\(\int \frac {(d x)^m}{(a x^n+b x^{1+n}+c x^{2+n})^2} \, dx\) [86]

Optimal result

Integrand size = 28, antiderivative size = 300 \[ \int \frac {(d x)^m}{\left (a x^n+b x^{1+n}+c x^{2+n}\right )^2} \, dx=\frac {x^{1-2 n} (d x)^m \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {c \left (b \left (b+\sqrt {b^2-4 a c}\right ) (m-2 n)+4 a c (1-m+2 n)\right ) x^{1-2 n} (d x)^m \operatorname {Hypergeometric2F1}\left (1,1+m-2 n,2+m-2 n,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) (1+m-2 n)}+\frac {c \left (b \left (b-\sqrt {b^2-4 a c}\right ) (m-2 n)+4 a c (1-m+2 n)\right ) x^{1-2 n} (d x)^m \operatorname {Hypergeometric2F1}\left (1,1+m-2 n,2+m-2 n,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) (1+m-2 n)} \] Output:

x^(1-2*n)*(d*x)^m*(b*c*x-2*a*c+b^2)/a/(-4*a*c+b^2)/(c*x^2+b*x+a)-c*(b*(b+( 
-4*a*c+b^2)^(1/2))*(m-2*n)+4*a*c*(1-m+2*n))*x^(1-2*n)*(d*x)^m*hypergeom([1 
, 1+m-2*n],[2+m-2*n],-2*c*x/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/( 
b-(-4*a*c+b^2)^(1/2))/(1+m-2*n)+c*(b*(b-(-4*a*c+b^2)^(1/2))*(m-2*n)+4*a*c* 
(1-m+2*n))*x^(1-2*n)*(d*x)^m*hypergeom([1, 1+m-2*n],[2+m-2*n],-2*c*x/(b+(- 
4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/(b+(-4*a*c+b^2)^(1/2))/(1+m-2*n)
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.88 \[ \int \frac {(d x)^m}{\left (a x^n+b x^{1+n}+c x^{2+n}\right )^2} \, dx=\frac {x^{1-2 n} (d x)^m \left (\frac {b^2-2 a c+b c x}{a+x (b+c x)}-\frac {c \left (\frac {-4 a c (-1+m-2 n)+b^2 (m-2 n)}{\sqrt {b^2-4 a c}}+b (m-2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+m-2 n,2+m-2 n,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) (1+m-2 n)}+\frac {c \left (-4 a c (-1+m-2 n)+b \left (b-\sqrt {b^2-4 a c}\right ) (m-2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+m-2 n,2+m-2 n,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m-2 n)}\right )}{a \left (b^2-4 a c\right )} \] Input:

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

Output:

(x^(1 - 2*n)*(d*x)^m*((b^2 - 2*a*c + b*c*x)/(a + x*(b + c*x)) - (c*((-4*a* 
c*(-1 + m - 2*n) + b^2*(m - 2*n))/Sqrt[b^2 - 4*a*c] + b*(m - 2*n))*Hyperge 
ometric2F1[1, 1 + m - 2*n, 2 + m - 2*n, (2*c*x)/(-b + Sqrt[b^2 - 4*a*c])]) 
/((b - Sqrt[b^2 - 4*a*c])*(1 + m - 2*n)) + (c*(-4*a*c*(-1 + m - 2*n) + b*( 
b - Sqrt[b^2 - 4*a*c])*(m - 2*n))*Hypergeometric2F1[1, 1 + m - 2*n, 2 + m 
- 2*n, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 
 - 4*a*c])*(1 + m - 2*n))))/(a*(b^2 - 4*a*c))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2028, 30, 1165, 1200, 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.

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

\(\Big \downarrow \) 2028

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

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 1165

\(\displaystyle x^{-m} (d x)^m \left (\frac {x^{m-2 n+1} \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int \frac {x^{m-2 n} \left ((m-2 n) b^2+c (m-2 n) x b+2 a c (-m+2 n+1)\right )}{c x^2+b x+a}dx}{a \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 1200

\(\displaystyle x^{-m} (d x)^m \left (\frac {x^{m-2 n+1} \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {\left (b c (m-2 n)+\frac {c \left (m b^2-2 n b^2+4 a c-4 a c m+8 a c n\right )}{\sqrt {b^2-4 a c}}\right ) x^{m-2 n}}{b+2 c x-\sqrt {b^2-4 a c}}+\frac {\left (b c (m-2 n)-\frac {c \left (m b^2-2 n b^2+4 a c-4 a c m+8 a c n\right )}{\sqrt {b^2-4 a c}}\right ) x^{m-2 n}}{b+2 c x+\sqrt {b^2-4 a c}}\right )dx}{a \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((d*x)^m*((x^(1 + m - 2*n)*(b^2 - 2*a*c + b*c*x))/(a*(b^2 - 4*a*c)*(a + b* 
x + c*x^2)) - ((c*(b*(m - 2*n) + (b^2*(m - 2*n) + 4*a*c*(1 - m + 2*n))/Sqr 
t[b^2 - 4*a*c])*x^(1 + m - 2*n)*Hypergeometric2F1[1, 1 + m - 2*n, 2 + m - 
2*n, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/((b - Sqrt[b^2 - 4*a*c])*(1 + m - 
2*n)) - (c*(b*(b - Sqrt[b^2 - 4*a*c])*(m - 2*n) + 4*a*c*(1 - m + 2*n))*x^( 
1 + m - 2*n)*Hypergeometric2F1[1, 1 + m - 2*n, 2 + m - 2*n, (-2*c*x)/(b + 
Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m - 2 
*n)))/(a*(b^2 - 4*a*c))))/x^m
 

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1200

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

rule 2009

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

rule 2028

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)*(x_)^(t_.))^(p_.), 
x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r))^p*Fx, x] /; FreeQ[ 
{a, b, c, r, s, t}, x] && IntegerQ[p] && PosQ[s - r] && PosQ[t - r] &&  !(E 
qQ[p, 1] && EqQ[u, 1])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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