\(\int (a+b x^3)^m (c+d x^3)^p \, dx\) [183]

Optimal result

Integrand size = 19, antiderivative size = 79 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^p \, dx=x \left (a+b x^3\right )^m \left (1+\frac {b x^3}{a}\right )^{-m} \left (c+d x^3\right )^p \left (1+\frac {d x^3}{c}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{3},-m,-p,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right ) \] Output:

x*(b*x^3+a)^m*(d*x^3+c)^p*AppellF1(1/3,-m,-p,4/3,-b*x^3/a,-d*x^3/c)/((1+b* 
x^3/a)^m)/((1+d*x^3/c)^p)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(172\) vs. \(2(79)=158\).

Time = 0.33 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.18 \[ \int \left (a+b x^3\right )^m \left (c+d x^3\right )^p \, dx=\frac {4 a c x \left (a+b x^3\right )^m \left (c+d x^3\right )^p \operatorname {AppellF1}\left (\frac {1}{3},-m,-p,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{4 a c \operatorname {AppellF1}\left (\frac {1}{3},-m,-p,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+3 x^3 \left (b c m \operatorname {AppellF1}\left (\frac {4}{3},1-m,-p,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+a d p \operatorname {AppellF1}\left (\frac {4}{3},-m,1-p,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )} \] Input:

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

Output:

(4*a*c*x*(a + b*x^3)^m*(c + d*x^3)^p*AppellF1[1/3, -m, -p, 4/3, -((b*x^3)/ 
a), -((d*x^3)/c)])/(4*a*c*AppellF1[1/3, -m, -p, 4/3, -((b*x^3)/a), -((d*x^ 
3)/c)] + 3*x^3*(b*c*m*AppellF1[4/3, 1 - m, -p, 7/3, -((b*x^3)/a), -((d*x^3 
)/c)] + a*d*p*AppellF1[4/3, -m, 1 - p, 7/3, -((b*x^3)/a), -((d*x^3)/c)]))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 79, 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 = {937, 937, 936}

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.

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 937

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

\(\Big \downarrow \) 936

\(\displaystyle x \left (a+b x^3\right )^m \left (\frac {b x^3}{a}+1\right )^{-m} \left (c+d x^3\right )^p \left (\frac {d x^3}{c}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{3},-m,-p,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\)

Input:

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

Output:

(x*(a + b*x^3)^m*(c + d*x^3)^p*AppellF1[1/3, -m, -p, 4/3, -((b*x^3)/a), -( 
(d*x^3)/c)])/((1 + (b*x^3)/a)^m*(1 + (d*x^3)/c)^p)
 

Defintions of rubi rules used

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 937

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), 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*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q 
}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((c + d*x**3)**p*(a + b*x**3)**m*x + 9*int(((c + d*x**3)**p*(a + b*x**3)** 
m*x**3)/(3*a*c*m + 3*a*c*p + a*c + 3*a*d*m*x**3 + 3*a*d*p*x**3 + a*d*x**3 
+ 3*b*c*m*x**3 + 3*b*c*p*x**3 + b*c*x**3 + 3*b*d*m*x**6 + 3*b*d*p*x**6 + b 
*d*x**6),x)*a*d*m**2 + 9*int(((c + d*x**3)**p*(a + b*x**3)**m*x**3)/(3*a*c 
*m + 3*a*c*p + a*c + 3*a*d*m*x**3 + 3*a*d*p*x**3 + a*d*x**3 + 3*b*c*m*x**3 
 + 3*b*c*p*x**3 + b*c*x**3 + 3*b*d*m*x**6 + 3*b*d*p*x**6 + b*d*x**6),x)*a* 
d*m*p + 3*int(((c + d*x**3)**p*(a + b*x**3)**m*x**3)/(3*a*c*m + 3*a*c*p + 
a*c + 3*a*d*m*x**3 + 3*a*d*p*x**3 + a*d*x**3 + 3*b*c*m*x**3 + 3*b*c*p*x**3 
 + b*c*x**3 + 3*b*d*m*x**6 + 3*b*d*p*x**6 + b*d*x**6),x)*a*d*m + 9*int(((c 
 + d*x**3)**p*(a + b*x**3)**m*x**3)/(3*a*c*m + 3*a*c*p + a*c + 3*a*d*m*x** 
3 + 3*a*d*p*x**3 + a*d*x**3 + 3*b*c*m*x**3 + 3*b*c*p*x**3 + b*c*x**3 + 3*b 
*d*m*x**6 + 3*b*d*p*x**6 + b*d*x**6),x)*b*c*m*p + 9*int(((c + d*x**3)**p*( 
a + b*x**3)**m*x**3)/(3*a*c*m + 3*a*c*p + a*c + 3*a*d*m*x**3 + 3*a*d*p*x** 
3 + a*d*x**3 + 3*b*c*m*x**3 + 3*b*c*p*x**3 + b*c*x**3 + 3*b*d*m*x**6 + 3*b 
*d*p*x**6 + b*d*x**6),x)*b*c*p**2 + 3*int(((c + d*x**3)**p*(a + b*x**3)**m 
*x**3)/(3*a*c*m + 3*a*c*p + a*c + 3*a*d*m*x**3 + 3*a*d*p*x**3 + a*d*x**3 + 
 3*b*c*m*x**3 + 3*b*c*p*x**3 + b*c*x**3 + 3*b*d*m*x**6 + 3*b*d*p*x**6 + b* 
d*x**6),x)*b*c*p + 9*int(((c + d*x**3)**p*(a + b*x**3)**m)/(3*a*c*m + 3*a* 
c*p + a*c + 3*a*d*m*x**3 + 3*a*d*p*x**3 + a*d*x**3 + 3*b*c*m*x**3 + 3*b*c* 
p*x**3 + b*c*x**3 + 3*b*d*m*x**6 + 3*b*d*p*x**6 + b*d*x**6),x)*a*c*m**2...