\(\int \frac {(c x)^{10/3}}{(a+b x^2)^{2/3}} \, dx\) [827]

Optimal result

Integrand size = 19, antiderivative size = 421 \[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=-\frac {7 a c^3 \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{9 b^2}+\frac {c (c x)^{7/3} \sqrt [3]{a+b x^2}}{3 b}+\frac {7 a c^{7/3} \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{18 \sqrt [4]{3} b^2 \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}} \] Output:

-7/9*a*c^3*(c*x)^(1/3)*(b*x^2+a)^(1/3)/b^2+1/3*c*(c*x)^(7/3)*(b*x^2+a)^(1/ 
3)/b+7/54*a*c^(7/3)*(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))*3^(3/4)/b^2/(-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 = 10.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.21 \[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {c^3 \sqrt [3]{c x} \left (-7 a^2-4 a b x^2+3 b^2 x^4+7 a^2 \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {2}{3},\frac {7}{6},-\frac {b x^2}{a}\right )\right )}{9 b^2 \left (a+b x^2\right )^{2/3}} \] Input:

Integrate[(c*x)^(10/3)/(a + b*x^2)^(2/3),x]
 

Output:

(c^3*(c*x)^(1/3)*(-7*a^2 - 4*a*b*x^2 + 3*b^2*x^4 + 7*a^2*(1 + (b*x^2)/a)^( 
2/3)*Hypergeometric2F1[1/6, 2/3, 7/6, -((b*x^2)/a)]))/(9*b^2*(a + b*x^2)^( 
2/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 365, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {262, 262, 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.

\(\displaystyle \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {c (c x)^{7/3} \sqrt [3]{a+b x^2}}{3 b}-\frac {7 a c^2 \int \frac {(c x)^{4/3}}{\left (b x^2+a\right )^{2/3}}dx}{9 b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {c (c x)^{7/3} \sqrt [3]{a+b x^2}}{3 b}-\frac {7 a c^2 \left (\frac {c \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{b}-\frac {a c^2 \int \frac {1}{(c x)^{2/3} \left (b x^2+a\right )^{2/3}}dx}{3 b}\right )}{9 b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {c (c x)^{7/3} \sqrt [3]{a+b x^2}}{3 b}-\frac {7 a c^2 \left (\frac {c \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{b}-\frac {a c \int \frac {1}{\left (b x^2+a\right )^{2/3}}d\sqrt [3]{c x}}{b}\right )}{9 b}\)

\(\Big \downarrow \) 771

\(\displaystyle \frac {c (c x)^{7/3} \sqrt [3]{a+b x^2}}{3 b}-\frac {7 a c^2 \left (\frac {c \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{b}-\frac {a c \int \frac {1}{\sqrt {1-b x^2}}d\frac {\sqrt [3]{c x}}{\sqrt [6]{b x^2+a}}}{b \sqrt {a+b x^2} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}\right )}{9 b}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {c (c x)^{7/3} \sqrt [3]{a+b x^2}}{3 b}-\frac {7 a c^2 \left (\frac {c \sqrt [3]{c x} \sqrt [3]{a+b x^2}}{b}-\frac {a \sqrt [3]{c} \sqrt [3]{c x} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right ) \sqrt {\frac {b^{2/3} (c x)^{4/3}+\sqrt [3]{b} c^{2/3} (c x)^{2/3}+c^{4/3}}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} b \sqrt {1-b x^2} \left (a+b x^2\right )^{2/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right )}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}\right )}{9 b}\)

Input:

Int[(c*x)^(10/3)/(a + b*x^2)^(2/3),x]
 

Output:

(c*(c*x)^(7/3)*(a + b*x^2)^(1/3))/(3*b) - (7*a*c^2*((c*(c*x)^(1/3)*(a + b* 
x^2)^(1/3))/b - (a*c^(1/3)*(c*x)^(1/3)*(c^(2/3) - b^(1/3)*(c*x)^(2/3))*Sqr 
t[(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 - Sq 
rt[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])/(2*3^(1/4)*b*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*b)
 

Defintions of rubi rules used

rule 262

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

Maple [F]

\[\int \frac {\left (c x \right )^{\frac {10}{3}}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x\]

Input:

int((c*x)^(10/3)/(b*x^2+a)^(2/3),x)
 

Output:

int((c*x)^(10/3)/(b*x^2+a)^(2/3),x)
 

Fricas [F]

\[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {\left (c x\right )^{\frac {10}{3}}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate((c*x)^(10/3)/(b*x^2+a)^(2/3),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 74.79 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.10 \[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {c^{\frac {10}{3}} x^{\frac {13}{3}} \Gamma \left (\frac {13}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {13}{6} \\ \frac {19}{6} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {2}{3}} \Gamma \left (\frac {19}{6}\right )} \] Input:

integrate((c*x)**(10/3)/(b*x**2+a)**(2/3),x)
 

Output:

c**(10/3)*x**(13/3)*gamma(13/6)*hyper((2/3, 13/6), (19/6,), b*x**2*exp_pol 
ar(I*pi)/a)/(2*a**(2/3)*gamma(19/6))
 

Maxima [F]

\[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {\left (c x\right )^{\frac {10}{3}}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate((c*x)^(10/3)/(b*x^2+a)^(2/3),x, algorithm="maxima")
 

Output:

integrate((c*x)^(10/3)/(b*x^2 + a)^(2/3), x)
 

Giac [F]

\[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {\left (c x\right )^{\frac {10}{3}}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate((c*x)^(10/3)/(b*x^2+a)^(2/3),x, algorithm="giac")
 

Output:

integrate((c*x)^(10/3)/(b*x^2 + a)^(2/3), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=\int \frac {{\left (c\,x\right )}^{10/3}}{{\left (b\,x^2+a\right )}^{2/3}} \,d x \] Input:

int((c*x)^(10/3)/(a + b*x^2)^(2/3),x)
 

Output:

int((c*x)^(10/3)/(a + b*x^2)^(2/3), x)
 

Reduce [F]

\[ \int \frac {(c x)^{10/3}}{\left (a+b x^2\right )^{2/3}} \, dx=c^{\frac {10}{3}} \left (\int \frac {x^{\frac {10}{3}}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) \] Input:

int((c*x)^(10/3)/(b*x^2+a)^(2/3),x)
 

Output:

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