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]

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

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Rubi [A]  time = 0.0955535, antiderivative size = 38, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [F]  time = 0.274566, size = 0, normalized size = 0. \[ \int \frac{e-2 f (-1+n) x^n}{e^2+4 d f x^2+4 e f x^n+4 f^2 x^{2 n}} \, dx \]

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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Maple [B]  time = 0.049, size = 78, normalized size = 2.1 \begin{align*} -{\frac{1}{4}\ln \left ({x}^{n}+{\frac{1}{2\,f} \left ( 2\,dfx+e\sqrt{-df} \right ){\frac{1}{\sqrt{-df}}}} \right ){\frac{1}{\sqrt{-df}}}}+{\frac{1}{4}\ln \left ({x}^{n}+{\frac{1}{2\,f} \left ( -2\,dfx+e\sqrt{-df} \right ){\frac{1}{\sqrt{-df}}}} \right ){\frac{1}{\sqrt{-df}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

-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., size = 0, normalized size = 0. \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.54707, size = 325, normalized size = 8.55 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 126.247, size = 139, normalized size = 3.66 \begin{align*} \begin{cases} \frac{x}{e} & \text{for}\: f = 0 \wedge \left (d = 0 \vee f = 0\right ) \\\frac{x}{e + 2 f x^{n}} & \text{for}\: d = 0 \\- \frac{i \log{\left (- \frac{i e \sqrt{\frac{1}{d}} \sqrt{\frac{1}{f}}}{2} - i f x^{n} \sqrt{\frac{1}{d}} \sqrt{\frac{1}{f}} + x \right )}}{4 d f \sqrt{\frac{1}{d}} \sqrt{\frac{1}{f}}} + \frac{i \log{\left (\frac{i e \sqrt{\frac{1}{d}} \sqrt{\frac{1}{f}}}{2} + i f x^{n} \sqrt{\frac{1}{d}} \sqrt{\frac{1}{f}} + x \right )}}{4 d f \sqrt{\frac{1}{d}} \sqrt{\frac{1}{f}}} & \text{otherwise} \end{cases} \end{align*}

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]

Piecewise((x/e, Eq(f, 0) & (Eq(d, 0) | Eq(f, 0))), (x/(e + 2*f*x**n), Eq(d, 0)), (-I*log(-I*e*sqrt(1/d)*sqrt(1
/f)/2 - I*f*x**n*sqrt(1/d)*sqrt(1/f) + x)/(4*d*f*sqrt(1/d)*sqrt(1/f)) + I*log(I*e*sqrt(1/d)*sqrt(1/f)/2 + I*f*
x**n*sqrt(1/d)*sqrt(1/f) + x)/(4*d*f*sqrt(1/d)*sqrt(1/f)), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)