\(\int \sqrt [3]{a+b x^3} (c+d x^3)^2 \, dx\) [95]

Optimal result

Integrand size = 21, antiderivative size = 128 \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\frac {d (4 b c-a d) x \left (a+b x^3\right )^{4/3}}{10 b^2}+\frac {d^2 x^4 \left (a+b x^3\right )^{4/3}}{8 b}+\frac {\left (10 b^2 c^2-4 a b c d+a^2 d^2\right ) x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{10 b^2 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:

1/10*d*(-a*d+4*b*c)*x*(b*x^3+a)^(4/3)/b^2+1/8*d^2*x^4*(b*x^3+a)^(4/3)/b+1/ 
10*(a^2*d^2-4*a*b*c*d+10*b^2*c^2)*x*(b*x^3+a)^(1/3)*hypergeom([-1/3, 1/3], 
[4/3],-b*x^3/a)/b^2/(1+b*x^3/a)^(1/3)
 

Mathematica [A] (verified)

Time = 8.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.40 \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\frac {x \sqrt [3]{a+b x^3} \left (20 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \operatorname {Gamma}\left (-\frac {1}{3}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {10}{3},-\frac {b x^3}{a}\right )-3 b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \operatorname {Gamma}\left (\frac {2}{3}\right ) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {13}{3},-\frac {b x^3}{a}\right )-9 b x^3 \left (c+d x^3\right )^2 \operatorname {Gamma}\left (\frac {2}{3}\right ) \, _3F_2\left (\frac {2}{3},\frac {4}{3},2;1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{280 a \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Gamma}\left (-\frac {1}{3}\right )} \] Input:

Integrate[(a + b*x^3)^(1/3)*(c + d*x^3)^2,x]
 

Output:

(x*(a + b*x^3)^(1/3)*(20*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)*Gamma[-1/3]*Hy 
pergeometric2F1[-1/3, 1/3, 10/3, -((b*x^3)/a)] - 3*b*x^3*(11*c^2 + 16*c*d* 
x^3 + 5*d^2*x^6)*Gamma[2/3]*Hypergeometric2F1[2/3, 4/3, 13/3, -((b*x^3)/a) 
] - 9*b*x^3*(c + d*x^3)^2*Gamma[2/3]*HypergeometricPFQ[{2/3, 4/3, 2}, {1, 
13/3}, -((b*x^3)/a)]))/(280*a*(1 + (b*x^3)/a)^(1/3)*Gamma[-1/3])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {933, 913, 779, 778}

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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int \sqrt [3]{b x^3+a} \left (d (11 b c-4 a d) x^3+c (8 b c-a d)\right )dx}{8 b}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {\frac {4 \left (a^2 d^2-4 a b c d+10 b^2 c^2\right ) \int \sqrt [3]{b x^3+a}dx}{5 b}+\frac {d x \left (a+b x^3\right )^{4/3} (11 b c-4 a d)}{5 b}}{8 b}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {\frac {4 \sqrt [3]{a+b x^3} \left (a^2 d^2-4 a b c d+10 b^2 c^2\right ) \int \sqrt [3]{\frac {b x^3}{a}+1}dx}{5 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{4/3} (11 b c-4 a d)}{5 b}}{8 b}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {\frac {4 x \sqrt [3]{a+b x^3} \left (a^2 d^2-4 a b c d+10 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{5 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{4/3} (11 b c-4 a d)}{5 b}}{8 b}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}\)

Input:

Int[(a + b*x^3)^(1/3)*(c + d*x^3)^2,x]
 

Output:

(d*x*(a + b*x^3)^(4/3)*(c + d*x^3))/(8*b) + ((d*(11*b*c - 4*a*d)*x*(a + b* 
x^3)^(4/3))/(5*b) + (4*(10*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*x*(a + b*x^3)^(1 
/3)*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)])/(5*b*(1 + (b*x^3)/a)^ 
(1/3)))/(8*b)
 

Defintions of rubi rules used

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x 
^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(1 + b*(x^n/a))^p, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

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 [F]

\[\int \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (d \,x^{3}+c \right )^{2}d x\]

Input:

int((b*x^3+a)^(1/3)*(d*x^3+c)^2,x)
 

Output:

int((b*x^3+a)^(1/3)*(d*x^3+c)^2,x)
 

Fricas [F]

\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x^{3} + c\right )}^{2} \,d x } \] Input:

integrate((b*x^3+a)^(1/3)*(d*x^3+c)^2,x, algorithm="fricas")
 

Output:

integral((d^2*x^6 + 2*c*d*x^3 + c^2)*(b*x^3 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.80 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.02 \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\frac {\sqrt [3]{a} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 \sqrt [3]{a} c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:

integrate((b*x**3+a)**(1/3)*(d*x**3+c)**2,x)
 

Output:

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

Maxima [F]

\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x^{3} + c\right )}^{2} \,d x } \] Input:

integrate((b*x^3+a)^(1/3)*(d*x^3+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(1/3)*(d*x^3 + c)^2, x)
 

Giac [F]

\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\int { {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (d x^{3} + c\right )}^{2} \,d x } \] Input:

integrate((b*x^3+a)^(1/3)*(d*x^3+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(1/3)*(d*x^3 + c)^2, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\int {\left (b\,x^3+a\right )}^{1/3}\,{\left (d\,x^3+c\right )}^2 \,d x \] Input:

int((a + b*x^3)^(1/3)*(c + d*x^3)^2,x)
 

Output:

int((a + b*x^3)^(1/3)*(c + d*x^3)^2, x)
 

Reduce [F]

\[ \int \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx=\frac {-2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d^{2} x +8 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c d x +\left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,d^{2} x^{4}+20 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c^{2} x +16 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c d \,x^{4}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} d^{2} x^{7}+2 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{3} d^{2}-8 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b c d +20 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a \,b^{2} c^{2}}{40 b^{2}} \] Input:

int((b*x^3+a)^(1/3)*(d*x^3+c)^2,x)
 

Output:

( - 2*(a + b*x**3)**(1/3)*a**2*d**2*x + 8*(a + b*x**3)**(1/3)*a*b*c*d*x + 
(a + b*x**3)**(1/3)*a*b*d**2*x**4 + 20*(a + b*x**3)**(1/3)*b**2*c**2*x + 1 
6*(a + b*x**3)**(1/3)*b**2*c*d*x**4 + 5*(a + b*x**3)**(1/3)*b**2*d**2*x**7 
 + 2*int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**3*d**2 - 8*int((a + b*x**3 
)**(1/3)/(a + b*x**3),x)*a**2*b*c*d + 20*int((a + b*x**3)**(1/3)/(a + b*x* 
*3),x)*a*b**2*c**2)/(40*b**2)