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

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

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 61, normalized size of antiderivative = 1.45, 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, 208} \[ \frac {\tanh ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} \left (1-m^2\right ) x^{m+1}}{(1-m) (m+1) \left (e+2 f x^2\right )}\right )}{2 \sqrt {d} \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 42, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {2 \sqrt {d} \sqrt {f} x^{m+1}}{e+2 f x^2}\right )}{2 \sqrt {d} \sqrt {f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 146, normalized size = 3.48 \[ \left [\frac {\sqrt {d f} \log \left (-\frac {4 \, f^{2} x^{4} + 4 \, d f x^{2} x^{2 \, m} + 4 \, e f x^{2} + 4 \, {\left (2 \, f x^{3} + e x\right )} \sqrt {d f} x^{m} + e^{2}}{4 \, f^{2} x^{4} - 4 \, d f x^{2} x^{2 \, m} + 4 \, e f x^{2} + e^{2}}\right )}{4 \, d f}, -\frac {\sqrt {-d f} \arctan \left (\frac {{\left (2 \, f x^{2} + e\right )} \sqrt {-d f}}{2 \, d f x x^{m}}\right )}{2 \, d f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.09, size = 74, normalized size = 1.76 \[ -\frac {\ln \left (x^{m}-\frac {\left (2 f \,x^{2}+e \right ) \sqrt {d f}}{2 d f x}\right )}{4 \sqrt {d f}}+\frac {\ln \left (x^{m}+\frac {\left (2 f \,x^{2}+e \right ) \sqrt {d f}}{2 d f x}\right )}{4 \sqrt {d f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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