\(\int \frac {(d+e x^n)^2}{(a+b x^n+c x^{2 n})^2} \, dx\) [70]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 498 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=-\frac {2 e (b d-a e) x}{a \left (b^2-4 a c\right ) n}-\frac {b e^2 x^{1+n}}{a \left (b^2-4 a c\right ) n}+\frac {x \left (b^2-2 a c+b c x^n\right ) \left (d+e x^n\right )^2}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {\left (4 a c \left (e \left (a e-\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )+b^2 \left (c d^2-a e^2\right ) (1-n)+b \left (a \sqrt {b^2-4 a c} e^2 (1-n)+c d \left (\sqrt {b^2-4 a c} d (1-n)-4 a e n\right )\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}+\frac {\left (4 a c \left (e \left (a e+\sqrt {b^2-4 a c} d (1-n)\right )-c d^2 (1-2 n)\right )+b^2 \left (c d^2-a e^2\right ) (1-n)-b \left (a \sqrt {b^2-4 a c} e^2 (1-n)+c d \left (\sqrt {b^2-4 a c} d (1-n)+4 a e n\right )\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2980\) vs. \(2(498)=996\).

Time = 6.09 (sec) , antiderivative size = 2980, normalized size of antiderivative = 5.98 \[ \int \frac {\left (d+e x^n\right )^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

-((x*(-(a*Sqrt[b^2 - 4*a*c]*(b^2*d^2 + 2*a^2*e^2 + b*c*d^2*x^n + a*b*e*(-2 
*d + e*x^n) - 2*a*c*d*(d + 2*e*x^n))) + (a*b*c*d^2*(a + x^n*(b + c*x^n))*( 
Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b 
 - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/((c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n 
))^n^(-1) - Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b + Sqrt[b^2 
- 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/((c*x^n)/(b + Sqrt[b^2 - 4*a* 
c] + 2*c*x^n))^n^(-1)))/2^n^(-1) - 2^(2 - n^(-1))*a^2*c*d*e*(a + x^n*(b + 
c*x^n))*(Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b - Sqrt[b^2 - 4 
*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/((c*x^n)/(b - Sqrt[b^2 - 4*a*c] 
+ 2*c*x^n))^n^(-1) - Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b + 
Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/((c*x^n)/(b + Sqrt[b 
^2 - 4*a*c] + 2*c*x^n))^n^(-1)) + (a^2*b*e^2*(a + x^n*(b + c*x^n))*(Hyperg 
eometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqr 
t[b^2 - 4*a*c] + 2*c*x^n)]/((c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^n^( 
-1) - Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b + Sqrt[b^2 - 4*a* 
c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2 
*c*x^n))^n^(-1)))/2^n^(-1) - (a*b*c*d^2*n*(a + x^n*(b + c*x^n))*(Hypergeom 
etric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b 
^2 - 4*a*c] + 2*c*x^n)]/((c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^n^(-1) 
 - Hypergeometric2F1[-n^(-1), -n^(-1), (-1 + n)/n, (b + Sqrt[b^2 - 4*a*...
 

Rubi [A] (verified)

Time = 1.62 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.09, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1766, 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.

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

\(\Big \downarrow \) 1766

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(x*(b^2*d^2 - 2*a*b*d*e - 2*a*(c*d^2 - a*e^2) + (b*c*d^2 - 4*a*c*d*e + a*b 
*e^2)*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) - (2*e^2*x*Hyperge 
ometric2F1[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]) - (((b*c*d^2 - 4*a*c*d*e + a*b*e^2)*(1 - 
n) - (b^2*(a*e^2*(1 - 3*n) - c*d^2*(1 - n)) + 4*a*c*(c*d^2 - a*e^2)*(1 - 2 
*n) + 4*a*b*c*d*e*n)/Sqrt[b^2 - 4*a*c])*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 - Sqrt[b 
^2 - 4*a*c])*n) - (2*e^2*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]) - (((b* 
c*d^2 - 4*a*c*d*e + a*b*e^2)*(1 - n) + (b^2*(a*e^2*(1 - 3*n) - c*d^2*(1 - 
n)) + 4*a*c*(c*d^2 - a*e^2)*(1 - 2*n) + 4*a*b*c*d*e*n)/Sqrt[b^2 - 4*a*c])* 
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 + Sqrt[b^2 - 4*a*c])*n)
 

Defintions of rubi rules used

rule 1766
Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ 
))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q*(a + b*x^n + c*x^(2 
*n))^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ 
[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ((IntegersQ[p, q] && 
!IntegerQ[n]) || IGtQ[p, 0] || (IGtQ[q, 0] &&  !IntegerQ[n]))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((e^2*x^(2*n) + 2*d*e*x^n + d^2)/(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)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n - x**(4*n)*c**2 + 2*x**(3*n)*b*c*n 
 - 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n - 2*x**(2*n)*a*c + x**(2*n)*b**2*n - 
x**(2*n)*b**2 + 2*x**n*a*b*n - 2*x**n*a*b + a**2*n - a**2),x)*b*c*e**2*n** 
2 - 2*x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n - x**(4*n)*c**2 + 2*x**(3*n)* 
b*c*n - 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n - 2*x**(2*n)*a*c + x**(2*n)*b**2 
*n - x**(2*n)*b**2 + 2*x**n*a*b*n - 2*x**n*a*b + a**2*n - a**2),x)*b*c*e** 
2*n + x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n - x**(4*n)*c**2 + 2*x**(3*n)* 
b*c*n - 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n - 2*x**(2*n)*a*c + x**(2*n)*b**2 
*n - x**(2*n)*b**2 + 2*x**n*a*b*n - 2*x**n*a*b + a**2*n - a**2),x)*b*c*e** 
2 - 4*x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n - x**(4*n)*c**2 + 2*x**(3*n)* 
b*c*n - 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n - 2*x**(2*n)*a*c + x**(2*n)*b**2 
*n - x**(2*n)*b**2 + 2*x**n*a*b*n - 2*x**n*a*b + a**2*n - a**2),x)*c**2*d* 
e*n**2 + 6*x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n - x**(4*n)*c**2 + 2*x**( 
3*n)*b*c*n - 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n - 2*x**(2*n)*a*c + x**(2*n) 
*b**2*n - x**(2*n)*b**2 + 2*x**n*a*b*n - 2*x**n*a*b + a**2*n - a**2),x)*c* 
*2*d*e*n - 2*x**(2*n)*int(x**(2*n)/(x**(4*n)*c**2*n - x**(4*n)*c**2 + 2*x* 
*(3*n)*b*c*n - 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n - 2*x**(2*n)*a*c + x**(2* 
n)*b**2*n - x**(2*n)*b**2 + 2*x**n*a*b*n - 2*x**n*a*b + a**2*n - a**2),x)* 
c**2*d*e + 2*x**(2*n)*int(1/(x**(4*n)*c**2*n - x**(4*n)*c**2 + 2*x**(3*n)* 
b*c*n - 2*x**(3*n)*b*c + 2*x**(2*n)*a*c*n - 2*x**(2*n)*a*c + x**(2*n)*b...