\(\int \frac {x^7}{(a+b x^3)^{2/3} (c+d x^3)} \, dx\) [753]

Optimal result

Integrand size = 24, antiderivative size = 279 \[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}+\frac {(3 b c+2 a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d^2}-\frac {c^{5/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2 (b c-a d)^{2/3}}+\frac {c^{5/3} \log \left (c+d x^3\right )}{6 d^2 (b c-a d)^{2/3}}+\frac {(3 b c+2 a d) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{5/3} d^2}-\frac {c^{5/3} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2 (b c-a d)^{2/3}} \] Output:

1/3*x^2*(b*x^3+a)^(1/3)/b/d+1/9*(2*a*d+3*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b 
*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(5/3)/d^2-1/3*c^(5/3)*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^2/(-a*d+b*c)^ 
(2/3)+1/6*c^(5/3)*ln(d*x^3+c)/d^2/(-a*d+b*c)^(2/3)+1/6*(2*a*d+3*b*c)*ln(b^ 
(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)/d^2-1/2*c^(5/3)*ln((-a*d+b*c)^(1/3)*x/c^( 
1/3)-(b*x^3+a)^(1/3))/d^2/(-a*d+b*c)^(2/3)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.85 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.69 \[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {\frac {12 d x^2 \sqrt [3]{a+b x^3}}{b}+\frac {4 \sqrt {3} (3 b c+2 a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{b^{5/3}}+\frac {6 \sqrt {-6-6 i \sqrt {3}} c^{5/3} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{(b c-a d)^{2/3}}+\frac {4 (3 b c+2 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{b^{5/3}}+\frac {6 \left (1-i \sqrt {3}\right ) c^{5/3} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{2/3}}-\frac {2 (3 b c+2 a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{b^{5/3}}+\frac {3 i \left (i+\sqrt {3}\right ) c^{5/3} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{2/3}}}{36 d^2} \] Input:

Integrate[x^7/((a + b*x^3)^(2/3)*(c + d*x^3)),x]
 

Output:

((12*d*x^2*(a + b*x^3)^(1/3))/b + (4*Sqrt[3]*(3*b*c + 2*a*d)*ArcTan[(Sqrt[ 
3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/b^(5/3) + (6*Sqrt[-6 - ( 
6*I)*Sqrt[3]]*c^(5/3)*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))])/(b*c - a*d)^(2/3) + 
 (4*(3*b*c + 2*a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/b^(5/3) + (6*(1 
 - I*Sqrt[3])*c^(5/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)* 
(a + b*x^3)^(1/3)])/(b*c - a*d)^(2/3) - (2*(3*b*c + 2*a*d)*Log[b^(2/3)*x^2 
 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/b^(5/3) + ((3*I)*(I + 
 Sqrt[3])*c^(5/3)*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 + Sqrt[3])*c^(2/3)*(a + b*x^3) 
^(2/3)])/(b*c - a*d)^(2/3))/(36*d^2)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {979, 1054, 2009}

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.

\(\displaystyle \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx\)

\(\Big \downarrow \) 979

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}-\frac {-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right ) (2 a d+3 b c)}{\sqrt {3} b^{2/3} d}+\frac {\sqrt {3} b c^{5/3} \arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{d (b c-a d)^{2/3}}-\frac {(2 a d+3 b c) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3} d}-\frac {b c^{5/3} \log \left (c+d x^3\right )}{2 d (b c-a d)^{2/3}}+\frac {3 b c^{5/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d (b c-a d)^{2/3}}}{3 b d}\)

Input:

Int[x^7/((a + b*x^3)^(2/3)*(c + d*x^3)),x]
 

Output:

(x^2*(a + b*x^3)^(1/3))/(3*b*d) - (-(((3*b*c + 2*a*d)*ArcTan[(1 + (2*b^(1/ 
3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3)*d)) + (Sqrt[3]*b*c^(5/ 
3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3 
]])/(d*(b*c - a*d)^(2/3)) - (b*c^(5/3)*Log[c + d*x^3])/(2*d*(b*c - a*d)^(2 
/3)) - ((3*b*c + 2*a*d)*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*b^(2/3)*d) 
+ (3*b*c^(5/3)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2* 
d*(b*c - a*d)^(2/3)))/(3*b*d)
 

Defintions of rubi rules used

rule 979

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 
 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q) + 1))), x] - Simp[e^(2*n)/(b*d 
*(m + n*(p + q) + 1))   Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Sim 
p[a*c*(m - 2*n + 1) + (a*d*(m + n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x 
^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && I 
GtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x 
]
 

rule 1054

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.18

method	result	size
pseudoelliptic	\(\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{2} d \,b^{\frac {2}{3}} \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}+c \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right ) b^{\frac {5}{3}}-\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} \left (2 a d +3 b c \right ) \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{3}}{6 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} b^{\frac {5}{3}} d^{2}}\)	\(329\)

Input:

int(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x,method=_RETURNVERBOSE)
 

Output:

1/6/((a*d-b*c)/c)^(2/3)*(2*(b*x^3+a)^(1/3)*x^2*d*b^(2/3)*((a*d-b*c)/c)^(2/ 
3)+c*(2*3^(1/2)*arctan(1/3*3^(1/2)*(((a*d-b*c)/c)^(1/3)*x-2*(b*x^3+a)^(1/3 
))/((a*d-b*c)/c)^(1/3)/x)+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)-2*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3 
+a)^(1/3))/x))*b^(5/3)-1/3*((a*d-b*c)/c)^(2/3)*(2*a*d+3*b*c)*(2*3^(1/2)*ar 
ctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)+ln((b^(2/3)*x^2+ 
b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2*ln((-b^(1/3)*x+(b*x^3+a) 
^(1/3))/x)))/b^(5/3)/d^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 554 vs. \(2 (225) = 450\).

Time = 0.53 (sec) , antiderivative size = 554, normalized size of antiderivative = 1.99 \[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\frac {6 \, \sqrt {3} b^{3} c \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}} + \sqrt {3} c x}{3 \, c x}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} d x^{2} + 6 \, b^{3} c \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left (\frac {{\left (b c - a d\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}} c}{x}\right ) - 3 \, b^{3} c \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left (\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} c^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c^{2} - a c d\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x}{x^{2}}\right ) - 6 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} c + 2 \, a b d\right )} {\left (b^{2}\right )}^{\frac {1}{6}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left ({\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{b^{2} x}\right ) + 2 \, {\left (b^{2}\right )}^{\frac {2}{3}} {\left (3 \, b c + 2 \, a d\right )} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) - {\left (b^{2}\right )}^{\frac {2}{3}} {\left (3 \, b c + 2 \, a d\right )} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{3} d^{2}} \] Input:

integrate(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")
 

Output:

1/18*(6*sqrt(3)*b^3*c*(-c^2/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))^(1/3)*arctan( 
-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*(b*c - a*d)*(-c^2/(b^2*c^2 - 2*a*b*c*d + 
 a^2*d^2))^(2/3) + sqrt(3)*c*x)/(c*x)) + 6*(b*x^3 + a)^(1/3)*b^2*d*x^2 + 6 
*b^3*c*(-c^2/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))^(1/3)*log(((b*c - a*d)*(-c^2 
/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))^(1/3)*x + (b*x^3 + a)^(1/3)*c)/x) - 3*b^ 
3*c*(-c^2/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))^(1/3)*log(((b^2*c^2 - 2*a*b*c*d 
 + a^2*d^2)*(-c^2/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))^(2/3)*x^2 + (b*x^3 + a) 
^(2/3)*c^2 - (b*x^3 + a)^(1/3)*(b*c^2 - a*c*d)*(-c^2/(b^2*c^2 - 2*a*b*c*d 
+ a^2*d^2))^(1/3)*x)/x^2) - 6*sqrt(1/3)*(3*b^2*c + 2*a*b*d)*(b^2)^(1/6)*ar 
ctan(sqrt(1/3)*((b^2)^(1/3)*b*x + 2*(b*x^3 + a)^(1/3)*(b^2)^(2/3))*(b^2)^( 
1/6)/(b^2*x)) + 2*(b^2)^(2/3)*(3*b*c + 2*a*d)*log(-((b^2)^(2/3)*x - (b*x^3 
 + a)^(1/3)*b)/x) - (b^2)^(2/3)*(3*b*c + 2*a*d)*log(((b^2)^(1/3)*b*x^2 + ( 
b*x^3 + a)^(1/3)*(b^2)^(2/3)*x + (b*x^3 + a)^(2/3)*b)/x^2))/(b^3*d^2)
 

Sympy [F]

\[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{7}}{\left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \] Input:

integrate(x**7/(b*x**3+a)**(2/3)/(d*x**3+c),x)
 

Output:

Integral(x**7/((a + b*x**3)**(2/3)*(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {x^{7}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:

integrate(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate(x^7/((b*x^3 + a)^(2/3)*(d*x^3 + c)), x)
 

Giac [F]

\[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int { \frac {x^{7}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}} \,d x } \] Input:

integrate(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")
 

Output:

integrate(x^7/((b*x^3 + a)^(2/3)*(d*x^3 + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^7}{{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \] Input:

int(x^7/((a + b*x^3)^(2/3)*(c + d*x^3)),x)
 

Output:

int(x^7/((a + b*x^3)^(2/3)*(c + d*x^3)), x)
 

Reduce [F]

\[ \int \frac {x^7}{\left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx=\int \frac {x^{7}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} c +\left (b \,x^{3}+a \right )^{\frac {2}{3}} d \,x^{3}}d x \] Input:

int(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x)
                                                                                    
                                                                                    
 

Output:

int(x**7/((a + b*x**3)**(2/3)*c + (a + b*x**3)**(2/3)*d*x**3),x)