\(\int \frac {1}{(d+e x^n)^3 (a+b x^n+c x^{2 n})} \, dx\) [68]

Optimal result

Integrand size = 26, antiderivative size = 653 \[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {e^2 x}{2 d \left (c d^2-b d e+a e^2\right ) n \left (d+e x^n\right )^2}+\frac {e^2 \left (e (b d (1-4 n)-a e (1-2 n))-c d^2 (1-6 n)\right ) x}{2 d^2 \left (c d^2-b d e+a e^2\right )^2 n^2 \left (d+e x^n\right )}-\frac {c \left (2 c^3 d^3-b^2 \left (b+\sqrt {b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d+\sqrt {b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d+a \sqrt {b^2-4 a c} e+3 b \left (\sqrt {b^2-4 a c} d+a e\right )\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}-\frac {c \left (2 c^3 d^3-b^2 \left (b-\sqrt {b^2-4 a c}\right ) e^3-3 c^2 d e \left (b d-\sqrt {b^2-4 a c} d+2 a e\right )+c e^2 \left (3 b^2 d-3 b \sqrt {b^2-4 a c} d+3 a b e-a \sqrt {b^2-4 a c} e\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^3}+\frac {e^2 \left (c^2 d^4 \left (1-7 n+12 n^2\right )+e^2 \left (a^2 e^2 \left (1-3 n+2 n^2\right )-2 a b d e \left (1-4 n+3 n^2\right )+b^2 d^2 \left (1-5 n+6 n^2\right )\right )+2 c d^2 e \left (a e \left (1-5 n+3 n^2\right )-b d \left (1-6 n+8 n^2\right )\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{2 d^3 \left (c d^2-b d e+a e^2\right )^3 n^2} \] Output:

1/2*e^2*x/d/(a*e^2-b*d*e+c*d^2)/n/(d+e*x^n)^2+1/2*e^2*(e*(b*d*(1-4*n)-a*e* 
(1-2*n))-c*d^2*(1-6*n))*x/d^2/(a*e^2-b*d*e+c*d^2)^2/n^2/(d+e*x^n)-c*(2*c^3 
*d^3-b^2*(b+(-4*a*c+b^2)^(1/2))*e^3-3*c^2*d*e*(b*d+(-4*a*c+b^2)^(1/2)*d+2* 
a*e)+c*e^2*(3*b^2*d+a*(-4*a*c+b^2)^(1/2)*e+3*b*((-4*a*c+b^2)^(1/2)*d+a*e)) 
)*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c 
-b*(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)^3-c*(2*c^3*d^3-b^2*(b-(-4*a*c+b 
^2)^(1/2))*e^3-3*c^2*d*e*(b*d-(-4*a*c+b^2)^(1/2)*d+2*a*e)+c*e^2*(3*b^2*d-3 
*b*(-4*a*c+b^2)^(1/2)*d+3*a*b*e-a*(-4*a*c+b^2)^(1/2)*e))*x*hypergeom([1, 1 
/n],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b 
^2)/(a*e^2-b*d*e+c*d^2)^3+1/2*e^2*(c^2*d^4*(12*n^2-7*n+1)+e^2*(a^2*e^2*(2* 
n^2-3*n+1)-2*a*b*d*e*(3*n^2-4*n+1)+b^2*d^2*(6*n^2-5*n+1))+2*c*d^2*e*(a*e*( 
3*n^2-5*n+1)-b*d*(8*n^2-6*n+1)))*x*hypergeom([1, 1/n],[1+1/n],-e*x^n/d)/d^ 
3/(a*e^2-b*d*e+c*d^2)^3/n^2
 

Mathematica [A] (verified)

Time = 2.62 (sec) , antiderivative size = 509, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {x \left (\frac {c \left (-2 c^3 d^3+b^2 \left (b+\sqrt {b^2-4 a c}\right ) e^3+3 c^2 d e \left (b d+\sqrt {b^2-4 a c} d+2 a e\right )-c e^2 \left (3 b^2 d+a \sqrt {b^2-4 a c} e+3 b \left (\sqrt {b^2-4 a c} d+a e\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {c \left (2 c^3 d^3+b^2 \left (-b+\sqrt {b^2-4 a c}\right ) e^3+3 c^2 d e \left (-b d+\sqrt {b^2-4 a c} d-2 a e\right )+c e^2 \left (3 b^2 d-3 b \sqrt {b^2-4 a c} d+3 a b e-a \sqrt {b^2-4 a c} e\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}+\frac {e^2 \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d}+\frac {e^2 (2 c d-b e) \left (c d^2+e (-b d+a e)\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2}+\frac {e^2 \left (c d^2+e (-b d+a e)\right )^2 \operatorname {Hypergeometric2F1}\left (3,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^3}\right )}{\left (c d^2+e (-b d+a e)\right )^3} \] Input:

Integrate[1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x]
 

Output:

(x*((c*(-2*c^3*d^3 + b^2*(b + Sqrt[b^2 - 4*a*c])*e^3 + 3*c^2*d*e*(b*d + Sq 
rt[b^2 - 4*a*c]*d + 2*a*e) - c*e^2*(3*b^2*d + a*Sqrt[b^2 - 4*a*c]*e + 3*b* 
(Sqrt[b^2 - 4*a*c]*d + a*e)))*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (2* 
c*x^n)/(-b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) - (c 
*(2*c^3*d^3 + b^2*(-b + Sqrt[b^2 - 4*a*c])*e^3 + 3*c^2*d*e*(-(b*d) + Sqrt[ 
b^2 - 4*a*c]*d - 2*a*e) + c*e^2*(3*b^2*d - 3*b*Sqrt[b^2 - 4*a*c]*d + 3*a*b 
*e - a*Sqrt[b^2 - 4*a*c]*e))*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2* 
c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]) + (e^ 
2*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*Hypergeometric2F1[1, n^(-1), 1 
 + n^(-1), -((e*x^n)/d)])/d + (e^2*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e) 
)*Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/d^2 + (e^2*(c*d^ 
2 + e*(-(b*d) + a*e))^2*Hypergeometric2F1[3, n^(-1), 1 + n^(-1), -((e*x^n) 
/d)])/d^3))/(c*d^2 + e*(-(b*d) + a*e))^3
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 552, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1754, 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.

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

\(\Big \downarrow \) 1754

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^2 x \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^3}-\frac {c x \left (-3 c^2 d e \left (d \sqrt {b^2-4 a c}+2 a e+b d\right )+c e^2 \left (3 b \left (d \sqrt {b^2-4 a c}+a e\right )+a e \sqrt {b^2-4 a c}+3 b^2 d\right )-b^2 e^3 \left (\sqrt {b^2-4 a c}+b\right )+2 c^3 d^3\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}-\frac {c x \left (-3 c^2 d e \left (-d \sqrt {b^2-4 a c}+2 a e+b d\right )+c e^2 \left (-3 b d \sqrt {b^2-4 a c}-a e \sqrt {b^2-4 a c}+3 a b e+3 b^2 d\right )-b^2 e^3 \left (b-\sqrt {b^2-4 a c}\right )+2 c^3 d^3\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^3}+\frac {e^2 x (2 c d-b e) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2 \left (a e^2-b d e+c d^2\right )^2}+\frac {e^2 x \operatorname {Hypergeometric2F1}\left (3,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^3 \left (a e^2-b d e+c d^2\right )}\)

Input:

Int[1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x]
 

Output:

-((c*(2*c^3*d^3 - b^2*(b + Sqrt[b^2 - 4*a*c])*e^3 - 3*c^2*d*e*(b*d + Sqrt[ 
b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d + a*Sqrt[b^2 - 4*a*c]*e + 3*b*(Sq 
rt[b^2 - 4*a*c]*d + a*e)))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2* 
c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(c*d 
^2 - b*d*e + a*e^2)^3)) - (c*(2*c^3*d^3 - b^2*(b - Sqrt[b^2 - 4*a*c])*e^3 
- 3*c^2*d*e*(b*d - Sqrt[b^2 - 4*a*c]*d + 2*a*e) + c*e^2*(3*b^2*d - 3*b*Sqr 
t[b^2 - 4*a*c]*d + 3*a*b*e - a*Sqrt[b^2 - 4*a*c]*e))*x*Hypergeometric2F1[1 
, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c + 
 b*Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^3) + (e^2*(3*c^2*d^2 + b^2*e 
^2 - c*e*(3*b*d + a*e))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^ 
n)/d)])/(d*(c*d^2 - b*d*e + a*e^2)^3) + (e^2*(2*c*d - b*e)*x*Hypergeometri 
c2F1[2, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d^2*(c*d^2 - b*d*e + a*e^2)^2) 
 + (e^2*x*Hypergeometric2F1[3, n^(-1), 1 + n^(-1), -((e*x^n)/d)])/(d^3*(c* 
d^2 - b*d*e + a*e^2))
 

Defintions of rubi rules used

rule 1754

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ 
)), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q/(a + b*x^n + c*x^(2*n)), 
 x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 
 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]
 

rule 2009

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

Maple [F]

\[\int \frac {1}{\left (d +e \,x^{n}\right )^{3} \left (a +b \,x^{n}+c \,x^{2 n}\right )}d x\]

Input:

int(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x)
 

Output:

int(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x)
 

Fricas [F]

\[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} {\left (e x^{n} + d\right )}^{3}} \,d x } \] Input:

integrate(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")
 

Output:

integral(1/(b*e^3*x^(4*n) + a*d^3 + (3*b*d*e^2 + a*e^3)*x^(3*n) + (c*e^3*x 
^(3*n) + 3*c*d*e^2*x^(2*n) + 3*c*d^2*e*x^n + c*d^3)*x^(2*n) + 3*(b*d^2*e + 
 a*d*e^2)*x^(2*n) + (b*d^3 + 3*a*d^2*e)*x^n), x)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate(1/(d+e*x**n)**3/(a+b*x**n+c*x**(2*n)),x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} {\left (e x^{n} + d\right )}^{3}} \,d x } \] Input:

integrate(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")
 

Output:

((12*n^2 - 7*n + 1)*c^2*d^4*e^2 - 2*(8*n^2 - 6*n + 1)*b*c*d^3*e^3 + (6*n^2 
 - 5*n + 1)*b^2*d^2*e^4 + (2*n^2 - 3*n + 1)*a^2*e^6 + 2*((3*n^2 - 5*n + 1) 
*c*d^2*e^4 - (3*n^2 - 4*n + 1)*b*d*e^5)*a)*integrate(1/2/(c^3*d^9*n^2 - 3* 
b*c^2*d^8*e*n^2 + 3*b^2*c*d^7*e^2*n^2 - b^3*d^6*e^3*n^2 + a^3*d^3*e^6*n^2 
+ 3*(c*d^5*e^4*n^2 - b*d^4*e^5*n^2)*a^2 + 3*(c^2*d^7*e^2*n^2 - 2*b*c*d^6*e 
^3*n^2 + b^2*d^5*e^4*n^2)*a + (c^3*d^8*e*n^2 - 3*b*c^2*d^7*e^2*n^2 + 3*b^2 
*c*d^6*e^3*n^2 - b^3*d^5*e^4*n^2 + a^3*d^2*e^7*n^2 + 3*(c*d^4*e^5*n^2 - b* 
d^3*e^6*n^2)*a^2 + 3*(c^2*d^6*e^3*n^2 - 2*b*c*d^5*e^4*n^2 + b^2*d^4*e^5*n^ 
2)*a)*x^n), x) + 1/2*((c*d^2*e^3*(6*n - 1) - b*d*e^4*(4*n - 1) + a*e^5*(2* 
n - 1))*x*x^n + (c*d^3*e^2*(7*n - 1) - b*d^2*e^3*(5*n - 1) + a*d*e^4*(3*n 
- 1))*x)/(c^2*d^8*n^2 - 2*b*c*d^7*e*n^2 + b^2*d^6*e^2*n^2 + a^2*d^4*e^4*n^ 
2 + 2*(c*d^6*e^2*n^2 - b*d^5*e^3*n^2)*a + (c^2*d^6*e^2*n^2 - 2*b*c*d^5*e^3 
*n^2 + b^2*d^4*e^4*n^2 + a^2*d^2*e^6*n^2 + 2*(c*d^4*e^4*n^2 - b*d^3*e^5*n^ 
2)*a)*x^(2*n) + 2*(c^2*d^7*e*n^2 - 2*b*c*d^6*e^2*n^2 + b^2*d^5*e^3*n^2 + a 
^2*d^3*e^5*n^2 + 2*(c*d^5*e^3*n^2 - b*d^4*e^4*n^2)*a)*x^n) + integrate((c^ 
3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3 - (3*c^2*d*e^2 - 2*b*c*e^3 
)*a - (3*c^3*d^2*e - 3*b*c^2*d*e^2 + b^2*c*e^3 - a*c^2*e^3)*x^n)/(a^4*e^6 
+ 3*(c*d^2*e^4 - b*d*e^5)*a^3 + 3*(c^2*d^4*e^2 - 2*b*c*d^3*e^3 + b^2*d^2*e 
^4)*a^2 + (c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*a + (c 
^4*d^6 - 3*b*c^3*d^5*e + 3*b^2*c^2*d^4*e^2 - b^3*c*d^3*e^3 + a^3*c*e^6 ...
 

Giac [F]

\[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )} {\left (e x^{n} + d\right )}^{3}} \,d x } \] Input:

integrate(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")
 

Output:

integrate(1/((c*x^(2*n) + b*x^n + a)*(e*x^n + d)^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {1}{{\left (d+e\,x^n\right )}^3\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \] Input:

int(1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))),x)
 

Output:

int(1/((d + e*x^n)^3*(a + b*x^n + c*x^(2*n))), x)
 

Reduce [F]

\[ \int \frac {1}{\left (d+e x^n\right )^3 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {1}{x^{5 n} c \,e^{3}+x^{4 n} b \,e^{3}+3 x^{4 n} c d \,e^{2}+x^{3 n} a \,e^{3}+3 x^{3 n} b d \,e^{2}+3 x^{3 n} c \,d^{2} e +3 x^{2 n} a d \,e^{2}+3 x^{2 n} b \,d^{2} e +x^{2 n} c \,d^{3}+3 x^{n} a \,d^{2} e +x^{n} b \,d^{3}+a \,d^{3}}d x \] Input:

int(1/(d+e*x^n)^3/(a+b*x^n+c*x^(2*n)),x)
 

Output:

int(1/(x**(5*n)*c*e**3 + x**(4*n)*b*e**3 + 3*x**(4*n)*c*d*e**2 + x**(3*n)* 
a*e**3 + 3*x**(3*n)*b*d*e**2 + 3*x**(3*n)*c*d**2*e + 3*x**(2*n)*a*d*e**2 + 
 3*x**(2*n)*b*d**2*e + x**(2*n)*c*d**3 + 3*x**n*a*d**2*e + x**n*b*d**3 + a 
*d**3),x)