Integrand size = 31, antiderivative size = 418 \[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=-\frac {d^2 x}{e \left (c d^2-b d e+a e^2\right ) n \left (d+e x^n\right )}+\frac {\left (b^2 d^2-2 a b d e-a \left (c d^2-a e^2\right )-\frac {b^3 d^2-2 a b^2 d e+4 a^2 c d e-a b \left (3 c d^2-a e^2\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}+\frac {\left (b^2 d^2-2 a b d e-a \left (c d^2-a e^2\right )+\frac {b^3 d^2-2 a b^2 d e+4 a^2 c d e-a b \left (3 c d^2-a e^2\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}+\frac {d \left (c d^2-e (b d (1+n)-a e (1+2 n))\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{e \left (c d^2-b d e+a e^2\right )^2 n} \] Output:
-d^2*x/e/(a*e^2-b*d*e+c*d^2)/n/(d+e*x^n)+(b^2*d^2-2*a*b*d*e-a*(-a*e^2+c*d^ 2)-(b^3*d^2-2*a*b^2*d*e+4*a^2*c*d*e-a*b*(-a*e^2+3*c*d^2))/(-4*a*c+b^2)^(1/ 2))*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))/(a*e^2-b*d*e+c*d^2)^2+(b^2*d^2-2*a*b*d*e-a*(-a*e^2+c*d^2)+( b^3*d^2-2*a*b^2*d*e+4*a^2*c*d*e-a*b*(-a*e^2+3*c*d^2))/(-4*a*c+b^2)^(1/2))* 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))/(a*e^2-b*d*e+c*d^2)^2+d*(c*d^2-e*(b*d*(1+n)-a*e*(1+2*n)))*x*hyp ergeom([1, 1/n],[1+1/n],-e*x^n/d)/e/(a*e^2-b*d*e+c*d^2)^2/n
Time = 2.89 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.79 \[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\frac {x^{1+3 n} \left (-\frac {c \left (2 c d e-b e^2-\frac {2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,3+\frac {1}{n},4+\frac {1}{n},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}-\frac {c \left (2 c d e-b e^2+\frac {2 c^2 d^2+b^2 e^2-2 c e (b d+a e)}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,3+\frac {1}{n},4+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}}+\frac {e^2 (2 c d-b e) \operatorname {Hypergeometric2F1}\left (1,3+\frac {1}{n},4+\frac {1}{n},-\frac {e x^n}{d}\right )}{d}+\frac {e^2 \left (c d^2+e (-b d+a e)\right ) \operatorname {Hypergeometric2F1}\left (2,3+\frac {1}{n},4+\frac {1}{n},-\frac {e x^n}{d}\right )}{d^2}\right )}{\left (c d^2+e (-b d+a e)\right )^2 (1+3 n)} \] Input:
Integrate[x^(3*n)/((d + e*x^n)^2*(a + b*x^n + c*x^(2*n))),x]
Output:
(x^(1 + 3*n)*(-((c*(2*c*d*e - b*e^2 - (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 3 + n^(-1), 4 + n^(-1), (2*c *x^n)/(-b + Sqrt[b^2 - 4*a*c])])/(b - Sqrt[b^2 - 4*a*c])) - (c*(2*c*d*e - b*e^2 + (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))/Sqrt[b^2 - 4*a*c])*Hyper geometric2F1[1, 3 + n^(-1), 4 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]) ])/(b + Sqrt[b^2 - 4*a*c]) + (e^2*(2*c*d - b*e)*Hypergeometric2F1[1, 3 + n ^(-1), 4 + n^(-1), -((e*x^n)/d)])/d + (e^2*(c*d^2 + e*(-(b*d) + a*e))*Hype rgeometric2F1[2, 3 + n^(-1), 4 + n^(-1), -((e*x^n)/d)])/d^2))/((c*d^2 + e* (-(b*d) + a*e))^2*(1 + 3*n))
Time = 1.33 (sec) , antiderivative size = 585, normalized size of antiderivative = 1.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1880, 1006, 1067, 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.
\(\displaystyle \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx\) |
\(\Big \downarrow \) 1880 |
\(\displaystyle \frac {2 c \int \frac {x^{3 n}}{\left (2 c x^n+b-\sqrt {b^2-4 a c}\right ) \left (e x^n+d\right )^2}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {x^{3 n}}{\left (2 c x^n+b+\sqrt {b^2-4 a c}\right ) \left (e x^n+d\right )^2}dx}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 1006 |
\(\displaystyle \frac {2 c \left (\frac {\int \frac {x^{3 n} \left (2 c e (2 n+1) x^n+2 c d n-\sqrt {b^2-4 a c} e (2 n+1)+b (2 n e+e)\right )}{\left (2 c x^n+b-\sqrt {b^2-4 a c}\right ) \left (e x^n+d\right )}dx}{d n \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )}-\frac {e x^{3 n+1}}{d n \left (d+e x^n\right ) \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {\int \frac {x^{3 n} \left (2 c e (2 n+1) x^n+2 c d n+\sqrt {b^2-4 a c} e (2 n+1)+b (2 n e+e)\right )}{\left (2 c x^n+b+\sqrt {b^2-4 a c}\right ) \left (e x^n+d\right )}dx}{d n \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {e x^{3 n+1}}{d n \left (d+e x^n\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 1067 |
\(\displaystyle \frac {2 c \left (\frac {\int \left (\frac {4 c^2 d n x^{3 n}}{\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (2 c x^n+b-\sqrt {b^2-4 a c}\right )}+\frac {e \left (2 c d (n+1)-\left (b-\sqrt {b^2-4 a c}\right ) e (2 n+1)\right ) x^{3 n}}{\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (e x^n+d\right )}\right )dx}{d n \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )}-\frac {e x^{3 n+1}}{d n \left (d+e x^n\right ) \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {\int \left (\frac {4 c^2 d n x^{3 n}}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (2 c x^n+b+\sqrt {b^2-4 a c}\right )}+\frac {e \left (2 c d (n+1)-\left (b+\sqrt {b^2-4 a c}\right ) e (2 n+1)\right ) x^{3 n}}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (e x^n+d\right )}\right )dx}{d n \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {e x^{3 n+1}}{d n \left (d+e x^n\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )}{\sqrt {b^2-4 a c}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 c \left (\frac {\frac {4 c^2 d n x^{3 n+1} \operatorname {Hypergeometric2F1}\left (1,3+\frac {1}{n},4+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{(3 n+1) \left (b-\sqrt {b^2-4 a c}\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {e x^{3 n+1} \operatorname {Hypergeometric2F1}\left (1,3+\frac {1}{n},4+\frac {1}{n},-\frac {e x^n}{d}\right ) \left (2 c d (n+1)-e (2 n+1) \left (b-\sqrt {b^2-4 a c}\right )\right )}{d (3 n+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}}{d n \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )}-\frac {e x^{3 n+1}}{d n \left (d+e x^n\right ) \left (e \sqrt {b^2-4 a c}-b e+2 c d\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {\frac {4 c^2 d n x^{3 n+1} \operatorname {Hypergeometric2F1}\left (1,3+\frac {1}{n},4+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{(3 n+1) \left (\sqrt {b^2-4 a c}+b\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}+\frac {e x^{3 n+1} \operatorname {Hypergeometric2F1}\left (1,3+\frac {1}{n},4+\frac {1}{n},-\frac {e x^n}{d}\right ) \left (2 c d (n+1)-e (2 n+1) \left (\sqrt {b^2-4 a c}+b\right )\right )}{d (3 n+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}}{d n \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {e x^{3 n+1}}{d n \left (d+e x^n\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )}{\sqrt {b^2-4 a c}}\) |
Input:
Int[x^(3*n)/((d + e*x^n)^2*(a + b*x^n + c*x^(2*n))),x]
Output:
(2*c*(-((e*x^(1 + 3*n))/(d*(2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)*n*(d + e*x^ n))) + ((4*c^2*d*n*x^(1 + 3*n)*Hypergeometric2F1[1, 3 + n^(-1), 4 + n^(-1) , (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/((b - Sqrt[b^2 - 4*a*c])*(2*c*d - ( b - Sqrt[b^2 - 4*a*c])*e)*(1 + 3*n)) + (e*(2*c*d*(1 + n) - (b - Sqrt[b^2 - 4*a*c])*e*(1 + 2*n))*x^(1 + 3*n)*Hypergeometric2F1[1, 3 + n^(-1), 4 + n^( -1), -((e*x^n)/d)])/(d*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + 3*n)))/(d* (2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)*n)))/Sqrt[b^2 - 4*a*c] - (2*c*(-((e*x^ (1 + 3*n))/(d*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*n*(d + e*x^n))) + ((4*c^ 2*d*n*x^(1 + 3*n)*Hypergeometric2F1[1, 3 + n^(-1), 4 + n^(-1), (-2*c*x^n)/ (b + Sqrt[b^2 - 4*a*c])])/((b + Sqrt[b^2 - 4*a*c])*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + 3*n)) + (e*(2*c*d*(1 + n) - (b + Sqrt[b^2 - 4*a*c])*e*(1 + 2*n))*x^(1 + 3*n)*Hypergeometric2F1[1, 3 + n^(-1), 4 + n^(-1), -((e*x^n )/d)])/(d*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + 3*n)))/(d*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*n)))/Sqrt[b^2 - 4*a*c]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x ^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*( b*c - a*d)*(p + 1) + d*b*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{ a, b, c, d, e, m, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomi alQ[a, b, c, d, e, m, n, p, q, x]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x]
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^( n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[ 2*(c/r) Int[(f*x)^m*((d + e*x^n)^q/(b - r + 2*c*x^n)), x], x] - Simp[2*(c /r) Int[(f*x)^m*((d + e*x^n)^q/(b + r + 2*c*x^n)), x], x]] /; FreeQ[{a, b , c, d, e, f, m, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && !Rati onalQ[n]
\[\int \frac {x^{3 n}}{\left (d +e \,x^{n}\right )^{2} \left (a +b \,x^{n}+c \,x^{2 n}\right )}d x\]
Input:
int(x^(3*n)/(d+e*x^n)^2/(a+b*x^n+c*x^(2*n)),x)
Output:
int(x^(3*n)/(d+e*x^n)^2/(a+b*x^n+c*x^(2*n)),x)
\[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {x^{3 \, n}}{{\left (c x^{2 \, n} + b x^{n} + a\right )} {\left (e x^{n} + d\right )}^{2}} \,d x } \] Input:
integrate(x^(3*n)/(d+e*x^n)^2/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")
Output:
integral(x^(3*n)/(b*e^2*x^(3*n) + a*d^2 + (c*e^2*x^(2*n) + 2*c*d*e*x^n + c *d^2)*x^(2*n) + (2*b*d*e + a*e^2)*x^(2*n) + (b*d^2 + 2*a*d*e)*x^n), x)
Exception generated. \[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:
integrate(x**(3*n)/(d+e*x**n)**2/(a+b*x**n+c*x**(2*n)),x)
Output:
Exception raised: HeuristicGCDFailed >> no luck
\[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {x^{3 \, n}}{{\left (c x^{2 \, n} + b x^{n} + a\right )} {\left (e x^{n} + d\right )}^{2}} \,d x } \] Input:
integrate(x^(3*n)/(d+e*x^n)^2/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")
Output:
-d^2*x/(c*d^3*e*n - b*d^2*e^2*n + a*d*e^3*n + (c*d^2*e^2*n - b*d*e^3*n + a *e^4*n)*x^n) + (a*d^2*e^2*(2*n + 1) - b*d^3*e*(n + 1) + c*d^4)*integrate(1 /(c^2*d^5*e*n - 2*b*c*d^4*e^2*n + b^2*d^3*e^3*n + a^2*d*e^5*n + 2*(c*d^3*e ^3*n - b*d^2*e^4*n)*a + (c^2*d^4*e^2*n - 2*b*c*d^3*e^3*n + b^2*d^2*e^4*n + a^2*e^6*n + 2*(c*d^2*e^4*n - b*d*e^5*n)*a)*x^n), x) + integrate((a*b*d^2 - 2*a^2*d*e + (b^2*d^2 + a^2*e^2 - (c*d^2 + 2*b*d*e)*a)*x^n)/(a^3*e^4 + 2* (c*d^2*e^2 - b*d*e^3)*a^2 + (c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*a + (c^3 *d^4 - 2*b*c^2*d^3*e + b^2*c*d^2*e^2 + a^2*c*e^4 + 2*(c^2*d^2*e^2 - b*c*d* e^3)*a)*x^(2*n) + (b*c^2*d^4 - 2*b^2*c*d^3*e + b^3*d^2*e^2 + a^2*b*e^4 + 2 *(b*c*d^2*e^2 - b^2*d*e^3)*a)*x^n), x)
\[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int { \frac {x^{3 \, n}}{{\left (c x^{2 \, n} + b x^{n} + a\right )} {\left (e x^{n} + d\right )}^{2}} \,d x } \] Input:
integrate(x^(3*n)/(d+e*x^n)^2/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")
Output:
integrate(x^(3*n)/((c*x^(2*n) + b*x^n + a)*(e*x^n + d)^2), x)
Timed out. \[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {x^{3\,n}}{{\left (d+e\,x^n\right )}^2\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \] Input:
int(x^(3*n)/((d + e*x^n)^2*(a + b*x^n + c*x^(2*n))),x)
Output:
int(x^(3*n)/((d + e*x^n)^2*(a + b*x^n + c*x^(2*n))), x)
\[ \int \frac {x^{3 n}}{\left (d+e x^n\right )^2 \left (a+b x^n+c x^{2 n}\right )} \, dx=\int \frac {x^{3 n}}{x^{4 n} c \,e^{2}+x^{3 n} b \,e^{2}+2 x^{3 n} c d e +x^{2 n} a \,e^{2}+2 x^{2 n} b d e +x^{2 n} c \,d^{2}+2 x^{n} a d e +x^{n} b \,d^{2}+a \,d^{2}}d x \] Input:
int(x^(3*n)/(d+e*x^n)^2/(a+b*x^n+c*x^(2*n)),x)
Output:
int(x**(3*n)/(x**(4*n)*c*e**2 + x**(3*n)*b*e**2 + 2*x**(3*n)*c*d*e + x**(2 *n)*a*e**2 + 2*x**(2*n)*b*d*e + x**(2*n)*c*d**2 + 2*x**n*a*d*e + x**n*b*d* *2 + a*d**2),x)