\(\int \frac {x^6 (a+b x^3)^{2/3}}{c+d x^3} \, dx\) [699]

Optimal result

Integrand size = 24, antiderivative size = 334 \[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=-\frac {(3 b c-a d) x \left (a+b x^3\right )^{2/3}}{9 b d^2}+\frac {x^4 \left (a+b x^3\right )^{2/3}}{6 d}+\frac {\left (9 b^2 c^2-6 a b c d-a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{4/3} d^3}-\frac {c^{4/3} (b c-a d)^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^3}-\frac {c^{4/3} (b c-a d)^{2/3} \log \left (c+d x^3\right )}{6 d^3}+\frac {c^{4/3} (b c-a d)^{2/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^3}-\frac {\left (9 b^2 c^2-6 a b c d-a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{4/3} d^3} \] Output:

-1/9*(-a*d+3*b*c)*x*(b*x^3+a)^(2/3)/b/d^2+1/6*x^4*(b*x^3+a)^(2/3)/d+1/27*( 
-a^2*d^2-6*a*b*c*d+9*b^2*c^2)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3 
^(1/2))*3^(1/2)/b^(4/3)/d^3-1/3*c^(4/3)*(-a*d+b*c)^(2/3)*arctan(1/3*(1+2*( 
-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/d^3-1/6*c^(4/3 
)*(-a*d+b*c)^(2/3)*ln(d*x^3+c)/d^3+1/2*c^(4/3)*(-a*d+b*c)^(2/3)*ln((-a*d+b 
*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/d^3-1/18*(-a^2*d^2-6*a*b*c*d+9*b^2*c^ 
2)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)/d^3
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.82 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.58 \[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {\frac {6 d \left (a+b x^3\right )^{2/3} \left (-6 b c x+2 a d x+3 b d x^4\right )}{b}+\frac {4 \sqrt {3} \left (9 b^2 c^2-6 a b c d-a^2 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{b^{4/3}}+18 \sqrt {-6+6 i \sqrt {3}} c^{4/3} (b c-a d)^{2/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+\frac {4 \left (-9 b^2 c^2+6 a b c d+a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{b^{4/3}}-18 i \left (-i+\sqrt {3}\right ) c^{4/3} (b c-a d)^{2/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+\frac {2 \left (9 b^2 c^2-6 a b c d-a^2 d^2\right ) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{b^{4/3}}+9 \left (1+i \sqrt {3}\right ) c^{4/3} (b c-a d)^{2/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{108 d^3} \] Input:

Integrate[(x^6*(a + b*x^3)^(2/3))/(c + d*x^3),x]
 

Output:

((6*d*(a + b*x^3)^(2/3)*(-6*b*c*x + 2*a*d*x + 3*b*d*x^4))/b + (4*Sqrt[3]*( 
9*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2 
*(a + b*x^3)^(1/3))])/b^(4/3) + 18*Sqrt[-6 + (6*I)*Sqrt[3]]*c^(4/3)*(b*c - 
 a*d)^(2/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - 
(3*I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))] + (4*(-9*b^2*c^2 + 6*a*b*c*d + 
 a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/b^(4/3) - (18*I)*(-I + Sq 
rt[3])*c^(4/3)*(b*c - a*d)^(2/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3 
])*c^(1/3)*(a + b*x^3)^(1/3)] + (2*(9*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*Log[b 
^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/b^(4/3) + 9 
*(1 + I*Sqrt[3])*c^(4/3)*(b*c - a*d)^(2/3)*Log[2*(b*c - a*d)^(2/3)*x^2 + ( 
-1 - I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(1/3) + I*(I + Sqr 
t[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(108*d^3)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {978, 27, 1052, 1026, 769, 901}

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.

\(\displaystyle \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx\)

\(\Big \downarrow \) 978

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1052

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

\(\Big \downarrow \) 1026

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

\(\Big \downarrow \) 769

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

\(\Big \downarrow \) 901

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

Input:

Int[(x^6*(a + b*x^3)^(2/3))/(c + d*x^3),x]
 

Output:

(x^4*(a + b*x^3)^(2/3))/(6*d) - (((3*b*c - a*d)*x*(a + b*x^3)^(2/3))/(3*b* 
d) - ((-9*b*c^2*(b*c - a*d)*(ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)* 
(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(1/3)) + Log[c + 
 d*x^3]/(6*c^(2/3)*(b*c - a*d)^(1/3)) - Log[((b*c - a*d)^(1/3)*x)/c^(1/3) 
- (a + b*x^3)^(1/3)]/(2*c^(2/3)*(b*c - a*d)^(1/3))))/d + ((9*b^2*c^2 - 6*a 
*b*c*d - a^2*d^2)*(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/( 
Sqrt[3]*b^(1/3)) - Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3))))/d)/ 
(3*b*d))/(3*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 769

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

rule 901

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] 
 + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0]
 

rule 978

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* 
((c + d*x^n)^q/(b*(m + n*(p + q) + 1))), x] - Simp[e^n/(b*(m + n*(p + q) + 
1))   Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 
 1) + (a*d*(m - n + 1) - n*q*(b*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c 
, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] && GtQ[m - n 
 + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
 

rule 1026

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)* 
(x_)^(n_)), x_Symbol] :> Simp[f/d   Int[(a + b*x^n)^p, x], x] + Simp[(d*e - 
 c*f)/d   Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, 
 p, n}, x]
 

rule 1052

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m 
- n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q + 1) + 
 1))), x] - Simp[g^n/(b*d*(m + n*(p + q + 1) + 1))   Int[(g*x)^(m - n)*(a + 
 b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*( 
f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x] /; FreeQ[ 
{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]
 

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(-\frac {\left (-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} d \left (\left (\frac {3 d \,x^{3}}{2}-3 c \right ) b^{\frac {7}{3}}+b^{\frac {4}{3}} a d \right ) x}{3}-\frac {\left (a^{2} d^{2}+6 a b c d -9 b^{2} c^{2}\right ) \left (\arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) b}{9}\right ) \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+c \left (b^{\frac {10}{3}} c -b^{\frac {7}{3}} a d \right ) \left (\arctan \left (\frac {\sqrt {3}\, \left (-\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} b^{\frac {7}{3}} d^{3}}\)	\(336\)

Input:

int(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

-1/3/((a*d-b*c)/c)^(1/3)*((-1/3*(b*x^3+a)^(2/3)*d*((3/2*d*x^3-3*c)*b^(7/3) 
+b^(4/3)*a*d)*x-1/9*(a^2*d^2+6*a*b*c*d-9*b^2*c^2)*(arctan(1/3*3^(1/2)*(2*( 
b*x^3+a)^(1/3)/b^(1/3)+x)/x)*3^(1/2)+ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)-1/ 
2*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))*b)*((a* 
d-b*c)/c)^(1/3)+c*(b^(10/3)*c-b^(7/3)*a*d)*(arctan(1/3*3^(1/2)*(-2/((a*d-b 
*c)/c)^(1/3)*(b*x^3+a)^(1/3)+x)/x)*3^(1/2)+ln((((a*d-b*c)/c)^(1/3)*x+(b*x^ 
3+a)^(1/3))/x)-1/2*ln((((a*d-b*c)/c)^(2/3)*x^2-((a*d-b*c)/c)^(1/3)*(b*x^3+ 
a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)))/b^(7/3)/d^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (276) = 552\).

Time = 1.21 (sec) , antiderivative size = 1164, normalized size of antiderivative = 3.49 \[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\text {Too large to display} \] Input:

integrate(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

[-1/54*(18*sqrt(3)*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*b^2*c*arctan( 
-1/3*(sqrt(3)*(b*c - a*d)*x + 2*sqrt(3)*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 
)^(1/3)*(b*x^3 + a)^(1/3))/((b*c - a*d)*x)) - 18*(b^2*c^3 - 2*a*b*c^2*d + 
a^2*c*d^2)^(1/3)*b^2*c*log(((b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(2/3)*x - 
(b*x^3 + a)^(1/3)*(b*c^2 - a*c*d))/x) + 9*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d 
^2)^(1/3)*b^2*c*log(-((b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*(b*c - a*d 
)*x^2 + (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(2/3)*(b*x^3 + a)^(1/3)*x + (b 
*x^3 + a)^(2/3)*(b*c^2 - a*c*d))/x^2) + 3*sqrt(1/3)*(9*b^3*c^2 - 6*a*b^2*c 
*d - a^2*b*d^2)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3) 
*x^2 - 3*sqrt(1/3)*(b^(4/3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^ 
(2/3)*b^(2/3)*x)*sqrt(-1/b^(2/3)) + 2*a) + 2*(9*b^2*c^2 - 6*a*b*c*d - a^2* 
d^2)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (9*b^2*c^2 - 6*a*b* 
c*d - a^2*d^2)*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b 
*x^3 + a)^(2/3))/x^2) - 3*(3*b^2*d^2*x^4 - 2*(3*b^2*c*d - a*b*d^2)*x)*(b*x 
^3 + a)^(2/3))/(b^2*d^3), -1/54*(18*sqrt(3)*(b^2*c^3 - 2*a*b*c^2*d + a^2*c 
*d^2)^(1/3)*b^2*c*arctan(-1/3*(sqrt(3)*(b*c - a*d)*x + 2*sqrt(3)*(b^2*c^3 
- 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*(b*x^3 + a)^(1/3))/((b*c - a*d)*x)) - 18* 
(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*b^2*c*log(((b^2*c^3 - 2*a*b*c^2* 
d + a^2*c*d^2)^(2/3)*x - (b*x^3 + a)^(1/3)*(b*c^2 - a*c*d))/x) + 9*(b^2*c^ 
3 - 2*a*b*c^2*d + a^2*c*d^2)^(1/3)*b^2*c*log(-((b^2*c^3 - 2*a*b*c^2*d +...
 

Sympy [F]

\[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\int \frac {x^{6} \left (a + b x^{3}\right )^{\frac {2}{3}}}{c + d x^{3}}\, dx \] Input:

integrate(x**6*(b*x**3+a)**(2/3)/(d*x**3+c),x)
 

Output:

Integral(x**6*(a + b*x**3)**(2/3)/(c + d*x**3), x)
 

Maxima [F]

\[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{6}}{d x^{3} + c} \,d x } \] Input:

integrate(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(2/3)*x^6/(d*x^3 + c), x)
 

Giac [F]

\[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{6}}{d x^{3} + c} \,d x } \] Input:

integrate(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(2/3)*x^6/(d*x^3 + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\int \frac {x^6\,{\left (b\,x^3+a\right )}^{2/3}}{d\,x^3+c} \,d x \] Input:

int((x^6*(a + b*x^3)^(2/3))/(c + d*x^3),x)
 

Output:

int((x^6*(a + b*x^3)^(2/3))/(c + d*x^3), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {x^6 \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx=\frac {2 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a d x -6 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b c x +3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b d \,x^{4}-2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} c d +6 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b \,c^{2}-2 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a^{2} d^{2}-12 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a b c d +18 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{3}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b^{2} c^{2}}{18 b \,d^{2}} \] Input:

int(x^6*(b*x^3+a)^(2/3)/(d*x^3+c),x)
 

Output:

(2*(a + b*x**3)**(2/3)*a*d*x - 6*(a + b*x**3)**(2/3)*b*c*x + 3*(a + b*x**3 
)**(2/3)*b*d*x**4 - 2*int((a + b*x**3)**(2/3)/(a*c + a*d*x**3 + b*c*x**3 + 
 b*d*x**6),x)*a**2*c*d + 6*int((a + b*x**3)**(2/3)/(a*c + a*d*x**3 + b*c*x 
**3 + b*d*x**6),x)*a*b*c**2 - 2*int(((a + b*x**3)**(2/3)*x**3)/(a*c + a*d* 
x**3 + b*c*x**3 + b*d*x**6),x)*a**2*d**2 - 12*int(((a + b*x**3)**(2/3)*x** 
3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*a*b*c*d + 18*int(((a + b*x**3 
)**(2/3)*x**3)/(a*c + a*d*x**3 + b*c*x**3 + b*d*x**6),x)*b**2*c**2)/(18*b* 
d**2)