3.2.33 \(\int \frac {(d+e x^2)^2}{d^2-e^2 x^4} \, dx\)

Optimal. Leaf size=29 \[ \frac {2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-x \]

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1150, 388, 208} \begin {gather*} \frac {2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)^2/(d^2 - e^2*x^4),x]

[Out]

-x + (2*Sqrt[d]*ArcTanh[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e]

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 1150

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(d + e*x^2)^(p + q)*(a/d + (c*x^
2)/e)^p, x] /; FreeQ[{a, c, d, e, q}, x] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x + (2*Sqrt[d]*ArcTanh[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x^2)^2/(d^2 - e^2*x^4),x]

[Out]

IntegrateAlgebraic[(d + e*x^2)^2/(d^2 - e^2*x^4), x]

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fricas [A]  time = 1.68, size = 73, normalized size = 2.52 \begin {gather*} \left [\sqrt {\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {\frac {d}{e}} + d}{e x^{2} - d}\right ) - x, -2 \, \sqrt {-\frac {d}{e}} \arctan \left (\frac {e x \sqrt {-\frac {d}{e}}}{d}\right ) - x\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[sqrt(d/e)*log((e*x^2 + 2*e*x*sqrt(d/e) + d)/(e*x^2 - d)) - x, -2*sqrt(-d/e)*arctan(e*x*sqrt(-d/e)/d) - x]

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giac [B]  time = 0.21, size = 118, normalized size = 4.07 \begin {gather*} \frac {{\left ({\left (d^{2}\right )}^{\frac {1}{4}} d e^{\frac {7}{2}} - {\left (d^{2}\right )}^{\frac {1}{4}} {\left | d \right |} e^{\frac {7}{2}}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{{\left (d^{2}\right )}^{\frac {1}{4}}}\right ) e^{\left (-4\right )}}{d} + \frac {{\left ({\left (d^{2}\right )}^{\frac {1}{4}} d e^{\frac {11}{2}} + {\left (d^{2}\right )}^{\frac {3}{4}} e^{\frac {11}{2}}\right )} e^{\left (-6\right )} \log \left ({\left | {\left (d^{2}\right )}^{\frac {1}{4}} e^{\left (-\frac {1}{2}\right )} + x \right |}\right )}{2 \, d} - \frac {{\left ({\left (d^{2}\right )}^{\frac {1}{4}} d e^{\frac {7}{2}} + {\left (d^{2}\right )}^{\frac {1}{4}} {\left | d \right |} e^{\frac {7}{2}}\right )} e^{\left (-4\right )} \log \left ({\left | -{\left (d^{2}\right )}^{\frac {1}{4}} e^{\left (-\frac {1}{2}\right )} + x \right |}\right )}{2 \, d} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^2)^(1/4)*d*e^(7/2) - (d^2)^(1/4)*abs(d)*e^(7/2))*arctan(x*e^(1/2)/(d^2)^(1/4))*e^(-4)/d + 1/2*((d^2)^(1/4)
*d*e^(11/2) + (d^2)^(3/4)*e^(11/2))*e^(-6)*log(abs((d^2)^(1/4)*e^(-1/2) + x))/d - 1/2*((d^2)^(1/4)*d*e^(7/2) +
(d^2)^(1/4)*abs(d)*e^(7/2))*e^(-4)*log(abs(-(d^2)^(1/4)*e^(-1/2) + x))/d - x

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maple [A]  time = 0.00, size = 22, normalized size = 0.76 \begin {gather*} \frac {2 d \arctanh \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.45, size = 36, normalized size = 1.24 \begin {gather*} -\frac {d \log \left (\frac {e x - \sqrt {d e}}{e x + \sqrt {d e}}\right )}{\sqrt {d e}} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*log((e*x - sqrt(d*e))/(e*x + sqrt(d*e)))/sqrt(d*e) - x

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mupad [B]  time = 4.43, size = 21, normalized size = 0.72 \begin {gather*} \frac {2\,\sqrt {d}\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )}{\sqrt {e}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)^2/(d^2 - e^2*x^4),x)

[Out]

(2*d^(1/2)*atanh((e^(1/2)*x)/d^(1/2)))/e^(1/2) - x

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sympy [A]  time = 0.18, size = 34, normalized size = 1.17 \begin {gather*} - x - \sqrt {\frac {d}{e}} \log {\left (x - \sqrt {\frac {d}{e}} \right )} + \sqrt {\frac {d}{e}} \log {\left (x + \sqrt {\frac {d}{e}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - sqrt(d/e)*log(x - sqrt(d/e)) + sqrt(d/e)*log(x + sqrt(d/e))

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