3.8.64 \(\int \frac {(a+b x^2)^{4/3}}{(c x)^{2/3}} \, dx\) [764]

3.8.64.1 Optimal result

Integrand size = 19, antiderivative size = 414 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx=\frac {8 a \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{9 c}+\frac {\sqrt [3]{c x} \left (a+b x^2\right )^{4/3}}{3 c}+\frac {8 a \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{9 \sqrt [4]{3} c^{5/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}} \]

output

8/9*a*(c*x)^(1/3)*(b*x^2+a)^(1/3)/c+1/3*(c*x)^(1/3)*(b*x^2+a)^(4/3)/c+8/27 
*a*(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)
 

3.8.64.2 Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.13 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx=\frac {3 a x \sqrt [3]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{6},\frac {7}{6},-\frac {b x^2}{a}\right )}{(c x)^{2/3} \sqrt [3]{1+\frac {b x^2}{a}}} \]

input

Integrate[(a + b*x^2)^(4/3)/(c*x)^(2/3),x]
 

output

(3*a*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, 1/6, 7/6, -((b*x^2)/a)])/ 
((c*x)^(2/3)*(1 + (b*x^2)/a)^(1/3))
 

3.8.64.3 Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 351, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {248, 248, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {8}{9} a \int \frac {\sqrt [3]{b x^2+a}}{(c x)^{2/3}}dx+\frac {\sqrt [3]{c x} \left (a+b x^2\right )^{4/3}}{3 c}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {8}{9} a \left (\frac {2}{3} a \int \frac {1}{(c x)^{2/3} \left (b x^2+a\right )^{2/3}}dx+\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}\right )+\frac {\sqrt [3]{c x} \left (a+b x^2\right )^{4/3}}{3 c}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {8}{9} a \left (\frac {2 a \int \frac {1}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{c}+\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}\right )+\frac {\sqrt [3]{c x} \left (a+b x^2\right )^{4/3}}{3 c}\)

\(\Big \downarrow \) 771

\(\displaystyle \frac {8}{9} a \left (\frac {2 a \int \frac {1}{\sqrt {1-b x^2}}d\frac {\sqrt [3]{c x}}{\sqrt [6]{b x^2+a}}}{c \sqrt {a+b x^2} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}+\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}\right )+\frac {\sqrt [3]{c x} \left (a+b x^2\right )^{4/3}}{3 c}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {8}{9} a \left (\frac {a \sqrt [3]{c x} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right ) \sqrt {\frac {b^{2/3} (c x)^{4/3}+\sqrt [3]{b} c^{2/3} (c x)^{2/3}+c^{4/3}}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} c^{5/3} \sqrt {1-b x^2} \left (a+b x^2\right )^{2/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right )}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}+\frac {\sqrt [3]{c x} \sqrt [3]{a+b x^2}}{c}\right )+\frac {\sqrt [3]{c x} \left (a+b x^2\right )^{4/3}}{3 c}\)

input

Int[(a + b*x^2)^(4/3)/(c*x)^(2/3),x]
 

output

((c*x)^(1/3)*(a + b*x^2)^(4/3))/(3*c) + (8*a*(((c*x)^(1/3)*(a + b*x^2)^(1/ 
3))/c + (a*(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])/(3^(1/4)*c^(5/3)*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)])))/9
 

3.8.64.3.1 Defintions of rubi rules used

rule 248

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) 
  Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ 
p, 0] && NeQ[m + 2*p + 1, 0] && 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]]
 

3.8.64.4 Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{\left (c x \right )^{\frac {2}{3}}}d x\]

input

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

output

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

3.8.64.5 Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {2}{3}}} \,d x } \]

input

integrate((b*x^2+a)^(4/3)/(c*x)^(2/3),x, algorithm="fricas")
 

output

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

3.8.64.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.54 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.11 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx=\frac {a^{\frac {4}{3}} \sqrt [3]{x} \Gamma \left (\frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{6} \\ \frac {7}{6} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {2}{3}} \Gamma \left (\frac {7}{6}\right )} \]

input

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

output

a**(4/3)*x**(1/3)*gamma(1/6)*hyper((-4/3, 1/6), (7/6,), b*x**2*exp_polar(I 
*pi)/a)/(2*c**(2/3)*gamma(7/6))
 

3.8.64.7 Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {2}{3}}} \,d x } \]

input

integrate((b*x^2+a)^(4/3)/(c*x)^(2/3),x, algorithm="maxima")
 

output

integrate((b*x^2 + a)^(4/3)/(c*x)^(2/3), x)
 

3.8.64.8 Giac [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {2}{3}}} \,d x } \]

input

integrate((b*x^2+a)^(4/3)/(c*x)^(2/3),x, algorithm="giac")
 

output

integrate((b*x^2 + a)^(4/3)/(c*x)^(2/3), x)
 

3.8.64.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{2/3}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{4/3}}{{\left (c\,x\right )}^{2/3}} \,d x \]

input

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

output

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