}{a+b x^2}} \sqrt {a+b x^2}}\\ &=-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}-\frac {6 b \sqrt [3]{a+b x^2}}{55 a c^3 (c x)^{5/3}}-\frac {3\ 3^{3/4} b^2 \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 )}{55 a^2 c^{17/3} \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*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 56, normalized size = 0.13 \begin {gather*} -\frac {3 x \sqrt [3]{a+b x^2} \, _2F_1\left (-\frac {11}{6},-\frac {1}{3};-\frac {5}{6};-\frac {b x^2}{a}\right )}{11 (c x)^{14/3} \sqrt [3]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/3)/(c*x)^(14/3),x]

[Out]

(-3*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-11/6, -1/3, -5/6, -((b*x^2)/a)])/(11*(c*x)^(14/3)*(1 + (b*x^2)/a)^(
1/3))

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{\left (c x \right )^{\frac {14}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(1/3)/(c*x)^(14/3),x)

[Out]

int((b*x^2+a)^(1/3)/(c*x)^(14/3),x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(14/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(14/3), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(14/3),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/3)*(c*x)^(1/3)/(c^5*x^5), x)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(1/3)/(c*x)**(14/3),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3278 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(1/3)/(c*x)^(14/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/3)/(c*x)^(14/3), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{1/3}}{{\left (c\,x\right )}^{14/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/3)/(c*x)^(14/3),x)

[Out]

int((a + b*x^2)^(1/3)/(c*x)^(14/3), x)

________________________________________________________________________________________