\(\int \frac {(a+b x^2)^{4/3}}{(c x)^{11/3}} \, dx\) [806]

Optimal result

Integrand size = 19, antiderivative size = 157 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=-\frac {3 b \sqrt [3]{a+b x^2}}{2 c^3 (c x)^{2/3}}-\frac {3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}-\frac {\sqrt {3} b^{4/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} (c x)^{2/3}}{c^{2/3} \sqrt [3]{a+b x^2}}}{\sqrt {3}}\right )}{2 c^{11/3}}-\frac {3 b^{4/3} \log \left (\sqrt [3]{b} (c x)^{2/3}-c^{2/3} \sqrt [3]{a+b x^2}\right )}{4 c^{11/3}} \] Output:

-3/2*b*(b*x^2+a)^(1/3)/c^3/(c*x)^(2/3)-3/8*(b*x^2+a)^(4/3)/c/(c*x)^(8/3)-1 
/2*3^(1/2)*b^(4/3)*arctan(1/3*(1+2*b^(1/3)*(c*x)^(2/3)/c^(2/3)/(b*x^2+a)^( 
1/3))*3^(1/2))/c^(11/3)-3/4*b^(4/3)*ln(b^(1/3)*(c*x)^(2/3)-c^(2/3)*(b*x^2+ 
a)^(1/3))/c^(11/3)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=-\frac {x \left (3 a \sqrt [3]{a+b x^2}+15 b x^2 \sqrt [3]{a+b x^2}+4 \sqrt {3} b^{4/3} x^{8/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x^{2/3}}{\sqrt [3]{b} x^{2/3}+2 \sqrt [3]{a+b x^2}}\right )+4 b^{4/3} x^{8/3} \log \left (-\sqrt [3]{b} x^{2/3}+\sqrt [3]{a+b x^2}\right )-2 b^{4/3} x^{8/3} \log \left (b^{2/3} x^{4/3}+\sqrt [3]{b} x^{2/3} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )\right )}{8 (c x)^{11/3}} \] Input:

Integrate[(a + b*x^2)^(4/3)/(c*x)^(11/3),x]
 

Output:

-1/8*(x*(3*a*(a + b*x^2)^(1/3) + 15*b*x^2*(a + b*x^2)^(1/3) + 4*Sqrt[3]*b^ 
(4/3)*x^(8/3)*ArcTan[(Sqrt[3]*b^(1/3)*x^(2/3))/(b^(1/3)*x^(2/3) + 2*(a + b 
*x^2)^(1/3))] + 4*b^(4/3)*x^(8/3)*Log[-(b^(1/3)*x^(2/3)) + (a + b*x^2)^(1/ 
3)] - 2*b^(4/3)*x^(8/3)*Log[b^(2/3)*x^(4/3) + b^(1/3)*x^(2/3)*(a + b*x^2)^ 
(1/3) + (a + b*x^2)^(2/3)]))/(c*x)^(11/3)
 

Rubi [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {247, 247, 266, 807, 853}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {b \int \frac {\sqrt [3]{b x^2+a}}{(c x)^{5/3}}dx}{c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{8 c (c x)^{8/3}}\)

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 853

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

Input:

Int[(a + b*x^2)^(4/3)/(c*x)^(11/3),x]
 

Output:

(-3*(a + b*x^2)^(4/3))/(8*c*(c*x)^(8/3)) + (b*((-3*(a + b*x^2)^(1/3))/(2*c 
*(c*x)^(2/3)) + (3*b*(-((c^(4/3)*ArcTan[(1 + (2*b^(1/3)*(c*x)^(2/3))/(c^(2 
/3)*(a + (b*x)/c)^(1/3)))/Sqrt[3]])/(Sqrt[3]*b^(2/3))) - (c^(4/3)*Log[(b^( 
1/3)*(c*x)^(2/3))/c^(2/3) - (a + (b*x)/c)^(1/3)])/(2*b^(2/3))))/(2*c^3)))/ 
c^2
 

Defintions of rubi rules used

rule 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 853

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim 
p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp 
[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{\left (c x \right )^{\frac {11}{3}}}d x\]

Input:

int((b*x^2+a)^(4/3)/(c*x)^(11/3),x)
 

Output:

int((b*x^2+a)^(4/3)/(c*x)^(11/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=\text {Timed out} \] Input:

integrate((b*x^2+a)^(4/3)/(c*x)^(11/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 33.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=\frac {a^{\frac {4}{3}} \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {4}{3} \\ - \frac {1}{3} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {11}{3}} x^{\frac {8}{3}} \Gamma \left (- \frac {1}{3}\right )} \] Input:

integrate((b*x**2+a)**(4/3)/(c*x)**(11/3),x)
 

Output:

a**(4/3)*gamma(-4/3)*hyper((-4/3, -4/3), (-1/3,), b*x**2*exp_polar(I*pi)/a 
)/(2*c**(11/3)*x**(8/3)*gamma(-1/3))
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {11}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(4/3)/(c*x)^(11/3),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(4/3)/(c*x)^(11/3), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {11}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(4/3)/(c*x)^(11/3),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(4/3)/(c*x)^(11/3), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{4/3}}{{\left (c\,x\right )}^{11/3}} \,d x \] Input:

int((a + b*x^2)^(4/3)/(c*x)^(11/3),x)
 

Output:

int((a + b*x^2)^(4/3)/(c*x)^(11/3), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{11/3}} \, dx=\frac {-3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a -3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{2}+8 x^{\frac {8}{3}} \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{x^{\frac {5}{3}}}d x \right ) b}{8 x^{\frac {8}{3}} c^{\frac {11}{3}}} \] Input:

int((b*x^2+a)^(4/3)/(c*x)^(11/3),x)
 

Output:

( - 3*(a + b*x**2)**(1/3)*a - 3*(a + b*x**2)**(1/3)*b*x**2 + 8*x**(2/3)*in 
t((a + b*x**2)**(1/3)/(x**(2/3)*x),x)*b*x**2)/(8*x**(2/3)*c**(2/3)*c**3*x* 
*2)