\(\int (a+b x^3)^m (c+d x^3)^3 \, dx\) [184]

Optimal result

Integrand size = 19, antiderivative size = 283 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^3 \, dx=\frac {d \left (28 a^2 d^2-12 a b c d (10+3 m)+3 b^2 c^2 \left (70+51 m+9 m^2\right )\right ) x \left (a+b x^3\right )^{1+m}}{b^3 (4+3 m) (7+3 m) (10+3 m)}-\frac {d^2 (7 a d-3 b c (10+3 m)) x^4 \left (a+b x^3\right )^{1+m}}{b^2 (7+3 m) (10+3 m)}+\frac {d^3 x^7 \left (a+b x^3\right )^{1+m}}{b (10+3 m)}+\frac {\left (b^3 c^3 \left (70+51 m+9 m^2\right )-\frac {a d \left (28 a^2 d^2-12 a b c d (10+3 m)+3 b^2 c^2 \left (70+51 m+9 m^2\right )\right )}{4+3 m}\right ) x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )}{b^3 (7+3 m) (10+3 m)} \] Output:

d*(28*a^2*d^2-12*a*b*c*d*(10+3*m)+3*b^2*c^2*(9*m^2+51*m+70))*x*(b*x^3+a)^( 
1+m)/b^3/(4+3*m)/(7+3*m)/(10+3*m)-d^2*(7*a*d-3*b*c*(10+3*m))*x^4*(b*x^3+a) 
^(1+m)/b^2/(7+3*m)/(10+3*m)+d^3*x^7*(b*x^3+a)^(1+m)/b/(10+3*m)+(b^3*c^3*(9 
*m^2+51*m+70)-a*d*(28*a^2*d^2-12*a*b*c*d*(10+3*m)+3*b^2*c^2*(9*m^2+51*m+70 
))/(4+3*m))*x*(b*x^3+a)^m*hypergeom([1/3, -m],[4/3],-b*x^3/a)/b^3/(7+3*m)/ 
(10+3*m)/((1+b*x^3/a)^m)
 

Mathematica [A] (verified)

Time = 7.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.48 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^3 \, dx=\frac {1}{140} x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \left (140 c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )+d x^3 \left (105 c^2 \operatorname {Hypergeometric2F1}\left (\frac {4}{3},-m,\frac {7}{3},-\frac {b x^3}{a}\right )+2 d x^3 \left (30 c \operatorname {Hypergeometric2F1}\left (\frac {7}{3},-m,\frac {10}{3},-\frac {b x^3}{a}\right )+7 d x^3 \operatorname {Hypergeometric2F1}\left (\frac {10}{3},-m,\frac {13}{3},-\frac {b x^3}{a}\right )\right )\right )\right ) \] Input:

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

Output:

(x*(a + b*x^3)^m*(140*c^3*Hypergeometric2F1[1/3, -m, 4/3, -((b*x^3)/a)] + 
d*x^3*(105*c^2*Hypergeometric2F1[4/3, -m, 7/3, -((b*x^3)/a)] + 2*d*x^3*(30 
*c*Hypergeometric2F1[7/3, -m, 10/3, -((b*x^3)/a)] + 7*d*x^3*Hypergeometric 
2F1[10/3, -m, 13/3, -((b*x^3)/a)]))))/(140*(1 + (b*x^3)/a)^m)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {933, 25, 1025, 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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\left (b x^3+a\right )^m \left (d x^3+c\right ) \left (d (7 a d-b c (3 m+16)) x^3+c (a d-b c (3 m+10))\right )dx}{b (3 m+10)}+\frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)}-\frac {\int \left (b x^3+a\right )^m \left (d x^3+c\right ) \left (d (7 a d-b c (3 m+16)) x^3+c (a d-b c (3 m+10))\right )dx}{b (3 m+10)}\)

\(\Big \downarrow \) 1025

\(\displaystyle \frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)}-\frac {\frac {\int -\left (b x^3+a\right )^m \left (d \left (b^2 \left (9 m^2+60 m+118\right ) c^2-a b d (15 m+92) c+28 a^2 d^2\right ) x^3+c \left (b^2 \left (9 m^2+51 m+70\right ) c^2-a b d (6 m+23) c+7 a^2 d^2\right )\right )dx}{b (3 m+7)}+\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1} (7 a d-b c (3 m+16))}{b (3 m+7)}}{b (3 m+10)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)}-\frac {\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1} (7 a d-b c (3 m+16))}{b (3 m+7)}-\frac {\int \left (b x^3+a\right )^m \left (d \left (b^2 \left (9 m^2+60 m+118\right ) c^2-a b d (15 m+92) c+28 a^2 d^2\right ) x^3+c \left (b^2 \left (9 m^2+51 m+70\right ) c^2-a b d (6 m+23) c+7 a^2 d^2\right )\right )dx}{b (3 m+7)}}{b (3 m+10)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)}-\frac {\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1} (7 a d-b c (3 m+16))}{b (3 m+7)}-\frac {\frac {d x \left (a+b x^3\right )^{m+1} \left (28 a^2 d^2-a b c d (15 m+92)+b^2 c^2 \left (9 m^2+60 m+118\right )\right )}{b (3 m+4)}-\frac {\left (28 a^3 d^3-12 a^2 b c d^2 (3 m+10)+3 a b^2 c^2 d \left (9 m^2+51 m+70\right )-b^3 c^3 \left (27 m^3+189 m^2+414 m+280\right )\right ) \int \left (b x^3+a\right )^mdx}{b (3 m+4)}}{b (3 m+7)}}{b (3 m+10)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)}-\frac {\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1} (7 a d-b c (3 m+16))}{b (3 m+7)}-\frac {\frac {d x \left (a+b x^3\right )^{m+1} \left (28 a^2 d^2-a b c d (15 m+92)+b^2 c^2 \left (9 m^2+60 m+118\right )\right )}{b (3 m+4)}-\frac {\left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (28 a^3 d^3-12 a^2 b c d^2 (3 m+10)+3 a b^2 c^2 d \left (9 m^2+51 m+70\right )-b^3 c^3 \left (27 m^3+189 m^2+414 m+280\right )\right ) \int \left (\frac {b x^3}{a}+1\right )^mdx}{b (3 m+4)}}{b (3 m+7)}}{b (3 m+10)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {d x \left (c+d x^3\right )^2 \left (a+b x^3\right )^{m+1}}{b (3 m+10)}-\frac {\frac {d x \left (c+d x^3\right ) \left (a+b x^3\right )^{m+1} (7 a d-b c (3 m+16))}{b (3 m+7)}-\frac {\frac {d x \left (a+b x^3\right )^{m+1} \left (28 a^2 d^2-a b c d (15 m+92)+b^2 c^2 \left (9 m^2+60 m+118\right )\right )}{b (3 m+4)}-\frac {x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (28 a^3 d^3-12 a^2 b c d^2 (3 m+10)+3 a b^2 c^2 d \left (9 m^2+51 m+70\right )-b^3 c^3 \left (27 m^3+189 m^2+414 m+280\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-m,\frac {4}{3},-\frac {b x^3}{a}\right )}{b (3 m+4)}}{b (3 m+7)}}{b (3 m+10)}\)

Input:

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

Output:

(d*x*(a + b*x^3)^(1 + m)*(c + d*x^3)^2)/(b*(10 + 3*m)) - ((d*(7*a*d - b*c* 
(16 + 3*m))*x*(a + b*x^3)^(1 + m)*(c + d*x^3))/(b*(7 + 3*m)) - ((d*(28*a^2 
*d^2 - a*b*c*d*(92 + 15*m) + b^2*c^2*(118 + 60*m + 9*m^2))*x*(a + b*x^3)^( 
1 + m))/(b*(4 + 3*m)) - ((28*a^3*d^3 - 12*a^2*b*c*d^2*(10 + 3*m) + 3*a*b^2 
*c^2*d*(70 + 51*m + 9*m^2) - b^3*c^3*(280 + 414*m + 189*m^2 + 27*m^3))*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)))/(b*(10 + 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]
 

rule 1025

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + ( 
f_.)*(x_)^(n_)), x_Symbol] :> Simp[f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/( 
b*(n*(p + q + 1) + 1))), x] + Simp[1/(b*(n*(p + q + 1) + 1))   Int[(a + b*x 
^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e 
- a*f) + f*n*q*(b*c - a*d) + b*d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[ 
{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^9 + 3*c*d^2*x^6 + 3*c^2*d*x^3 + c^3)*(b*x^3 + a)^m, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(84*(a + b*x**3)**m*a**3*d**3*m*x - 108*(a + b*x**3)**m*a**2*b*c*d**2*m**2 
*x - 360*(a + b*x**3)**m*a**2*b*c*d**2*m*x - 63*(a + b*x**3)**m*a**2*b*d** 
3*m**2*x**4 - 21*(a + b*x**3)**m*a**2*b*d**3*m*x**4 + 81*(a + b*x**3)**m*a 
*b**2*c**2*d*m**3*x + 459*(a + b*x**3)**m*a*b**2*c**2*d*m**2*x + 630*(a + 
b*x**3)**m*a*b**2*c**2*d*m*x + 81*(a + b*x**3)**m*a*b**2*c*d**2*m**3*x**4 
+ 297*(a + b*x**3)**m*a*b**2*c*d**2*m**2*x**4 + 90*(a + b*x**3)**m*a*b**2* 
c*d**2*m*x**4 + 27*(a + b*x**3)**m*a*b**2*d**3*m**3*x**7 + 45*(a + b*x**3) 
**m*a*b**2*d**3*m**2*x**7 + 12*(a + b*x**3)**m*a*b**2*d**3*m*x**7 + 27*(a 
+ b*x**3)**m*b**3*c**3*m**3*x + 189*(a + b*x**3)**m*b**3*c**3*m**2*x + 414 
*(a + b*x**3)**m*b**3*c**3*m*x + 280*(a + b*x**3)**m*b**3*c**3*x + 81*(a + 
 b*x**3)**m*b**3*c**2*d*m**3*x**4 + 486*(a + b*x**3)**m*b**3*c**2*d*m**2*x 
**4 + 783*(a + b*x**3)**m*b**3*c**2*d*m*x**4 + 210*(a + b*x**3)**m*b**3*c* 
*2*d*x**4 + 81*(a + b*x**3)**m*b**3*c*d**2*m**3*x**7 + 405*(a + b*x**3)**m 
*b**3*c*d**2*m**2*x**7 + 486*(a + b*x**3)**m*b**3*c*d**2*m*x**7 + 120*(a + 
 b*x**3)**m*b**3*c*d**2*x**7 + 27*(a + b*x**3)**m*b**3*d**3*m**3*x**10 + 1 
08*(a + b*x**3)**m*b**3*d**3*m**2*x**10 + 117*(a + b*x**3)**m*b**3*d**3*m* 
x**10 + 28*(a + b*x**3)**m*b**3*d**3*x**10 - 6804*int((a + b*x**3)**m/(81* 
a*m**4 + 594*a*m**3 + 1431*a*m**2 + 1254*a*m + 280*a + 81*b*m**4*x**3 + 59 
4*b*m**3*x**3 + 1431*b*m**2*x**3 + 1254*b*m*x**3 + 280*b*x**3),x)*a**4*d** 
3*m**5 - 49896*int((a + b*x**3)**m/(81*a*m**4 + 594*a*m**3 + 1431*a*m**...