\(\int \frac {(a+b x^2)^{4/3}}{(c x)^{20/3}} \, dx\) [815]

Optimal result

Integrand size = 19, antiderivative size = 450 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=-\frac {24 b \sqrt [3]{a+b x^2}}{187 c^3 (c x)^{11/3}}-\frac {48 b^2 \sqrt [3]{a+b x^2}}{935 a c^5 (c x)^{5/3}}-\frac {3 \left (a+b x^2\right )^{4/3}}{17 c (c x)^{17/3}}-\frac {24\ 3^{3/4} b^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{935 a^2 c^{23/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}} \] Output:

-24/187*b*(b*x^2+a)^(1/3)/c^3/(c*x)^(11/3)-48/935*b^2*(b*x^2+a)^(1/3)/a/c^ 
5/(c*x)^(5/3)-3/17*(b*x^2+a)^(4/3)/c/(c*x)^(17/3)-24/935*3^(3/4)*b^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^( 
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^2/c^(23/3)/(-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 = 57, normalized size of antiderivative = 0.13 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=-\frac {3 a x \sqrt [3]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {17}{6},-\frac {4}{3},-\frac {11}{6},-\frac {b x^2}{a}\right )}{17 (c x)^{20/3} \sqrt [3]{1+\frac {b x^2}{a}}} \] Input:

Integrate[(a + b*x^2)^(4/3)/(c*x)^(20/3),x]
 

Output:

(-3*a*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-17/6, -4/3, -11/6, -((b*x^2)/ 
a)])/(17*(c*x)^(20/3)*(1 + (b*x^2)/a)^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 400, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {247, 247, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {8 b \int \frac {\sqrt [3]{b x^2+a}}{(c x)^{14/3}}dx}{17 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{17 c (c x)^{17/3}}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {8 b \left (\frac {2 b \int \frac {1}{(c x)^{8/3} \left (b x^2+a\right )^{2/3}}dx}{11 c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}\right )}{17 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{17 c (c x)^{17/3}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {8 b \left (\frac {2 b \left (-\frac {3 b \int \frac {1}{(c x)^{2/3} \left (b x^2+a\right )^{2/3}}dx}{5 a c^2}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}\right )}{17 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{17 c (c x)^{17/3}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {8 b \left (\frac {2 b \left (-\frac {9 b \int \frac {1}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{5 a c^3}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}\right )}{17 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{17 c (c x)^{17/3}}\)

\(\Big \downarrow \) 771

\(\displaystyle \frac {8 b \left (\frac {2 b \left (-\frac {9 b \int \frac {1}{\sqrt {1-b x^2}}d\frac {\sqrt [3]{c x}}{\sqrt [6]{b x^2+a}}}{5 a c^3 \sqrt {a+b x^2} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}\right )}{17 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{17 c (c x)^{17/3}}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {8 b \left (\frac {2 b \left (-\frac {3\ 3^{3/4} b \sqrt [3]{c x} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right ) \sqrt {\frac {b^{2/3} (c x)^{4/3}+\sqrt [3]{b} c^{2/3} (c x)^{2/3}+c^{4/3}}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{10 a c^{11/3} \sqrt {1-b x^2} \left (a+b x^2\right )^{2/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right )}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}-\frac {3 \sqrt [3]{a+b x^2}}{5 a c (c x)^{5/3}}\right )}{11 c^2}-\frac {3 \sqrt [3]{a+b x^2}}{11 c (c x)^{11/3}}\right )}{17 c^2}-\frac {3 \left (a+b x^2\right )^{4/3}}{17 c (c x)^{17/3}}\)

Input:

Int[(a + b*x^2)^(4/3)/(c*x)^(20/3),x]
 

Output:

(-3*(a + b*x^2)^(4/3))/(17*c*(c*x)^(17/3)) + (8*b*((-3*(a + b*x^2)^(1/3))/ 
(11*c*(c*x)^(11/3)) + (2*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*c^2)))/(17*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 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]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {4}{3}}}{\left (c x \right )^{\frac {20}{3}}}d x\]

Input:

int((b*x^2+a)^(4/3)/(c*x)^(20/3),x)
 

Output:

int((b*x^2+a)^(4/3)/(c*x)^(20/3),x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {20}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(4/3)/(c*x)^(20/3),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(4/3)*(c*x)^(1/3)/(c^7*x^7), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**(4/3)/(c*x)**(20/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {20}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(4/3)/(c*x)^(20/3),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(4/3)/(c*x)^(20/3), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {4}{3}}}{\left (c x\right )^{\frac {20}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(4/3)/(c*x)^(20/3),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(4/3)/(c*x)^(20/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{4/3}}{{\left (c\,x\right )}^{20/3}} \,d x \] Input:

int((a + b*x^2)^(4/3)/(c*x)^(20/3),x)
 

Output:

int((a + b*x^2)^(4/3)/(c*x)^(20/3), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{(c x)^{20/3}} \, dx=\frac {-27 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a -51 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{2}-16 x^{\frac {17}{3}} \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{x^{\frac {14}{3}} a +x^{\frac {20}{3}} b}d x \right ) a b}{153 x^{\frac {17}{3}} c^{\frac {20}{3}}} \] Input:

int((b*x^2+a)^(4/3)/(c*x)^(20/3),x)
 

Output:

( - 27*(a + b*x**2)**(1/3)*a - 51*(a + b*x**2)**(1/3)*b*x**2 - 16*x**(2/3) 
*int((a + b*x**2)**(1/3)/(x**(2/3)*a*x**4 + x**(2/3)*b*x**6),x)*a*b*x**5)/ 
(153*x**(2/3)*c**(2/3)*c**6*x**5)