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

Optimal. Leaf size=38 \[ \frac {\tan ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x}{e+2 f x^n}\right )}{2 \sqrt {d} \sqrt {f}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2093, 205} \[ \frac {\tan ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x}{e+2 f x^n}\right )}{2 \sqrt {d} \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 2093

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

Rubi steps

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

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Mathematica [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \frac {e-2 f (-1+n) x^n}{e^2+4 d f x^2+4 e f x^n+4 f^2 x^{2 n}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 144, normalized size = 3.79 \[ \left [-\frac {\sqrt {-d f} \log \left (-\frac {4 \, d f x^{2} - 4 \, f^{2} x^{2 \, n} - 4 \, \sqrt {-d f} e x - e^{2} - 4 \, {\left (2 \, \sqrt {-d f} f x + e f\right )} x^{n}}{4 \, d f x^{2} + 4 \, f^{2} x^{2 \, n} + 4 \, e f x^{n} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {d f} \arctan \left (\frac {2 \, \sqrt {d f} f x^{n} + \sqrt {d f} e}{2 \, d f x}\right )}{2 \, d f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 78, normalized size = 2.05 \[ \frac {\ln \left (x^{n}+\frac {-2 d f x +\sqrt {-d f}\, e}{2 \sqrt {-d f}\, f}\right )}{4 \sqrt {-d f}}-\frac {\ln \left (x^{n}+\frac {2 d f x +\sqrt {-d f}\, e}{2 \sqrt {-d f}\, f}\right )}{4 \sqrt {-d f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.42, size = 196, normalized size = 5.16 \[ \frac {\ln \left (-\frac {e+2\,f\,x^n-2\,f\,n\,x^n}{4\,f^2}-\frac {e^2\,n-4\,d\,f\,x^2+4\,d\,f\,n\,x^2+2\,e\,f\,n\,x^n}{8\,\sqrt {-d}\,f^{5/2}\,x}\right )}{4\,\sqrt {-d}\,\sqrt {f}}-\frac {\mathrm {atan}\left (\frac {x\,\left (8\,d\,f\,n^2-16\,d\,f\,n+8\,d\,f\right )}{4\,\sqrt {d}\,\sqrt {f}\,\left (e\,n-e\,n^2\right )}\right )}{2\,\sqrt {d}\,\sqrt {f}}-\frac {\ln \left (\frac {e^2\,n-4\,d\,f\,x^2+4\,d\,f\,n\,x^2+2\,e\,f\,n\,x^n}{8\,\sqrt {-d}\,f^{5/2}\,x}-\frac {e+2\,f\,x^n-2\,f\,n\,x^n}{4\,f^2}\right )}{4\,\sqrt {-d}\,\sqrt {f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(- (e + 2*f*x^n - 2*f*n*x^n)/(4*f^2) - (e^2*n - 4*d*f*x^2 + 4*d*f*n*x^2 + 2*e*f*n*x^n)/(8*(-d)^(1/2)*f^(5/2
)*x))/(4*(-d)^(1/2)*f^(1/2)) - atan((x*(8*d*f - 16*d*f*n + 8*d*f*n^2))/(4*d^(1/2)*f^(1/2)*(e*n - e*n^2)))/(2*d
^(1/2)*f^(1/2)) - log((e^2*n - 4*d*f*x^2 + 4*d*f*n*x^2 + 2*e*f*n*x^n)/(8*(-d)^(1/2)*f^(5/2)*x) - (e + 2*f*x^n
- 2*f*n*x^n)/(4*f^2))/(4*(-d)^(1/2)*f^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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