\(\int (a+b x^3)^{2/3} (c+d x^3)^2 \, dx\) [123]

Optimal result

Integrand size = 21, antiderivative size = 212 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )^2 \, dx=\frac {1}{81} \left (27 c^2-\frac {a d (9 b c-2 a d)}{b^2}\right ) x \left (a+b x^3\right )^{2/3}+\frac {d (9 b c-2 a d) x \left (a+b x^3\right )^{5/3}}{27 b^2}+\frac {d^2 x^4 \left (a+b x^3\right )^{5/3}}{9 b}+\frac {2 a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{81 \sqrt {3} b^{7/3}}-\frac {a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{81 b^{7/3}} \] Output:

1/81*(27*c^2-a*d*(-2*a*d+9*b*c)/b^2)*x*(b*x^3+a)^(2/3)+1/27*d*(-2*a*d+9*b* 
c)*x*(b*x^3+a)^(5/3)/b^2+1/9*d^2*x^4*(b*x^3+a)^(5/3)/b+2/243*a*(2*a^2*d^2- 
9*a*b*c*d+27*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^(7/3)-1/81*a*(2*a^2*d^2-9*a*b*c*d+27*b^2*c^2)*ln(-b^(1/3)*x+(b*x 
^3+a)^(1/3))/b^(7/3)
 

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.21 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )^2 \, dx=\frac {3 \sqrt [3]{b} x \left (a+b x^3\right )^{2/3} \left (-4 a^2 d^2+3 a b d \left (6 c+d x^3\right )+9 b^2 \left (3 c^2+3 c d x^3+d^2 x^6\right )\right )+2 \sqrt {3} a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+a \left (27 b^2 c^2-9 a b c d+2 a^2 d^2\right ) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{243 b^{7/3}} \] Input:

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

Output:

(3*b^(1/3)*x*(a + b*x^3)^(2/3)*(-4*a^2*d^2 + 3*a*b*d*(6*c + d*x^3) + 9*b^2 
*(3*c^2 + 3*c*d*x^3 + d^2*x^6)) + 2*Sqrt[3]*a*(27*b^2*c^2 - 9*a*b*c*d + 2* 
a^2*d^2)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] - 2 
*a*(27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/ 
3)] + a*(27*b^2*c^2 - 9*a*b*c*d + 2*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)])/(243*b^(7/3))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {933, 913, 748, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int \left (b x^3+a\right )^{2/3} \left (4 d (3 b c-a d) x^3+c (9 b c-a d)\right )dx}{9 b}+\frac {d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {\left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right ) \int \left (b x^3+a\right )^{2/3}dx}{3 b}+\frac {2 d x \left (a+b x^3\right )^{5/3} (3 b c-a d)}{3 b}}{9 b}+\frac {d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}\)

\(\Big \downarrow \) 748

\(\displaystyle \frac {\frac {\left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right ) \left (\frac {2}{3} a \int \frac {1}{\sqrt [3]{b x^3+a}}dx+\frac {1}{3} x \left (a+b x^3\right )^{2/3}\right )}{3 b}+\frac {2 d x \left (a+b x^3\right )^{5/3} (3 b c-a d)}{3 b}}{9 b}+\frac {d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {\frac {\left (2 a^2 d^2-9 a b c d+27 b^2 c^2\right ) \left (\frac {2}{3} a \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )+\frac {1}{3} x \left (a+b x^3\right )^{2/3}\right )}{3 b}+\frac {2 d x \left (a+b x^3\right )^{5/3} (3 b c-a d)}{3 b}}{9 b}+\frac {d x \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )}{9 b}\)

Input:

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

Output:

(d*x*(a + b*x^3)^(5/3)*(c + d*x^3))/(9*b) + ((2*d*(3*b*c - a*d)*x*(a + b*x 
^3)^(5/3))/(3*b) + ((27*b^2*c^2 - 9*a*b*c*d + 2*a^2*d^2)*((x*(a + b*x^3)^( 
2/3))/3 + (2*a*(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqr 
t[3]*b^(1/3)) - Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3))))/3))/(3 
*b))/(9*b)
 

Defintions of rubi rules used

rule 748

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p 
+ 1)), x] + Simp[a*n*(p/(n*p + 1))   Int[(a + b*x^n)^(p - 1), x], x] /; Fre 
eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat 
or[p + 1/n], Denominator[p]])
 

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 913

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

rule 933

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

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(\frac {\frac {2 a d x \left (\frac {d \,x^{3}}{6}+c \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {4}{3}}}{9}+\frac {\left (\frac {1}{3} d^{2} x^{6}+c d \,x^{3}+c^{2}\right ) x \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{\frac {7}{3}}}{3}+\frac {2 a \left (-6 a \,d^{2} x \,b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right ) \left (a^{2} d^{2}-\frac {9}{2} a b c d +\frac {27}{2} b^{2} c^{2}\right )\right )}{243}}{b^{\frac {7}{3}}}\)	\(204\)

Input:

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

Output:

2/243/b^(7/3)*(27*a*d*x*(1/6*d*x^3+c)*(b*x^3+a)^(2/3)*b^(4/3)+81/2*(1/3*d^ 
2*x^6+c*d*x^3+c^2)*x*(b*x^3+a)^(2/3)*b^(7/3)+a*(-6*a*d^2*x*b^(1/3)*(b*x^3+ 
a)^(2/3)+(-2*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^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2*l 
n((-b^(1/3)*x+(b*x^3+a)^(1/3))/x))*(a^2*d^2-9/2*a*b*c*d+27/2*b^2*c^2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 634, normalized size of antiderivative = 2.99 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

[1/243*(3*sqrt(1/3)*(27*a*b^3*c^2 - 9*a^2*b^2*c*d + 2*a^3*b*d^2)*sqrt((-b) 
^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*( 
(-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^(2/3 
)*x)*sqrt((-b)^(1/3)/b) + 2*a) - 2*(27*a*b^2*c^2 - 9*a^2*b*c*d + 2*a^3*d^2 
)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + (27*a*b^2*c^2 - 9 
*a^2*b*c*d + 2*a^3*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*(9*b^3*d^2*x^7 + 3*(9*b^3*c*d 
+ a*b^2*d^2)*x^4 + (27*b^3*c^2 + 18*a*b^2*c*d - 4*a^2*b*d^2)*x)*(b*x^3 + a 
)^(2/3))/b^3, -1/243*(6*sqrt(1/3)*(27*a*b^3*c^2 - 9*a^2*b^2*c*d + 2*a^3*b* 
d^2)*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a)^( 
1/3))*sqrt(-(-b)^(1/3)/b)/x) + 2*(27*a*b^2*c^2 - 9*a^2*b*c*d + 2*a^3*d^2)* 
(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - (27*a*b^2*c^2 - 9*a 
^2*b*c*d + 2*a^3*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*(9*b^3*d^2*x^7 + 3*(9*b^3*c*d + 
a*b^2*d^2)*x^4 + (27*b^3*c^2 + 18*a*b^2*c*d - 4*a^2*b*d^2)*x)*(b*x^3 + a)^ 
(2/3))/b^3]
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.56 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.62 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )^2 \, dx=\frac {a^{\frac {2}{3}} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 a^{\frac {2}{3}} c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {2}{3}} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

a**(2/3)*c**2*x*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), b*x**3*exp_polar(I*p 
i)/a)/(3*gamma(4/3)) + 2*a**(2/3)*c*d*x**4*gamma(4/3)*hyper((-2/3, 4/3), ( 
7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + a**(2/3)*d**2*x**7*gamma 
(7/3)*hyper((-2/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3) 
)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 552 vs. \(2 (181) = 362\).

Time = 0.11 (sec) , antiderivative size = 552, normalized size of antiderivative = 2.60 \[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

-1/9*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^( 
1/3))/b^(1/3) - a*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^ 
(2/3)/x^2)/b^(1/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3) + 3*( 
b*x^3 + a)^(2/3)*a/((b - (b*x^3 + a)/x^3)*x^2))*c^2 + 1/27*(2*sqrt(3)*a^2* 
arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) - 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^(4/ 
3) + 2*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) + 3*((b*x^3 + a)^(2 
/3)*a^2*b/x^2 + 2*(b*x^3 + a)^(5/3)*a^2/x^5)/(b^3 - 2*(b*x^3 + a)*b^2/x^3 
+ (b*x^3 + a)^2*b/x^6))*c*d - 1/243*(4*sqrt(3)*a^3*arctan(1/3*sqrt(3)*(b^( 
1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) - 2*a^3*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^3*log(-b^(1/ 
3) + (b*x^3 + a)^(1/3)/x)/b^(7/3) + 3*(2*(b*x^3 + a)^(2/3)*a^3*b^2/x^2 + 1 
1*(b*x^3 + a)^(5/3)*a^3*b/x^5 - 4*(b*x^3 + a)^(8/3)*a^3/x^8)/(b^5 - 3*(b*x 
^3 + a)*b^4/x^3 + 3*(b*x^3 + a)^2*b^3/x^6 - (b*x^3 + a)^3*b^2/x^9))*d^2
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )^2 \, dx=\frac {-4 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} d^{2} x +18 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b c d x +3 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,d^{2} x^{4}+27 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} c^{2} x +27 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} c d \,x^{4}+9 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} d^{2} x^{7}+4 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{3} d^{2}-18 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{2} b c d +54 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a \,b^{2} c^{2}}{81 b^{2}} \] Input:

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

Output:

( - 4*(a + b*x**3)**(2/3)*a**2*d**2*x + 18*(a + b*x**3)**(2/3)*a*b*c*d*x + 
 3*(a + b*x**3)**(2/3)*a*b*d**2*x**4 + 27*(a + b*x**3)**(2/3)*b**2*c**2*x 
+ 27*(a + b*x**3)**(2/3)*b**2*c*d*x**4 + 9*(a + b*x**3)**(2/3)*b**2*d**2*x 
**7 + 4*int((a + b*x**3)**(2/3)/(a + b*x**3),x)*a**3*d**2 - 18*int((a + b* 
x**3)**(2/3)/(a + b*x**3),x)*a**2*b*c*d + 54*int((a + b*x**3)**(2/3)/(a + 
b*x**3),x)*a*b**2*c**2)/(81*b**2)