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

Optimal. Leaf size=78 \[ \frac{2 d (d g+e f)^2}{e^3 (d-e x)}+\frac{g x (3 d g+2 e f)}{e^2}+\frac{(5 d g+e f) (d g+e f) \log (d-e x)}{e^3}+\frac{g^2 x^2}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.0961631, antiderivative size = 78, 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, 77} \[ \frac{2 d (d g+e f)^2}{e^3 (d-e x)}+\frac{g x (3 d g+2 e f)}{e^2}+\frac{(5 d g+e f) (d g+e f) \log (d-e x)}{e^3}+\frac{g^2 x^2}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(g*(2*e*f + 3*d*g)*x)/e^2 + (g^2*x^2)/(2*e) + (2*d*(e*f + d*g)^2)/(e^3*(d - e*x)) + ((e*f + d*g)*(e*f + 5*d*g)
*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0633919, size = 83, normalized size = 1.06 \[ \frac{2 \left (5 d^2 g^2+6 d e f g+e^2 f^2\right ) \log (d-e x)+\frac{4 d (d g+e f)^2}{d-e x}+2 e g x (3 d g+2 e f)+e^2 g^2 x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 138, normalized size = 1.8 \begin{align*}{\frac{{g}^{2}{x}^{2}}{2\,e}}+3\,{\frac{d{g}^{2}x}{{e}^{2}}}+2\,{\frac{fgx}{e}}+5\,{\frac{\ln \left ( ex-d \right ){d}^{2}{g}^{2}}{{e}^{3}}}+6\,{\frac{\ln \left ( ex-d \right ) dfg}{{e}^{2}}}+{\frac{\ln \left ( ex-d \right ){f}^{2}}{e}}-2\,{\frac{{d}^{3}{g}^{2}}{{e}^{3} \left ( ex-d \right ) }}-4\,{\frac{{d}^{2}fg}{{e}^{2} \left ( ex-d \right ) }}-2\,{\frac{d{f}^{2}}{e \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.00677, size = 140, normalized size = 1.79 \begin{align*} -\frac{2 \,{\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac{e g^{2} x^{2} + 2 \,{\left (2 \, e f g + 3 \, d g^{2}\right )} x}{2 \, e^{2}} + \frac{{\left (e^{2} f^{2} + 6 \, d e f g + 5 \, d^{2} g^{2}\right )} \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.71681, size = 321, normalized size = 4.12 \begin{align*} \frac{e^{3} g^{2} x^{3} - 4 \, d e^{2} f^{2} - 8 \, d^{2} e f g - 4 \, d^{3} g^{2} +{\left (4 \, e^{3} f g + 5 \, d e^{2} g^{2}\right )} x^{2} - 2 \,{\left (2 \, d e^{2} f g + 3 \, d^{2} e g^{2}\right )} x - 2 \,{\left (d e^{2} f^{2} + 6 \, d^{2} e f g + 5 \, d^{3} g^{2} -{\left (e^{3} f^{2} + 6 \, d e^{2} f g + 5 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{2 \,{\left (e^{4} x - d 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.723033, size = 92, normalized size = 1.18 \begin{align*} - \frac{2 d^{3} g^{2} + 4 d^{2} e f g + 2 d e^{2} f^{2}}{- d e^{3} + e^{4} x} + \frac{g^{2} x^{2}}{2 e} + \frac{x \left (3 d g^{2} + 2 e f g\right )}{e^{2}} + \frac{\left (d g + e f\right ) \left (5 d g + e f\right ) \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)**2,x)

[Out]

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

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Giac [B]  time = 1.24317, size = 286, normalized size = 3.67 \begin{align*} \frac{1}{2} \,{\left (5 \, d^{2} g^{2} e^{3} + 6 \, d f g e^{4} + f^{2} e^{5}\right )} e^{\left (-6\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) + \frac{1}{2} \,{\left (g^{2} x^{2} e^{7} + 6 \, d g^{2} x e^{6} + 4 \, f g x e^{7}\right )} e^{\left (-8\right )} + \frac{{\left (5 \, d^{3} g^{2} e^{2} + 6 \, d^{2} f g e^{3} + d f^{2} e^{4}\right )} e^{\left (-5\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{2 \,{\left (d^{4} g^{2} e^{3} + 2 \, d^{3} f g e^{4} + d^{2} f^{2} e^{5} +{\left (d^{3} g^{2} e^{4} + 2 \, d^{2} f g e^{5} + d f^{2} e^{6}\right )} x\right )} e^{\left (-6\right )}}{x^{2} e^{2} - d^{2}} \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)^2,x, algorithm="giac")

[Out]

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