\(\int \frac {(f+g x)^2}{(d+e x)^3 (d^2-e^2 x^2)} \, dx\) [555]

Optimal result

Integrand size = 29, antiderivative size = 113 \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=-\frac {(e f-d g)^2}{6 d e^3 (d+e x)^3}-\frac {(e f-d g) (e f+3 d g)}{8 d^2 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{8 d^3 e^3 (d+e x)}+\frac {(e f+d g)^2 \text {arctanh}\left (\frac {e x}{d}\right )}{8 d^4 e^3} \]

[Out]

-1/6*(-d*g+e*f)^2/d/e^3/(e*x+d)^3-1/8*(-d*g+e*f)*(3*d*g+e*f)/d^2/e^3/(e*x+d)^2-1/8*(d*g+e*f)^2/d^3/e^3/(e*x+d)
+1/8*(d*g+e*f)^2*arctanh(e*x/d)/d^4/e^3

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 113, 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} \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=\frac {\text {arctanh}\left (\frac {e x}{d}\right ) (d g+e f)^2}{8 d^4 e^3}-\frac {(d g+e f)^2}{8 d^3 e^3 (d+e x)}-\frac {(3 d g+e f) (e f-d g)}{8 d^2 e^3 (d+e x)^2}-\frac {(e f-d g)^2}{6 d e^3 (d+e x)^3} \]

[In]

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

[Out]

-1/6*(e*f - d*g)^2/(d*e^3*(d + e*x)^3) - ((e*f - d*g)*(e*f + 3*d*g))/(8*d^2*e^3*(d + e*x)^2) - (e*f + d*g)^2/(
8*d^3*e^3*(d + e*x)) + ((e*f + d*g)^2*ArcTanh[(e*x)/d])/(8*d^4*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) (d+e x)^4} \, dx \\ & = \int \left (\frac {(-e f+d g)^2}{2 d e^2 (d+e x)^4}+\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^2 (d+e x)^3}+\frac {(e f+d g)^2}{8 d^3 e^2 (d+e x)^2}+\frac {(e f+d g)^2}{8 d^3 e^2 \left (d^2-e^2 x^2\right )}\right ) \, dx \\ & = -\frac {(e f-d g)^2}{6 d e^3 (d+e x)^3}-\frac {(e f-d g) (e f+3 d g)}{8 d^2 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{8 d^3 e^3 (d+e x)}+\frac {(e f+d g)^2 \int \frac {1}{d^2-e^2 x^2} \, dx}{8 d^3 e^2} \\ & = -\frac {(e f-d g)^2}{6 d e^3 (d+e x)^3}-\frac {(e f-d g) (e f+3 d g)}{8 d^2 e^3 (d+e x)^2}-\frac {(e f+d g)^2}{8 d^3 e^3 (d+e x)}+\frac {(e f+d g)^2 \tanh ^{-1}\left (\frac {e x}{d}\right )}{8 d^4 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.08 \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=\frac {-\frac {8 d^3 (e f-d g)^2}{(d+e x)^3}+\frac {6 d^2 \left (-e^2 f^2-2 d e f g+3 d^2 g^2\right )}{(d+e x)^2}-\frac {6 d (e f+d g)^2}{d+e x}-3 (e f+d g)^2 \log (d-e x)+3 (e f+d g)^2 \log (d+e x)}{48 d^4 e^3} \]

[In]

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

[Out]

((-8*d^3*(e*f - d*g)^2)/(d + e*x)^3 + (6*d^2*(-(e^2*f^2) - 2*d*e*f*g + 3*d^2*g^2))/(d + e*x)^2 - (6*d*(e*f + d
*g)^2)/(d + e*x) - 3*(e*f + d*g)^2*Log[d - e*x] + 3*(e*f + d*g)^2*Log[d + e*x])/(48*d^4*e^3)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.57

method	result	size
norman	\(\frac {-\frac {\left (d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}\right ) x^{3}}{12 d^{4}}-\frac {\left (d^{2} g^{2}+2 d e f g -7 e^{2} f^{2}\right ) x}{8 d^{2} e^{2}}-\frac {\left (3 d^{2} g^{2}-2 d e f g -9 e^{2} f^{2}\right ) x^{2}}{8 d^{3} e}}{\left (e x +d \right )^{3}}-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 e^{3} d^{4}}+\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 e^{3} d^{4}}\)	\(177\)
default	\(\frac {\left (-d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{16 e^{3} d^{4}}-\frac {-3 d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{8 d^{2} e^{3} \left (e x +d \right )^{2}}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{6 e^{3} d \left (e x +d \right )^{3}}+\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{16 e^{3} d^{4}}-\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{8 e^{3} d^{3} \left (e x +d \right )}\)	\(184\)
risch	\(\frac {-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) x^{2}}{8 d^{3} e}+\frac {\left (d^{2} g^{2}-6 d e f g -3 e^{2} f^{2}\right ) x}{8 d^{2} e^{2}}+\frac {d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}}{12 d \,e^{3}}}{\left (e x +d \right )^{3}}-\frac {\ln \left (-e x +d \right ) g^{2}}{16 e^{3} d^{2}}-\frac {\ln \left (-e x +d \right ) f g}{8 e^{2} d^{3}}-\frac {\ln \left (-e x +d \right ) f^{2}}{16 e \,d^{4}}+\frac {\ln \left (e x +d \right ) g^{2}}{16 e^{3} d^{2}}+\frac {\ln \left (e x +d \right ) f g}{8 e^{2} d^{3}}+\frac {\ln \left (e x +d \right ) f^{2}}{16 e \,d^{4}}\)	\(207\)
parallelrisch	\(-\frac {-6 \ln \left (e x +d \right ) d^{4} e f g +3 \ln \left (e x -d \right ) x^{3} d^{2} e^{3} g^{2}-18 \ln \left (e x +d \right ) x \,d^{3} e^{2} f g +18 \ln \left (e x -d \right ) x \,d^{3} e^{2} f g +18 \ln \left (e x -d \right ) x^{2} d^{2} e^{3} f g -18 \ln \left (e x +d \right ) x^{2} d^{2} e^{3} f g -6 \ln \left (e x +d \right ) x^{3} d \,e^{4} f g +6 \ln \left (e x -d \right ) x^{3} d \,e^{4} f g -9 \ln \left (e x +d \right ) x \,d^{2} e^{3} f^{2}+6 x \,d^{4} e \,g^{2}-42 x \,d^{2} e^{3} f^{2}+4 x^{3} d^{2} e^{3} g^{2}+18 x^{2} d^{3} e^{2} g^{2}-54 x^{2} d \,e^{4} f^{2}+3 \ln \left (e x -d \right ) d^{3} e^{2} f^{2}+3 \ln \left (e x -d \right ) x^{3} e^{5} f^{2}-12 x^{2} d^{2} e^{3} f g +6 \ln \left (e x -d \right ) d^{4} e f g +12 x \,d^{3} e^{2} f g -3 \ln \left (e x +d \right ) d^{5} g^{2}-3 \ln \left (e x +d \right ) x^{3} d^{2} e^{3} g^{2}+9 \ln \left (e x -d \right ) x^{2} d^{3} e^{2} g^{2}+9 \ln \left (e x -d \right ) x^{2} d \,e^{4} f^{2}-9 \ln \left (e x +d \right ) x^{2} d^{3} e^{2} g^{2}-9 \ln \left (e x +d \right ) x^{2} d \,e^{4} f^{2}+9 \ln \left (e x -d \right ) x \,d^{4} e \,g^{2}+9 \ln \left (e x -d \right ) x \,d^{2} e^{3} f^{2}-9 \ln \left (e x +d \right ) x \,d^{4} e \,g^{2}-8 x^{3} d \,e^{4} f g -3 \ln \left (e x +d \right ) d^{3} e^{2} f^{2}-3 \ln \left (e x +d \right ) x^{3} e^{5} f^{2}-20 x^{3} e^{5} f^{2}+3 \ln \left (e x -d \right ) d^{5} g^{2}}{48 e^{3} d^{4} \left (e x +d \right )^{3}}\)	\(563\)

[In]

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

[Out]

(-1/12*(d^2*g^2-2*d*e*f*g-5*e^2*f^2)/d^4*x^3-1/8*(d^2*g^2+2*d*e*f*g-7*e^2*f^2)/d^2/e^2*x-1/8*(3*d^2*g^2-2*d*e*
f*g-9*e^2*f^2)/d^3/e*x^2)/(e*x+d)^3-1/16*(d^2*g^2+2*d*e*f*g+e^2*f^2)/e^3/d^4*ln(-e*x+d)+1/16*(d^2*g^2+2*d*e*f*
g+e^2*f^2)/e^3/d^4*ln(e*x+d)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (105) = 210\).

Time = 0.28 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.54 \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=-\frac {20 \, d^{3} e^{2} f^{2} + 8 \, d^{4} e f g - 4 \, d^{5} g^{2} + 6 \, {\left (d e^{4} f^{2} + 2 \, d^{2} e^{3} f g + d^{3} e^{2} g^{2}\right )} x^{2} + 6 \, {\left (3 \, d^{2} e^{3} f^{2} + 6 \, d^{3} e^{2} f g - d^{4} e g^{2}\right )} x - 3 \, {\left (d^{3} e^{2} f^{2} + 2 \, d^{4} e f g + d^{5} g^{2} + {\left (e^{5} f^{2} + 2 \, d e^{4} f g + d^{2} e^{3} g^{2}\right )} x^{3} + 3 \, {\left (d e^{4} f^{2} + 2 \, d^{2} e^{3} f g + d^{3} e^{2} g^{2}\right )} x^{2} + 3 \, {\left (d^{2} e^{3} f^{2} + 2 \, d^{3} e^{2} f g + d^{4} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + 3 \, {\left (d^{3} e^{2} f^{2} + 2 \, d^{4} e f g + d^{5} g^{2} + {\left (e^{5} f^{2} + 2 \, d e^{4} f g + d^{2} e^{3} g^{2}\right )} x^{3} + 3 \, {\left (d e^{4} f^{2} + 2 \, d^{2} e^{3} f g + d^{3} e^{2} g^{2}\right )} x^{2} + 3 \, {\left (d^{2} e^{3} f^{2} + 2 \, d^{3} e^{2} f g + d^{4} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{48 \, {\left (d^{4} e^{6} x^{3} + 3 \, d^{5} e^{5} x^{2} + 3 \, d^{6} e^{4} x + d^{7} e^{3}\right )}} \]

[In]

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

[Out]

-1/48*(20*d^3*e^2*f^2 + 8*d^4*e*f*g - 4*d^5*g^2 + 6*(d*e^4*f^2 + 2*d^2*e^3*f*g + d^3*e^2*g^2)*x^2 + 6*(3*d^2*e
^3*f^2 + 6*d^3*e^2*f*g - d^4*e*g^2)*x - 3*(d^3*e^2*f^2 + 2*d^4*e*f*g + d^5*g^2 + (e^5*f^2 + 2*d*e^4*f*g + d^2*
e^3*g^2)*x^3 + 3*(d*e^4*f^2 + 2*d^2*e^3*f*g + d^3*e^2*g^2)*x^2 + 3*(d^2*e^3*f^2 + 2*d^3*e^2*f*g + d^4*e*g^2)*x
)*log(e*x + d) + 3*(d^3*e^2*f^2 + 2*d^4*e*f*g + d^5*g^2 + (e^5*f^2 + 2*d*e^4*f*g + d^2*e^3*g^2)*x^3 + 3*(d*e^4
*f^2 + 2*d^2*e^3*f*g + d^3*e^2*g^2)*x^2 + 3*(d^2*e^3*f^2 + 2*d^3*e^2*f*g + d^4*e*g^2)*x)*log(e*x - d))/(d^4*e^
6*x^3 + 3*d^5*e^5*x^2 + 3*d^6*e^4*x + d^7*e^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (99) = 198\).

Time = 0.59 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.19 \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=- \frac {- 2 d^{4} g^{2} + 4 d^{3} e f g + 10 d^{2} e^{2} f^{2} + x^{2} \cdot \left (3 d^{2} e^{2} g^{2} + 6 d e^{3} f g + 3 e^{4} f^{2}\right ) + x \left (- 3 d^{3} e g^{2} + 18 d^{2} e^{2} f g + 9 d e^{3} f^{2}\right )}{24 d^{6} e^{3} + 72 d^{5} e^{4} x + 72 d^{4} e^{5} x^{2} + 24 d^{3} e^{6} x^{3}} - \frac {\left (d g + e f\right )^{2} \log {\left (- \frac {d \left (d g + e f\right )^{2}}{e \left (d^{2} g^{2} + 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{16 d^{4} e^{3}} + \frac {\left (d g + e f\right )^{2} \log {\left (\frac {d \left (d g + e f\right )^{2}}{e \left (d^{2} g^{2} + 2 d e f g + e^{2} f^{2}\right )} + x \right )}}{16 d^{4} e^{3}} \]

[In]

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

[Out]

-(-2*d**4*g**2 + 4*d**3*e*f*g + 10*d**2*e**2*f**2 + x**2*(3*d**2*e**2*g**2 + 6*d*e**3*f*g + 3*e**4*f**2) + x*(
-3*d**3*e*g**2 + 18*d**2*e**2*f*g + 9*d*e**3*f**2))/(24*d**6*e**3 + 72*d**5*e**4*x + 72*d**4*e**5*x**2 + 24*d*
*3*e**6*x**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)/(16*d**4*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)/(16*d**4*e**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.82 \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=-\frac {10 \, d^{2} e^{2} f^{2} + 4 \, d^{3} e f g - 2 \, d^{4} g^{2} + 3 \, {\left (e^{4} f^{2} + 2 \, d e^{3} f g + d^{2} e^{2} g^{2}\right )} x^{2} + 3 \, {\left (3 \, d e^{3} f^{2} + 6 \, d^{2} e^{2} f g - d^{3} e g^{2}\right )} x}{24 \, {\left (d^{3} e^{6} x^{3} + 3 \, d^{4} e^{5} x^{2} + 3 \, d^{5} e^{4} x + d^{6} e^{3}\right )}} + \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{16 \, d^{4} e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{16 \, d^{4} e^{3}} \]

[In]

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

[Out]

-1/24*(10*d^2*e^2*f^2 + 4*d^3*e*f*g - 2*d^4*g^2 + 3*(e^4*f^2 + 2*d*e^3*f*g + d^2*e^2*g^2)*x^2 + 3*(3*d*e^3*f^2
+ 6*d^2*e^2*f*g - d^3*e*g^2)*x)/(d^3*e^6*x^3 + 3*d^4*e^5*x^2 + 3*d^5*e^4*x + d^6*e^3) + 1/16*(e^2*f^2 + 2*d*e
*f*g + d^2*g^2)*log(e*x + d)/(d^4*e^3) - 1/16*(e^2*f^2 + 2*d*e*f*g + d^2*g^2)*log(e*x - d)/(d^4*e^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.65 \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=\frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{16 \, d^{4} e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{16 \, d^{4} e^{3}} - \frac {10 \, d^{3} e^{2} f^{2} + 4 \, d^{4} e f g - 2 \, d^{5} g^{2} + 3 \, {\left (d e^{4} f^{2} + 2 \, d^{2} e^{3} f g + d^{3} e^{2} g^{2}\right )} x^{2} + 3 \, {\left (3 \, d^{2} e^{3} f^{2} + 6 \, d^{3} e^{2} f g - d^{4} e g^{2}\right )} x}{24 \, {\left (e x + d\right )}^{3} d^{4} e^{3}} \]

[In]

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

[Out]

1/16*(e^2*f^2 + 2*d*e*f*g + d^2*g^2)*log(abs(e*x + d))/(d^4*e^3) - 1/16*(e^2*f^2 + 2*d*e*f*g + d^2*g^2)*log(ab
s(e*x - d))/(d^4*e^3) - 1/24*(10*d^3*e^2*f^2 + 4*d^4*e*f*g - 2*d^5*g^2 + 3*(d*e^4*f^2 + 2*d^2*e^3*f*g + d^3*e^
2*g^2)*x^2 + 3*(3*d^2*e^3*f^2 + 6*d^3*e^2*f*g - d^4*e*g^2)*x)/((e*x + d)^3*d^4*e^3)

Mupad [B] (verification not implemented)

Time = 12.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.35 \[ \int \frac {(f+g x)^2}{(d+e x)^3 \left (d^2-e^2 x^2\right )} \, dx=\frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,{\left (d\,g+e\,f\right )}^2}{8\,d^4\,e^3}-\frac {\frac {-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2}{12\,d\,e^3}+\frac {x\,\left (-d^2\,g^2+6\,d\,e\,f\,g+3\,e^2\,f^2\right )}{8\,d^2\,e^2}+\frac {x^2\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{8\,d^3\,e}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \]

[In]

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

[Out]

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