\(\int (a+b x^3)^m (c+d x^3)^2 \, dx\) [185]

Optimal result

Integrand size = 19, antiderivative size = 173 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^2 \, dx=-\frac {2 d (2 a d-b c (7+3 m)) x \left (a+b x^3\right )^{1+m}}{b^2 (4+3 m) (7+3 m)}+\frac {d^2 x^4 \left (a+b x^3\right )^{1+m}}{b (7+3 m)}+\frac {\left (4 a^2 d^2-2 a b c d (7+3 m)+b^2 c^2 \left (28+33 m+9 m^2\right )\right ) x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )}{b^2 (4+3 m) (7+3 m)} \] Output:

-2*d*(2*a*d-b*c*(7+3*m))*x*(b*x^3+a)^(1+m)/b^2/(4+3*m)/(7+3*m)+d^2*x^4*(b* 
x^3+a)^(1+m)/b/(7+3*m)+(4*a^2*d^2-2*a*b*c*d*(7+3*m)+b^2*c^2*(9*m^2+33*m+28 
))*x*(b*x^3+a)^m*hypergeom([1/3, -m],[4/3],-b*x^3/a)/b^2/(4+3*m)/(7+3*m)/( 
(1+b*x^3/a)^m)
 

Mathematica [A] (verified)

Time = 5.22 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.61 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^2 \, dx=\frac {1}{14} x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \left (14 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )+d x^3 \left (7 c \operatorname {Hypergeometric2F1}\left (\frac {4}{3},-m,\frac {7}{3},-\frac {b x^3}{a}\right )+2 d x^3 \operatorname {Hypergeometric2F1}\left (\frac {7}{3},-m,\frac {10}{3},-\frac {b x^3}{a}\right )\right )\right ) \] Input:

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

Output:

(x*(a + b*x^3)^m*(14*c^2*Hypergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)] + d 
*x^3*(7*c*Hypergeometric2F1[4/3, -m, 7/3, -((b*x^3)/a)] + 2*d*x^3*Hypergeo 
metric2F1[7/3, -m, 10/3, -((b*x^3)/a)])))/(14*(1 + (b*x^3)/a)^m)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {933, 25, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\left (b x^3+a\right )^m \left (d (4 a d-b c (3 m+10)) x^3+c (a d-b c (3 m+7))\right )dx}{b (3 m+7)}+\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1}}{b (3 m+7)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1}}{b (3 m+7)}-\frac {\int \left (b x^3+a\right )^m \left (d (4 a d-b c (3 m+10)) x^3+c (a d-b c (3 m+7))\right )dx}{b (3 m+7)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1}}{b (3 m+7)}-\frac {\frac {d x \left (a+b x^3\right )^{m+1} (4 a d-b c (3 m+10))}{b (3 m+4)}-\frac {\left (4 a^2 d^2-2 a b c d (3 m+7)+b^2 c^2 \left (9 m^2+33 m+28\right )\right ) \int \left (b x^3+a\right )^mdx}{b (3 m+4)}}{b (3 m+7)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1}}{b (3 m+7)}-\frac {\frac {d x \left (a+b x^3\right )^{m+1} (4 a d-b c (3 m+10))}{b (3 m+4)}-\frac {\left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (4 a^2 d^2-2 a b c d (3 m+7)+b^2 c^2 \left (9 m^2+33 m+28\right )\right ) \int \left (\frac {b x^3}{a}+1\right )^mdx}{b (3 m+4)}}{b (3 m+7)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1}}{b (3 m+7)}-\frac {\frac {d x \left (a+b x^3\right )^{m+1} (4 a d-b c (3 m+10))}{b (3 m+4)}-\frac {x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (4 a^2 d^2-2 a b c d (3 m+7)+b^2 c^2 \left (9 m^2+33 m+28\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )}{b (3 m+4)}}{b (3 m+7)}\)

Input:

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

Output:

(d*x*(a + b*x^3)^(1 + m)*(c + d*x^3))/(b*(7 + 3*m)) - ((d*(4*a*d - b*c*(10 
 + 3*m))*x*(a + b*x^3)^(1 + m))/(b*(4 + 3*m)) - ((4*a^2*d^2 - 2*a*b*c*d*(7 
 + 3*m) + b^2*c^2*(28 + 33*m + 9*m^2))*x*(a + b*x^3)^m*Hypergeometric2F1[1 
/3, -m, 4/3, -((b*x^3)/a)])/(b*(4 + 3*m)*(1 + (b*x^3)/a)^m))/(b*(7 + 3*m))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 92.41 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.70 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^2 \, dx=\frac {a^{m} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, - m \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 a^{m} c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, - m \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{m} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, - m \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

( - 12*(a + b*x**3)**m*a**2*d**2*m*x + 18*(a + b*x**3)**m*a*b*c*d*m**2*x + 
 42*(a + b*x**3)**m*a*b*c*d*m*x + 9*(a + b*x**3)**m*a*b*d**2*m**2*x**4 + 3 
*(a + b*x**3)**m*a*b*d**2*m*x**4 + 9*(a + b*x**3)**m*b**2*c**2*m**2*x + 33 
*(a + b*x**3)**m*b**2*c**2*m*x + 28*(a + b*x**3)**m*b**2*c**2*x + 18*(a + 
b*x**3)**m*b**2*c*d*m**2*x**4 + 48*(a + b*x**3)**m*b**2*c*d*m*x**4 + 14*(a 
 + b*x**3)**m*b**2*c*d*x**4 + 9*(a + b*x**3)**m*b**2*d**2*m**2*x**7 + 15*( 
a + b*x**3)**m*b**2*d**2*m*x**7 + 4*(a + b*x**3)**m*b**2*d**2*x**7 + 324*i 
nt((a + b*x**3)**m/(27*a*m**3 + 108*a*m**2 + 117*a*m + 28*a + 27*b*m**3*x* 
*3 + 108*b*m**2*x**3 + 117*b*m*x**3 + 28*b*x**3),x)*a**3*d**2*m**4 + 1296* 
int((a + b*x**3)**m/(27*a*m**3 + 108*a*m**2 + 117*a*m + 28*a + 27*b*m**3*x 
**3 + 108*b*m**2*x**3 + 117*b*m*x**3 + 28*b*x**3),x)*a**3*d**2*m**3 + 1404 
*int((a + b*x**3)**m/(27*a*m**3 + 108*a*m**2 + 117*a*m + 28*a + 27*b*m**3* 
x**3 + 108*b*m**2*x**3 + 117*b*m*x**3 + 28*b*x**3),x)*a**3*d**2*m**2 + 336 
*int((a + b*x**3)**m/(27*a*m**3 + 108*a*m**2 + 117*a*m + 28*a + 27*b*m**3* 
x**3 + 108*b*m**2*x**3 + 117*b*m*x**3 + 28*b*x**3),x)*a**3*d**2*m - 486*in 
t((a + b*x**3)**m/(27*a*m**3 + 108*a*m**2 + 117*a*m + 28*a + 27*b*m**3*x** 
3 + 108*b*m**2*x**3 + 117*b*m*x**3 + 28*b*x**3),x)*a**2*b*c*d*m**5 - 3078* 
int((a + b*x**3)**m/(27*a*m**3 + 108*a*m**2 + 117*a*m + 28*a + 27*b*m**3*x 
**3 + 108*b*m**2*x**3 + 117*b*m*x**3 + 28*b*x**3),x)*a**2*b*c*d*m**4 - 664 
2*int((a + b*x**3)**m/(27*a*m**3 + 108*a*m**2 + 117*a*m + 28*a + 27*b*m...