\(\int \frac {x^{2-3 p} (a+b x^3)^p}{(c+d x^3)^3} \, dx\) [889]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {x^{2-3 p} \left (a+b x^3\right )^p}{\left (c+d x^3\right )^3} \, dx=\frac {x^{3-3 p} \left (a+b x^3\right )^{1+p}}{3 a c (1-p) \left (c+d x^3\right )^2}-\frac {a (2 b c-a d (1+p)) x^{6-3 p} \left (a+b x^3\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (3,2-p,3-p,\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{3 c^4 (1-p) (2-p)} \] Output:

1/3*x^(3-3*p)*(b*x^3+a)^(p+1)/a/c/(1-p)/(d*x^3+c)^2-1/3*a*(2*b*c-a*d*(p+1) 
)*x^(6-3*p)*(b*x^3+a)^(-2+p)*hypergeom([3, 2-p],[3-p],(-a*d+b*c)*x^3/c/(b* 
x^3+a))/c^4/(1-p)/(2-p)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 0.17 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.19 \[ \int \frac {x^{2-3 p} \left (a+b x^3\right )^p}{\left (c+d x^3\right )^3} \, dx=\frac {p x^{3-3 p} \left (a+b x^3\right )^p \left (2 \left (c+d p x^3\right ) \Phi \left (\frac {(b c-a d) x^3}{c \left (a+b x^3\right )},1,1-p\right )-d (-1+p) x^3 \Phi \left (\frac {(b c-a d) x^3}{c \left (a+b x^3\right )},1,2-p\right )-\left (2 c+d (1+p) x^3\right ) \Phi \left (\frac {(b c-a d) x^3}{c \left (a+b x^3\right )},1,-p\right )\right )}{6 c^2 \left (c+d x^3\right )^2} \] Input:

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

Output:

(p*x^(3 - 3*p)*(a + b*x^3)^p*(2*(c + d*p*x^3)*HurwitzLerchPhi[((b*c - a*d) 
*x^3)/(c*(a + b*x^3)), 1, 1 - p] - d*(-1 + p)*x^3*HurwitzLerchPhi[((b*c - 
a*d)*x^3)/(c*(a + b*x^3)), 1, 2 - p] - (2*c + d*(1 + p)*x^3)*HurwitzLerchP 
hi[((b*c - a*d)*x^3)/(c*(a + b*x^3)), 1, -p]))/(6*c^2*(c + d*x^3)^2)
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 0.66 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1013, 1012}

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.

\(\displaystyle \int \frac {x^{2-3 p} \left (a+b x^3\right )^p}{\left (c+d x^3\right )^3} \, dx\)

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {p x^{3-3 p} \left (a+b x^3\right )^p \left (d (1-p) x^3 \Phi \left (\frac {(b c-a d) x^3}{c \left (b x^3+a\right )},1,2-p\right )+2 \left (c+d p x^3\right ) \Phi \left (\frac {(b c-a d) x^3}{c \left (b x^3+a\right )},1,1-p\right )-\left (2 c+d (p+1) x^3\right ) \Phi \left (\frac {(b c-a d) x^3}{c \left (b x^3+a\right )},1,-p\right )\right )}{6 c^2 \left (c+d x^3\right )^2}\)

Input:

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

Output:

(p*x^(3 - 3*p)*(a + b*x^3)^p*(2*(c + d*p*x^3)*HurwitzLerchPhi[((b*c - a*d) 
*x^3)/(c*(a + b*x^3)), 1, 1 - p] + d*(1 - p)*x^3*HurwitzLerchPhi[((b*c - a 
*d)*x^3)/(c*(a + b*x^3)), 1, 2 - p] - (2*c + d*(1 + p)*x^3)*HurwitzLerchPh 
i[((b*c - a*d)*x^3)/(c*(a + b*x^3)), 1, -p]))/(6*c^2*(c + d*x^3)^2)
 

Defintions of rubi rules used

rule 1012

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

rule 1013

Int[((e_.)*(x_))^(m_.)*((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[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

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

Input:

int(x^(2-3*p)*(b*x^3+a)^p/(d*x^3+c)^3,x)
 

Output:

int(x^(2-3*p)*(b*x^3+a)^p/(d*x^3+c)^3,x)
 

Fricas [F]

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

integrate(x^(2-3*p)*(b*x^3+a)^p/(d*x^3+c)^3,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate(x**(2-3*p)*(b*x**3+a)**p/(d*x**3+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((b*x^3 + a)^p*x^(-3*p + 2)/(d*x^3 + c)^3, x)
 

Giac [F]

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

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

Output:

integrate((b*x^3 + a)^p*x^(-3*p + 2)/(d*x^3 + c)^3, x)
 

Mupad [F(-1)]

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

int((x^(2 - 3*p)*(a + b*x^3)^p)/(c + d*x^3)^3,x)
 

Output:

int((x^(2 - 3*p)*(a + b*x^3)^p)/(c + d*x^3)^3, x)
 

Reduce [F]

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

int(x^(2-3*p)*(b*x^3+a)^p/(d*x^3+c)^3,x)
 

Output:

( - (a + b*x**3)**p*a**3*b*d**3*p**4*x**3 - 5*(a + b*x**3)**p*a**3*b*d**3* 
p**3*x**3 - 8*(a + b*x**3)**p*a**3*b*d**3*p**2*x**3 - 4*(a + b*x**3)**p*a* 
*3*b*d**3*p*x**3 + 10*(a + b*x**3)**p*a**2*b**2*c*d**2*p**3*x**3 + 26*(a + 
 b*x**3)**p*a**2*b**2*c*d**2*p**2*x**3 + 28*(a + b*x**3)**p*a**2*b**2*c*d* 
*2*p*x**3 + 8*(a + b*x**3)**p*a**2*b**2*c*d**2*x**3 + (a + b*x**3)**p*a**2 
*b**2*d**3*p**3*x**6 + 5*(a + b*x**3)**p*a**2*b**2*d**3*p**2*x**6 + 8*(a + 
 b*x**3)**p*a**2*b**2*d**3*p*x**6 + 4*(a + b*x**3)**p*a**2*b**2*d**3*x**6 
- 26*(a + b*x**3)**p*a*b**3*c**2*d*p**2*x**3 - 40*(a + b*x**3)**p*a*b**3*c 
**2*d*p*x**3 - 24*(a + b*x**3)**p*a*b**3*c**2*d*x**3 - 8*(a + b*x**3)**p*a 
*b**3*c*d**2*p**2*x**6 - 16*(a + b*x**3)**p*a*b**3*c*d**2*p*x**6 - 12*(a + 
 b*x**3)**p*a*b**3*c*d**2*x**6 + 20*(a + b*x**3)**p*b**4*c**3*p*x**3 + 16* 
(a + b*x**3)**p*b**4*c**3*x**3 + 10*(a + b*x**3)**p*b**4*c**2*d*p*x**6 + 8 
*(a + b*x**3)**p*b**4*c**2*d*x**6 - 12*x**(3*p)*int(((a + b*x**3)**p*x**5) 
/(x**(3*p)*a**5*c**3*d**4*p**4 + 6*x**(3*p)*a**5*c**3*d**4*p**3 + 13*x**(3 
*p)*a**5*c**3*d**4*p**2 + 12*x**(3*p)*a**5*c**3*d**4*p + 4*x**(3*p)*a**5*c 
**3*d**4 + 3*x**(3*p)*a**5*c**2*d**5*p**4*x**3 + 18*x**(3*p)*a**5*c**2*d** 
5*p**3*x**3 + 39*x**(3*p)*a**5*c**2*d**5*p**2*x**3 + 36*x**(3*p)*a**5*c**2 
*d**5*p*x**3 + 12*x**(3*p)*a**5*c**2*d**5*x**3 + 3*x**(3*p)*a**5*c*d**6*p* 
*4*x**6 + 18*x**(3*p)*a**5*c*d**6*p**3*x**6 + 39*x**(3*p)*a**5*c*d**6*p**2 
*x**6 + 36*x**(3*p)*a**5*c*d**6*p*x**6 + 12*x**(3*p)*a**5*c*d**6*x**6 +...