\(\int \frac {(d+e x)^4 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\) [560]

Optimal result

Integrand size = 29, antiderivative size = 107 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {\left (e^2 f^2+8 d e f g+8 d^2 g^2\right ) x}{e^2}+\frac {g (e f+2 d g) x^2}{e}+\frac {g^2 x^3}{3}+\frac {4 d^2 (e f+d g)^2}{e^3 (d-e x)}+\frac {4 d (e f+d g) (e f+3 d g) \log (d-e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 107, 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} \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {4 d^2 (d g+e f)^2}{e^3 (d-e x)}+\frac {x \left (8 d^2 g^2+8 d e f g+e^2 f^2\right )}{e^2}+\frac {4 d (d g+e f) (3 d g+e f) \log (d-e x)}{e^3}+\frac {g x^2 (2 d g+e f)}{e}+\frac {g^2 x^3}{3} \]

[In]

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

[Out]

((e^2*f^2 + 8*d*e*f*g + 8*d^2*g^2)*x)/e^2 + (g*(e*f + 2*d*g)*x^2)/e + (g^2*x^3)/3 + (4*d^2*(e*f + d*g)^2)/(e^3
*(d - e*x)) + (4*d*(e*f + d*g)*(e*f + 3*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)^2 (f+g x)^2}{(d-e x)^2} \, dx \\ & = \int \left (\frac {e^2 f^2+8 d e f g+8 d^2 g^2}{e^2}+\frac {2 g (e f+2 d g) x}{e}+g^2 x^2+\frac {4 d (-e f-3 d g) (e f+d g)}{e^2 (d-e x)}+\frac {4 d^2 (e f+d g)^2}{e^2 (-d+e x)^2}\right ) \, dx \\ & = \frac {\left (e^2 f^2+8 d e f g+8 d^2 g^2\right ) x}{e^2}+\frac {g (e f+2 d g) x^2}{e}+\frac {g^2 x^3}{3}+\frac {4 d^2 (e f+d g)^2}{e^3 (d-e x)}+\frac {4 d (e f+d g) (e f+3 d g) \log (d-e x)}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {\left (e^2 f^2+8 d e f g+8 d^2 g^2\right ) x}{e^2}+\frac {g (e f+2 d g) x^2}{e}+\frac {g^2 x^3}{3}-\frac {4 d^2 (e f+d g)^2}{e^3 (-d+e x)}+\frac {4 d \left (e^2 f^2+4 d e f g+3 d^2 g^2\right ) \log (d-e x)}{e^3} \]

[In]

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

[Out]

((e^2*f^2 + 8*d*e*f*g + 8*d^2*g^2)*x)/e^2 + (g*(e*f + 2*d*g)*x^2)/e + (g^2*x^3)/3 - (4*d^2*(e*f + d*g)^2)/(e^3
*(-d + e*x)) + (4*d*(e^2*f^2 + 4*d*e*f*g + 3*d^2*g^2)*Log[d - e*x])/e^3

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {\frac {1}{3} g^{2} x^{3} e^{2}+2 d e \,g^{2} x^{2}+e^{2} f g \,x^{2}+8 d^{2} g^{2} x +8 d e f g x +e^{2} f^{2} x}{e^{2}}+\frac {4 d \left (3 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {4 d^{2} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}\)	\(133\)
risch	\(\frac {g^{2} x^{3}}{3}+\frac {2 d \,g^{2} x^{2}}{e}+f g \,x^{2}+\frac {8 d^{2} g^{2} x}{e^{2}}+\frac {8 d f g x}{e}+f^{2} x +\frac {12 d^{3} \ln \left (-e x +d \right ) g^{2}}{e^{3}}+\frac {16 d^{2} \ln \left (-e x +d \right ) f g}{e^{2}}+\frac {4 d \ln \left (-e x +d \right ) f^{2}}{e}+\frac {4 d^{4} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {8 d^{3} f g}{e^{2} \left (-e x +d \right )}+\frac {4 d^{2} f^{2}}{e \left (-e x +d \right )}\)	\(161\)
norman	\(\frac {\left (-\frac {23}{3} d^{2} g^{2}-8 d e f g -e^{2} f^{2}\right ) x^{3}+\frac {d^{2} \left (6 d^{3} g^{2}+9 d^{2} e f g +4 d \,e^{2} f^{2}\right )}{e^{3}}+\frac {d^{2} \left (12 d^{2} g^{2}+16 d e f g +5 e^{2} f^{2}\right ) x}{e^{2}}-\frac {e^{2} g^{2} x^{5}}{3}-e g \left (2 d g +e f \right ) x^{4}}{-e^{2} x^{2}+d^{2}}+\frac {4 d \left (3 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(170\)
parallelrisch	\(\frac {g^{2} e^{4} x^{4}+5 x^{3} d \,e^{3} g^{2}+3 x^{3} e^{4} f g +36 \ln \left (e x -d \right ) x \,d^{3} e \,g^{2}+48 \ln \left (e x -d \right ) x \,d^{2} e^{2} f g +12 \ln \left (e x -d \right ) x d \,e^{3} f^{2}+18 x^{2} d^{2} e^{2} g^{2}+21 x^{2} d \,e^{3} f g +3 x^{2} e^{4} f^{2}-36 \ln \left (e x -d \right ) d^{4} g^{2}-48 \ln \left (e x -d \right ) d^{3} e f g -12 \ln \left (e x -d \right ) d^{2} e^{2} f^{2}-36 d^{4} g^{2}-48 f g e \,d^{3}-15 d^{2} e^{2} f^{2}}{3 e^{3} \left (e x -d \right )}\)	\(217\)

[In]

int((e*x+d)^4*(g*x+f)^2/(-e^2*x^2+d^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/e^2*(1/3*g^2*x^3*e^2+2*d*e*g^2*x^2+e^2*f*g*x^2+8*d^2*g^2*x+8*d*e*f*g*x+e^2*f^2*x)+4*d/e^3*(3*d^2*g^2+4*d*e*f
*g+e^2*f^2)*ln(-e*x+d)+4*d^2*(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 = 0.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.93 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {e^{4} g^{2} x^{4} - 12 \, d^{2} e^{2} f^{2} - 24 \, d^{3} e f g - 12 \, d^{4} g^{2} + {\left (3 \, e^{4} f g + 5 \, d e^{3} g^{2}\right )} x^{3} + 3 \, {\left (e^{4} f^{2} + 7 \, d e^{3} f g + 6 \, d^{2} e^{2} g^{2}\right )} x^{2} - 3 \, {\left (d e^{3} f^{2} + 8 \, d^{2} e^{2} f g + 8 \, d^{3} e g^{2}\right )} x - 12 \, {\left (d^{2} e^{2} f^{2} + 4 \, d^{3} e f g + 3 \, d^{4} g^{2} - {\left (d e^{3} f^{2} + 4 \, d^{2} e^{2} f g + 3 \, d^{3} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{3 \, {\left (e^{4} x - d e^{3}\right )}} \]

[In]

integrate((e*x+d)^4*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="fricas")

[Out]

1/3*(e^4*g^2*x^4 - 12*d^2*e^2*f^2 - 24*d^3*e*f*g - 12*d^4*g^2 + (3*e^4*f*g + 5*d*e^3*g^2)*x^3 + 3*(e^4*f^2 + 7
*d*e^3*f*g + 6*d^2*e^2*g^2)*x^2 - 3*(d*e^3*f^2 + 8*d^2*e^2*f*g + 8*d^3*e*g^2)*x - 12*(d^2*e^2*f^2 + 4*d^3*e*f*
g + 3*d^4*g^2 - (d*e^3*f^2 + 4*d^2*e^2*f*g + 3*d^3*e*g^2)*x)*log(e*x - d))/(e^4*x - d*e^3)

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.11 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {4 d \left (d g + e f\right ) \left (3 d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} + \frac {g^{2} x^{3}}{3} + x^{2} \cdot \left (\frac {2 d g^{2}}{e} + f g\right ) + x \left (\frac {8 d^{2} g^{2}}{e^{2}} + \frac {8 d f g}{e} + f^{2}\right ) + \frac {- 4 d^{4} g^{2} - 8 d^{3} e f g - 4 d^{2} e^{2} f^{2}}{- d e^{3} + e^{4} x} \]

[In]

integrate((e*x+d)**4*(g*x+f)**2/(-e**2*x**2+d**2)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.32 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=-\frac {4 \, {\left (d^{2} e^{2} f^{2} + 2 \, d^{3} e f g + d^{4} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac {e^{2} g^{2} x^{3} + 3 \, {\left (e^{2} f g + 2 \, d e g^{2}\right )} x^{2} + 3 \, {\left (e^{2} f^{2} + 8 \, d e f g + 8 \, d^{2} g^{2}\right )} x}{3 \, e^{2}} + \frac {4 \, {\left (d e^{2} f^{2} + 4 \, d^{2} e f g + 3 \, d^{3} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

[In]

integrate((e*x+d)^4*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.38 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {4 \, {\left (d e^{2} f^{2} + 4 \, d^{2} e f g + 3 \, d^{3} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {4 \, {\left (d^{2} e^{2} f^{2} + 2 \, d^{3} e f g + d^{4} g^{2}\right )}}{{\left (e x - d\right )} e^{3}} + \frac {e^{6} g^{2} x^{3} + 3 \, e^{6} f g x^{2} + 6 \, d e^{5} g^{2} x^{2} + 3 \, e^{6} f^{2} x + 24 \, d e^{5} f g x + 24 \, d^{2} e^{4} g^{2} x}{3 \, e^{6}} \]

[In]

integrate((e*x+d)^4*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="giac")

[Out]

4*(d*e^2*f^2 + 4*d^2*e*f*g + 3*d^3*g^2)*log(abs(e*x - d))/e^3 - 4*(d^2*e^2*f^2 + 2*d^3*e*f*g + d^4*g^2)/((e*x
- d)*e^3) + 1/3*(e^6*g^2*x^3 + 3*e^6*f*g*x^2 + 6*d*e^5*g^2*x^2 + 3*e^6*f^2*x + 24*d*e^5*f*g*x + 24*d^2*e^4*g^2
*x)/e^6

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.73 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=x^2\,\left (\frac {g\,\left (d\,g+e\,f\right )}{e}+\frac {d\,g^2}{e}\right )+x\,\left (\frac {d^2\,g^2+4\,d\,e\,f\,g+e^2\,f^2}{e^2}+\frac {2\,d\,\left (\frac {2\,g\,\left (d\,g+e\,f\right )}{e}+\frac {2\,d\,g^2}{e}\right )}{e}-\frac {d^2\,g^2}{e^2}\right )+\frac {g^2\,x^3}{3}+\frac {4\,\left (d^4\,g^2+2\,d^3\,e\,f\,g+d^2\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {\ln \left (e\,x-d\right )\,\left (12\,d^3\,g^2+16\,d^2\,e\,f\,g+4\,d\,e^2\,f^2\right )}{e^3} \]

[In]

int(((f + g*x)^2*(d + e*x)^4)/(d^2 - e^2*x^2)^2,x)

[Out]

x^2*((g*(d*g + e*f))/e + (d*g^2)/e) + x*((d^2*g^2 + e^2*f^2 + 4*d*e*f*g)/e^2 + (2*d*((2*g*(d*g + e*f))/e + (2*
d*g^2)/e))/e - (d^2*g^2)/e^2) + (g^2*x^3)/3 + (4*(d^4*g^2 + d^2*e^2*f^2 + 2*d^3*e*f*g))/(e*(d*e^2 - e^3*x)) +
(log(e*x - d)*(12*d^3*g^2 + 4*d*e^2*f^2 + 16*d^2*e*f*g))/e^3