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

Optimal. Leaf size=146 \[ -\frac{e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac{(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}+\frac{(d g+e f)^2}{16 d^4 e^3 (d-e x)}-\frac{(3 e f-d g) (d g+e f)}{16 d^4 e^3 (d+e x)}+\frac{f (d g+e f) \tanh ^{-1}\left (\frac{e x}{d}\right )}{4 d^5 e^2} \]

[Out]

(e*f + d*g)^2/(16*d^4*e^3*(d - e*x)) - (e*f - d*g)^2/(12*d^2*e^3*(d + e*x)^3) - (e^2*f^2 - d^2*g^2)/(8*d^3*e^3
*(d + e*x)^2) - ((3*e*f - d*g)*(e*f + d*g))/(16*d^4*e^3*(d + e*x)) + (f*(e*f + d*g)*ArcTanh[(e*x)/d])/(4*d^5*e
^2)

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Rubi [A]  time = 0.155293, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {848, 88, 208} \[ -\frac{e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac{(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}+\frac{(d g+e f)^2}{16 d^4 e^3 (d-e x)}-\frac{(3 e f-d g) (d g+e f)}{16 d^4 e^3 (d+e x)}+\frac{f (d g+e f) \tanh ^{-1}\left (\frac{e x}{d}\right )}{4 d^5 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*f + d*g)^2/(16*d^4*e^3*(d - e*x)) - (e*f - d*g)^2/(12*d^2*e^3*(d + e*x)^3) - (e^2*f^2 - d^2*g^2)/(8*d^3*e^3
*(d + e*x)^2) - ((3*e*f - d*g)*(e*f + d*g))/(16*d^4*e^3*(d + e*x)) + (f*(e*f + d*g)*ArcTanh[(e*x)/d])/(4*d^5*e
^2)

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx &=\int \frac{(f+g x)^2}{(d-e x)^2 (d+e x)^4} \, dx\\ &=\int \left (\frac{(e f+d g)^2}{16 d^4 e^2 (d-e x)^2}+\frac{(-e f+d g)^2}{4 d^2 e^2 (d+e x)^4}+\frac{e^2 f^2-d^2 g^2}{4 d^3 e^2 (d+e x)^3}+\frac{(3 e f-d g) (e f+d g)}{16 d^4 e^2 (d+e x)^2}+\frac{f (e f+d g)}{4 d^4 e \left (d^2-e^2 x^2\right )}\right ) \, dx\\ &=\frac{(e f+d g)^2}{16 d^4 e^3 (d-e x)}-\frac{(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}-\frac{e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac{(3 e f-d g) (e f+d g)}{16 d^4 e^3 (d+e x)}+\frac{(f (e f+d g)) \int \frac{1}{d^2-e^2 x^2} \, dx}{4 d^4 e}\\ &=\frac{(e f+d g)^2}{16 d^4 e^3 (d-e x)}-\frac{(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}-\frac{e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac{(3 e f-d g) (e f+d g)}{16 d^4 e^3 (d+e x)}+\frac{f (e f+d g) \tanh ^{-1}\left (\frac{e x}{d}\right )}{4 d^5 e^2}\\ \end{align*}

Mathematica [A]  time = 0.0990623, size = 171, normalized size = 1.17 \[ \frac{2 d \left (d^3 e^2 f (g x-4 f)+d^2 e^3 f x (f+6 g x)+2 d^4 e g (f+2 g x)+2 d^5 g^2+3 d e^4 f x^2 (2 f+g x)+3 e^5 f^2 x^3\right )+3 e f (e x-d) (d+e x)^3 (d g+e f) \log (d-e x)+3 e f (d-e x) (d+e x)^3 (d g+e f) \log (d+e x)}{24 d^5 e^3 (d-e x) (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(2*d^5*g^2 + 3*e^5*f^2*x^3 + d^3*e^2*f*(-4*f + g*x) + 3*d*e^4*f*x^2*(2*f + g*x) + 2*d^4*e*g*(f + 2*g*x) +
d^2*e^3*f*x*(f + 6*g*x)) + 3*e*f*(e*f + d*g)*(-d + e*x)*(d + e*x)^3*Log[d - e*x] + 3*e*f*(e*f + d*g)*(d - e*x
)*(d + e*x)^3*Log[d + e*x])/(24*d^5*e^3*(d - e*x)*(d + e*x)^3)

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Maple [A]  time = 0.056, size = 270, normalized size = 1.9 \begin{align*} -{\frac{{g}^{2}}{16\,{d}^{2}{e}^{3} \left ( ex-d \right ) }}-{\frac{fg}{8\,{e}^{2}{d}^{3} \left ( ex-d \right ) }}-{\frac{{f}^{2}}{16\,e{d}^{4} \left ( ex-d \right ) }}-{\frac{\ln \left ( ex-d \right ) fg}{8\,{e}^{2}{d}^{4}}}-{\frac{\ln \left ( ex-d \right ){f}^{2}}{8\,{d}^{5}e}}+{\frac{{g}^{2}}{8\,d{e}^{3} \left ( ex+d \right ) ^{2}}}-{\frac{{f}^{2}}{8\,e{d}^{3} \left ( ex+d \right ) ^{2}}}+{\frac{{g}^{2}}{16\,{d}^{2}{e}^{3} \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}}{12\,{e}^{3} \left ( ex+d \right ) ^{3}}}+{\frac{fg}{6\,d{e}^{2} \left ( ex+d \right ) ^{3}}}-{\frac{{f}^{2}}{12\,e{d}^{2} \left ( ex+d \right ) ^{3}}}+{\frac{\ln \left ( ex+d \right ) fg}{8\,{e}^{2}{d}^{4}}}+{\frac{\ln \left ( ex+d \right ){f}^{2}}{8\,{d}^{5}e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/e^3/d^2/(e*x-d)*g^2-1/8/e^2/d^3/(e*x-d)*f*g-1/16/e/d^4/(e*x-d)*f^2-1/8/e^2/d^4*ln(e*x-d)*f*g-1/8/e/d^5*l
n(e*x-d)*f^2+1/8/d/e^3/(e*x+d)^2*g^2-1/8/d^3/e/(e*x+d)^2*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^2-1/12/e^3/(e*x+d)^3*g^2+1/6/d/e^2/(e*x+d)^3*f*g-1/12/d^2/e/(e*x+d)^3*f^2+1/8/e^2/d^4*ln(
e*x+d)*f*g+1/8/e/d^5*ln(e*x+d)*f^2

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Maxima [A]  time = 1.01457, size = 266, normalized size = 1.82 \begin{align*} \frac{4 \, d^{3} e^{2} f^{2} - 2 \, d^{4} e f g - 2 \, d^{5} g^{2} - 3 \,{\left (e^{5} f^{2} + d e^{4} f g\right )} x^{3} - 6 \,{\left (d e^{4} f^{2} + d^{2} e^{3} f g\right )} x^{2} -{\left (d^{2} e^{3} f^{2} + d^{3} e^{2} f g + 4 \, d^{4} e g^{2}\right )} x}{12 \,{\left (d^{4} e^{7} x^{4} + 2 \, d^{5} e^{6} x^{3} - 2 \, d^{7} e^{4} x - d^{8} e^{3}\right )}} + \frac{{\left (e f^{2} + d f g\right )} \log \left (e x + d\right )}{8 \, d^{5} e^{2}} - \frac{{\left (e f^{2} + d f g\right )} \log \left (e x - d\right )}{8 \, d^{5} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*d^3*e^2*f^2 - 2*d^4*e*f*g - 2*d^5*g^2 - 3*(e^5*f^2 + d*e^4*f*g)*x^3 - 6*(d*e^4*f^2 + d^2*e^3*f*g)*x^2
- (d^2*e^3*f^2 + d^3*e^2*f*g + 4*d^4*e*g^2)*x)/(d^4*e^7*x^4 + 2*d^5*e^6*x^3 - 2*d^7*e^4*x - d^8*e^3) + 1/8*(e*
f^2 + d*f*g)*log(e*x + d)/(d^5*e^2) - 1/8*(e*f^2 + d*f*g)*log(e*x - d)/(d^5*e^2)

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Fricas [B]  time = 1.74351, size = 674, normalized size = 4.62 \begin{align*} \frac{8 \, d^{4} e^{2} f^{2} - 4 \, d^{5} e f g - 4 \, d^{6} g^{2} - 6 \,{\left (d e^{5} f^{2} + d^{2} e^{4} f g\right )} x^{3} - 12 \,{\left (d^{2} e^{4} f^{2} + d^{3} e^{3} f g\right )} x^{2} - 2 \,{\left (d^{3} e^{3} f^{2} + d^{4} e^{2} f g + 4 \, d^{5} e g^{2}\right )} x - 3 \,{\left (d^{4} e^{2} f^{2} + d^{5} e f g -{\left (e^{6} f^{2} + d e^{5} f g\right )} x^{4} - 2 \,{\left (d e^{5} f^{2} + d^{2} e^{4} f g\right )} x^{3} + 2 \,{\left (d^{3} e^{3} f^{2} + d^{4} e^{2} f g\right )} x\right )} \log \left (e x + d\right ) + 3 \,{\left (d^{4} e^{2} f^{2} + d^{5} e f g -{\left (e^{6} f^{2} + d e^{5} f g\right )} x^{4} - 2 \,{\left (d e^{5} f^{2} + d^{2} e^{4} f g\right )} x^{3} + 2 \,{\left (d^{3} e^{3} f^{2} + d^{4} e^{2} f g\right )} x\right )} \log \left (e x - d\right )}{24 \,{\left (d^{5} e^{7} x^{4} + 2 \, d^{6} e^{6} x^{3} - 2 \, d^{8} e^{4} x - d^{9} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*d^4*e^2*f^2 - 4*d^5*e*f*g - 4*d^6*g^2 - 6*(d*e^5*f^2 + d^2*e^4*f*g)*x^3 - 12*(d^2*e^4*f^2 + d^3*e^3*f*
g)*x^2 - 2*(d^3*e^3*f^2 + d^4*e^2*f*g + 4*d^5*e*g^2)*x - 3*(d^4*e^2*f^2 + d^5*e*f*g - (e^6*f^2 + d*e^5*f*g)*x^
4 - 2*(d*e^5*f^2 + d^2*e^4*f*g)*x^3 + 2*(d^3*e^3*f^2 + d^4*e^2*f*g)*x)*log(e*x + d) + 3*(d^4*e^2*f^2 + d^5*e*f
*g - (e^6*f^2 + d*e^5*f*g)*x^4 - 2*(d*e^5*f^2 + d^2*e^4*f*g)*x^3 + 2*(d^3*e^3*f^2 + d^4*e^2*f*g)*x)*log(e*x -
d))/(d^5*e^7*x^4 + 2*d^6*e^6*x^3 - 2*d^8*e^4*x - d^9*e^3)

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Sympy [A]  time = 1.62728, size = 236, normalized size = 1.62 \begin{align*} - \frac{2 d^{5} g^{2} + 2 d^{4} e f g - 4 d^{3} e^{2} f^{2} + x^{3} \left (3 d e^{4} f g + 3 e^{5} f^{2}\right ) + x^{2} \left (6 d^{2} e^{3} f g + 6 d e^{4} f^{2}\right ) + x \left (4 d^{4} e g^{2} + d^{3} e^{2} f g + d^{2} e^{3} f^{2}\right )}{- 12 d^{8} e^{3} - 24 d^{7} e^{4} x + 24 d^{5} e^{6} x^{3} + 12 d^{4} e^{7} x^{4}} - \frac{f \left (d g + e f\right ) \log{\left (- \frac{d f \left (d g + e f\right )}{e \left (d f g + e f^{2}\right )} + x \right )}}{8 d^{5} e^{2}} + \frac{f \left (d g + e f\right ) \log{\left (\frac{d f \left (d g + e f\right )}{e \left (d f g + e f^{2}\right )} + x \right )}}{8 d^{5} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*d**5*g**2 + 2*d**4*e*f*g - 4*d**3*e**2*f**2 + x**3*(3*d*e**4*f*g + 3*e**5*f**2) + x**2*(6*d**2*e**3*f*g +
6*d*e**4*f**2) + x*(4*d**4*e*g**2 + d**3*e**2*f*g + d**2*e**3*f**2))/(-12*d**8*e**3 - 24*d**7*e**4*x + 24*d**5
*e**6*x**3 + 12*d**4*e**7*x**4) - f*(d*g + e*f)*log(-d*f*(d*g + e*f)/(e*(d*f*g + e*f**2)) + x)/(8*d**5*e**2) +
f*(d*g + e*f)*log(d*f*(d*g + e*f)/(e*(d*f*g + e*f**2)) + x)/(8*d**5*e**2)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError