\(\int \frac {(d+e x)^6 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\) [558]

Optimal result

Integrand size = 29, antiderivative size = 177 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right ) x}{e^2}+\frac {d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x^2}{e}+\frac {1}{3} \left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^3+\frac {1}{2} e g (e f+3 d g) x^4+\frac {1}{5} e^2 g^2 x^5+\frac {16 d^4 (e f+d g)^2}{e^3 (d-e x)}+\frac {32 d^3 (e f+d g) (e f+2 d g) \log (d-e x)}{e^3} \]

[Out]

d^2*(48*d^2*g^2+64*d*e*f*g+17*e^2*f^2)*x/e^2+d*(16*d^2*g^2+17*d*e*f*g+3*e^2*f^2)*x^2/e+1/3*(17*d^2*g^2+12*d*e*
f*g+e^2*f^2)*x^3+1/2*e*g*(3*d*g+e*f)*x^4+1/5*e^2*g^2*x^5+16*d^4*(d*g+e*f)^2/e^3/(-e*x+d)+32*d^3*(d*g+e*f)*(2*d
*g+e*f)*ln(-e*x+d)/e^3

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {862, 90} \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {16 d^4 (d g+e f)^2}{e^3 (d-e x)}+\frac {32 d^3 (d g+e f) (2 d g+e f) \log (d-e x)}{e^3}+\frac {1}{3} x^3 \left (17 d^2 g^2+12 d e f g+e^2 f^2\right )+\frac {d x^2 \left (16 d^2 g^2+17 d e f g+3 e^2 f^2\right )}{e}+\frac {d^2 x \left (48 d^2 g^2+64 d e f g+17 e^2 f^2\right )}{e^2}+\frac {1}{2} e g x^4 (3 d g+e f)+\frac {1}{5} e^2 g^2 x^5 \]

[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 90

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 862

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/e)*x)^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]))

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.07 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d^2 \left (17 e^2 f^2+64 d e f g+48 d^2 g^2\right ) x}{e^2}+\frac {d \left (3 e^2 f^2+17 d e f g+16 d^2 g^2\right ) x^2}{e}+\frac {1}{3} \left (e^2 f^2+12 d e f g+17 d^2 g^2\right ) x^3+\frac {1}{2} e g (e f+3 d g) x^4+\frac {1}{5} e^2 g^2 x^5-\frac {16 d^4 (e f+d g)^2}{e^3 (-d+e x)}+\frac {32 d^3 \left (e^2 f^2+3 d e f g+2 d^2 g^2\right ) \log (d-e x)}{e^3} \]

[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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {\frac {1}{5} g^{2} e^{4} x^{5}+\frac {3}{2} d \,e^{3} g^{2} x^{4}+\frac {1}{2} e^{4} f g \,x^{4}+\frac {17}{3} d^{2} e^{2} g^{2} x^{3}+4 d \,e^{3} f g \,x^{3}+\frac {1}{3} e^{4} f^{2} x^{3}+16 d^{3} e \,g^{2} x^{2}+17 d^{2} e^{2} f g \,x^{2}+3 d \,e^{3} f^{2} x^{2}+48 d^{4} g^{2} x +64 d^{3} e f g x +17 d^{2} e^{2} f^{2} x}{e^{2}}+\frac {32 d^{3} \left (2 d^{2} g^{2}+3 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {16 d^{4} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}\)	\(217\)
risch	\(\frac {e^{2} g^{2} x^{5}}{5}+\frac {3 e d \,g^{2} x^{4}}{2}+\frac {e^{2} f g \,x^{4}}{2}+\frac {17 d^{2} g^{2} x^{3}}{3}+4 e d f g \,x^{3}+\frac {e^{2} f^{2} x^{3}}{3}+\frac {16 d^{3} g^{2} x^{2}}{e}+17 d^{2} f g \,x^{2}+3 e d \,f^{2} x^{2}+\frac {48 d^{4} g^{2} x}{e^{2}}+\frac {64 d^{3} f g x}{e}+17 d^{2} f^{2} x +\frac {64 d^{5} \ln \left (-e x +d \right ) g^{2}}{e^{3}}+\frac {96 d^{4} \ln \left (-e x +d \right ) f g}{e^{2}}+\frac {32 d^{3} \ln \left (-e x +d \right ) f^{2}}{e}+\frac {16 d^{6} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {32 d^{5} f g}{e^{2} \left (-e x +d \right )}+\frac {16 d^{4} f^{2}}{e \left (-e x +d \right )}\)	\(239\)
norman	\(\frac {\left (-\frac {127}{3} d^{4} g^{2}-60 f g e \,d^{3}-\frac {50}{3} d^{2} e^{2} f^{2}\right ) x^{3}+\left (-\frac {82}{15} d^{2} g^{2} e^{2}-4 d f g \,e^{3}-\frac {1}{3} f^{2} e^{4}\right ) x^{5}+\left (-\frac {29}{2} g^{2} e \,d^{3}-\frac {33}{2} e^{2} f g \,d^{2}-3 e^{3} f^{2} d \right ) x^{4}+\frac {d^{2} \left (32 g^{2} d^{5}+49 f g \,d^{4} e +19 f^{2} d^{3} e^{2}\right )}{e^{3}}+\frac {d^{4} \left (64 d^{2} g^{2}+96 d e f g +33 e^{2} f^{2}\right ) x}{e^{2}}-\frac {g^{2} e^{4} x^{7}}{5}-\frac {e^{3} g \left (3 d g +e f \right ) x^{6}}{2}}{-e^{2} x^{2}+d^{2}}+\frac {32 d^{3} \left (2 d^{2} g^{2}+3 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(246\)
parallelrisch	\(\frac {6 g^{2} e^{6} x^{6}+39 x^{5} d \,e^{5} g^{2}+15 x^{5} e^{6} f g +125 x^{4} d^{2} e^{4} g^{2}+105 d \,e^{5} f g \,x^{4}+10 e^{6} f^{2} x^{4}+310 d^{3} e^{3} g^{2} x^{3}+390 d^{2} e^{4} f g \,x^{3}+80 d \,e^{5} f^{2} x^{3}+1920 \ln \left (e x -d \right ) x \,d^{5} e \,g^{2}+2880 \ln \left (e x -d \right ) x \,d^{4} e^{2} f g +960 \ln \left (e x -d \right ) x \,d^{3} e^{3} f^{2}+960 d^{4} e^{2} g^{2} x^{2}+1410 d^{3} e^{3} f g \,x^{2}+420 d^{2} e^{4} f^{2} x^{2}-1920 \ln \left (e x -d \right ) d^{6} g^{2}-2880 \ln \left (e x -d \right ) d^{5} e f g -960 \ln \left (e x -d \right ) d^{4} e^{2} f^{2}-1920 g^{2} d^{6}-2880 f g e \,d^{5}-990 e^{2} f^{2} d^{4}}{30 e^{3} \left (e x -d \right )}\)	\(300\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.63 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {6 \, e^{6} g^{2} x^{6} - 480 \, d^{4} e^{2} f^{2} - 960 \, d^{5} e f g - 480 \, d^{6} g^{2} + 3 \, {\left (5 \, e^{6} f g + 13 \, d e^{5} g^{2}\right )} x^{5} + 5 \, {\left (2 \, e^{6} f^{2} + 21 \, d e^{5} f g + 25 \, d^{2} e^{4} g^{2}\right )} x^{4} + 10 \, {\left (8 \, d e^{5} f^{2} + 39 \, d^{2} e^{4} f g + 31 \, d^{3} e^{3} g^{2}\right )} x^{3} + 30 \, {\left (14 \, d^{2} e^{4} f^{2} + 47 \, d^{3} e^{3} f g + 32 \, d^{4} e^{2} g^{2}\right )} x^{2} - 30 \, {\left (17 \, d^{3} e^{3} f^{2} + 64 \, d^{4} e^{2} f g + 48 \, d^{5} e g^{2}\right )} x - 960 \, {\left (d^{4} e^{2} f^{2} + 3 \, d^{5} e f g + 2 \, d^{6} g^{2} - {\left (d^{3} e^{3} f^{2} + 3 \, d^{4} e^{2} f g + 2 \, d^{5} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{30 \, {\left (e^{4} x - d e^{3}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {32 d^{3} \left (d g + e f\right ) \left (2 d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} + \frac {e^{2} g^{2} x^{5}}{5} + x^{4} \cdot \left (\frac {3 d e g^{2}}{2} + \frac {e^{2} f g}{2}\right ) + x^{3} \cdot \left (\frac {17 d^{2} g^{2}}{3} + 4 d e f g + \frac {e^{2} f^{2}}{3}\right ) + x^{2} \cdot \left (\frac {16 d^{3} g^{2}}{e} + 17 d^{2} f g + 3 d e f^{2}\right ) + x \left (\frac {48 d^{4} g^{2}}{e^{2}} + \frac {64 d^{3} f g}{e} + 17 d^{2} f^{2}\right ) + \frac {- 16 d^{6} g^{2} - 32 d^{5} e f g - 16 d^{4} e^{2} f^{2}}{- d e^{3} + e^{4} x} \]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.23 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=-\frac {16 \, {\left (d^{4} e^{2} f^{2} + 2 \, d^{5} e f g + d^{6} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac {6 \, e^{4} g^{2} x^{5} + 15 \, {\left (e^{4} f g + 3 \, d e^{3} g^{2}\right )} x^{4} + 10 \, {\left (e^{4} f^{2} + 12 \, d e^{3} f g + 17 \, d^{2} e^{2} g^{2}\right )} x^{3} + 30 \, {\left (3 \, d e^{3} f^{2} + 17 \, d^{2} e^{2} f g + 16 \, d^{3} e g^{2}\right )} x^{2} + 30 \, {\left (17 \, d^{2} e^{2} f^{2} + 64 \, d^{3} e f g + 48 \, d^{4} g^{2}\right )} x}{30 \, e^{2}} + \frac {32 \, {\left (d^{3} e^{2} f^{2} + 3 \, d^{4} e f g + 2 \, d^{5} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {32 \, {\left (d^{3} e^{2} f^{2} + 3 \, d^{4} e f g + 2 \, d^{5} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {16 \, {\left (d^{4} e^{2} f^{2} + 2 \, d^{5} e f g + d^{6} g^{2}\right )}}{{\left (e x - d\right )} e^{3}} + \frac {6 \, e^{12} g^{2} x^{5} + 15 \, e^{12} f g x^{4} + 45 \, d e^{11} g^{2} x^{4} + 10 \, e^{12} f^{2} x^{3} + 120 \, d e^{11} f g x^{3} + 170 \, d^{2} e^{10} g^{2} x^{3} + 90 \, d e^{11} f^{2} x^{2} + 510 \, d^{2} e^{10} f g x^{2} + 480 \, d^{3} e^{9} g^{2} x^{2} + 510 \, d^{2} e^{10} f^{2} x + 1920 \, d^{3} e^{9} f g x + 1440 \, d^{4} e^{8} g^{2} x}{30 \, e^{10}} \]

[In]

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

[Out]

32*(d^3*e^2*f^2 + 3*d^4*e*f*g + 2*d^5*g^2)*log(abs(e*x - d))/e^3 - 16*(d^4*e^2*f^2 + 2*d^5*e*f*g + d^6*g^2)/((
e*x - d)*e^3) + 1/30*(6*e^12*g^2*x^5 + 15*e^12*f*g*x^4 + 45*d*e^11*g^2*x^4 + 10*e^12*f^2*x^3 + 120*d*e^11*f*g*
x^3 + 170*d^2*e^10*g^2*x^3 + 90*d*e^11*f^2*x^2 + 510*d^2*e^10*f*g*x^2 + 480*d^3*e^9*g^2*x^2 + 510*d^2*e^10*f^2
*x + 1920*d^3*e^9*f*g*x + 1440*d^4*e^8*g^2*x)/e^10

Mupad [B] (verification not implemented)

Time = 11.85 (sec) , antiderivative size = 565, normalized size of antiderivative = 3.19 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=x^2\,\left (\frac {2\,d\,\left (d^2\,g^2+3\,d\,e\,f\,g+e^2\,f^2\right )}{e}-\frac {d^2\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{2\,e^2}+\frac {d\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e}\right )+x^4\,\left (\frac {e\,g\,\left (2\,d\,g+e\,f\right )}{2}+\frac {d\,e\,g^2}{2}\right )+x\,\left (\frac {d^4\,g^2+8\,d^3\,e\,f\,g+6\,d^2\,e^2\,f^2}{e^2}-\frac {d^2\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e^2}+\frac {2\,d\,\left (\frac {4\,d\,\left (d^2\,g^2+3\,d\,e\,f\,g+e^2\,f^2\right )}{e}-\frac {d^2\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e^2}+\frac {2\,d\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^2}-d^2\,g^2+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{e}\right )}{e}\right )}{e}\right )+x^3\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{3\,e^2}-\frac {d^2\,g^2}{3}+\frac {2\,d\,\left (2\,e\,g\,\left (2\,d\,g+e\,f\right )+2\,d\,e\,g^2\right )}{3\,e}\right )+\frac {\ln \left (e\,x-d\right )\,\left (64\,d^5\,g^2+96\,d^4\,e\,f\,g+32\,d^3\,e^2\,f^2\right )}{e^3}+\frac {16\,\left (d^6\,g^2+2\,d^5\,e\,f\,g+d^4\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {e^2\,g^2\,x^5}{5} \]

[In]

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

[Out]

x^2*((2*d*(d^2*g^2 + e^2*f^2 + 3*d*e*f*g))/e - (d^2*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/(2*e^2) + (d*((e^4*f^2
+ 6*d^2*e^2*g^2 + 8*d*e^3*f*g)/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e) + x^4*((e*g*(2*d
*g + e*f))/2 + (d*e*g^2)/2) + x*((d^4*g^2 + 6*d^2*e^2*f^2 + 8*d^3*e*f*g)/e^2 - (d^2*((e^4*f^2 + 6*d^2*e^2*g^2
+ 8*d*e^3*f*g)/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e^2 + (2*d*((4*d*(d^2*g^2 + e^2*f^2
+ 3*d*e*f*g))/e - (d^2*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e^2 + (2*d*((e^4*f^2 + 6*d^2*e^2*g^2 + 8*d*e^3*f*g)
/e^2 - d^2*g^2 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/e))/e))/e) + x^3*((e^4*f^2 + 6*d^2*e^2*g^2 + 8*d*e^3*
f*g)/(3*e^2) - (d^2*g^2)/3 + (2*d*(2*e*g*(2*d*g + e*f) + 2*d*e*g^2))/(3*e)) + (log(e*x - d)*(64*d^5*g^2 + 32*d
^3*e^2*f^2 + 96*d^4*e*f*g))/e^3 + (16*(d^6*g^2 + d^4*e^2*f^2 + 2*d^5*e*f*g))/(e*(d*e^2 - e^3*x)) + (e^2*g^2*x^
5)/5