\(\int (a+b x^3)^{4/3} (c+d x^3) \, dx\) [117]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) \, dx=\frac {d x \left (a+b x^3\right )^{7/3}}{8 b}+\frac {a (8 b c-a d) x \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{8 b \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:

1/8*d*x*(b*x^3+a)^(7/3)/b+1/8*a*(-a*d+8*b*c)*x*(b*x^3+a)^(1/3)*hypergeom([ 
-4/3, 1/3],[4/3],-b*x^3/a)/b/(1+b*x^3/a)^(1/3)
 

Mathematica [A] (verified)

Time = 7.46 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) \, dx=\frac {x \sqrt [3]{a+b x^3} \left (d \left (a+b x^3\right )^2-\frac {a (-8 b c+a d) \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{1+\frac {b x^3}{a}}}\right )}{8 b} \] Input:

Integrate[(a + b*x^3)^(4/3)*(c + d*x^3),x]
 

Output:

(x*(a + b*x^3)^(1/3)*(d*(a + b*x^3)^2 - (a*(-8*b*c + a*d)*Hypergeometric2F 
1[-4/3, 1/3, 4/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^(1/3)))/(8*b)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.

\(\displaystyle \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) \, dx\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {(8 b c-a d) \int \left (b x^3+a\right )^{4/3}dx}{8 b}+\frac {d x \left (a+b x^3\right )^{7/3}}{8 b}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {a \sqrt [3]{a+b x^3} (8 b c-a d) \int \left (\frac {b x^3}{a}+1\right )^{4/3}dx}{8 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3}}{8 b}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {a x \sqrt [3]{a+b x^3} (8 b c-a d) \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{8 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3}}{8 b}\)

Input:

Int[(a + b*x^3)^(4/3)*(c + d*x^3),x]
 

Output:

(d*x*(a + b*x^3)^(7/3))/(8*b) + (a*(8*b*c - a*d)*x*(a + b*x^3)^(1/3)*Hyper 
geometric2F1[-4/3, 1/3, 4/3, -((b*x^3)/a)])/(8*b*(1 + (b*x^3)/a)^(1/3))
 

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]
 

Maple [F]

\[\int \left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )d x\]

Input:

int((b*x^3+a)^(4/3)*(d*x^3+c),x)
 

Output:

int((b*x^3+a)^(4/3)*(d*x^3+c),x)
 

Fricas [F]

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

integrate((b*x^3+a)^(4/3)*(d*x^3+c),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.05 \[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) \, dx=\frac {a^{\frac {4}{3}} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {a^{\frac {4}{3}} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} b c x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} b d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:

integrate((b*x**3+a)**(4/3)*(d*x**3+c),x)
 

Output:

a**(4/3)*c*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/ 
a)/(3*gamma(4/3)) + a**(4/3)*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)*b*c*x**4*gamma(4/3)*hy 
per((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + a**(1/ 
3)*b*d*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]

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

integrate((b*x^3+a)^(4/3)*(d*x^3+c),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a)^(4/3)*(d*x^3 + c), x)
 

Giac [F]

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

integrate((b*x^3+a)^(4/3)*(d*x^3+c),x, algorithm="giac")
 

Output:

integrate((b*x^3 + a)^(4/3)*(d*x^3 + c), x)
 

Mupad [F(-1)]

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

int((a + b*x^3)^(4/3)*(c + d*x^3),x)
 

Output:

int((a + b*x^3)^(4/3)*(c + d*x^3), x)
 

Reduce [F]

\[ \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) \, dx=\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} d x +24 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b c x +9 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b d \,x^{4}+8 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} c \,x^{4}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} d \,x^{7}-2 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{3} d +16 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{2} b c}{40 b} \] Input:

int((b*x^3+a)^(4/3)*(d*x^3+c),x)
 

Output:

(2*(a + b*x**3)**(1/3)*a**2*d*x + 24*(a + b*x**3)**(1/3)*a*b*c*x + 9*(a + 
b*x**3)**(1/3)*a*b*d*x**4 + 8*(a + b*x**3)**(1/3)*b**2*c*x**4 + 5*(a + b*x 
**3)**(1/3)*b**2*d*x**7 - 2*int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**3*d 
 + 16*int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**2*b*c)/(40*b)