\(\int \frac {x^6 (A+B x^3)}{(a+b x^3)^{4/3}} \, dx\) [365]

Optimal result

Integrand size = 22, antiderivative size = 168 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {a (A b-a B) x}{b^3 \sqrt [3]{a+b x^3}}+\frac {(3 A b-5 a B) x \left (a+b x^3\right )^{2/3}}{9 b^3}+\frac {B x^4 \left (a+b x^3\right )^{2/3}}{6 b^2}-\frac {2 a (6 A b-7 a B) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{10/3}}+\frac {a (6 A b-7 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{9 b^{10/3}} \] Output:

a*(A*b-B*a)*x/b^3/(b*x^3+a)^(1/3)+1/9*(3*A*b-5*B*a)*x*(b*x^3+a)^(2/3)/b^3+ 
1/6*B*x^4*(b*x^3+a)^(2/3)/b^2-2/27*a*(6*A*b-7*B*a)*arctan(1/3*(1+2*b^(1/3) 
*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(10/3)+1/9*a*(6*A*b-7*B*a)*ln(-b^(1 
/3)*x+(b*x^3+a)^(1/3))/b^(10/3)
 

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.22 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {\frac {3 \sqrt [3]{b} \left (-28 a^2 B x+a b x \left (24 A-7 B x^3\right )+3 b^2 x^4 \left (2 A+B x^3\right )\right )}{\sqrt [3]{a+b x^3}}+4 \sqrt {3} a (-6 A b+7 a B) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-4 a (-6 A b+7 a B) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+2 a (-6 A b+7 a B) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{10/3}} \] Input:

Integrate[(x^6*(A + B*x^3))/(a + b*x^3)^(4/3),x]
 

Output:

((3*b^(1/3)*(-28*a^2*B*x + a*b*x*(24*A - 7*B*x^3) + 3*b^2*x^4*(2*A + B*x^3 
)))/(a + b*x^3)^(1/3) + 4*Sqrt[3]*a*(-6*A*b + 7*a*B)*ArcTan[(Sqrt[3]*b^(1/ 
3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] - 4*a*(-6*A*b + 7*a*B)*Log[-(b^(1 
/3)*x) + (a + b*x^3)^(1/3)] + 2*a*(-6*A*b + 7*a*B)*Log[b^(2/3)*x^2 + b^(1/ 
3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(54*b^(10/3))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {957, 843, 843, 769}

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.

\(\displaystyle \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^7 (A b-a B)}{a b \sqrt [3]{a+b x^3}}-\frac {(6 A b-7 a B) \int \frac {x^6}{\sqrt [3]{b x^3+a}}dx}{a b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^7 (A b-a B)}{a b \sqrt [3]{a+b x^3}}-\frac {(6 A b-7 a B) \left (\frac {x^4 \left (a+b x^3\right )^{2/3}}{6 b}-\frac {2 a \int \frac {x^3}{\sqrt [3]{b x^3+a}}dx}{3 b}\right )}{a b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^7 (A b-a B)}{a b \sqrt [3]{a+b x^3}}-\frac {(6 A b-7 a B) \left (\frac {x^4 \left (a+b x^3\right )^{2/3}}{6 b}-\frac {2 a \left (\frac {x \left (a+b x^3\right )^{2/3}}{3 b}-\frac {a \int \frac {1}{\sqrt [3]{b x^3+a}}dx}{3 b}\right )}{3 b}\right )}{a b}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {x^7 (A b-a B)}{a b \sqrt [3]{a+b x^3}}-\frac {(6 A b-7 a B) \left (\frac {x^4 \left (a+b x^3\right )^{2/3}}{6 b}-\frac {2 a \left (\frac {x \left (a+b x^3\right )^{2/3}}{3 b}-\frac {a \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{3 b}\right )}{3 b}\right )}{a b}\)

Input:

Int[(x^6*(A + B*x^3))/(a + b*x^3)^(4/3),x]
 

Output:

((A*b - a*B)*x^7)/(a*b*(a + b*x^3)^(1/3)) - ((6*A*b - 7*a*B)*((x^4*(a + b* 
x^3)^(2/3))/(6*b) - (2*a*((x*(a + b*x^3)^(2/3))/(3*b) - (a*(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))))/(3*b)))/(3*b)))/(a*b)
 

Defintions of rubi rules used

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 843

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

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(\frac {\frac {4 a \left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) \left (A b -\frac {7 B a}{6}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{9}+\frac {4 \left (a \left (-\frac {7 B \,x^{3}}{24}+A \right ) b^{\frac {4}{3}}+\frac {\left (\frac {B \,x^{3}}{2}+A \right ) x^{3} b^{\frac {7}{3}}}{4}-\frac {7 B \,a^{2} b^{\frac {1}{3}}}{6}\right ) x}{3}}{b^{\frac {10}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}\)	\(170\)

Input:

int(x^6*(B*x^3+A)/(b*x^3+a)^(4/3),x,method=_RETURNVERBOSE)
 

Output:

4/9/b^(10/3)*(a*(3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/ 
b^(1/3)/x)+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))*(A*b-7/6*B*a)*(b*x^3+a)^(1/3)+3*( 
a*(-7/24*B*x^3+A)*b^(4/3)+1/4*(1/2*B*x^3+A)*x^3*b^(7/3)-7/6*B*a^2*b^(1/3)) 
*x)/(b*x^3+a)^(1/3)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (140) = 280\).

Time = 0.11 (sec) , antiderivative size = 624, normalized size of antiderivative = 3.71 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx =\text {Too large to display} \] Input:

integrate(x^6*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="fricas")
 

Output:

[-1/54*(6*sqrt(1/3)*(7*B*a^3*b - 6*A*a^2*b^2 + (7*B*a^2*b^2 - 6*A*a*b^3)*x 
^3)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 - 3*sqr 
t(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) + 4*(7*B*a^3 - 6*A*a^2*b + (7*B*a^2*b - 6*A*a 
*b^2)*x^3)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - 2*(7*B*a^3 - 
6*A*a^2*b + (7*B*a^2*b - 6*A*a*b^2)*x^3)*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*b^3*x^7 - (7*B*a* 
b^2 - 6*A*b^3)*x^4 - 4*(7*B*a^2*b - 6*A*a*b^2)*x)*(b*x^3 + a)^(2/3))/(b^5* 
x^3 + a*b^4), -1/54*(4*(7*B*a^3 - 6*A*a^2*b + (7*B*a^2*b - 6*A*a*b^2)*x^3) 
*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - 2*(7*B*a^3 - 6*A*a^2*b 
+ (7*B*a^2*b - 6*A*a*b^2)*x^3)*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) + 12*sqrt(1/3)*(7*B*a^3*b - 6*A*a^2* 
b^2 + (7*B*a^2*b^2 - 6*A*a*b^3)*x^3)*arctan(sqrt(1/3)*(b^(1/3)*x + 2*(b*x^ 
3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3) - 3*(3*B*b^3*x^7 - (7*B*a*b^2 - 6*A*b^3 
)*x^4 - 4*(7*B*a^2*b - 6*A*a*b^2)*x)*(b*x^3 + a)^(2/3))/(b^5*x^3 + a*b^4)]
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 15.55 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.48 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {A x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {4}{3}} \Gamma \left (\frac {10}{3}\right )} + \frac {B x^{10} \Gamma \left (\frac {10}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {10}{3} \\ \frac {13}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {4}{3}} \Gamma \left (\frac {13}{3}\right )} \] Input:

integrate(x**6*(B*x**3+A)/(b*x**3+a)**(4/3),x)
 

Output:

A*x**7*gamma(7/3)*hyper((4/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3* 
a**(4/3)*gamma(10/3)) + B*x**10*gamma(10/3)*hyper((4/3, 10/3), (13/3,), b* 
x**3*exp_polar(I*pi)/a)/(3*a**(4/3)*gamma(13/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (140) = 280\).

Time = 0.12 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.21 \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=-\frac {1}{54} \, B {\left (\frac {28 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {10}{3}}} + \frac {3 \, {\left (18 \, a^{2} b^{2} - \frac {49 \, {\left (b x^{3} + a\right )} a^{2} b}{x^{3}} + \frac {28 \, {\left (b x^{3} + a\right )}^{2} a^{2}}{x^{6}}\right )}}{\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{5}}{x} - \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{4}}{x^{4}} + \frac {{\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{3}}{x^{7}}} - \frac {14 \, a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {10}{3}}} + \frac {28 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {10}{3}}}\right )} + \frac {1}{9} \, A {\left (\frac {4 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {7}{3}}} + \frac {3 \, {\left (3 \, a b - \frac {4 \, {\left (b x^{3} + a\right )} a}{x^{3}}\right )}}{\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{3}}{x} - \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{2}}{x^{4}}} - \frac {2 \, a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {7}{3}}} + \frac {4 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {7}{3}}}\right )} \] Input:

integrate(x^6*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="maxima")
 

Output:

-1/54*B*(28*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/ 
x)/b^(1/3))/b^(10/3) + 3*(18*a^2*b^2 - 49*(b*x^3 + a)*a^2*b/x^3 + 28*(b*x^ 
3 + a)^2*a^2/x^6)/((b*x^3 + a)^(1/3)*b^5/x - 2*(b*x^3 + a)^(4/3)*b^4/x^4 + 
 (b*x^3 + a)^(7/3)*b^3/x^7) - 14*a^2*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/ 
3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(10/3) + 28*a^2*log(-b^(1/3) + (b*x^3 + a) 
^(1/3)/x)/b^(10/3)) + 1/9*A*(4*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*( 
b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) + 3*(3*a*b - 4*(b*x^3 + a)*a/x^3)/((b 
*x^3 + a)^(1/3)*b^3/x - (b*x^3 + a)^(4/3)*b^2/x^4) - 2*a*log(b^(2/3) + (b* 
x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(7/3) + 4*a*log(-b^(1/ 
3) + (b*x^3 + a)^(1/3)/x)/b^(7/3))
 

Giac [F]

\[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{6}}{{\left (b x^{3} + a\right )}^{\frac {4}{3}}} \,d x } \] Input:

integrate(x^6*(B*x^3+A)/(b*x^3+a)^(4/3),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*x^6/(b*x^3 + a)^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int \frac {x^6\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{4/3}} \,d x \] Input:

int((x^6*(A + B*x^3))/(a + b*x^3)^(4/3),x)
 

Output:

int((x^6*(A + B*x^3))/(a + b*x^3)^(4/3), x)
 

Reduce [F]

\[ \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^{4/3}} \, dx=\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \] Input:

int(x^6*(B*x^3+A)/(b*x^3+a)^(4/3),x)
 

Output:

int(x**6/(a + b*x**3)**(1/3),x)