3.576 \(\int \frac{(f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\)

Optimal. Leaf size=127 \[ \frac{x \left (3 e^2 f^2-d^2 g^2\right )+2 d^2 f g}{8 d^4 e^2 \left (d^2-e^2 x^2\right )}+\frac{\left (3 e^2 f^2-d^2 g^2\right ) \tanh ^{-1}\left (\frac{e x}{d}\right )}{8 d^5 e^3}+\frac{(f+g x) \left (d^2 g+e^2 f x\right )}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2} \]

[Out]

((d^2*g + e^2*f*x)*(f + g*x))/(4*d^2*e^2*(d^2 - e^2*x^2)^2) + (2*d^2*f*g + (3*e^2*f^2 - d^2*g^2)*x)/(8*d^4*e^2
*(d^2 - e^2*x^2)) + ((3*e^2*f^2 - d^2*g^2)*ArcTanh[(e*x)/d])/(8*d^5*e^3)

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Rubi [A]  time = 0.0614481, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {739, 639, 208} \[ \frac{x \left (3 e^2 f^2-d^2 g^2\right )+2 d^2 f g}{8 d^4 e^2 \left (d^2-e^2 x^2\right )}+\frac{\left (3 e^2 f^2-d^2 g^2\right ) \tanh ^{-1}\left (\frac{e x}{d}\right )}{8 d^5 e^3}+\frac{(f+g x) \left (d^2 g+e^2 f x\right )}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^2/(d^2 - e^2*x^2)^3,x]

[Out]

((d^2*g + e^2*f*x)*(f + g*x))/(4*d^2*e^2*(d^2 - e^2*x^2)^2) + (2*d^2*f*g + (3*e^2*f^2 - d^2*g^2)*x)/(8*d^4*e^2
*(d^2 - e^2*x^2)) + ((3*e^2*f^2 - d^2*g^2)*ArcTanh[(e*x)/d])/(8*d^5*e^3)

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

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]

Rubi steps

\begin{align*} \int \frac{(f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx &=\frac{\left (d^2 g+e^2 f x\right ) (f+g x)}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}-\frac{\int \frac{-3 e^2 f^2+d^2 g^2-2 e^2 f g x}{\left (d^2-e^2 x^2\right )^2} \, dx}{4 d^2 e^2}\\ &=\frac{\left (d^2 g+e^2 f x\right ) (f+g x)}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}+\frac{2 d^2 f g+\left (3 e^2 f^2-d^2 g^2\right ) x}{8 d^4 e^2 \left (d^2-e^2 x^2\right )}-\frac{\left (-\frac{3 e^2 f^2}{d^2}+g^2\right ) \int \frac{1}{d^2-e^2 x^2} \, dx}{8 d^2 e^2}\\ &=\frac{\left (d^2 g+e^2 f x\right ) (f+g x)}{4 d^2 e^2 \left (d^2-e^2 x^2\right )^2}+\frac{2 d^2 f g+\left (3 e^2 f^2-d^2 g^2\right ) x}{8 d^4 e^2 \left (d^2-e^2 x^2\right )}+\frac{\left (3 e^2 f^2-d^2 g^2\right ) \tanh ^{-1}\left (\frac{e x}{d}\right )}{8 d^5 e^3}\\ \end{align*}

Mathematica [A]  time = 0.044333, size = 110, normalized size = 0.87 \[ \frac{d^3 e^3 x \left (5 f^2+g^2 x^2\right )+\left (d^2-e^2 x^2\right )^2 \left (3 e^2 f^2-d^2 g^2\right ) \tanh ^{-1}\left (\frac{e x}{d}\right )+d^5 e g (4 f+g x)-3 d e^5 f^2 x^3}{8 d^5 e^3 \left (d^2-e^2 x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^2/(d^2 - e^2*x^2)^3,x]

[Out]

(-3*d*e^5*f^2*x^3 + d^5*e*g*(4*f + g*x) + d^3*e^3*x*(5*f^2 + g^2*x^2) + (3*e^2*f^2 - d^2*g^2)*(d^2 - e^2*x^2)^
2*ArcTanh[(e*x)/d])/(8*d^5*e^3*(d^2 - e^2*x^2)^2)

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Maple [B]  time = 0.057, size = 298, normalized size = 2.4 \begin{align*}{\frac{\ln \left ( ex-d \right ){g}^{2}}{16\,{e}^{3}{d}^{3}}}-{\frac{3\,\ln \left ( ex-d \right ){f}^{2}}{16\,e{d}^{5}}}+{\frac{{g}^{2}}{16\,{e}^{3}d \left ( ex-d \right ) ^{2}}}+{\frac{fg}{8\,{d}^{2}{e}^{2} \left ( ex-d \right ) ^{2}}}+{\frac{{f}^{2}}{16\,e{d}^{3} \left ( ex-d \right ) ^{2}}}+{\frac{{g}^{2}}{16\,{e}^{3}{d}^{2} \left ( ex-d \right ) }}-{\frac{fg}{8\,{e}^{2}{d}^{3} \left ( ex-d \right ) }}-{\frac{3\,{f}^{2}}{16\,e{d}^{4} \left ( ex-d \right ) }}-{\frac{\ln \left ( ex+d \right ){g}^{2}}{16\,{e}^{3}{d}^{3}}}+{\frac{3\,\ln \left ( ex+d \right ){f}^{2}}{16\,e{d}^{5}}}+{\frac{{g}^{2}}{16\,{e}^{3}{d}^{2} \left ( ex+d \right ) }}+{\frac{fg}{8\,{e}^{2}{d}^{3} \left ( ex+d \right ) }}-{\frac{3\,{f}^{2}}{16\,e{d}^{4} \left ( ex+d \right ) }}-{\frac{{g}^{2}}{16\,{e}^{3}d \left ( ex+d \right ) ^{2}}}+{\frac{fg}{8\,{d}^{2}{e}^{2} \left ( ex+d \right ) ^{2}}}-{\frac{{f}^{2}}{16\,e{d}^{3} \left ( ex+d \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^2/(-e^2*x^2+d^2)^3,x)

[Out]

1/16/e^3/d^3*ln(e*x-d)*g^2-3/16/e/d^5*ln(e*x-d)*f^2+1/16/e^3/d/(e*x-d)^2*g^2+1/8/e^2/d^2/(e*x-d)^2*f*g+1/16/e/
d^3/(e*x-d)^2*f^2+1/16/e^3/d^2/(e*x-d)*g^2-1/8/e^2/d^3/(e*x-d)*f*g-3/16/e/d^4/(e*x-d)*f^2-1/16/e^3/d^3*ln(e*x+
d)*g^2+3/16/e/d^5*ln(e*x+d)*f^2+1/16/d^2/e^3/(e*x+d)*g^2+1/8/d^3/e^2/(e*x+d)*f*g-3/16/d^4/e/(e*x+d)*f^2-1/16/d
/e^3/(e*x+d)^2*g^2+1/8/d^2/e^2/(e*x+d)^2*f*g-1/16/d^3/e/(e*x+d)^2*f^2

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Maxima [A]  time = 0.987638, size = 205, normalized size = 1.61 \begin{align*} \frac{4 \, d^{4} f g -{\left (3 \, e^{4} f^{2} - d^{2} e^{2} g^{2}\right )} x^{3} +{\left (5 \, d^{2} e^{2} f^{2} + d^{4} g^{2}\right )} x}{8 \,{\left (d^{4} e^{6} x^{4} - 2 \, d^{6} e^{4} x^{2} + d^{8} e^{2}\right )}} + \frac{{\left (3 \, e^{2} f^{2} - d^{2} g^{2}\right )} \log \left (e x + d\right )}{16 \, d^{5} e^{3}} - \frac{{\left (3 \, e^{2} f^{2} - d^{2} g^{2}\right )} \log \left (e x - d\right )}{16 \, d^{5} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

1/8*(4*d^4*f*g - (3*e^4*f^2 - d^2*e^2*g^2)*x^3 + (5*d^2*e^2*f^2 + d^4*g^2)*x)/(d^4*e^6*x^4 - 2*d^6*e^4*x^2 + d
^8*e^2) + 1/16*(3*e^2*f^2 - d^2*g^2)*log(e*x + d)/(d^5*e^3) - 1/16*(3*e^2*f^2 - d^2*g^2)*log(e*x - d)/(d^5*e^3
)

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Fricas [B]  time = 1.66353, size = 475, normalized size = 3.74 \begin{align*} \frac{8 \, d^{5} e f g - 2 \,{\left (3 \, d e^{5} f^{2} - d^{3} e^{3} g^{2}\right )} x^{3} + 2 \,{\left (5 \, d^{3} e^{3} f^{2} + d^{5} e g^{2}\right )} x +{\left (3 \, d^{4} e^{2} f^{2} - d^{6} g^{2} +{\left (3 \, e^{6} f^{2} - d^{2} e^{4} g^{2}\right )} x^{4} - 2 \,{\left (3 \, d^{2} e^{4} f^{2} - d^{4} e^{2} g^{2}\right )} x^{2}\right )} \log \left (e x + d\right ) -{\left (3 \, d^{4} e^{2} f^{2} - d^{6} g^{2} +{\left (3 \, e^{6} f^{2} - d^{2} e^{4} g^{2}\right )} x^{4} - 2 \,{\left (3 \, d^{2} e^{4} f^{2} - d^{4} e^{2} g^{2}\right )} x^{2}\right )} \log \left (e x - d\right )}{16 \,{\left (d^{5} e^{7} x^{4} - 2 \, d^{7} e^{5} x^{2} + d^{9} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

1/16*(8*d^5*e*f*g - 2*(3*d*e^5*f^2 - d^3*e^3*g^2)*x^3 + 2*(5*d^3*e^3*f^2 + d^5*e*g^2)*x + (3*d^4*e^2*f^2 - d^6
*g^2 + (3*e^6*f^2 - d^2*e^4*g^2)*x^4 - 2*(3*d^2*e^4*f^2 - d^4*e^2*g^2)*x^2)*log(e*x + d) - (3*d^4*e^2*f^2 - d^
6*g^2 + (3*e^6*f^2 - d^2*e^4*g^2)*x^4 - 2*(3*d^2*e^4*f^2 - d^4*e^2*g^2)*x^2)*log(e*x - d))/(d^5*e^7*x^4 - 2*d^
7*e^5*x^2 + d^9*e^3)

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Sympy [A]  time = 1.13298, size = 143, normalized size = 1.13 \begin{align*} \frac{4 d^{4} f g + x^{3} \left (d^{2} e^{2} g^{2} - 3 e^{4} f^{2}\right ) + x \left (d^{4} g^{2} + 5 d^{2} e^{2} f^{2}\right )}{8 d^{8} e^{2} - 16 d^{6} e^{4} x^{2} + 8 d^{4} e^{6} x^{4}} + \frac{\left (d^{2} g^{2} - 3 e^{2} f^{2}\right ) \log{\left (- \frac{d}{e} + x \right )}}{16 d^{5} e^{3}} - \frac{\left (d^{2} g^{2} - 3 e^{2} f^{2}\right ) \log{\left (\frac{d}{e} + x \right )}}{16 d^{5} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**2/(-e**2*x**2+d**2)**3,x)

[Out]

(4*d**4*f*g + x**3*(d**2*e**2*g**2 - 3*e**4*f**2) + x*(d**4*g**2 + 5*d**2*e**2*f**2))/(8*d**8*e**2 - 16*d**6*e
**4*x**2 + 8*d**4*e**6*x**4) + (d**2*g**2 - 3*e**2*f**2)*log(-d/e + x)/(16*d**5*e**3) - (d**2*g**2 - 3*e**2*f*
*2)*log(d/e + x)/(16*d**5*e**3)

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Giac [A]  time = 1.13916, size = 171, normalized size = 1.35 \begin{align*} \frac{{\left (d^{2} g^{2} - 3 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (\frac{{\left | 2 \, x e^{2} - 2 \,{\left | d \right |} e \right |}}{{\left | 2 \, x e^{2} + 2 \,{\left | d \right |} e \right |}}\right )}{16 \, d^{4}{\left | d \right |}} + \frac{{\left (d^{2} g^{2} x^{3} e^{2} + d^{4} g^{2} x + 4 \, d^{4} f g - 3 \, f^{2} x^{3} e^{4} + 5 \, d^{2} f^{2} x e^{2}\right )} e^{\left (-2\right )}}{8 \,{\left (x^{2} e^{2} - d^{2}\right )}^{2} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="giac")

[Out]

1/16*(d^2*g^2 - 3*f^2*e^2)*e^(-3)*log(abs(2*x*e^2 - 2*abs(d)*e)/abs(2*x*e^2 + 2*abs(d)*e))/(d^4*abs(d)) + 1/8*
(d^2*g^2*x^3*e^2 + d^4*g^2*x + 4*d^4*f*g - 3*f^2*x^3*e^4 + 5*d^2*f^2*x*e^2)*e^(-2)/((x^2*e^2 - d^2)^2*d^4)