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

Optimal. Leaf size=61 \[ -\frac{2 g (d g+e f)}{e^3 (d-e x)}+\frac{(d g+e f)^2}{2 e^3 (d-e x)^2}-\frac{g^2 \log (d-e x)}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.0597248, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {848, 43} \[ -\frac{2 g (d g+e f)}{e^3 (d-e x)}+\frac{(d g+e f)^2}{2 e^3 (d-e x)^2}-\frac{g^2 \log (d-e x)}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.025836, size = 49, normalized size = 0.8 \[ \frac{\frac{(d g+e f) (e (f+4 g x)-3 d g)}{(d-e x)^2}-2 g^2 \log (d-e x)}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 105, normalized size = 1.7 \begin{align*} -{\frac{\ln \left ( ex-d \right ){g}^{2}}{{e}^{3}}}+{\frac{{d}^{2}{g}^{2}}{2\,{e}^{3} \left ( ex-d \right ) ^{2}}}+{\frac{dfg}{{e}^{2} \left ( ex-d \right ) ^{2}}}+{\frac{{f}^{2}}{2\,e \left ( ex-d \right ) ^{2}}}+2\,{\frac{{g}^{2}d}{{e}^{3} \left ( ex-d \right ) }}+2\,{\frac{fg}{{e}^{2} \left ( ex-d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00053, size = 109, normalized size = 1.79 \begin{align*} \frac{e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2} + 4 \,{\left (e^{2} f g + d e g^{2}\right )} x}{2 \,{\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac{g^{2} \log \left (e x - d\right )}{e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.72297, size = 205, normalized size = 3.36 \begin{align*} \frac{e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2} + 4 \,{\left (e^{2} f g + d e g^{2}\right )} x - 2 \,{\left (e^{2} g^{2} x^{2} - 2 \, d e g^{2} x + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \,{\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.708844, size = 80, normalized size = 1.31 \begin{align*} \frac{- 3 d^{2} g^{2} - 2 d e f g + e^{2} f^{2} + x \left (4 d e g^{2} + 4 e^{2} f g\right )}{2 d^{2} e^{3} - 4 d e^{4} x + 2 e^{5} x^{2}} - \frac{g^{2} \log{\left (- d + e x \right )}}{e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17624, size = 263, normalized size = 4.31 \begin{align*} -\frac{d g^{2} 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 )}{2 \,{\left | d \right |}} - \frac{1}{2} \, g^{2} e^{\left (-3\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) + \frac{{\left (4 \,{\left (d^{2} g^{2} e^{4} + d f g e^{5}\right )} x^{3} +{\left (5 \, d^{3} g^{2} e^{3} + 6 \, d^{2} f g e^{4} + d f^{2} e^{5}\right )} x^{2} - 2 \,{\left (d^{4} g^{2} e^{2} - d^{2} f^{2} e^{4}\right )} x -{\left (3 \, d^{5} g^{2} e^{3} + 2 \, d^{4} f g e^{4} - d^{3} f^{2} e^{5}\right )} e^{\left (-2\right )}\right )} e^{\left (-4\right )}}{2 \,{\left (x^{2} e^{2} - d^{2}\right )}^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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