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

Optimal. Leaf size=42 \[ \frac {\tanh ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x^{m+1}}{e+2 f x^n}\right )}{2 \sqrt {d} \sqrt {f}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2094, 208} \[ \frac {\tanh ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x^{m+1}}{e+2 f x^n}\right )}{2 \sqrt {d} \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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]

[Out]

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

Rule 208

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

Rule 2094

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_
Symbol] :> Dist[(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)] && Eq
Q[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]

Rubi steps

\begin {align*} \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 (e^2 (1+m) (1+m-n)\right ) \operatorname {Subst}\left (\int \frac {1}{e^2-4 d e^2 f (1+m)^2 (1+m-n)^2 x^2} \, dx,x,\frac {x^{1+m}}{e (1+m) (1+m-n)+2 f (1+m) (1+m-n) x^n}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x^{1+m}}{e+2 f x^n}\right )}{2 \sqrt {d} \sqrt {f}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \frac {x^m \left (e (1+m)+2 f (1+m-n) x^n\right )}{e^2-4 d f x^{2+2 m}+4 e f x^n+4 f^2 x^{2 n}} \, dx \]

Verification is Not applicable to the result.

[In]

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]

[Out]

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]

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 165, normalized size = 3.93 \[ \left [\frac {\sqrt {d f} \log \left (-\frac {4 \, d f x^{2} x^{2 \, m} + 4 \, \sqrt {d f} e x x^{m} + 4 \, f^{2} x^{2 \, n} + e^{2} + 4 \, {\left (2 \, \sqrt {d f} f x x^{m} + e f\right )} x^{n}}{4 \, d f x^{2} x^{2 \, m} - 4 \, f^{2} x^{2 \, n} - 4 \, e f x^{n} - e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {-d f} \arctan \left (\frac {2 \, \sqrt {-d f} f x^{n} + \sqrt {-d f} e}{2 \, d f x x^{m}}\right )}{2 \, d f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

[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)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (2 \, f {\left (m - n + 1\right )} x^{n} + e {\left (m + 1\right )}\right )} x^{m}}{4 \, d f x^{2 \, m + 2} - 4 \, f^{2} x^{2 \, n} - 4 \, e f x^{n} - e^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

________________________________________________________________________________________

maple [B]  time = 0.08, size = 78, normalized size = 1.86 \[ -\frac {\ln \left (x^{n}+\frac {-2 d f x \,x^{m}+\sqrt {d f}\, e}{2 \sqrt {d f}\, f}\right )}{4 \sqrt {d f}}+\frac {\ln \left (x^{n}+\frac {2 d f x \,x^{m}+\sqrt {d f}\, e}{2 \sqrt {d f}\, f}\right )}{4 \sqrt {d f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(e*(m+1)+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)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (2 \, f {\left (m - n + 1\right )} x^{n} + e {\left (m + 1\right )}\right )} x^{m}}{4 \, d f x^{2 \, m + 2} - 4 \, f^{2} x^{2 \, n} - 4 \, e f x^{n} - e^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

-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]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^m\,\left (e\,\left (m+1\right )+2\,f\,x^n\,\left (m-n+1\right )\right )}{e^2+4\,f^2\,x^{2\,n}-4\,d\,f\,x^{2\,m+2}+4\,e\,f\,x^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {e x^{m}}{4 d f x^{2} x^{2 m} - e^{2} - 4 e f x^{n} - 4 f^{2} x^{2 n}}\, dx - \int \frac {e m x^{m}}{4 d f x^{2} x^{2 m} - e^{2} - 4 e f x^{n} - 4 f^{2} x^{2 n}}\, dx - \int \frac {2 f x^{m} x^{n}}{4 d f x^{2} x^{2 m} - e^{2} - 4 e f x^{n} - 4 f^{2} x^{2 n}}\, dx - \int \frac {2 f m x^{m} x^{n}}{4 d f x^{2} x^{2 m} - e^{2} - 4 e f x^{n} - 4 f^{2} x^{2 n}}\, dx - \int \left (- \frac {2 f n x^{m} x^{n}}{4 d f x^{2} x^{2 m} - e^{2} - 4 e f x^{n} - 4 f^{2} x^{2 n}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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

________________________________________________________________________________________