3.558 \(\int \frac{(d+e x)^6 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\)

Optimal. Leaf size=177 \[ \frac{1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac{d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac{d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac{16 d^4 (d g+e f)^2}{e^3 (d-e x)}+\frac{32 d^3 (d g+e f) (2 d g+e f) \log (d-e x)}{e^3}+\frac{1}{2} e g x^4 (3 d g+e f)+\frac{1}{5} e^2 g^2 x^5 \]

[Out]

(d^2*(17*e^2*f^2 + 64*d*e*f*g + 48*d^2*g^2)*x)/e^2 + (d*(3*e^2*f^2 + 17*d*e*f*g + 16*d^2*g^2)*x^2)/e + ((e^2*f
^2 + 12*d*e*f*g + 17*d^2*g^2)*x^3)/3 + (e*g*(e*f + 3*d*g)*x^4)/2 + (e^2*g^2*x^5)/5 + (16*d^4*(e*f + d*g)^2)/(e
^3*(d - e*x)) + (32*d^3*(e*f + d*g)*(e*f + 2*d*g)*Log[d - e*x])/e^3

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Rubi [A]  time = 0.232196, antiderivative size = 177, 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, 88} \[ \frac{1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac{d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac{d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac{16 d^4 (d g+e f)^2}{e^3 (d-e x)}+\frac{32 d^3 (d g+e f) (2 d g+e f) \log (d-e x)}{e^3}+\frac{1}{2} e g x^4 (3 d g+e f)+\frac{1}{5} e^2 g^2 x^5 \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^6*(f + g*x)^2)/(d^2 - e^2*x^2)^2,x]

[Out]

(d^2*(17*e^2*f^2 + 64*d*e*f*g + 48*d^2*g^2)*x)/e^2 + (d*(3*e^2*f^2 + 17*d*e*f*g + 16*d^2*g^2)*x^2)/e + ((e^2*f
^2 + 12*d*e*f*g + 17*d^2*g^2)*x^3)/3 + (e*g*(e*f + 3*d*g)*x^4)/2 + (e^2*g^2*x^5)/5 + (16*d^4*(e*f + d*g)^2)/(e
^3*(d - e*x)) + (32*d^3*(e*f + d*g)*(e*f + 2*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 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)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx &=\int \frac{(d+e x)^4 (f+g x)^2}{(d-e x)^2} \, dx\\ &=\int \left (\frac{d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right )}{e^2}+\frac{2 d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x}{e}+\left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^2+2 e g (e f+3 d g) x^3+e^2 g^2 x^4+\frac{32 d^3 (-e f-2 d g) (e f+d g)}{e^2 (d-e x)}+\frac{16 d^4 (e f+d g)^2}{e^2 (-d+e x)^2}\right ) \, dx\\ &=\frac{d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right ) x}{e^2}+\frac{d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x^2}{e}+\frac{1}{3} \left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^3+\frac{1}{2} e g (e f+3 d g) x^4+\frac{1}{5} e^2 g^2 x^5+\frac{16 d^4 (e f+d g)^2}{e^3 (d-e x)}+\frac{32 d^3 (e f+d g) (e f+2 d g) \log (d-e x)}{e^3}\\ \end{align*}

Mathematica [A]  time = 0.120351, size = 185, normalized size = 1.05 \[ \frac{1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac{d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac{d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac{32 d^3 \left (2 d^2 g^2+3 d e f g+e^2 f^2\right ) \log (d-e x)}{e^3}-\frac{16 d^4 (d g+e f)^2}{e^3 (e x-d)}+\frac{1}{2} e g x^4 (3 d g+e f)+\frac{1}{5} e^2 g^2 x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^6*(f + g*x)^2)/(d^2 - e^2*x^2)^2,x]

[Out]

(d^2*(17*e^2*f^2 + 64*d*e*f*g + 48*d^2*g^2)*x)/e^2 + (d*(3*e^2*f^2 + 17*d*e*f*g + 16*d^2*g^2)*x^2)/e + ((e^2*f
^2 + 12*d*e*f*g + 17*d^2*g^2)*x^3)/3 + (e*g*(e*f + 3*d*g)*x^4)/2 + (e^2*g^2*x^5)/5 - (16*d^4*(e*f + d*g)^2)/(e
^3*(-d + e*x)) + (32*d^3*(e^2*f^2 + 3*d*e*f*g + 2*d^2*g^2)*Log[d - e*x])/e^3

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Maple [A]  time = 0.049, size = 245, normalized size = 1.4 \begin{align*}{\frac{{e}^{2}{g}^{2}{x}^{5}}{5}}+{\frac{3\,e{x}^{4}d{g}^{2}}{2}}+{\frac{{e}^{2}{x}^{4}fg}{2}}+{\frac{17\,{x}^{3}{d}^{2}{g}^{2}}{3}}+4\,e{x}^{3}dfg+{\frac{{e}^{2}{x}^{3}{f}^{2}}{3}}+16\,{\frac{{x}^{2}{d}^{3}{g}^{2}}{e}}+17\,{x}^{2}{d}^{2}fg+3\,e{x}^{2}d{f}^{2}+48\,{\frac{{d}^{4}{g}^{2}x}{{e}^{2}}}+64\,{\frac{{d}^{3}fgx}{e}}+17\,{d}^{2}{f}^{2}x+64\,{\frac{{d}^{5}\ln \left ( ex-d \right ){g}^{2}}{{e}^{3}}}+96\,{\frac{{d}^{4}\ln \left ( ex-d \right ) fg}{{e}^{2}}}+32\,{\frac{{d}^{3}\ln \left ( ex-d \right ){f}^{2}}{e}}-16\,{\frac{{d}^{6}{g}^{2}}{{e}^{3} \left ( ex-d \right ) }}-32\,{\frac{{d}^{5}fg}{{e}^{2} \left ( ex-d \right ) }}-16\,{\frac{{d}^{4}{f}^{2}}{e \left ( ex-d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x)

[Out]

1/5*e^2*g^2*x^5+3/2*e*x^4*d*g^2+1/2*e^2*x^4*f*g+17/3*x^3*d^2*g^2+4*e*x^3*d*f*g+1/3*e^2*x^3*f^2+16/e*x^2*d^3*g^
2+17*x^2*d^2*f*g+3*e*x^2*d*f^2+48/e^2*d^4*g^2*x+64/e*d^3*f*g*x+17*d^2*f^2*x+64*d^5/e^3*ln(e*x-d)*g^2+96*d^4/e^
2*ln(e*x-d)*f*g+32*d^3/e*ln(e*x-d)*f^2-16*d^6/e^3/(e*x-d)*g^2-32*d^5/e^2/(e*x-d)*f*g-16*d^4/e/(e*x-d)*f^2

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Maxima [A]  time = 0.98052, size = 294, normalized size = 1.66 \begin{align*} -\frac{16 \,{\left (d^{4} e^{2} f^{2} + 2 \, d^{5} e f g + d^{6} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac{6 \, e^{4} g^{2} x^{5} + 15 \,{\left (e^{4} f g + 3 \, d e^{3} g^{2}\right )} x^{4} + 10 \,{\left (e^{4} f^{2} + 12 \, d e^{3} f g + 17 \, d^{2} e^{2} g^{2}\right )} x^{3} + 30 \,{\left (3 \, d e^{3} f^{2} + 17 \, d^{2} e^{2} f g + 16 \, d^{3} e g^{2}\right )} x^{2} + 30 \,{\left (17 \, d^{2} e^{2} f^{2} + 64 \, d^{3} e f g + 48 \, d^{4} g^{2}\right )} x}{30 \, e^{2}} + \frac{32 \,{\left (d^{3} e^{2} f^{2} + 3 \, d^{4} e f g + 2 \, d^{5} 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)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="maxima")

[Out]

-16*(d^4*e^2*f^2 + 2*d^5*e*f*g + d^6*g^2)/(e^4*x - d*e^3) + 1/30*(6*e^4*g^2*x^5 + 15*(e^4*f*g + 3*d*e^3*g^2)*x
^4 + 10*(e^4*f^2 + 12*d*e^3*f*g + 17*d^2*e^2*g^2)*x^3 + 30*(3*d*e^3*f^2 + 17*d^2*e^2*f*g + 16*d^3*e*g^2)*x^2 +
30*(17*d^2*e^2*f^2 + 64*d^3*e*f*g + 48*d^4*g^2)*x)/e^2 + 32*(d^3*e^2*f^2 + 3*d^4*e*f*g + 2*d^5*g^2)*log(e*x -
d)/e^3

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Fricas [A]  time = 1.76268, size = 612, normalized size = 3.46 \begin{align*} \frac{6 \, e^{6} g^{2} x^{6} - 480 \, d^{4} e^{2} f^{2} - 960 \, d^{5} e f g - 480 \, d^{6} g^{2} + 3 \,{\left (5 \, e^{6} f g + 13 \, d e^{5} g^{2}\right )} x^{5} + 5 \,{\left (2 \, e^{6} f^{2} + 21 \, d e^{5} f g + 25 \, d^{2} e^{4} g^{2}\right )} x^{4} + 10 \,{\left (8 \, d e^{5} f^{2} + 39 \, d^{2} e^{4} f g + 31 \, d^{3} e^{3} g^{2}\right )} x^{3} + 30 \,{\left (14 \, d^{2} e^{4} f^{2} + 47 \, d^{3} e^{3} f g + 32 \, d^{4} e^{2} g^{2}\right )} x^{2} - 30 \,{\left (17 \, d^{3} e^{3} f^{2} + 64 \, d^{4} e^{2} f g + 48 \, d^{5} e g^{2}\right )} x - 960 \,{\left (d^{4} e^{2} f^{2} + 3 \, d^{5} e f g + 2 \, d^{6} g^{2} -{\left (d^{3} e^{3} f^{2} + 3 \, d^{4} e^{2} f g + 2 \, d^{5} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{30 \,{\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)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="fricas")

[Out]

1/30*(6*e^6*g^2*x^6 - 480*d^4*e^2*f^2 - 960*d^5*e*f*g - 480*d^6*g^2 + 3*(5*e^6*f*g + 13*d*e^5*g^2)*x^5 + 5*(2*
e^6*f^2 + 21*d*e^5*f*g + 25*d^2*e^4*g^2)*x^4 + 10*(8*d*e^5*f^2 + 39*d^2*e^4*f*g + 31*d^3*e^3*g^2)*x^3 + 30*(14
*d^2*e^4*f^2 + 47*d^3*e^3*f*g + 32*d^4*e^2*g^2)*x^2 - 30*(17*d^3*e^3*f^2 + 64*d^4*e^2*f*g + 48*d^5*e*g^2)*x -
960*(d^4*e^2*f^2 + 3*d^5*e*f*g + 2*d^6*g^2 - (d^3*e^3*f^2 + 3*d^4*e^2*f*g + 2*d^5*e*g^2)*x)*log(e*x - d))/(e^4
*x - d*e^3)

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Sympy [A]  time = 1.08256, size = 204, normalized size = 1.15 \begin{align*} \frac{32 d^{3} \left (d g + e f\right ) \left (2 d g + e f\right ) \log{\left (- d + e x \right )}}{e^{3}} + \frac{e^{2} g^{2} x^{5}}{5} + x^{4} \left (\frac{3 d e g^{2}}{2} + \frac{e^{2} f g}{2}\right ) + x^{3} \left (\frac{17 d^{2} g^{2}}{3} + 4 d e f g + \frac{e^{2} f^{2}}{3}\right ) - \frac{16 d^{6} g^{2} + 32 d^{5} e f g + 16 d^{4} e^{2} f^{2}}{- d e^{3} + e^{4} x} + \frac{x^{2} \left (16 d^{3} g^{2} + 17 d^{2} e f g + 3 d e^{2} f^{2}\right )}{e} + \frac{x \left (48 d^{4} g^{2} + 64 d^{3} e f g + 17 d^{2} e^{2} f^{2}\right )}{e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**6*(g*x+f)**2/(-e**2*x**2+d**2)**2,x)

[Out]

32*d**3*(d*g + e*f)*(2*d*g + e*f)*log(-d + e*x)/e**3 + e**2*g**2*x**5/5 + x**4*(3*d*e*g**2/2 + e**2*f*g/2) + x
**3*(17*d**2*g**2/3 + 4*d*e*f*g + e**2*f**2/3) - (16*d**6*g**2 + 32*d**5*e*f*g + 16*d**4*e**2*f**2)/(-d*e**3 +
e**4*x) + x**2*(16*d**3*g**2 + 17*d**2*e*f*g + 3*d*e**2*f**2)/e + x*(48*d**4*g**2 + 64*d**3*e*f*g + 17*d**2*e
**2*f**2)/e**2

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Giac [A]  time = 1.18565, size = 441, normalized size = 2.49 \begin{align*} 16 \,{\left (2 \, d^{5} g^{2} e^{5} + 3 \, d^{4} f g e^{6} + d^{3} f^{2} e^{7}\right )} e^{\left (-8\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) + \frac{1}{30} \,{\left (6 \, g^{2} x^{5} e^{22} + 45 \, d g^{2} x^{4} e^{21} + 170 \, d^{2} g^{2} x^{3} e^{20} + 480 \, d^{3} g^{2} x^{2} e^{19} + 1440 \, d^{4} g^{2} x e^{18} + 15 \, f g x^{4} e^{22} + 120 \, d f g x^{3} e^{21} + 510 \, d^{2} f g x^{2} e^{20} + 1920 \, d^{3} f g x e^{19} + 10 \, f^{2} x^{3} e^{22} + 90 \, d f^{2} x^{2} e^{21} + 510 \, d^{2} f^{2} x e^{20}\right )} e^{\left (-20\right )} + \frac{16 \,{\left (2 \, d^{6} g^{2} e^{6} + 3 \, d^{5} f g e^{7} + d^{4} f^{2} e^{8}\right )} e^{\left (-9\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 )}{{\left | d \right |}} - \frac{16 \,{\left (d^{7} g^{2} e^{5} + 2 \, d^{6} f g e^{6} + d^{5} f^{2} e^{7} +{\left (d^{6} g^{2} e^{6} + 2 \, d^{5} f g e^{7} + d^{4} f^{2} e^{8}\right )} x\right )} e^{\left (-8\right )}}{x^{2} e^{2} - d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="giac")

[Out]

16*(2*d^5*g^2*e^5 + 3*d^4*f*g*e^6 + d^3*f^2*e^7)*e^(-8)*log(abs(x^2*e^2 - d^2)) + 1/30*(6*g^2*x^5*e^22 + 45*d*
g^2*x^4*e^21 + 170*d^2*g^2*x^3*e^20 + 480*d^3*g^2*x^2*e^19 + 1440*d^4*g^2*x*e^18 + 15*f*g*x^4*e^22 + 120*d*f*g
*x^3*e^21 + 510*d^2*f*g*x^2*e^20 + 1920*d^3*f*g*x*e^19 + 10*f^2*x^3*e^22 + 90*d*f^2*x^2*e^21 + 510*d^2*f^2*x*e
^20)*e^(-20) + 16*(2*d^6*g^2*e^6 + 3*d^5*f*g*e^7 + d^4*f^2*e^8)*e^(-9)*log(abs(2*x*e^2 - 2*abs(d)*e)/abs(2*x*e
^2 + 2*abs(d)*e))/abs(d) - 16*(d^7*g^2*e^5 + 2*d^6*f*g*e^6 + d^5*f^2*e^7 + (d^6*g^2*e^6 + 2*d^5*f*g*e^7 + d^4*
f^2*e^8)*x)*e^(-8)/(x^2*e^2 - d^2)