\(\int \frac {(d+e x)^2 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [574]

Optimal result

Integrand size = 29, antiderivative size = 88 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {(e f+d g)^2}{4 d e^3 (d-e x)^2}+\frac {(e f-3 d g) (e f+d g)}{4 d^2 e^3 (d-e x)}+\frac {(e f-d g)^2 \text {arctanh}\left (\frac {e x}{d}\right )}{4 d^3 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 88, 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} \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {\text {arctanh}\left (\frac {e x}{d}\right ) (e f-d g)^2}{4 d^3 e^3}+\frac {(e f-3 d g) (d g+e f)}{4 d^2 e^3 (d-e x)}+\frac {(d g+e f)^2}{4 d e^3 (d-e x)^2} \]

[In]

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

[Out]

(e*f + d*g)^2/(4*d*e^3*(d - e*x)^2) + ((e*f - 3*d*g)*(e*f + d*g))/(4*d^2*e^3*(d - e*x)) + ((e*f - d*g)^2*ArcTa
nh[(e*x)/d])/(4*d^3*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 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)^3 (d+e x)} \, dx \\ & = \int \left (\frac {(e f+d g)^2}{2 d e^2 (d-e x)^3}+\frac {(e f-3 d g) (e f+d g)}{4 d^2 e^2 (d-e x)^2}+\frac {(-e f+d g)^2}{4 d^2 e^2 \left (d^2-e^2 x^2\right )}\right ) \, dx \\ & = \frac {(e f+d g)^2}{4 d e^3 (d-e x)^2}+\frac {(e f-3 d g) (e f+d g)}{4 d^2 e^3 (d-e x)}+\frac {(e f-d g)^2 \int \frac {1}{d^2-e^2 x^2} \, dx}{4 d^2 e^2} \\ & = \frac {(e f+d g)^2}{4 d e^3 (d-e x)^2}+\frac {(e f-3 d g) (e f+d g)}{4 d^2 e^3 (d-e x)}+\frac {(e f-d g)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{4 d^3 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {-\frac {2 d (e f+d g) \left (2 d^2 g+e^2 f x-d e (2 f+3 g x)\right )}{(d-e x)^2}-(e f-d g)^2 \log (d-e x)+(e f-d g)^2 \log (d+e x)}{8 d^3 e^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {-3 d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{4 e^{3} d^{2} \left (-e x +d \right )}-\frac {-d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{4 d \,e^{3} \left (-e x +d \right )^{2}}+\frac {\left (-d^{2} g^{2}+2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{8 e^{3} d^{3}}+\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{8 e^{3} d^{3}}\)	\(152\)
risch	\(\frac {\frac {\left (3 d^{2} g^{2}+2 d e f g -e^{2} f^{2}\right ) x}{4 d^{2} e^{2}}-\frac {d^{2} g^{2}-e^{2} f^{2}}{2 d \,e^{3}}}{\left (-e x +d \right )^{2}}-\frac {\ln \left (-e x +d \right ) g^{2}}{8 e^{3} d}+\frac {\ln \left (-e x +d \right ) f g}{4 e^{2} d^{2}}-\frac {\ln \left (-e x +d \right ) f^{2}}{8 e \,d^{3}}+\frac {\ln \left (e x +d \right ) g^{2}}{8 e^{3} d}-\frac {\ln \left (e x +d \right ) f g}{4 e^{2} d^{2}}+\frac {\ln \left (e x +d \right ) f^{2}}{8 e \,d^{3}}\)	\(171\)
norman	\(\frac {\frac {d \left (-d^{2} g^{2} e +f^{2} e^{3}\right )}{2 e^{4}}-\frac {\left (d^{2} g^{2}-2 d e f g -3 e^{2} f^{2}\right ) x}{4 e^{2}}+\frac {\left (3 d^{2} g^{2}+2 d e f g -e^{2} f^{2}\right ) x^{3}}{4 d^{2}}-\frac {\left (-d e \,g^{2}-e^{2} f g \right ) x^{2}}{e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{8 e^{3} d^{3}}+\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{8 e^{3} d^{3}}\)	\(194\)
parallelrisch	\(-\frac {-4 d^{2} e^{2} f^{2}+4 d^{4} g^{2}-\ln \left (e x +d \right ) d^{2} e^{2} f^{2}+\ln \left (e x -d \right ) x^{2} e^{4} f^{2}-\ln \left (e x +d \right ) x^{2} e^{4} f^{2}+\ln \left (e x -d \right ) d^{2} e^{2} f^{2}-6 x \,d^{3} e \,g^{2}+2 x d \,e^{3} f^{2}-2 \ln \left (e x -d \right ) x^{2} d \,e^{3} f g +2 \ln \left (e x +d \right ) x^{2} d \,e^{3} f g +4 \ln \left (e x -d \right ) x \,d^{2} e^{2} f g -4 \ln \left (e x +d \right ) x \,d^{2} e^{2} f g +2 \ln \left (e x +d \right ) x d \,e^{3} f^{2}+2 \ln \left (e x +d \right ) d^{3} e f g +\ln \left (e x -d \right ) d^{4} g^{2}+\ln \left (e x -d \right ) x^{2} d^{2} e^{2} g^{2}-\ln \left (e x +d \right ) x^{2} d^{2} e^{2} g^{2}-2 \ln \left (e x -d \right ) x \,d^{3} e \,g^{2}-2 \ln \left (e x -d \right ) x d \,e^{3} f^{2}+2 \ln \left (e x +d \right ) x \,d^{3} e \,g^{2}-2 \ln \left (e x -d \right ) d^{3} e f g -4 x \,d^{2} e^{2} f g -\ln \left (e x +d \right ) d^{4} g^{2}}{8 e^{3} d^{3} \left (e x -d \right )^{2}}\)	\(379\)

[In]

int((e*x+d)^2*(g*x+f)^2/(-e^2*x^2+d^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (84) = 168\).

Time = 0.27 (sec) , antiderivative size = 271, normalized size of antiderivative = 3.08 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {4 \, d^{2} e^{2} f^{2} - 4 \, d^{4} g^{2} - 2 \, {\left (d e^{3} f^{2} - 2 \, d^{2} e^{2} f g - 3 \, d^{3} e g^{2}\right )} x + {\left (d^{2} e^{2} f^{2} - 2 \, d^{3} e f g + d^{4} g^{2} + {\left (e^{4} f^{2} - 2 \, d e^{3} f g + d^{2} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (d e^{3} f^{2} - 2 \, d^{2} e^{2} f g + d^{3} e g^{2}\right )} x\right )} \log \left (e x + d\right ) - {\left (d^{2} e^{2} f^{2} - 2 \, d^{3} e f g + d^{4} g^{2} + {\left (e^{4} f^{2} - 2 \, d e^{3} f g + d^{2} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (d e^{3} f^{2} - 2 \, d^{2} e^{2} f g + d^{3} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{8 \, {\left (d^{3} e^{5} x^{2} - 2 \, d^{4} e^{4} x + d^{5} e^{3}\right )}} \]

[In]

integrate((e*x+d)^2*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (75) = 150\).

Time = 0.46 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.10 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=- \frac {2 d^{3} g^{2} - 2 d e^{2} f^{2} + x \left (- 3 d^{2} e g^{2} - 2 d e^{2} f g + e^{3} f^{2}\right )}{4 d^{4} e^{3} - 8 d^{3} e^{4} x + 4 d^{2} e^{5} x^{2}} - \frac {\left (d g - e f\right )^{2} \log {\left (- \frac {d \left (d g - e f\right )^{2}}{e \left (d^{2} g^{2} - 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{8 d^{3} e^{3}} + \frac {\left (d g - e f\right )^{2} \log {\left (\frac {d \left (d g - e f\right )^{2}}{e \left (d^{2} g^{2} - 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{8 d^{3} e^{3}} \]

[In]

integrate((e*x+d)**2*(g*x+f)**2/(-e**2*x**2+d**2)**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.70 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {2 \, d e^{2} f^{2} - 2 \, d^{3} g^{2} - {\left (e^{3} f^{2} - 2 \, d e^{2} f g - 3 \, d^{2} e g^{2}\right )} x}{4 \, {\left (d^{2} e^{5} x^{2} - 2 \, d^{3} e^{4} x + d^{4} e^{3}\right )}} + \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{8 \, d^{3} e^{3}} - \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{8 \, d^{3} e^{3}} \]

[In]

integrate((e*x+d)^2*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.62 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{8 \, d^{3} e^{3}} - \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{8 \, d^{3} e^{3}} + \frac {2 \, d^{2} e^{2} f^{2} - 2 \, d^{4} g^{2} - {\left (d e^{3} f^{2} - 2 \, d^{2} e^{2} f g - 3 \, d^{3} e g^{2}\right )} x}{4 \, {\left (e x - d\right )}^{2} d^{3} e^{3}} \]

[In]

integrate((e*x+d)^2*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,{\left (d\,g-e\,f\right )}^2}{4\,d^3\,e^3}-\frac {\frac {d^2\,g^2-e^2\,f^2}{2\,d\,e^3}-\frac {x\,\left (3\,d^2\,g^2+2\,d\,e\,f\,g-e^2\,f^2\right )}{4\,d^2\,e^2}}{d^2-2\,d\,e\,x+e^2\,x^2} \]

[In]

int(((f + g*x)^2*(d + e*x)^2)/(d^2 - e^2*x^2)^3,x)

[Out]

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