\(\int \frac {(f+g x)^2}{(d+e x)^2 (d^2-e^2 x^2)^2} \, dx\) [566]

Optimal result

Integrand size = 29, antiderivative size = 146 \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=\frac {(e f+d g)^2}{16 d^4 e^3 (d-e x)}-\frac {(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}-\frac {e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac {(3 e f-d g) (e f+d g)}{16 d^4 e^3 (d+e x)}+\frac {f (e f+d g) \text {arctanh}\left (\frac {e x}{d}\right )}{4 d^5 e^2} \]

[Out]

1/16*(d*g+e*f)^2/d^4/e^3/(-e*x+d)-1/12*(-d*g+e*f)^2/d^2/e^3/(e*x+d)^3+1/8*(d^2*g^2-e^2*f^2)/d^3/e^3/(e*x+d)^2-
1/16*(-d*g+3*e*f)*(d*g+e*f)/d^4/e^3/(e*x+d)+1/4*f*(d*g+e*f)*arctanh(e*x/d)/d^5/e^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {862, 90, 214} \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=\frac {f \text {arctanh}\left (\frac {e x}{d}\right ) (d g+e f)}{4 d^5 e^2}+\frac {(d g+e f)^2}{16 d^4 e^3 (d-e x)}-\frac {(3 e f-d g) (d g+e f)}{16 d^4 e^3 (d+e x)}-\frac {(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}-\frac {e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2} \]

[In]

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

[Out]

(e*f + d*g)^2/(16*d^4*e^3*(d - e*x)) - (e*f - d*g)^2/(12*d^2*e^3*(d + e*x)^3) - (e^2*f^2 - d^2*g^2)/(8*d^3*e^3
*(d + e*x)^2) - ((3*e*f - d*g)*(e*f + d*g))/(16*d^4*e^3*(d + e*x)) + (f*(e*f + d*g)*ArcTanh[(e*x)/d])/(4*d^5*e
^2)

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

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 {(f+g x)^2}{(d-e x)^2 (d+e x)^4} \, dx \\ & = \int \left (\frac {(e f+d g)^2}{16 d^4 e^2 (d-e x)^2}+\frac {(-e f+d g)^2}{4 d^2 e^2 (d+e x)^4}+\frac {e^2 f^2-d^2 g^2}{4 d^3 e^2 (d+e x)^3}+\frac {(3 e f-d g) (e f+d g)}{16 d^4 e^2 (d+e x)^2}+\frac {f (e f+d g)}{4 d^4 e \left (d^2-e^2 x^2\right )}\right ) \, dx \\ & = \frac {(e f+d g)^2}{16 d^4 e^3 (d-e x)}-\frac {(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}-\frac {e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac {(3 e f-d g) (e f+d g)}{16 d^4 e^3 (d+e x)}+\frac {(f (e f+d g)) \int \frac {1}{d^2-e^2 x^2} \, dx}{4 d^4 e} \\ & = \frac {(e f+d g)^2}{16 d^4 e^3 (d-e x)}-\frac {(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}-\frac {e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac {(3 e f-d g) (e f+d g)}{16 d^4 e^3 (d+e x)}+\frac {f (e f+d g) \tanh ^{-1}\left (\frac {e x}{d}\right )}{4 d^5 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.17 \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=\frac {2 d \left (2 d^5 g^2+3 e^5 f^2 x^3+d^3 e^2 f (-4 f+g x)+3 d e^4 f x^2 (2 f+g x)+2 d^4 e g (f+2 g x)+d^2 e^3 f x (f+6 g x)\right )+3 e f (e f+d g) (-d+e x) (d+e x)^3 \log (d-e x)+3 e f (e f+d g) (d-e x) (d+e x)^3 \log (d+e x)}{24 d^5 e^3 (d-e x) (d+e x)^3} \]

[In]

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

[Out]

(2*d*(2*d^5*g^2 + 3*e^5*f^2*x^3 + d^3*e^2*f*(-4*f + g*x) + 3*d*e^4*f*x^2*(2*f + g*x) + 2*d^4*e*g*(f + 2*g*x) +
d^2*e^3*f*x*(f + 6*g*x)) + 3*e*f*(e*f + d*g)*(-d + e*x)*(d + e*x)^3*Log[d - e*x] + 3*e*f*(e*f + d*g)*(d - e*x
)*(d + e*x)^3*Log[d + e*x])/(24*d^5*e^3*(d - e*x)*(d + e*x)^3)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.11

method	result	size
norman	\(\frac {\frac {d f g -3 e \,f^{2}}{8 d \,e^{2}}-\frac {\left (-d^{2} g^{2}-d e f g -e^{2} f^{2}\right ) x^{3}}{3 d^{4}}+\frac {f \left (d g +e f \right ) x^{2}}{2 d^{3}}-\frac {e \left (-4 d^{2} g^{2}-d e f g -e^{2} f^{2}\right ) x^{4}}{24 d^{5}}}{\left (e x +d \right )^{3} \left (-e x +d \right )}-\frac {f \left (d g +e f \right ) \ln \left (-e x +d \right )}{8 d^{5} e^{2}}+\frac {f \left (d g +e f \right ) \ln \left (e x +d \right )}{8 d^{5} e^{2}}\)	\(162\)
risch	\(\frac {\frac {e f \left (d g +e f \right ) x^{3}}{4 d^{4}}+\frac {f \left (d g +e f \right ) x^{2}}{2 d^{3}}+\frac {\left (4 d^{2} g^{2}+d e f g +e^{2} f^{2}\right ) x}{12 d^{2} e^{2}}+\frac {d^{2} g^{2}+d e f g -2 e^{2} f^{2}}{6 d \,e^{3}}}{\left (e x +d \right )^{2} \left (-e^{2} x^{2}+d^{2}\right )}-\frac {f \ln \left (-e x +d \right ) g}{8 d^{4} e^{2}}-\frac {\ln \left (-e x +d \right ) f^{2}}{8 e \,d^{5}}+\frac {f \ln \left (e x +d \right ) g}{8 d^{4} e^{2}}+\frac {f^{2} \ln \left (e x +d \right )}{8 d^{5} e}\)	\(185\)
default	\(\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{16 e^{3} d^{4} \left (-e x +d \right )}-\frac {f \left (d g +e f \right ) \ln \left (-e x +d \right )}{8 d^{5} e^{2}}-\frac {-d^{2} g^{2}+e^{2} f^{2}}{8 e^{3} d^{3} \left (e x +d \right )^{2}}-\frac {-d^{2} g^{2}+2 d e f g +3 e^{2} f^{2}}{16 e^{3} d^{4} \left (e x +d \right )}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{12 d^{2} e^{3} \left (e x +d \right )^{3}}+\frac {f \left (d g +e f \right ) \ln \left (e x +d \right )}{8 d^{5} e^{2}}\)	\(189\)
parallelrisch	\(-\frac {3 \ln \left (e x -d \right ) x^{4} e^{5} f^{2}-3 \ln \left (e x +d \right ) x^{4} e^{5} f^{2}-3 \ln \left (e x -d \right ) d^{5} f g -3 \ln \left (e x -d \right ) d^{4} e \,f^{2}+3 \ln \left (e x +d \right ) d^{5} f g +3 \ln \left (e x +d \right ) d^{4} e \,f^{2}+18 f^{2} d^{3} e^{2} x +12 x^{2} d^{2} e^{3} f^{2}+4 x^{4} d^{2} e^{3} g^{2}+8 x^{3} d^{3} e^{2} g^{2}-10 x^{3} d \,e^{4} f^{2}+6 \ln \left (e x -d \right ) x^{3} d \,e^{4} f^{2}-6 \ln \left (e x +d \right ) x^{3} d \,e^{4} f^{2}-6 \ln \left (e x -d \right ) x \,d^{3} e^{2} f^{2}+6 \ln \left (e x +d \right ) x \,d^{3} e^{2} f^{2}+14 x^{3} d^{2} e^{3} f g +12 x^{2} d^{3} e^{2} f g -6 f g \,d^{4} e x +4 x^{4} d \,e^{4} f g -8 x^{4} e^{5} f^{2}-6 \ln \left (e x +d \right ) x^{3} d^{2} e^{3} f g -6 \ln \left (e x -d \right ) x \,d^{4} e f g +6 \ln \left (e x +d \right ) x \,d^{4} e f g +3 \ln \left (e x -d \right ) x^{4} d \,e^{4} f g -3 \ln \left (e x +d \right ) x^{4} d \,e^{4} f g +6 \ln \left (e x -d \right ) x^{3} d^{2} e^{3} f g}{24 d^{5} e^{2} \left (e^{2} x^{2}-d^{2}\right ) \left (e x +d \right )^{2}}\)	\(435\)

[In]

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

[Out]

(1/8*(d*f*g-3*e*f^2)/d/e^2-1/3*(-d^2*g^2-d*e*f*g-e^2*f^2)/d^4*x^3+1/2/d^3*f*(d*g+e*f)*x^2-1/24*e*(-4*d^2*g^2-d
*e*f*g-e^2*f^2)/d^5*x^4)/(e*x+d)^3/(-e*x+d)-1/8*f*(d*g+e*f)/d^5/e^2*ln(-e*x+d)+1/8*f*(d*g+e*f)/d^5/e^2*ln(e*x+
d)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (137) = 274\).

Time = 0.37 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.31 \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=\frac {8 \, d^{4} e^{2} f^{2} - 4 \, d^{5} e f g - 4 \, d^{6} g^{2} - 6 \, {\left (d e^{5} f^{2} + d^{2} e^{4} f g\right )} x^{3} - 12 \, {\left (d^{2} e^{4} f^{2} + d^{3} e^{3} f g\right )} x^{2} - 2 \, {\left (d^{3} e^{3} f^{2} + d^{4} e^{2} f g + 4 \, d^{5} e g^{2}\right )} x - 3 \, {\left (d^{4} e^{2} f^{2} + d^{5} e f g - {\left (e^{6} f^{2} + d e^{5} f g\right )} x^{4} - 2 \, {\left (d e^{5} f^{2} + d^{2} e^{4} f g\right )} x^{3} + 2 \, {\left (d^{3} e^{3} f^{2} + d^{4} e^{2} f g\right )} x\right )} \log \left (e x + d\right ) + 3 \, {\left (d^{4} e^{2} f^{2} + d^{5} e f g - {\left (e^{6} f^{2} + d e^{5} f g\right )} x^{4} - 2 \, {\left (d e^{5} f^{2} + d^{2} e^{4} f g\right )} x^{3} + 2 \, {\left (d^{3} e^{3} f^{2} + d^{4} e^{2} f g\right )} x\right )} \log \left (e x - d\right )}{24 \, {\left (d^{5} e^{7} x^{4} + 2 \, d^{6} e^{6} x^{3} - 2 \, d^{8} e^{4} x - d^{9} e^{3}\right )}} \]

[In]

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

[Out]

1/24*(8*d^4*e^2*f^2 - 4*d^5*e*f*g - 4*d^6*g^2 - 6*(d*e^5*f^2 + d^2*e^4*f*g)*x^3 - 12*(d^2*e^4*f^2 + d^3*e^3*f*
g)*x^2 - 2*(d^3*e^3*f^2 + d^4*e^2*f*g + 4*d^5*e*g^2)*x - 3*(d^4*e^2*f^2 + d^5*e*f*g - (e^6*f^2 + d*e^5*f*g)*x^
4 - 2*(d*e^5*f^2 + d^2*e^4*f*g)*x^3 + 2*(d^3*e^3*f^2 + d^4*e^2*f*g)*x)*log(e*x + d) + 3*(d^4*e^2*f^2 + d^5*e*f
*g - (e^6*f^2 + d*e^5*f*g)*x^4 - 2*(d*e^5*f^2 + d^2*e^4*f*g)*x^3 + 2*(d^3*e^3*f^2 + d^4*e^2*f*g)*x)*log(e*x -
d))/(d^5*e^7*x^4 + 2*d^6*e^6*x^3 - 2*d^8*e^4*x - d^9*e^3)

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.65 \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=\frac {- 2 d^{5} g^{2} - 2 d^{4} e f g + 4 d^{3} e^{2} f^{2} + x^{3} \left (- 3 d e^{4} f g - 3 e^{5} f^{2}\right ) + x^{2} \left (- 6 d^{2} e^{3} f g - 6 d e^{4} f^{2}\right ) + x \left (- 4 d^{4} e g^{2} - d^{3} e^{2} f g - d^{2} e^{3} f^{2}\right )}{- 12 d^{8} e^{3} - 24 d^{7} e^{4} x + 24 d^{5} e^{6} x^{3} + 12 d^{4} e^{7} x^{4}} - \frac {f \left (d g + e f\right ) \log {\left (- \frac {d f \left (d g + e f\right )}{e \left (d f g + e f^{2}\right )} + x \right )}}{8 d^{5} e^{2}} + \frac {f \left (d g + e f\right ) \log {\left (\frac {d f \left (d g + e f\right )}{e \left (d f g + e f^{2}\right )} + x \right )}}{8 d^{5} e^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.35 \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=\frac {4 \, d^{3} e^{2} f^{2} - 2 \, d^{4} e f g - 2 \, d^{5} g^{2} - 3 \, {\left (e^{5} f^{2} + d e^{4} f g\right )} x^{3} - 6 \, {\left (d e^{4} f^{2} + d^{2} e^{3} f g\right )} x^{2} - {\left (d^{2} e^{3} f^{2} + d^{3} e^{2} f g + 4 \, d^{4} e g^{2}\right )} x}{12 \, {\left (d^{4} e^{7} x^{4} + 2 \, d^{5} e^{6} x^{3} - 2 \, d^{7} e^{4} x - d^{8} e^{3}\right )}} + \frac {{\left (e f^{2} + d f g\right )} \log \left (e x + d\right )}{8 \, d^{5} e^{2}} - \frac {{\left (e f^{2} + d f g\right )} \log \left (e x - d\right )}{8 \, d^{5} e^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.55 \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=-\frac {{\left (e f^{2} + d f g\right )} \log \left ({\left | -\frac {2 \, d}{e x + d} + 1 \right |}\right )}{8 \, d^{5} e^{2}} + \frac {e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}}{32 \, d^{5} e^{3} {\left (\frac {2 \, d}{e x + d} - 1\right )}} - \frac {\frac {9 \, d^{2} e^{5} f^{2}}{e x + d} + \frac {6 \, d^{3} e^{5} f^{2}}{{\left (e x + d\right )}^{2}} + \frac {4 \, d^{4} e^{5} f^{2}}{{\left (e x + d\right )}^{3}} + \frac {6 \, d^{3} e^{4} f g}{e x + d} - \frac {8 \, d^{5} e^{4} f g}{{\left (e x + d\right )}^{3}} - \frac {3 \, d^{4} e^{3} g^{2}}{e x + d} - \frac {6 \, d^{5} e^{3} g^{2}}{{\left (e x + d\right )}^{2}} + \frac {4 \, d^{6} e^{3} g^{2}}{{\left (e x + d\right )}^{3}}}{48 \, d^{6} e^{6}} \]

[In]

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

[Out]

-1/8*(e*f^2 + d*f*g)*log(abs(-2*d/(e*x + d) + 1))/(d^5*e^2) + 1/32*(e^2*f^2 + 2*d*e*f*g + d^2*g^2)/(d^5*e^3*(2
*d/(e*x + d) - 1)) - 1/48*(9*d^2*e^5*f^2/(e*x + d) + 6*d^3*e^5*f^2/(e*x + d)^2 + 4*d^4*e^5*f^2/(e*x + d)^3 + 6
*d^3*e^4*f*g/(e*x + d) - 8*d^5*e^4*f*g/(e*x + d)^3 - 3*d^4*e^3*g^2/(e*x + d) - 6*d^5*e^3*g^2/(e*x + d)^2 + 4*d
^6*e^3*g^2/(e*x + d)^3)/(d^6*e^6)

Mupad [B] (verification not implemented)

Time = 12.06 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx=\frac {\frac {d^2\,g^2+d\,e\,f\,g-2\,e^2\,f^2}{6\,d\,e^3}+\frac {f\,x^2\,\left (d\,g+e\,f\right )}{2\,d^3}+\frac {x\,\left (4\,d^2\,g^2+d\,e\,f\,g+e^2\,f^2\right )}{12\,d^2\,e^2}+\frac {e\,f\,x^3\,\left (d\,g+e\,f\right )}{4\,d^4}}{d^4+2\,d^3\,e\,x-2\,d\,e^3\,x^3-e^4\,x^4}+\frac {f\,\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,\left (d\,g+e\,f\right )}{4\,d^5\,e^2} \]

[In]

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

[Out]

((d^2*g^2 - 2*e^2*f^2 + d*e*f*g)/(6*d*e^3) + (f*x^2*(d*g + e*f))/(2*d^3) + (x*(4*d^2*g^2 + e^2*f^2 + d*e*f*g))
/(12*d^2*e^2) + (e*f*x^3*(d*g + e*f))/(4*d^4))/(d^4 - e^4*x^4 - 2*d*e^3*x^3 + 2*d^3*e*x) + (f*atanh((e*x)/d)*(
d*g + e*f))/(4*d^5*e^2)