\(\int \frac {x^m (e (1+m)+2 f (1+m-n) x^n)}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx\) [40]

Optimal result

Integrand size = 56, antiderivative size = 42 \[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=\frac {\arctan \left (\frac {2 \sqrt {d} \sqrt {f} x^{1+m}}{e+2 f x^n}\right )}{2 \sqrt {d} \sqrt {f}} \] Output:

1/2*arctan(2*d^(1/2)*f^(1/2)*x^(1+m)/(e+2*f*x^n))/d^(1/2)/f^(1/2)
 

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=\frac {\arctan \left (\frac {2 \sqrt {d} \sqrt {f} x^{1+m}}{e+2 f x^n}\right )}{2 \sqrt {d} \sqrt {f}} \] Input:

Integrate[(x^m*(e*(1 + m) + 2*f*(1 + m - n)*x^n))/(e^2 + 4*d*f*x^(2 + 2*m) 
 + 4*e*f*x^n + 4*f^2*x^(2*n)),x]
 

Output:

ArcTan[(2*Sqrt[d]*Sqrt[f]*x^(1 + m))/(e + 2*f*x^n)]/(2*Sqrt[d]*Sqrt[f])
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.67, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2520, 218}

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.

\(\displaystyle \int \frac {x^m \left (e (m+1)+2 f (m-n+1) x^n\right )}{4 d f x^{2 m+2}+e^2+4 e f x^n+4 f^2 x^{2 n}} \, dx\)

\(\Big \downarrow \) 2520

\(\displaystyle e^2 (m+1) (m-n+1) \int \frac {1}{\frac {4 d e^2 f (m+1)^2 (m-n+1)^2 x^{2 m+2}}{\left (2 f (m+1) (m-n+1) x^n+e (m+1) (m-n+1)\right )^2}+e^2}d\frac {x^{m+1}}{2 f (m+1) (m-n+1) x^n+e (m+1) (m-n+1)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\arctan \left (\frac {2 \sqrt {d} \sqrt {f} (m+1) (m-n+1) x^{m+1}}{e (m+1) (m-n+1)+2 f (m+1) (m-n+1) x^n}\right )}{2 \sqrt {d} \sqrt {f}}\)

Input:

Int[(x^m*(e*(1 + m) + 2*f*(1 + m - n)*x^n))/(e^2 + 4*d*f*x^(2 + 2*m) + 4*e 
*f*x^n + 4*f^2*x^(2*n)),x]
 

Output:

ArcTan[(2*Sqrt[d]*Sqrt[f]*(1 + m)*(1 + m - n)*x^(1 + m))/(e*(1 + m)*(1 + m 
 - n) + 2*f*(1 + m)*(1 + m - n)*x^n)]/(2*Sqrt[d]*Sqrt[f])
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 2520

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.) 
*(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_Symbol] :> Simp[A^2*((m - n + 1)/(m + 1) 
)   Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x], x, x^(m + 1)/(A*(m - n + 
 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x] && EqQ[n2, 
2*n] && EqQ[k, 2*(m + 1)] && EqQ[a*B^2*(m + 1)^2 - A^2*d*(m - n + 1)^2, 0] 
&& EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(83\) vs. \(2(32)=64\).

Time = 2.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.00

method	result	size
risch	\(-\frac {\ln \left (x^{n}+\frac {2 x^{m} d f x +\sqrt {-d f}\, e}{2 \sqrt {-d f}\, f}\right )}{4 \sqrt {-d f}}+\frac {\ln \left (x^{n}+\frac {-2 x^{m} d f x +\sqrt {-d f}\, e}{2 \sqrt {-d f}\, f}\right )}{4 \sqrt {-d f}}\)	\(84\)

Input:

int(x^m*(e*(1+m)+2*f*(1+m-n)*x^n)/(e^2+4*d*f*x^(2+2*m)+4*e*f*x^n+4*f^2*x^( 
2*n)),x,method=_RETURNVERBOSE)
 

Output:

-1/4/(-d*f)^(1/2)*ln(x^n+1/2*(2*x^m*d*f*x+(-d*f)^(1/2)*e)/(-d*f)^(1/2)/f)+ 
1/4/(-d*f)^(1/2)*ln(x^n+1/2*(-2*x^m*d*f*x+(-d*f)^(1/2)*e)/(-d*f)^(1/2)/f)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 165, normalized size of antiderivative = 3.93 \[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=\left [-\frac {\sqrt {-d f} \log \left (-\frac {4 \, d f x^{2} x^{2 \, m} - 4 \, \sqrt {-d f} e x x^{m} - 4 \, f^{2} x^{2 \, n} - e^{2} - 4 \, {\left (2 \, \sqrt {-d f} f x x^{m} + e f\right )} x^{n}}{4 \, d f x^{2} x^{2 \, m} + 4 \, f^{2} x^{2 \, n} + 4 \, e f x^{n} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {d f} \arctan \left (\frac {2 \, \sqrt {d f} f x^{n} + \sqrt {d f} e}{2 \, d f x x^{m}}\right )}{2 \, d f}\right ] \] Input:

integrate(x^m*(e*(1+m)+2*f*(1+m-n)*x^n)/(e^2+4*d*f*x^(2+2*m)+4*e*f*x^n+4*f 
^2*x^(2*n)),x, algorithm="fricas")
 

Output:

[-1/4*sqrt(-d*f)*log(-(4*d*f*x^2*x^(2*m) - 4*sqrt(-d*f)*e*x*x^m - 4*f^2*x^ 
(2*n) - e^2 - 4*(2*sqrt(-d*f)*f*x*x^m + e*f)*x^n)/(4*d*f*x^2*x^(2*m) + 4*f 
^2*x^(2*n) + 4*e*f*x^n + e^2))/(d*f), -1/2*sqrt(d*f)*arctan(1/2*(2*sqrt(d* 
f)*f*x^n + sqrt(d*f)*e)/(d*f*x*x^m))/(d*f)]
 

Sympy [F]

\[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=\int \frac {x^{m} \left (e m + e + 2 f m x^{n} - 2 f n x^{n} + 2 f x^{n}\right )}{4 d f x^{2 m + 2} + e^{2} + 4 e f x^{n} + 4 f^{2} x^{2 n}}\, dx \] Input:

integrate(x**m*(e*(1+m)+2*f*(1+m-n)*x**n)/(e**2+4*d*f*x**(2+2*m)+4*e*f*x** 
n+4*f**2*x**(2*n)),x)
 

Output:

Integral(x**m*(e*m + e + 2*f*m*x**n - 2*f*n*x**n + 2*f*x**n)/(4*d*f*x**(2* 
m + 2) + e**2 + 4*e*f*x**n + 4*f**2*x**(2*n)), x)
 

Maxima [F]

\[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=\int { \frac {{\left (2 \, f {\left (m - n + 1\right )} x^{n} + e {\left (m + 1\right )}\right )} x^{m}}{4 \, d f x^{2 \, m + 2} + 4 \, f^{2} x^{2 \, n} + 4 \, e f x^{n} + e^{2}} \,d x } \] Input:

integrate(x^m*(e*(1+m)+2*f*(1+m-n)*x^n)/(e^2+4*d*f*x^(2+2*m)+4*e*f*x^n+4*f 
^2*x^(2*n)),x, algorithm="maxima")
 

Output:

integrate((2*f*(m - n + 1)*x^n + e*(m + 1))*x^m/(4*d*f*x^(2*m + 2) + 4*f^2 
*x^(2*n) + 4*e*f*x^n + e^2), x)
 

Giac [F]

\[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=\int { \frac {{\left (2 \, f {\left (m - n + 1\right )} x^{n} + e {\left (m + 1\right )}\right )} x^{m}}{4 \, d f x^{2 \, m + 2} + 4 \, f^{2} x^{2 \, n} + 4 \, e f x^{n} + e^{2}} \,d x } \] Input:

integrate(x^m*(e*(1+m)+2*f*(1+m-n)*x^n)/(e^2+4*d*f*x^(2+2*m)+4*e*f*x^n+4*f 
^2*x^(2*n)),x, algorithm="giac")
 

Output:

integrate((2*f*(m - n + 1)*x^n + e*(m + 1))*x^m/(4*d*f*x^(2*m + 2) + 4*f^2 
*x^(2*n) + 4*e*f*x^n + e^2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=\int \frac {x^m\,\left (e\,\left (m+1\right )+2\,f\,x^n\,\left (m-n+1\right )\right )}{e^2+4\,f^2\,x^{2\,n}+4\,d\,f\,x^{2\,m+2}+4\,e\,f\,x^n} \,d x \] Input:

int((x^m*(e*(m + 1) + 2*f*x^n*(m - n + 1)))/(e^2 + 4*f^2*x^(2*n) + 4*d*f*x 
^(2*m + 2) + 4*e*f*x^n),x)
 

Output:

int((x^m*(e*(m + 1) + 2*f*x^n*(m - n + 1)))/(e^2 + 4*f^2*x^(2*n) + 4*d*f*x 
^(2*m + 2) + 4*e*f*x^n), x)
 

Reduce [F]

\[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2+4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx=2 \left (\int \frac {x^{m +n}}{4 x^{2 m} d f \,x^{2}+4 x^{2 n} f^{2}+4 x^{n} e f +e^{2}}d x \right ) f m -2 \left (\int \frac {x^{m +n}}{4 x^{2 m} d f \,x^{2}+4 x^{2 n} f^{2}+4 x^{n} e f +e^{2}}d x \right ) f n +2 \left (\int \frac {x^{m +n}}{4 x^{2 m} d f \,x^{2}+4 x^{2 n} f^{2}+4 x^{n} e f +e^{2}}d x \right ) f +\left (\int \frac {x^{m}}{4 x^{2 m} d f \,x^{2}+4 x^{2 n} f^{2}+4 x^{n} e f +e^{2}}d x \right ) e m +\left (\int \frac {x^{m}}{4 x^{2 m} d f \,x^{2}+4 x^{2 n} f^{2}+4 x^{n} e f +e^{2}}d x \right ) e \] Input:

int(x^m*(e*(1+m)+2*f*(1+m-n)*x^n)/(e^2+4*d*f*x^(2+2*m)+4*e*f*x^n+4*f^2*x^( 
2*n)),x)
 

Output:

2*int(x**(m + n)/(4*x**(2*m)*d*f*x**2 + 4*x**(2*n)*f**2 + 4*x**n*e*f + e** 
2),x)*f*m - 2*int(x**(m + n)/(4*x**(2*m)*d*f*x**2 + 4*x**(2*n)*f**2 + 4*x* 
*n*e*f + e**2),x)*f*n + 2*int(x**(m + n)/(4*x**(2*m)*d*f*x**2 + 4*x**(2*n) 
*f**2 + 4*x**n*e*f + e**2),x)*f + int(x**m/(4*x**(2*m)*d*f*x**2 + 4*x**(2* 
n)*f**2 + 4*x**n*e*f + e**2),x)*e*m + int(x**m/(4*x**(2*m)*d*f*x**2 + 4*x* 
*(2*n)*f**2 + 4*x**n*e*f + e**2),x)*e