\(\int \frac {x^{2-3 p} (a+b x^3)^p}{(c+d x^3)^2} \, dx\) [888]

Optimal result

Integrand size = 26, antiderivative size = 69 \[ \int \frac {x^{2-3 p} \left (a+b x^3\right )^p}{\left (c+d x^3\right )^2} \, dx=\frac {a x^{3-3 p} \left (a+b x^3\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{3 c^2 (1-p)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.46 \[ \int \frac {x^{2-3 p} \left (a+b x^3\right )^p}{\left (c+d x^3\right )^2} \, dx=-\frac {x^{3-3 p} \left (a+b x^3\right )^p \left (1+\frac {b x^3}{a}\right )^{-p} \left (1+\frac {d x^3}{c}\right )^p \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{3 c (-1+p) \left (c+d x^3\right )} \] Input:

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

Output:

-1/3*(x^(3 - 3*p)*(a + b*x^3)^p*(1 + (d*x^3)/c)^p*Hypergeometric2F1[1 - p, 
 -p, 2 - p, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))])/(c*(-1 + p)*(1 + (b*x^3 
)/a)^p*(c + d*x^3))
 

Rubi [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.48, 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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {x^{3-3 p} \left (a+b x^3\right )^p \left (\frac {b x^3}{a}+1\right )^{-p} \left (\frac {d x^3}{c}+1\right )^{p-1} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {c \left (\frac {b x^3}{a}-\frac {d x^3}{c}\right )}{d x^3+c}\right )}{3 c^2 (1-p)}\)

Input:

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

Output:

(x^(3 - 3*p)*(a + b*x^3)^p*(1 + (d*x^3)/c)^(-1 + p)*Hypergeometric2F1[1 - 
p, -p, 2 - p, -((c*((b*x^3)/a - (d*x^3)/c))/(c + d*x^3))])/(3*c^2*(1 - p)* 
(1 + (b*x^3)/a)^p)
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - (a + b*x**3)**p*b*x**3 + 3*x**(3*p)*int(((a + b*x**3)**p*x**5)/(x**(3* 
p)*a**2*c**2*d*p + x**(3*p)*a**2*c**2*d + 2*x**(3*p)*a**2*c*d**2*p*x**3 + 
2*x**(3*p)*a**2*c*d**2*x**3 + x**(3*p)*a**2*d**3*p*x**6 + x**(3*p)*a**2*d* 
*3*x**6 - 2*x**(3*p)*a*b*c**3 + x**(3*p)*a*b*c**2*d*p*x**3 - 3*x**(3*p)*a* 
b*c**2*d*x**3 + 2*x**(3*p)*a*b*c*d**2*p*x**6 + x**(3*p)*a*b*d**3*p*x**9 + 
x**(3*p)*a*b*d**3*x**9 - 2*x**(3*p)*b**2*c**3*x**3 - 4*x**(3*p)*b**2*c**2* 
d*x**6 - 2*x**(3*p)*b**2*c*d**2*x**9),x)*a**2*b*c*d**2*p + 3*x**(3*p)*int( 
((a + b*x**3)**p*x**5)/(x**(3*p)*a**2*c**2*d*p + x**(3*p)*a**2*c**2*d + 2* 
x**(3*p)*a**2*c*d**2*p*x**3 + 2*x**(3*p)*a**2*c*d**2*x**3 + x**(3*p)*a**2* 
d**3*p*x**6 + x**(3*p)*a**2*d**3*x**6 - 2*x**(3*p)*a*b*c**3 + x**(3*p)*a*b 
*c**2*d*p*x**3 - 3*x**(3*p)*a*b*c**2*d*x**3 + 2*x**(3*p)*a*b*c*d**2*p*x**6 
 + x**(3*p)*a*b*d**3*p*x**9 + x**(3*p)*a*b*d**3*x**9 - 2*x**(3*p)*b**2*c** 
3*x**3 - 4*x**(3*p)*b**2*c**2*d*x**6 - 2*x**(3*p)*b**2*c*d**2*x**9),x)*a** 
2*b*c*d**2 + 3*x**(3*p)*int(((a + b*x**3)**p*x**5)/(x**(3*p)*a**2*c**2*d*p 
 + x**(3*p)*a**2*c**2*d + 2*x**(3*p)*a**2*c*d**2*p*x**3 + 2*x**(3*p)*a**2* 
c*d**2*x**3 + x**(3*p)*a**2*d**3*p*x**6 + x**(3*p)*a**2*d**3*x**6 - 2*x**( 
3*p)*a*b*c**3 + x**(3*p)*a*b*c**2*d*p*x**3 - 3*x**(3*p)*a*b*c**2*d*x**3 + 
2*x**(3*p)*a*b*c*d**2*p*x**6 + x**(3*p)*a*b*d**3*p*x**9 + x**(3*p)*a*b*d** 
3*x**9 - 2*x**(3*p)*b**2*c**3*x**3 - 4*x**(3*p)*b**2*c**2*d*x**6 - 2*x**(3 
*p)*b**2*c*d**2*x**9),x)*a**2*b*d**3*p*x**3 + 3*x**(3*p)*int(((a + b*x*...