\(\int (a+b x^3)^{5/3} (c+d x^3)^2 \, dx\) [122]

Optimal result

Integrand size = 21, antiderivative size = 259 \[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx=\frac {5 a \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) x \left (a+b x^3\right )^{2/3}}{486 b^2}+\frac {\left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) x \left (a+b x^3\right )^{5/3}}{162 b^2}+\frac {d (6 b c-a d) x \left (a+b x^3\right )^{8/3}}{27 b^2}+\frac {d^2 x^4 \left (a+b x^3\right )^{8/3}}{12 b}+\frac {5 a^2 \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{243 \sqrt {3} b^{7/3}}-\frac {5 a^2 \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{486 b^{7/3}} \] Output:

5/486*a*(a^2*d^2-6*a*b*c*d+27*b^2*c^2)*x*(b*x^3+a)^(2/3)/b^2+1/162*(a^2*d^ 
2-6*a*b*c*d+27*b^2*c^2)*x*(b*x^3+a)^(5/3)/b^2+1/27*d*(-a*d+6*b*c)*x*(b*x^3 
+a)^(8/3)/b^2+1/12*d^2*x^4*(b*x^3+a)^(8/3)/b+5/729*a^2*(a^2*d^2-6*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)-5/486*a^2*(a^2*d^2-6*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 = 1.99 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.13 \[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx=\frac {3 \sqrt [3]{b} x \left (a+b x^3\right )^{2/3} \left (-20 a^3 d^2+15 a^2 b d \left (8 c+d x^3\right )+27 b^3 x^3 \left (6 c^2+8 c d x^3+3 d^2 x^6\right )+18 a b^2 \left (24 c^2+22 c d x^3+7 d^2 x^6\right )\right )+20 \sqrt {3} a^2 \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-20 a^2 \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+10 a^2 \left (27 b^2 c^2-6 a b c d+a^2 d^2\right ) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{2916 b^{7/3}} \] Input:

Integrate[(a + b*x^3)^(5/3)*(c + d*x^3)^2,x]
 

Output:

(3*b^(1/3)*x*(a + b*x^3)^(2/3)*(-20*a^3*d^2 + 15*a^2*b*d*(8*c + d*x^3) + 2 
7*b^3*x^3*(6*c^2 + 8*c*d*x^3 + 3*d^2*x^6) + 18*a*b^2*(24*c^2 + 22*c*d*x^3 
+ 7*d^2*x^6)) + 20*Sqrt[3]*a^2*(27*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))] - 20*a^2*(27*b^2*c^2 
 - 6*a*b*c*d + a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + 10*a^2*(27 
*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)])/(2916*b^(7/3))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {933, 913, 748, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int \left (b x^3+a\right )^{5/3} \left (d (15 b c-4 a d) x^3+c (12 b c-a d)\right )dx}{12 b}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {4 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \int \left (b x^3+a\right )^{5/3}dx}{9 b}+\frac {d x \left (a+b x^3\right )^{8/3} (15 b c-4 a d)}{9 b}}{12 b}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}\)

\(\Big \downarrow \) 748

\(\displaystyle \frac {\frac {4 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \left (\frac {5}{6} a \int \left (b x^3+a\right )^{2/3}dx+\frac {1}{6} x \left (a+b x^3\right )^{5/3}\right )}{9 b}+\frac {d x \left (a+b x^3\right )^{8/3} (15 b c-4 a d)}{9 b}}{12 b}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}\)

\(\Big \downarrow \) 748

\(\displaystyle \frac {\frac {4 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \left (\frac {5}{6} a \left (\frac {2}{3} a \int \frac {1}{\sqrt [3]{b x^3+a}}dx+\frac {1}{3} x \left (a+b x^3\right )^{2/3}\right )+\frac {1}{6} x \left (a+b x^3\right )^{5/3}\right )}{9 b}+\frac {d x \left (a+b x^3\right )^{8/3} (15 b c-4 a d)}{9 b}}{12 b}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {\frac {4 \left (a^2 d^2-6 a b c d+27 b^2 c^2\right ) \left (\frac {5}{6} a \left (\frac {2}{3} a \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )+\frac {1}{3} x \left (a+b x^3\right )^{2/3}\right )+\frac {1}{6} x \left (a+b x^3\right )^{5/3}\right )}{9 b}+\frac {d x \left (a+b x^3\right )^{8/3} (15 b c-4 a d)}{9 b}}{12 b}+\frac {d x \left (a+b x^3\right )^{8/3} \left (c+d x^3\right )}{12 b}\)

Input:

Int[(a + b*x^3)^(5/3)*(c + d*x^3)^2,x]
 

Output:

(d*x*(a + b*x^3)^(8/3)*(c + d*x^3))/(12*b) + ((d*(15*b*c - 4*a*d)*x*(a + b 
*x^3)^(8/3))/(9*b) + (4*(27*b^2*c^2 - 6*a*b*c*d + a^2*d^2)*((x*(a + b*x^3) 
^(5/3))/6 + (5*a*((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]]/(Sqrt[3]*b^(1/3)) - Log[-(b^(1/3)*x) + (a + 
b*x^3)^(1/3)]/(2*b^(1/3))))/3))/6))/(9*b))/(12*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 = 1.41 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(-\frac {5 \left (-\frac {243 \left (\frac {1}{2} d^{2} x^{6}+\frac {4}{3} c d \,x^{3}+c^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{4} b^{\frac {10}{3}}}{10}+a \left (-18 a \left (b \,x^{3}+a \right )^{\frac {2}{3}} d \left (\frac {d \,x^{3}}{8}+c \right ) x \,b^{\frac {4}{3}}-\frac {324 \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\frac {7}{24} d^{2} x^{6}+\frac {11}{12} c d \,x^{3}+c^{2}\right ) x \,b^{\frac {7}{3}}}{5}+a \left (3 a \,d^{2} x \,b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {2}{3}}+\left (a^{2} d^{2}-6 a b c d +27 b^{2} c^{2}\right ) \left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right )\right )\right )\right )}{729 b^{\frac {7}{3}}}\)	\(243\)

Input:

int((b*x^3+a)^(5/3)*(d*x^3+c)^2,x,method=_RETURNVERBOSE)
 

Output:

-5/729*(-243/10*(1/2*d^2*x^6+4/3*c*d*x^3+c^2)*(b*x^3+a)^(2/3)*x^4*b^(10/3) 
+a*(-18*a*(b*x^3+a)^(2/3)*d*(1/8*d*x^3+c)*x*b^(4/3)-324/5*(b*x^3+a)^(2/3)* 
(7/24*d^2*x^6+11/12*c*d*x^3+c^2)*x*b^(7/3)+a*(3*a*d^2*x*b^(1/3)*(b*x^3+a)^ 
(2/3)+(a^2*d^2-6*a*b*c*d+27*b^2*c^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^(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^(7/3)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 717, normalized size of antiderivative = 2.77 \[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx =\text {Too large to display} \] Input:

integrate((b*x^3+a)^(5/3)*(d*x^3+c)^2,x, algorithm="fricas")
 

Output:

[1/2916*(30*sqrt(1/3)*(27*a^2*b^3*c^2 - 6*a^3*b^2*c*d + a^4*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) - 20*(27*a^2*b^2*c^2 - 6*a^3*b*c*d + a^4* 
d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + 10*(27*a^2*b^2 
*c^2 - 6*a^3*b*c*d + a^4*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*(81*b^4*d^2*x^10 + 18*(1 
2*b^4*c*d + 7*a*b^3*d^2)*x^7 + 3*(54*b^4*c^2 + 132*a*b^3*c*d + 5*a^2*b^2*d 
^2)*x^4 + 4*(108*a*b^3*c^2 + 30*a^2*b^2*c*d - 5*a^3*b*d^2)*x)*(b*x^3 + a)^ 
(2/3))/b^3, -1/2916*(60*sqrt(1/3)*(27*a^2*b^3*c^2 - 6*a^3*b^2*c*d + a^4*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) + 20*(27*a^2*b^2*c^2 - 6*a^3*b*c*d + a^4*d^2) 
*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - 10*(27*a^2*b^2*c^2 
 - 6*a^3*b*c*d + a^4*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*(81*b^4*d^2*x^10 + 18*(12*b^ 
4*c*d + 7*a*b^3*d^2)*x^7 + 3*(54*b^4*c^2 + 132*a*b^3*c*d + 5*a^2*b^2*d^2)* 
x^4 + 4*(108*a*b^3*c^2 + 30*a^2*b^2*c*d - 5*a^3*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 = 32.66 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.04 \[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx=\frac {a^{\frac {5}{3}} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 a^{\frac {5}{3}} c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {5}{3}} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {a^{\frac {2}{3}} b c^{2} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {2 a^{\frac {2}{3}} b c d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {a^{\frac {2}{3}} b d^{2} x^{10} \Gamma \left (\frac {10}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {10}{3} \\ \frac {13}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {13}{3}\right )} \] Input:

integrate((b*x**3+a)**(5/3)*(d*x**3+c)**2,x)
 

Output:

a**(5/3)*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**(5/3)*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**(5/3)*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) 
) + a**(2/3)*b*c**2*x**4*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), b*x**3*exp_ 
polar(I*pi)/a)/(3*gamma(7/3)) + 2*a**(2/3)*b*c*d*x**7*gamma(7/3)*hyper((-2 
/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3)) + a**(2/3)*b* 
d**2*x**10*gamma(10/3)*hyper((-2/3, 10/3), (13/3,), b*x**3*exp_polar(I*pi) 
/a)/(3*gamma(13/3))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 672 vs. \(2 (224) = 448\).

Time = 0.12 (sec) , antiderivative size = 672, normalized size of antiderivative = 2.59 \[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx =\text {Too large to display} \] Input:

integrate((b*x^3+a)^(5/3)*(d*x^3+c)^2,x, algorithm="maxima")
 

Output:

-1/54*(10*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x) 
/b^(1/3))/b^(1/3) - 5*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^(1/3) + 10*a^2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^ 
(1/3) + 3*(5*(b*x^3 + a)^(2/3)*a^2*b/x^2 - 8*(b*x^3 + a)^(5/3)*a^2/x^5)/(b 
^2 - 2*(b*x^3 + a)*b/x^3 + (b*x^3 + a)^2/x^6))*c^2 + 1/243*(10*sqrt(3)*a^3 
*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) - 5 
*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^ 
(4/3) + 10*a^3*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) + 3*(5*(b*x^3 + 
 a)^(2/3)*a^3*b^2/x^2 - 13*(b*x^3 + a)^(5/3)*a^3*b/x^5 - 10*(b*x^3 + a)^(8 
/3)*a^3/x^8)/(b^4 - 3*(b*x^3 + a)*b^3/x^3 + 3*(b*x^3 + a)^2*b^2/x^6 - (b*x 
^3 + a)^3*b/x^9))*c*d - 1/2916*(20*sqrt(3)*a^4*arctan(1/3*sqrt(3)*(b^(1/3) 
 + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(7/3) - 10*a^4*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) + 20*a^4*log(-b^(1/3) 
 + (b*x^3 + a)^(1/3)/x)/b^(7/3) + 3*(10*(b*x^3 + a)^(2/3)*a^4*b^3/x^2 - 36 
*(b*x^3 + a)^(5/3)*a^4*b^2/x^5 - 75*(b*x^3 + a)^(8/3)*a^4*b/x^8 + 20*(b*x^ 
3 + a)^(11/3)*a^4/x^11)/(b^6 - 4*(b*x^3 + a)*b^5/x^3 + 6*(b*x^3 + a)^2*b^4 
/x^6 - 4*(b*x^3 + a)^3*b^3/x^9 + (b*x^3 + a)^4*b^2/x^12))*d^2
 

Giac [F]

\[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}^{2} \,d x } \] Input:

integrate((b*x^3+a)^(5/3)*(d*x^3+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(5/3)*(d*x^3 + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx=\int {\left (b\,x^3+a\right )}^{5/3}\,{\left (d\,x^3+c\right )}^2 \,d x \] Input:

int((a + b*x^3)^(5/3)*(c + d*x^3)^2,x)
 

Output:

int((a + b*x^3)^(5/3)*(c + d*x^3)^2, x)
 

Reduce [F]

\[ \int \left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2 \, dx=\frac {-20 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{3} d^{2} x +120 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} b c d x +15 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} b \,d^{2} x^{4}+432 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,b^{2} c^{2} x +396 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,b^{2} c d \,x^{4}+126 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a \,b^{2} d^{2} x^{7}+162 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{3} c^{2} x^{4}+216 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{3} c d \,x^{7}+81 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{3} d^{2} x^{10}+20 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{4} d^{2}-120 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{3} b c d +540 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{2} b^{2} c^{2}}{972 b^{2}} \] Input:

int((b*x^3+a)^(5/3)*(d*x^3+c)^2,x)
 

Output:

( - 20*(a + b*x**3)**(2/3)*a**3*d**2*x + 120*(a + b*x**3)**(2/3)*a**2*b*c* 
d*x + 15*(a + b*x**3)**(2/3)*a**2*b*d**2*x**4 + 432*(a + b*x**3)**(2/3)*a* 
b**2*c**2*x + 396*(a + b*x**3)**(2/3)*a*b**2*c*d*x**4 + 126*(a + b*x**3)** 
(2/3)*a*b**2*d**2*x**7 + 162*(a + b*x**3)**(2/3)*b**3*c**2*x**4 + 216*(a + 
 b*x**3)**(2/3)*b**3*c*d*x**7 + 81*(a + b*x**3)**(2/3)*b**3*d**2*x**10 + 2 
0*int((a + b*x**3)**(2/3)/(a + b*x**3),x)*a**4*d**2 - 120*int((a + b*x**3) 
**(2/3)/(a + b*x**3),x)*a**3*b*c*d + 540*int((a + b*x**3)**(2/3)/(a + b*x* 
*3),x)*a**2*b**2*c**2)/(972*b**2)