Integrand size = 27, antiderivative size = 892 \[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\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 )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\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 )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {c e \left (8 a c (1-n)-b^2 (2-n)-b \sqrt {b^2-4 a c} (2-n)\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c e \left (8 a c (1-n)-b^2 (2-n)+b \sqrt {b^2-4 a c} (2-n)\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {c f \left (4 a c (3-2 n)-b^2 (3-n)-b \sqrt {b^2-4 a c} (3-n)\right ) x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {3+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c f \left (4 a c (3-2 n)-b^2 (3-n)+b \sqrt {b^2-4 a c} (3-n)\right ) x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {3+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n} \] Output:
d*x*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+e*x^2*(b^2-2* a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+f*x^3*(b^2-2*a*c+b*c*x^n )/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))-c*d*(4*a*c*(1-2*n)-b^2*(1-n)-b*(-4* a*c+b^2)^(1/2)*(1-n))*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2 )^(1/2)))/a/(-4*a*c+b^2)/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/n-c*d*(4*a*c*(1- 2*n)-b^2*(1-n)+b*(-4*a*c+b^2)^(1/2)*(1-n))*x*hypergeom([1, 1/n],[1+1/n],-2 *c*x^n/(b+(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/(b*(-4*a*c+b^2)^(1/2)-4*a*c+ b^2)/n-1/2*c*e*(8*a*c*(1-n)-b^2*(2-n)-b*(-4*a*c+b^2)^(1/2)*(2-n))*x^2*hype rgeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/( b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/n-1/2*c*e*(8*a*c*(1-n)-b^2*(2-n)+b*(-4*a*c +b^2)^(1/2)*(2-n))*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b+(-4*a*c+b^ 2)^(1/2)))/a/(-4*a*c+b^2)/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)/n-1/3*c*f*(4*a* c*(3-2*n)-b^2*(3-n)-b*(-4*a*c+b^2)^(1/2)*(3-n))*x^3*hypergeom([1, 3/n],[(3 +n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/(b^2-4*a*c-b*(-4*a* c+b^2)^(1/2))/n-1/3*c*f*(4*a*c*(3-2*n)-b^2*(3-n)+b*(-4*a*c+b^2)^(1/2)*(3-n ))*x^3*hypergeom([1, 3/n],[(3+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/a/(-4 *a*c+b^2)/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)/n
Leaf count is larger than twice the leaf count of optimal. \(6525\) vs. \(2(892)=1784\).
Time = 6.71 (sec) , antiderivative size = 6525, normalized size of antiderivative = 7.32 \[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \] Input:
Integrate[(d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x]
Output:
Result too large to show
Time = 1.97 (sec) , antiderivative size = 1194, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2328, 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.
\(\displaystyle \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\) |
\(\Big \downarrow \) 2328 |
\(\displaystyle \int \left (\frac {d}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {e x}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {f x^2}{\left (a+b x^n+c x^{2 n}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 b c^2 e (2-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+2)}+\frac {2 b c^2 e (2-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+2)}-\frac {2 b c^2 f (3-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+3}{n},2+\frac {3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+3)}+\frac {2 b c^2 f (3-n) \operatorname {Hypergeometric2F1}\left (1,\frac {n+3}{n},2+\frac {3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+3)}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {f \left (b c x^n+b^2-2 a c\right ) x^3}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {e \left (b c x^n+b^2-2 a c\right ) x^2}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c d \left (-\left ((1-n) b^2\right )-\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\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 ) x}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c d \left (-\left ((1-n) b^2\right )+\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\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 ) x}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {d \left (b c x^n+b^2-2 a c\right ) x}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}\) |
Input:
Int[(d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x]
Output:
(d*x*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (e*x^2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n ))) + (f*x^3*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^ (2*n))) - (c*d*(4*a*c*(1 - 2*n) - b^2*(1 - n) - b*Sqrt[b^2 - 4*a*c]*(1 - n ))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4 *a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c*d*(4 *a*c*(1 - 2*n) - b^2*(1 - n) + b*Sqrt[b^2 - 4*a*c]*(1 - n))*x*Hypergeometr ic2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n) - (c*e*(4*a*c*(1 - n) - b^ 2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b - Sqrt[b ^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - ( c*e*(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sq rt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*x^3*Hypergeom etric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b + S qrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c]) *n) - (2*b*c^2*e*(2 - n)*x^(2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b...
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && ILtQ[p, -1]
\[\int \frac {f \,x^{2}+e x +d}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]
Input:
int((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)
Output:
int((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)
\[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:
integrate((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")
Output:
integral((f*x^2 + e*x + d)/(c^2*x^(4*n) + b^2*x^(2*n) + 2*a*b*x^n + a^2 + 2*(b*c*x^n + a*c)*x^(2*n)), x)
Timed out. \[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \] Input:
integrate((f*x**2+e*x+d)/(a+b*x**n+c*x**(2*n))**2,x)
Output:
Timed out
\[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:
integrate((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")
Output:
((b^2*f - 2*a*c*f)*x^3 + (b^2*e - 2*a*c*e)*x^2 + (b*c*f*x^3 + b*c*e*x^2 + b*c*d*x)*x^n + (b^2*d - 2*a*c*d)*x)/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n) - integrate((2*a*c*d*( 2*n - 1) - b^2*d*(n - 1) + (2*a*c*f*(2*n - 3) - b^2*f*(n - 3))*x^2 - (b*c* f*(n - 3)*x^2 + b*c*e*(n - 2)*x + b*c*d*(n - 1))*x^n + (4*a*c*e*(n - 1) - b^2*e*(n - 2))*x)/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2* n) + (a*b^3*n - 4*a^2*b*c*n)*x^n), x)
\[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {f x^{2} + e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:
integrate((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")
Output:
integrate((f*x^2 + e*x + d)/(c*x^(2*n) + b*x^n + a)^2, x)
Timed out. \[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {f\,x^2+e\,x+d}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \] Input:
int((d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x)
Output:
int((d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2, x)
\[ \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\left (\int \frac {x^{2}}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) f +\left (\int \frac {x}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) e +\left (\int \frac {1}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) d \] Input:
int((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)
Output:
int(x**2/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c + x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)*f + int(x/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**( 2*n)*a*c + x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)*e + int(1/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c + x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)* d