\(\int \frac {1}{(c x)^{14/3} (a+b x^2)^{2/3}} \, dx\) [831]

Optimal result

Integrand size = 19, antiderivative size = 425 \[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=-\frac {3 \sqrt [3]{a+b x^2}}{11 a c (c x)^{11/3}}+\frac {27 b \sqrt [3]{a+b x^2}}{55 a^2 c^3 (c x)^{5/3}}+\frac {27\ 3^{3/4} b^2 \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{110 a^3 c^{17/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}} \] Output:

-3/11*(b*x^2+a)^(1/3)/a/c/(c*x)^(11/3)+27/55*b*(b*x^2+a)^(1/3)/a^2/c^3/(c* 
x)^(5/3)+27/110*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^(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)-(1+3^(1/2))*b^(1/3)* 
(c*x)^(2/3)/(b*x^2+a)^(1/3))^2)^(1/2)*InverseJacobiAM(arccos((c^(2/3)-(1-3 
^(1/2))*b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))/(c^(2/3)-(1+3^(1/2))*b^(1/3)* 
(c*x)^(2/3)/(b*x^2+a)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2))/a^3/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)-(1+3^(1/2))*b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.13 \[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=-\frac {3 x \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {2}{3},-\frac {5}{6},-\frac {b x^2}{a}\right )}{11 (c x)^{14/3} \left (a+b x^2\right )^{2/3}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.38 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {264, 264, 266, 771, 766}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {9 b \int \frac {1}{(c x)^{8/3} \left (b x^2+a\right )^{2/3}}dx}{11 a c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 a c (c x)^{11/3}}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {9 b \left (-\frac {3 b \int \frac {1}{(c x)^{2/3} \left (b x^2+a\right )^{2/3}}dx}{5 a c^2}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 a c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 a c (c x)^{11/3}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {9 b \left (-\frac {9 b \int \frac {1}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{5 a c^3}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 a c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 a c (c x)^{11/3}}\)

\(\Big \downarrow \) 771

\(\displaystyle -\frac {9 b \left (-\frac {9 b \int \frac {1}{\sqrt {1-b x^2}}d\frac {\sqrt [3]{c x}}{\sqrt [6]{b x^2+a}}}{5 a c^3 \sqrt {a+b x^2} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 a c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 a c (c x)^{11/3}}\)

\(\Big \downarrow \) 766

\(\displaystyle -\frac {9 b \left (-\frac {3\ 3^{3/4} b \sqrt [3]{c x} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right ) \sqrt {\frac {b^{2/3} (c x)^{4/3}+\sqrt [3]{b} c^{2/3} (c x)^{2/3}+c^{4/3}}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{10 a c^{11/3} \sqrt {1-b x^2} \left (a+b x^2\right )^{2/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right )}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 a c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 a c (c x)^{11/3}}\)

Input:

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

Output:

(-3*(a + b*x^2)^(1/3))/(11*a*c*(c*x)^(11/3)) - (9*b*((-3*(a + b*x^2)^(1/3) 
)/(5*a*c*(c*x)^(5/3)) - (3*3^(3/4)*b*(c*x)^(1/3)*(c^(2/3) - b^(1/3)*(c*x)^ 
(2/3))*Sqrt[(c^(4/3) + b^(1/3)*c^(2/3)*(c*x)^(2/3) + b^(2/3)*(c*x)^(4/3))/ 
(c^(2/3) - (1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))^2]*EllipticF[ArcCos[(c^(2/3) 
 - (1 - Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(c^(2/3) - (1 + Sqrt[3])*b^(1/3)*(c* 
x)^(2/3))], (2 + Sqrt[3])/4])/(10*a*c^(11/3)*Sqrt[1 - b*x^2]*(a + b*x^2)^( 
2/3)*Sqrt[(a*c^2)/(a*c^2 + b*c^2*x^2)]*Sqrt[-((b^(1/3)*(c*x)^(2/3)*(c^(2/3 
) - b^(1/3)*(c*x)^(2/3)))/(c^(2/3) - (1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))^2) 
])))/(11*a*c^2)
 

Defintions of rubi rules used

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 766

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 771

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a/(a + b*x^n))^(p + 1 
/n)*(a + b*x^n)^(p + 1/n)   Subst[Int[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]]
 

Maple [F]

\[\int \frac {1}{\left (c x \right )^{\frac {14}{3}} \left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {14}{3}}} \,d x } \] Input:

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

Output:

integral((b*x^2 + a)^(1/3)*(c*x)^(1/3)/(b*c^5*x^7 + a*c^5*x^5), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {14}{3}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {14}{3}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=\int \frac {1}{{\left (c\,x\right )}^{14/3}\,{\left (b\,x^2+a\right )}^{2/3}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(c x)^{14/3} \left (a+b x^2\right )^{2/3}} \, dx=\frac {\int \frac {1}{x^{\frac {14}{3}} \left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x}{c^{\frac {14}{3}}} \] Input:

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

Output:

int(1/(x**(2/3)*(a + b*x**2)**(2/3)*x**4),x)/(c**(2/3)*c**4)