\(\int (a+b x^3)^m (c+d x^3) \, dx\) [186]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right ) \, dx=\frac {d x \left (a+b x^3\right )^{1+m}}{b (4+3 m)}+\left (c-\frac {a d}{4 b+3 b m}\right ) x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right ) \, dx=\frac {x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \left (d \left (a+b x^3\right ) \left (1+\frac {b x^3}{a}\right )^m+(-a d+b c (4+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{b (4+3 m)} \] Input:

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

Output:

(x*(a + b*x^3)^m*(d*(a + b*x^3)*(1 + (b*x^3)/a)^m + (-(a*d) + b*c*(4 + 3*m 
))*Hypergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)]))/(b*(4 + 3*m)*(1 + (b*x^ 
3)/a)^m)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, 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.

\(\displaystyle \int \left (c+d x^3\right ) \left (a+b x^3\right )^m \, dx\)

\(\Big \downarrow \) 913

\(\displaystyle \left (c-\frac {a d}{3 b m+4 b}\right ) \int \left (b x^3+a\right )^mdx+\frac {d x \left (a+b x^3\right )^{m+1}}{b (3 m+4)}\)

\(\Big \downarrow \) 779

\(\displaystyle \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (c-\frac {a d}{3 b m+4 b}\right ) \int \left (\frac {b x^3}{a}+1\right )^mdx+\frac {d x \left (a+b x^3\right )^{m+1}}{b (3 m+4)}\)

\(\Big \downarrow \) 778

\(\displaystyle x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (c-\frac {a d}{3 b m+4 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )+\frac {d x \left (a+b x^3\right )^{m+1}}{b (3 m+4)}\)

Input:

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

Output:

(d*x*(a + b*x^3)^(1 + m))/(b*(4 + 3*m)) + ((c - (a*d)/(4*b + 3*b*m))*x*(a 
+ b*x^3)^m*Hypergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^ 
m
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d*x^3 + c)*(b*x^3 + a)^m, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 38.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right ) \, dx=\frac {a^{m} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, - m \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {a^{m} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, - m \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \] Input:

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

Output:

a**m*c*x*gamma(1/3)*hyper((1/3, -m), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3* 
gamma(4/3)) + a**m*d*x**4*gamma(4/3)*hyper((4/3, -m), (7/3,), b*x**3*exp_p 
olar(I*pi)/a)/(3*gamma(7/3))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b x^3\right )^m \left (c+d x^3\right ) \, dx=\frac {3 \left (b \,x^{3}+a \right )^{m} a d m x +3 \left (b \,x^{3}+a \right )^{m} b c m x +4 \left (b \,x^{3}+a \right )^{m} b c x +3 \left (b \,x^{3}+a \right )^{m} b d m \,x^{4}+\left (b \,x^{3}+a \right )^{m} b d \,x^{4}-27 \left (\int \frac {\left (b \,x^{3}+a \right )^{m}}{9 b \,m^{2} x^{3}+15 b m \,x^{3}+4 b \,x^{3}+9 a \,m^{2}+15 a m +4 a}d x \right ) a^{2} d \,m^{3}-45 \left (\int \frac {\left (b \,x^{3}+a \right )^{m}}{9 b \,m^{2} x^{3}+15 b m \,x^{3}+4 b \,x^{3}+9 a \,m^{2}+15 a m +4 a}d x \right ) a^{2} d \,m^{2}-12 \left (\int \frac {\left (b \,x^{3}+a \right )^{m}}{9 b \,m^{2} x^{3}+15 b m \,x^{3}+4 b \,x^{3}+9 a \,m^{2}+15 a m +4 a}d x \right ) a^{2} d m +81 \left (\int \frac {\left (b \,x^{3}+a \right )^{m}}{9 b \,m^{2} x^{3}+15 b m \,x^{3}+4 b \,x^{3}+9 a \,m^{2}+15 a m +4 a}d x \right ) a b c \,m^{4}+243 \left (\int \frac {\left (b \,x^{3}+a \right )^{m}}{9 b \,m^{2} x^{3}+15 b m \,x^{3}+4 b \,x^{3}+9 a \,m^{2}+15 a m +4 a}d x \right ) a b c \,m^{3}+216 \left (\int \frac {\left (b \,x^{3}+a \right )^{m}}{9 b \,m^{2} x^{3}+15 b m \,x^{3}+4 b \,x^{3}+9 a \,m^{2}+15 a m +4 a}d x \right ) a b c \,m^{2}+48 \left (\int \frac {\left (b \,x^{3}+a \right )^{m}}{9 b \,m^{2} x^{3}+15 b m \,x^{3}+4 b \,x^{3}+9 a \,m^{2}+15 a m +4 a}d x \right ) a b c m}{b \left (9 m^{2}+15 m +4\right )} \] Input:

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

Output:

(3*(a + b*x**3)**m*a*d*m*x + 3*(a + b*x**3)**m*b*c*m*x + 4*(a + b*x**3)**m 
*b*c*x + 3*(a + b*x**3)**m*b*d*m*x**4 + (a + b*x**3)**m*b*d*x**4 - 27*int( 
(a + b*x**3)**m/(9*a*m**2 + 15*a*m + 4*a + 9*b*m**2*x**3 + 15*b*m*x**3 + 4 
*b*x**3),x)*a**2*d*m**3 - 45*int((a + b*x**3)**m/(9*a*m**2 + 15*a*m + 4*a 
+ 9*b*m**2*x**3 + 15*b*m*x**3 + 4*b*x**3),x)*a**2*d*m**2 - 12*int((a + b*x 
**3)**m/(9*a*m**2 + 15*a*m + 4*a + 9*b*m**2*x**3 + 15*b*m*x**3 + 4*b*x**3) 
,x)*a**2*d*m + 81*int((a + b*x**3)**m/(9*a*m**2 + 15*a*m + 4*a + 9*b*m**2* 
x**3 + 15*b*m*x**3 + 4*b*x**3),x)*a*b*c*m**4 + 243*int((a + b*x**3)**m/(9* 
a*m**2 + 15*a*m + 4*a + 9*b*m**2*x**3 + 15*b*m*x**3 + 4*b*x**3),x)*a*b*c*m 
**3 + 216*int((a + b*x**3)**m/(9*a*m**2 + 15*a*m + 4*a + 9*b*m**2*x**3 + 1 
5*b*m*x**3 + 4*b*x**3),x)*a*b*c*m**2 + 48*int((a + b*x**3)**m/(9*a*m**2 + 
15*a*m + 4*a + 9*b*m**2*x**3 + 15*b*m*x**3 + 4*b*x**3),x)*a*b*c*m)/(b*(9*m 
**2 + 15*m + 4))