3.540 \(\int \frac{x^m (e (1+m)+2 f (-2+m) x^3)}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx\)

Optimal. Leaf size=42 \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^{m+1}}{e+2 f x^3}\right )}{2 \sqrt{d} \sqrt{f}} \]

[Out]

ArcTan[(2*Sqrt[d]*Sqrt[f]*x^(1 + m))/(e + 2*f*x^3)]/(2*Sqrt[d]*Sqrt[f])

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Rubi [A]  time = 0.22301, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 51, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.039, Rules used = {2094, 205} \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^{m+1}}{e+2 f x^3}\right )}{2 \sqrt{d} \sqrt{f}} \]

Antiderivative was successfully verified.

[In]

Int[(x^m*(e*(1 + m) + 2*f*(-2 + m)*x^3))/(e^2 + 4*e*f*x^3 + 4*f^2*x^6 + 4*d*f*x^(2 + 2*m)),x]

[Out]

ArcTan[(2*Sqrt[d]*Sqrt[f]*x^(1 + m))/(e + 2*f*x^3)]/(2*Sqrt[d]*Sqrt[f])

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]

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{x^m \left (e (1+m)+2 f (-2+m) x^3\right )}{e^2+4 e f x^3+4 f^2 x^6+4 d f x^{2+2 m}} \, dx &=-\left (\left (e^2 (2-m) (1+m)\right ) \operatorname{Subst}\left (\int \frac{1}{e^2+4 d e^2 f (-2+m)^2 (1+m)^2 x^2} \, dx,x,\frac{x^{1+m}}{e (-2+m) (1+m)+2 f (-2+m) (1+m) x^3}\right )\right )\\ &=\frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^{1+m}}{e+2 f x^3}\right )}{2 \sqrt{d} \sqrt{f}}\\ \end{align*}

Mathematica [A]  time = 0.317145, size = 42, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^{m+1}}{e+2 f x^3}\right )}{2 \sqrt{d} \sqrt{f}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^m*(e*(1 + m) + 2*f*(-2 + m)*x^3))/(e^2 + 4*e*f*x^3 + 4*f^2*x^6 + 4*d*f*x^(2 + 2*m)),x]

[Out]

ArcTan[(2*Sqrt[d]*Sqrt[f]*x^(1 + m))/(e + 2*f*x^3)]/(2*Sqrt[d]*Sqrt[f])

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Maple [B]  time = 0.035, size = 78, normalized size = 1.9 \begin{align*} -{\frac{1}{4}\ln \left ({x}^{m}+{\frac{2\,f{x}^{3}+e}{2\,dfx}\sqrt{-df}} \right ){\frac{1}{\sqrt{-df}}}}+{\frac{1}{4}\ln \left ({x}^{m}-{\frac{2\,f{x}^{3}+e}{2\,dfx}\sqrt{-df}} \right ){\frac{1}{\sqrt{-df}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2+2*m)),x)

[Out]

-1/4/(-d*f)^(1/2)*ln(x^m+1/2*(2*f*x^3+e)*(-d*f)^(1/2)/d/f/x)+1/4/(-d*f)^(1/2)*ln(x^m-1/2*(2*f*x^3+e)*(-d*f)^(1
/2)/d/f/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, f{\left (m - 2\right )} x^{3} + e{\left (m + 1\right )}\right )} x^{m}}{4 \, f^{2} x^{6} + 4 \, e f x^{3} + 4 \, d f x^{2 \, m + 2} + e^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2+2*m)),x, algorithm="maxima")

[Out]

integrate((2*f*(m - 2)*x^3 + e*(m + 1))*x^m/(4*f^2*x^6 + 4*e*f*x^3 + 4*d*f*x^(2*m + 2) + e^2), x)

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Fricas [A]  time = 1.27622, size = 321, normalized size = 7.64 \begin{align*} \left [-\frac{\sqrt{-d f} \log \left (-\frac{4 \, f^{2} x^{6} - 4 \, d f x^{2} x^{2 \, m} + 4 \, e f x^{3} + 4 \,{\left (2 \, f x^{4} + e x\right )} \sqrt{-d f} x^{m} + e^{2}}{4 \, f^{2} x^{6} + 4 \, d f x^{2} x^{2 \, m} + 4 \, e f x^{3} + e^{2}}\right )}{4 \, d f}, -\frac{\sqrt{d f} \arctan \left (\frac{{\left (2 \, f x^{3} + e\right )} \sqrt{d f}}{2 \, d f x x^{m}}\right )}{2 \, d f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2+2*m)),x, algorithm="fricas")

[Out]

[-1/4*sqrt(-d*f)*log(-(4*f^2*x^6 - 4*d*f*x^2*x^(2*m) + 4*e*f*x^3 + 4*(2*f*x^4 + e*x)*sqrt(-d*f)*x^m + e^2)/(4*
f^2*x^6 + 4*d*f*x^2*x^(2*m) + 4*e*f*x^3 + e^2))/(d*f), -1/2*sqrt(d*f)*arctan(1/2*(2*f*x^3 + e)*sqrt(d*f)/(d*f*
x*x^m))/(d*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(e*(1+m)+2*f*(-2+m)*x**3)/(e**2+4*e*f*x**3+4*f**2*x**6+4*d*f*x**(2+2*m)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, f{\left (m - 2\right )} x^{3} + e{\left (m + 1\right )}\right )} x^{m}}{4 \, f^{2} x^{6} + 4 \, e f x^{3} + 4 \, d f x^{2 \, m + 2} + e^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(e*(1+m)+2*f*(-2+m)*x^3)/(e^2+4*e*f*x^3+4*f^2*x^6+4*d*f*x^(2+2*m)),x, algorithm="giac")

[Out]

integrate((2*f*(m - 2)*x^3 + e*(m + 1))*x^m/(4*f^2*x^6 + 4*e*f*x^3 + 4*d*f*x^(2*m + 2) + e^2), x)