\(\int \frac {(a+b x^3+c x^6)^p}{x^4 (d+e x^3)^2} \, dx\) [104]

Optimal result

Integrand size = 27, antiderivative size = 738 \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=-\frac {4^p \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 d^2 (1-2 p) x^3}-\frac {4^p e \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c \left (d+e x^3\right )},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 \left (d+e x^3\right )}\right )}{3 d^2 (1-2 p) \left (d+e x^3\right )}-\frac {4^p e \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{3 d^3 p}+\frac {4^p e \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (a+b x^3+c x^6\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c \left (d+e x^3\right )},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 \left (d+e x^3\right )}\right )}{3 d^3 p} \] Output:

-1/3*4^p*(c*x^6+b*x^3+a)^p*AppellF1(1-2*p,-p,-p,2-2*p,-1/2*(b-(-4*a*c+b^2) 
^(1/2))/c/x^3,-1/2*(b+(-4*a*c+b^2)^(1/2))/c/x^3)/d^2/(1-2*p)/x^3/(((b-(-4* 
a*c+b^2)^(1/2)+2*c*x^3)/c/x^3)^p)/(((b+(-4*a*c+b^2)^(1/2)+2*c*x^3)/c/x^3)^ 
p)-1/3*4^p*e*(c*x^6+b*x^3+a)^p*AppellF1(1-2*p,-p,-p,2-2*p,(2*d-(b+(-4*a*c+ 
b^2)^(1/2))*e/c)/(2*e*x^3+2*d),1/2*(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)/c/(e*x 
^3+d))/d^2/(1-2*p)/((e*(b-(-4*a*c+b^2)^(1/2)+2*c*x^3)/c/(e*x^3+d))^p)/((e* 
(b+(-4*a*c+b^2)^(1/2)+2*c*x^3)/c/(e*x^3+d))^p)/(e*x^3+d)-1/3*4^p*e*(c*x^6+ 
b*x^3+a)^p*AppellF1(-2*p,-p,-p,1-2*p,-1/2*(b-(-4*a*c+b^2)^(1/2))/c/x^3,-1/ 
2*(b+(-4*a*c+b^2)^(1/2))/c/x^3)/d^3/p/(((b-(-4*a*c+b^2)^(1/2)+2*c*x^3)/c/x 
^3)^p)/(((b+(-4*a*c+b^2)^(1/2)+2*c*x^3)/c/x^3)^p)+1/3*4^p*e*(c*x^6+b*x^3+a 
)^p*AppellF1(-2*p,-p,-p,1-2*p,(2*d-(b+(-4*a*c+b^2)^(1/2))*e/c)/(2*e*x^3+2* 
d),1/2*(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)/c/(e*x^3+d))/d^3/p/((e*(b-(-4*a*c+ 
b^2)^(1/2)+2*c*x^3)/c/(e*x^3+d))^p)/((e*(b+(-4*a*c+b^2)^(1/2)+2*c*x^3)/c/( 
e*x^3+d))^p)
 

Mathematica [A] (warning: unable to verify)

Time = 3.01 (sec) , antiderivative size = 626, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=\frac {\left (1+\frac {b-\sqrt {b^2-4 a c}}{2 c x^3}\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}}{2 c}+x^3\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{c}\right )^p \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \left (d p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3},\frac {-b+\sqrt {b^2-4 a c}}{2 c x^3}\right )+e (1-2 p) x^3 \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3},\frac {-b+\sqrt {b^2-4 a c}}{2 c x^3}\right )\right )}{3 d^3 p (-1+2 p) x^3}+\frac {4^p e \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (a+b x^3+c x^6\right )^p \left (d p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c \left (d+e x^3\right )},\frac {2 c d-b e+\sqrt {b^2-4 a c} e}{2 c d+2 c e x^3}\right )+(-1+2 p) \left (d+e x^3\right ) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c \left (d+e x^3\right )},\frac {2 c d-b e+\sqrt {b^2-4 a c} e}{2 c d+2 c e x^3}\right )\right )}{3 d^3 p (-1+2 p) \left (d+e x^3\right )} \] Input:

Integrate[(a + b*x^3 + c*x^6)^p/(x^4*(d + e*x^3)^2),x]
 

Output:

(((b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/c)^p*(a + b*x^3 + c*x^6)^p*(d*p*Appell 
F1[1 - 2*p, -p, -p, 2 - 2*p, -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x^3), (-b + S 
qrt[b^2 - 4*a*c])/(2*c*x^3)] + e*(1 - 2*p)*x^3*AppellF1[-2*p, -p, -p, 1 - 
2*p, -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x^3), (-b + Sqrt[b^2 - 4*a*c])/(2*c*x 
^3)]))/(3*d^3*p*(-1 + 2*p)*(1 + (b - Sqrt[b^2 - 4*a*c])/(2*c*x^3))^p*x^3*( 
(b - Sqrt[b^2 - 4*a*c])/(2*c) + x^3)^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/ 
(c*x^3))^p) + (4^p*e*(a + b*x^3 + c*x^6)^p*(d*p*AppellF1[1 - 2*p, -p, -p, 
2 - 2*p, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x^3)), (2*c*d - b 
*e + Sqrt[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x^3)] + (-1 + 2*p)*(d + e*x^3)*Ap 
pellF1[-2*p, -p, -p, 1 - 2*p, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*(d 
+ e*x^3)), (2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x^3)]))/(3*d 
^3*p*(-1 + 2*p)*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3))/(c*(d + e*x^3)))^p* 
((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3))/(c*(d + e*x^3)))^p*(d + e*x^3))
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 733, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1802, 1289, 2009}

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.

\(\displaystyle \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx\)

\(\Big \downarrow \) 1802

\(\displaystyle \frac {1}{3} \int \frac {\left (c x^6+b x^3+a\right )^p}{x^6 \left (e x^3+d\right )^2}dx^3\)

\(\Big \downarrow \) 1289

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (-\frac {e 4^p \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{d^3 p}+\frac {e 4^p \left (a+b x^3+c x^6\right )^p \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c \left (e x^3+d\right )},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 \left (e x^3+d\right )}\right )}{d^3 p}-\frac {e 4^p \left (a+b x^3+c x^6\right )^p \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x^3\right )}{c \left (d+e x^3\right )}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c \left (e x^3+d\right )},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 \left (e x^3+d\right )}\right )}{d^2 (1-2 p) \left (d+e x^3\right )}-\frac {4^p \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \left (a+b x^3+c x^6\right )^p \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{c x^3}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),-\frac {b-\sqrt {b^2-4 a c}}{2 c x^3},-\frac {b+\sqrt {b^2-4 a c}}{2 c x^3}\right )}{d^2 (1-2 p) x^3}\right )\)

Input:

Int[(a + b*x^3 + c*x^6)^p/(x^4*(d + e*x^3)^2),x]
 

Output:

(-((4^p*(a + b*x^3 + c*x^6)^p*AppellF1[1 - 2*p, -p, -p, 2*(1 - p), -1/2*(b 
 - Sqrt[b^2 - 4*a*c])/(c*x^3), -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x^3)])/(d^2 
*(1 - 2*p)*x^3*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(c*x^3))^p*((b + Sqrt[b^ 
2 - 4*a*c] + 2*c*x^3)/(c*x^3))^p)) - (4^p*e*(a + b*x^3 + c*x^6)^p*AppellF1 
[1 - 2*p, -p, -p, 2*(1 - p), (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + 
 e*x^3)), (2*d - ((b + Sqrt[b^2 - 4*a*c])*e)/c)/(2*(d + e*x^3))])/(d^2*(1 
- 2*p)*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3))/(c*(d + e*x^3)))^p*((e*(b + 
Sqrt[b^2 - 4*a*c] + 2*c*x^3))/(c*(d + e*x^3)))^p*(d + e*x^3)) - (4^p*e*(a 
+ b*x^3 + c*x^6)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -1/2*(b - Sqrt[b^2 - 4* 
a*c])/(c*x^3), -1/2*(b + Sqrt[b^2 - 4*a*c])/(c*x^3)])/(d^3*p*((b - Sqrt[b^ 
2 - 4*a*c] + 2*c*x^3)/(c*x^3))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(c*x^3 
))^p) + (4^p*e*(a + b*x^3 + c*x^6)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, (2*c* 
d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x^3)), (2*d - ((b + Sqrt[b^2 - 
4*a*c])*e)/c)/(2*(d + e*x^3))])/(d^3*p*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x^ 
3))/(c*(d + e*x^3)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3))/(c*(d + e*x^ 
3)))^p))/3
 

Defintions of rubi rules used

rule 1289

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( 
IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0]))
 

rule 1802

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + ( 
e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1 
)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, 
c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{p}}{x^{4} \left (e \,x^{3}+d \right )^{2}}d x\]

Input:

int((c*x^6+b*x^3+a)^p/x^4/(e*x^3+d)^2,x)
 

Output:

int((c*x^6+b*x^3+a)^p/x^4/(e*x^3+d)^2,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{{\left (e x^{3} + d\right )}^{2} x^{4}} \,d x } \] Input:

integrate((c*x^6+b*x^3+a)^p/x^4/(e*x^3+d)^2,x, algorithm="fricas")
 

Output:

integral((c*x^6 + b*x^3 + a)^p/(e^2*x^10 + 2*d*e*x^7 + d^2*x^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=\text {Timed out} \] Input:

integrate((c*x**6+b*x**3+a)**p/x**4/(e*x**3+d)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{{\left (e x^{3} + d\right )}^{2} x^{4}} \,d x } \] Input:

integrate((c*x^6+b*x^3+a)^p/x^4/(e*x^3+d)^2,x, algorithm="maxima")
 

Output:

integrate((c*x^6 + b*x^3 + a)^p/((e*x^3 + d)^2*x^4), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=\int { \frac {{\left (c x^{6} + b x^{3} + a\right )}^{p}}{{\left (e x^{3} + d\right )}^{2} x^{4}} \,d x } \] Input:

integrate((c*x^6+b*x^3+a)^p/x^4/(e*x^3+d)^2,x, algorithm="giac")
 

Output:

integrate((c*x^6 + b*x^3 + a)^p/((e*x^3 + d)^2*x^4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=\int \frac {{\left (c\,x^6+b\,x^3+a\right )}^p}{x^4\,{\left (e\,x^3+d\right )}^2} \,d x \] Input:

int((a + b*x^3 + c*x^6)^p/(x^4*(d + e*x^3)^2),x)
 

Output:

int((a + b*x^3 + c*x^6)^p/(x^4*(d + e*x^3)^2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^3+c x^6\right )^p}{x^4 \left (d+e x^3\right )^2} \, dx=\int \frac {\left (c \,x^{6}+b \,x^{3}+a \right )^{p}}{e^{2} x^{10}+2 d e \,x^{7}+d^{2} x^{4}}d x \] Input:

int((c*x^6+b*x^3+a)^p/x^4/(e*x^3+d)^2,x)
 

Output:

int((a + b*x**3 + c*x**6)**p/(d**2*x**4 + 2*d*e*x**7 + e**2*x**10),x)