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

Optimal. Leaf size=86 \[ \frac{(e f-d g)^2 \log (d+e x)}{4 d^2 e^3}-\frac{(e f-3 d g) (d g+e f) \log (d-e x)}{4 d^2 e^3}+\frac{(d g+e f)^2}{2 d e^3 (d-e x)} \]

[Out]

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

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Rubi [A]  time = 0.0815934, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {799, 88} \[ \frac{(e f-d g)^2 \log (d+e x)}{4 d^2 e^3}-\frac{(e f-3 d g) (d g+e f) \log (d-e x)}{4 d^2 e^3}+\frac{(d g+e f)^2}{2 d e^3 (d-e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 799

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^m*(
f + g*x)^(p + 1)*(a/f + (c*x)/g)^p, x] /; FreeQ[{a, c, d, e, f, g, m}, x] && EqQ[c*f^2 + a*g^2, 0] && (Integer
Q[p] || (GtQ[a, 0] && GtQ[f, 0] && EqQ[p, -1]))

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]))

Rubi steps

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

Mathematica [A]  time = 0.0483859, size = 91, normalized size = 1.06 \[ \frac{(d-e x) \left (3 d^2 g^2+2 d e f g-e^2 f^2\right ) \log (d-e x)+(d-e x) (e f-d g)^2 \log (d+e x)+2 d (d g+e f)^2}{4 d^2 e^3 (d-e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.09, size = 156, normalized size = 1.8 \begin{align*} -{\frac{{g}^{2}d}{2\,{e}^{3} \left ( ex-d \right ) }}-{\frac{fg}{{e}^{2} \left ( ex-d \right ) }}-{\frac{{f}^{2}}{2\,de \left ( ex-d \right ) }}+{\frac{3\,\ln \left ( ex-d \right ){g}^{2}}{4\,{e}^{3}}}+{\frac{\ln \left ( ex-d \right ) fg}{2\,d{e}^{2}}}-{\frac{\ln \left ( ex-d \right ){f}^{2}}{4\,{d}^{2}e}}+{\frac{\ln \left ( ex+d \right ){g}^{2}}{4\,{e}^{3}}}-{\frac{\ln \left ( ex+d \right ) fg}{2\,d{e}^{2}}}+{\frac{\ln \left ( ex+d \right ){f}^{2}}{4\,{d}^{2}e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.975871, size = 154, normalized size = 1.79 \begin{align*} -\frac{e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}}{2 \,{\left (d e^{4} x - d^{2} e^{3}\right )}} + \frac{{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{4 \, d^{2} e^{3}} - \frac{{\left (e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2}\right )} \log \left (e x - d\right )}{4 \, d^{2} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.71683, size = 343, normalized size = 3.99 \begin{align*} -\frac{2 \, d e^{2} f^{2} + 4 \, d^{2} e f g + 2 \, d^{3} g^{2} +{\left (d e^{2} f^{2} - 2 \, d^{2} e f g + d^{3} g^{2} -{\left (e^{3} f^{2} - 2 \, d e^{2} f g + d^{2} e g^{2}\right )} x\right )} \log \left (e x + d\right ) -{\left (d e^{2} f^{2} - 2 \, d^{2} e f g - 3 \, d^{3} g^{2} -{\left (e^{3} f^{2} - 2 \, d e^{2} f g - 3 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{4 \,{\left (d^{2} e^{4} x - d^{3} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.14259, size = 180, normalized size = 2.09 \begin{align*} - \frac{d^{2} g^{2} + 2 d e f g + e^{2} f^{2}}{- 2 d^{2} e^{3} + 2 d e^{4} x} + \frac{\left (d g - e f\right )^{2} \log{\left (x + \frac{2 d^{3} g^{2} - d \left (d g - e f\right )^{2}}{d^{2} e g^{2} + 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} + \frac{\left (d g + e f\right ) \left (3 d g - e f\right ) \log{\left (x + \frac{2 d^{3} g^{2} - d \left (d g + e f\right ) \left (3 d g - e f\right )}{d^{2} e g^{2} + 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d**2*g**2 + 2*d*e*f*g + e**2*f**2)/(-2*d**2*e**3 + 2*d*e**4*x) + (d*g - e*f)**2*log(x + (2*d**3*g**2 - d*(d*
g - e*f)**2)/(d**2*e*g**2 + 2*d*e**2*f*g - e**3*f**2))/(4*d**2*e**3) + (d*g + e*f)*(3*d*g - e*f)*log(x + (2*d*
*3*g**2 - d*(d*g + e*f)*(3*d*g - e*f))/(d**2*e*g**2 + 2*d*e**2*f*g - e**3*f**2))/(4*d**2*e**3)

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Giac [A]  time = 1.14363, size = 215, normalized size = 2.5 \begin{align*} \frac{1}{2} \, g^{2} e^{\left (-3\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) + \frac{{\left (d^{2} g^{2} + 2 \, d f g e - 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 )}{4 \, d{\left | d \right |}} - \frac{{\left ({\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} x +{\left (d^{3} g^{2} e + 2 \, d^{2} f g e^{2} + d f^{2} e^{3}\right )} e^{\left (-2\right )}\right )} e^{\left (-2\right )}}{2 \,{\left (x^{2} e^{2} - d^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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