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

Optimal. Leaf size=188 \[ \frac {f (d g+e f)}{8 d^5 e^2 (d-e x)}-\frac {(d g+3 e f) (e f-d g)}{32 d^4 e^3 (d+e x)^2}+\frac {(d g+e f)^2}{32 d^4 e^3 (d-e x)^2}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}+\frac {\left (-d^2 g^2+2 d e f g+5 e^2 f^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^6 e^3}-\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)} \]

[Out]

1/32*(d*g+e*f)^2/d^4/e^3/(-e*x+d)^2+1/8*f*(d*g+e*f)/d^5/e^2/(-e*x+d)-1/24*(-d*g+e*f)^2/d^3/e^3/(e*x+d)^3-1/32*
(-d*g+e*f)*(d*g+3*e*f)/d^4/e^3/(e*x+d)^2+1/16*(d^2*g^2-3*e^2*f^2)/d^5/e^3/(e*x+d)+1/16*(-d^2*g^2+2*d*e*f*g+5*e
^2*f^2)*arctanh(e*x/d)/d^6/e^3

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Rubi [A]  time = 0.21, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {848, 88, 208} \[ -\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)}+\frac {\left (-d^2 g^2+2 d e f g+5 e^2 f^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^6 e^3}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}-\frac {(d g+3 e f) (e f-d g)}{32 d^4 e^3 (d+e x)^2}+\frac {f (d g+e f)}{8 d^5 e^2 (d-e x)}+\frac {(d g+e f)^2}{32 d^4 e^3 (d-e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*f + d*g)^2/(32*d^4*e^3*(d - e*x)^2) + (f*(e*f + d*g))/(8*d^5*e^2*(d - e*x)) - (e*f - d*g)^2/(24*d^3*e^3*(d
+ e*x)^3) - ((e*f - d*g)*(3*e*f + d*g))/(32*d^4*e^3*(d + e*x)^2) - (3*e^2*f^2 - d^2*g^2)/(16*d^5*e^3*(d + e*x)
) + ((5*e^2*f^2 + 2*d*e*f*g - d^2*g^2)*ArcTanh[(e*x)/d])/(16*d^6*e^3)

Rule 88

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 208

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

Rule 848

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*x)/e)^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*} \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^3} \, dx &=\int \frac {(f+g x)^2}{(d-e x)^3 (d+e x)^4} \, dx\\ &=\int \left (\frac {(e f+d g)^2}{16 d^4 e^2 (d-e x)^3}+\frac {f (e f+d g)}{8 d^5 e (d-e x)^2}+\frac {(-e f+d g)^2}{8 d^3 e^2 (d+e x)^4}+\frac {(e f-d g) (3 e f+d g)}{16 d^4 e^2 (d+e x)^3}+\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^2 (d+e x)^2}+\frac {-5 e^2 f^2-2 d e f g+d^2 g^2}{16 d^5 e^2 \left (-d^2+e^2 x^2\right )}\right ) \, dx\\ &=\frac {(e f+d g)^2}{32 d^4 e^3 (d-e x)^2}+\frac {f (e f+d g)}{8 d^5 e^2 (d-e x)}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}-\frac {(e f-d g) (3 e f+d g)}{32 d^4 e^3 (d+e x)^2}-\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)}-\frac {\left (5 e^2 f^2+2 d e f g-d^2 g^2\right ) \int \frac {1}{-d^2+e^2 x^2} \, dx}{16 d^5 e^2}\\ &=\frac {(e f+d g)^2}{32 d^4 e^3 (d-e x)^2}+\frac {f (e f+d g)}{8 d^5 e^2 (d-e x)}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}-\frac {(e f-d g) (3 e f+d g)}{32 d^4 e^3 (d+e x)^2}-\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)}+\frac {\left (5 e^2 f^2+2 d e f g-d^2 g^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^6 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 197, normalized size = 1.05 \[ \frac {-\frac {4 d^3 (e f-d g)^2}{(d+e x)^3}+\frac {3 d^2 \left (d^2 g^2+2 d e f g-3 e^2 f^2\right )}{(d+e x)^2}+\frac {6 d \left (d^2 g^2-3 e^2 f^2\right )}{d+e x}+3 \left (d^2 g^2-2 d e f g-5 e^2 f^2\right ) \log (d-e x)+3 \left (-d^2 g^2+2 d e f g+5 e^2 f^2\right ) \log (d+e x)+\frac {3 d^2 (d g+e f)^2}{(d-e x)^2}+\frac {12 d e f (d g+e f)}{d-e x}}{96 d^6 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*d^2*(e*f + d*g)^2)/(d - e*x)^2 + (12*d*e*f*(e*f + d*g))/(d - e*x) - (4*d^3*(e*f - d*g)^2)/(d + e*x)^3 + (3
*d^2*(-3*e^2*f^2 + 2*d*e*f*g + d^2*g^2))/(d + e*x)^2 + (6*d*(-3*e^2*f^2 + d^2*g^2))/(d + e*x) + 3*(-5*e^2*f^2
- 2*d*e*f*g + d^2*g^2)*Log[d - e*x] + 3*(5*e^2*f^2 + 2*d*e*f*g - d^2*g^2)*Log[d + e*x])/(96*d^6*e^3)

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fricas [B]  time = 0.68, size = 662, normalized size = 3.52 \[ -\frac {16 \, d^{5} e^{2} f^{2} - 32 \, d^{6} e f g - 8 \, d^{7} g^{2} + 6 \, {\left (5 \, d e^{6} f^{2} + 2 \, d^{2} e^{5} f g - d^{3} e^{4} g^{2}\right )} x^{4} + 6 \, {\left (5 \, d^{2} e^{5} f^{2} + 2 \, d^{3} e^{4} f g - d^{4} e^{3} g^{2}\right )} x^{3} - 10 \, {\left (5 \, d^{3} e^{4} f^{2} + 2 \, d^{4} e^{3} f g - d^{5} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (25 \, d^{4} e^{3} f^{2} + 10 \, d^{5} e^{2} f g + 7 \, d^{6} e g^{2}\right )} x - 3 \, {\left (5 \, d^{5} e^{2} f^{2} + 2 \, d^{6} e f g - d^{7} g^{2} + {\left (5 \, e^{7} f^{2} + 2 \, d e^{6} f g - d^{2} e^{5} g^{2}\right )} x^{5} + {\left (5 \, d e^{6} f^{2} + 2 \, d^{2} e^{5} f g - d^{3} e^{4} g^{2}\right )} x^{4} - 2 \, {\left (5 \, d^{2} e^{5} f^{2} + 2 \, d^{3} e^{4} f g - d^{4} e^{3} g^{2}\right )} x^{3} - 2 \, {\left (5 \, d^{3} e^{4} f^{2} + 2 \, d^{4} e^{3} f g - d^{5} e^{2} g^{2}\right )} x^{2} + {\left (5 \, d^{4} e^{3} f^{2} + 2 \, d^{5} e^{2} f g - d^{6} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + 3 \, {\left (5 \, d^{5} e^{2} f^{2} + 2 \, d^{6} e f g - d^{7} g^{2} + {\left (5 \, e^{7} f^{2} + 2 \, d e^{6} f g - d^{2} e^{5} g^{2}\right )} x^{5} + {\left (5 \, d e^{6} f^{2} + 2 \, d^{2} e^{5} f g - d^{3} e^{4} g^{2}\right )} x^{4} - 2 \, {\left (5 \, d^{2} e^{5} f^{2} + 2 \, d^{3} e^{4} f g - d^{4} e^{3} g^{2}\right )} x^{3} - 2 \, {\left (5 \, d^{3} e^{4} f^{2} + 2 \, d^{4} e^{3} f g - d^{5} e^{2} g^{2}\right )} x^{2} + {\left (5 \, d^{4} e^{3} f^{2} + 2 \, d^{5} e^{2} f g - d^{6} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{96 \, {\left (d^{6} e^{8} x^{5} + d^{7} e^{7} x^{4} - 2 \, d^{8} e^{6} x^{3} - 2 \, d^{9} e^{5} x^{2} + d^{10} e^{4} x + d^{11} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(16*d^5*e^2*f^2 - 32*d^6*e*f*g - 8*d^7*g^2 + 6*(5*d*e^6*f^2 + 2*d^2*e^5*f*g - d^3*e^4*g^2)*x^4 + 6*(5*d^
2*e^5*f^2 + 2*d^3*e^4*f*g - d^4*e^3*g^2)*x^3 - 10*(5*d^3*e^4*f^2 + 2*d^4*e^3*f*g - d^5*e^2*g^2)*x^2 - 2*(25*d^
4*e^3*f^2 + 10*d^5*e^2*f*g + 7*d^6*e*g^2)*x - 3*(5*d^5*e^2*f^2 + 2*d^6*e*f*g - d^7*g^2 + (5*e^7*f^2 + 2*d*e^6*
f*g - d^2*e^5*g^2)*x^5 + (5*d*e^6*f^2 + 2*d^2*e^5*f*g - d^3*e^4*g^2)*x^4 - 2*(5*d^2*e^5*f^2 + 2*d^3*e^4*f*g -
d^4*e^3*g^2)*x^3 - 2*(5*d^3*e^4*f^2 + 2*d^4*e^3*f*g - d^5*e^2*g^2)*x^2 + (5*d^4*e^3*f^2 + 2*d^5*e^2*f*g - d^6*
e*g^2)*x)*log(e*x + d) + 3*(5*d^5*e^2*f^2 + 2*d^6*e*f*g - d^7*g^2 + (5*e^7*f^2 + 2*d*e^6*f*g - d^2*e^5*g^2)*x^
5 + (5*d*e^6*f^2 + 2*d^2*e^5*f*g - d^3*e^4*g^2)*x^4 - 2*(5*d^2*e^5*f^2 + 2*d^3*e^4*f*g - d^4*e^3*g^2)*x^3 - 2*
(5*d^3*e^4*f^2 + 2*d^4*e^3*f*g - d^5*e^2*g^2)*x^2 + (5*d^4*e^3*f^2 + 2*d^5*e^2*f*g - d^6*e*g^2)*x)*log(e*x - d
))/(d^6*e^8*x^5 + d^7*e^7*x^4 - 2*d^8*e^6*x^3 - 2*d^9*e^5*x^2 + d^10*e^4*x + d^11*e^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -(d^2*exp(1)^4*g^2-2*d*exp(1)^5*g*f+exp(
1)^6*f^2)/(exp(2)^3*d^6*exp(1)-3*exp(2)^2*d^6*exp(1)^3+3*exp(2)*d^6*exp(1)^5-d^6*exp(1)^7)*ln(abs(x*exp(1)+d))
-(-d^2*exp(1)^3*g^2+2*d*exp(1)^4*g*f-exp(1)^5*f^2)/(2*exp(2)^3*d^6-6*exp(2)^2*d^6*exp(1)^2+6*exp(2)*d^6*exp(1)
^4-2*d^6*exp(1)^6)*ln(abs(-x^2*exp(2)+d^2))-(-3*exp(2)^3*f^2+exp(2)^2*d^2*g^2-2*exp(2)^2*d*exp(1)*g*f+10*exp(2
)^2*exp(1)^2*f^2-6*exp(2)*d^2*exp(1)^2*g^2+12*exp(2)*d*exp(1)^3*g*f-15*exp(2)*exp(1)^4*f^2-3*d^2*exp(1)^4*g^2+
6*d*exp(1)^5*g*f)*1/2/(8*exp(2)^3*d^5-24*exp(2)^2*d^5*exp(1)^2+24*exp(2)*d^5*exp(1)^4-8*d^5*exp(1)^6)/exp(1)/a
bs(d)*ln(abs(-2*x*exp(2)-2*exp(1)*abs(d))/abs(-2*x*exp(2)+2*exp(1)*abs(d)))-((3*exp(2)^5*d*f^2-exp(2)^4*d^3*g^
2+2*exp(2)^4*d^2*exp(1)*g*f-10*exp(2)^4*d*exp(1)^2*f^2-2*exp(2)^3*d^3*exp(1)^2*g^2+4*exp(2)^3*d^2*exp(1)^3*g*f
+7*exp(2)^3*d*exp(1)^4*f^2+3*exp(2)^2*d^3*exp(1)^4*g^2-6*exp(2)^2*d^2*exp(1)^5*g*f)*x^3+(4*exp(2)^3*d^4*exp(1)
*g^2-8*exp(2)^3*d^3*exp(1)^2*g*f+4*exp(2)^3*d^2*exp(1)^3*f^2-4*exp(2)^2*d^4*exp(1)^3*g^2+8*exp(2)^2*d^3*exp(1)
^4*g*f-4*exp(2)^2*d^2*exp(1)^5*f^2)*x^2+(-5*exp(2)^4*d^3*f^2-exp(2)^3*d^5*g^2+2*exp(2)^3*d^4*exp(1)*g*f+14*exp
(2)^3*d^3*exp(1)^2*f^2+6*exp(2)^2*d^5*exp(1)^2*g^2-12*exp(2)^2*d^4*exp(1)^3*g*f-9*exp(2)^2*d^3*exp(1)^4*f^2-5*
exp(2)*d^5*exp(1)^4*g^2+10*exp(2)*d^4*exp(1)^5*g*f)*x-4*exp(2)^3*d^5*g*f+2*exp(2)^3*d^4*exp(1)*f^2-2*exp(2)^2*
d^6*exp(1)*g^2+16*exp(2)^2*d^5*exp(1)^2*g*f-8*exp(2)^2*d^4*exp(1)^3*f^2-12*exp(2)*d^5*exp(1)^4*g*f+6*exp(2)*d^
4*exp(1)^5*f^2+2*d^6*exp(1)^5*g^2)/8/d^6/exp(2)/(exp(2)-exp(1)^2)^3/(-x^2*exp(2)+d^2)^2

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maple [A]  time = 0.02, size = 348, normalized size = 1.85 \[ -\frac {g^{2}}{24 \left (e x +d \right )^{3} d \,e^{3}}+\frac {f g}{12 \left (e x +d \right )^{3} d^{2} e^{2}}-\frac {f^{2}}{24 \left (e x +d \right )^{3} d^{3} e}+\frac {g^{2}}{32 \left (e x -d \right )^{2} d^{2} e^{3}}+\frac {g^{2}}{32 \left (e x +d \right )^{2} d^{2} e^{3}}+\frac {f g}{16 \left (e x -d \right )^{2} d^{3} e^{2}}+\frac {f g}{16 \left (e x +d \right )^{2} d^{3} e^{2}}+\frac {f^{2}}{32 \left (e x -d \right )^{2} d^{4} e}-\frac {3 f^{2}}{32 \left (e x +d \right )^{2} d^{4} e}+\frac {g^{2}}{16 \left (e x +d \right ) d^{3} e^{3}}-\frac {f g}{8 \left (e x -d \right ) d^{4} e^{2}}+\frac {g^{2} \ln \left (e x -d \right )}{32 d^{4} e^{3}}-\frac {g^{2} \ln \left (e x +d \right )}{32 d^{4} e^{3}}-\frac {f^{2}}{8 \left (e x -d \right ) d^{5} e}-\frac {3 f^{2}}{16 \left (e x +d \right ) d^{5} e}-\frac {f g \ln \left (e x -d \right )}{16 d^{5} e^{2}}+\frac {f g \ln \left (e x +d \right )}{16 d^{5} e^{2}}-\frac {5 f^{2} \ln \left (e x -d \right )}{32 d^{6} e}+\frac {5 f^{2} \ln \left (e x +d \right )}{32 d^{6} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/e^3/d^2/(e*x-d)^2*g^2+1/16/e^2/d^3/(e*x-d)^2*f*g+1/32/e/d^4/(e*x-d)^2*f^2+1/32/d^4/e^3*g^2*ln(e*x-d)-1/16
/d^5/e^2*f*g*ln(e*x-d)-5/32/d^6/e*f^2*ln(e*x-d)-1/8/(e*x-d)/d^4/e^2*f*g-1/8/(e*x-d)/d^5/e*f^2+1/16/e^3/d^3/(e*
x+d)*g^2-3/16/(e*x+d)/d^5/e*f^2+1/32/(e*x+d)^2/d^2/e^3*g^2+1/16/(e*x+d)^2/d^3/e^2*f*g-3/32/(e*x+d)^2/d^4/e*f^2
-1/32/d^4/e^3*g^2*ln(e*x+d)+1/16/d^5/e^2*f*g*ln(e*x+d)+5/32/d^6/e*f^2*ln(e*x+d)-1/24/(e*x+d)^3/d/e^3*g^2+1/12/
e^2/d^2/(e*x+d)^3*f*g-1/24/(e*x+d)^3/d^3/e*f^2

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maxima [A]  time = 0.50, size = 308, normalized size = 1.64 \[ -\frac {8 \, d^{4} e^{2} f^{2} - 16 \, d^{5} e f g - 4 \, d^{6} g^{2} + 3 \, {\left (5 \, e^{6} f^{2} + 2 \, d e^{5} f g - d^{2} e^{4} g^{2}\right )} x^{4} + 3 \, {\left (5 \, d e^{5} f^{2} + 2 \, d^{2} e^{4} f g - d^{3} e^{3} g^{2}\right )} x^{3} - 5 \, {\left (5 \, d^{2} e^{4} f^{2} + 2 \, d^{3} e^{3} f g - d^{4} e^{2} g^{2}\right )} x^{2} - {\left (25 \, d^{3} e^{3} f^{2} + 10 \, d^{4} e^{2} f g + 7 \, d^{5} e g^{2}\right )} x}{48 \, {\left (d^{5} e^{8} x^{5} + d^{6} e^{7} x^{4} - 2 \, d^{7} e^{6} x^{3} - 2 \, d^{8} e^{5} x^{2} + d^{9} e^{4} x + d^{10} e^{3}\right )}} + \frac {{\left (5 \, e^{2} f^{2} + 2 \, d e f g - d^{2} g^{2}\right )} \log \left (e x + d\right )}{32 \, d^{6} e^{3}} - \frac {{\left (5 \, e^{2} f^{2} + 2 \, d e f g - d^{2} g^{2}\right )} \log \left (e x - d\right )}{32 \, d^{6} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(8*d^4*e^2*f^2 - 16*d^5*e*f*g - 4*d^6*g^2 + 3*(5*e^6*f^2 + 2*d*e^5*f*g - d^2*e^4*g^2)*x^4 + 3*(5*d*e^5*f
^2 + 2*d^2*e^4*f*g - d^3*e^3*g^2)*x^3 - 5*(5*d^2*e^4*f^2 + 2*d^3*e^3*f*g - d^4*e^2*g^2)*x^2 - (25*d^3*e^3*f^2
+ 10*d^4*e^2*f*g + 7*d^5*e*g^2)*x)/(d^5*e^8*x^5 + d^6*e^7*x^4 - 2*d^7*e^6*x^3 - 2*d^8*e^5*x^2 + d^9*e^4*x + d^
10*e^3) + 1/32*(5*e^2*f^2 + 2*d*e*f*g - d^2*g^2)*log(e*x + d)/(d^6*e^3) - 1/32*(5*e^2*f^2 + 2*d*e*f*g - d^2*g^
2)*log(e*x - d)/(d^6*e^3)

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mupad [B]  time = 2.68, size = 249, normalized size = 1.32 \[ \frac {\frac {d^2\,g^2+4\,d\,e\,f\,g-2\,e^2\,f^2}{12\,d\,e^3}-\frac {x^3\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{16\,d^4}-\frac {e\,x^4\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{16\,d^5}+\frac {x\,\left (7\,d^2\,g^2+10\,d\,e\,f\,g+25\,e^2\,f^2\right )}{48\,d^2\,e^2}+\frac {5\,x^2\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{48\,d^3\,e}}{d^5+d^4\,e\,x-2\,d^3\,e^2\,x^2-2\,d^2\,e^3\,x^3+d\,e^4\,x^4+e^5\,x^5}+\frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{16\,d^6\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^2*g^2 - 2*e^2*f^2 + 4*d*e*f*g)/(12*d*e^3) - (x^3*(5*e^2*f^2 - d^2*g^2 + 2*d*e*f*g))/(16*d^4) - (e*x^4*(5*e
^2*f^2 - d^2*g^2 + 2*d*e*f*g))/(16*d^5) + (x*(7*d^2*g^2 + 25*e^2*f^2 + 10*d*e*f*g))/(48*d^2*e^2) + (5*x^2*(5*e
^2*f^2 - d^2*g^2 + 2*d*e*f*g))/(48*d^3*e))/(d^5 + e^5*x^5 + d*e^4*x^4 - 2*d^3*e^2*x^2 - 2*d^2*e^3*x^3 + d^4*e*
x) + (atanh((e*x)/d)*(5*e^2*f^2 - d^2*g^2 + 2*d*e*f*g))/(16*d^6*e^3)

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sympy [A]  time = 1.83, size = 321, normalized size = 1.71 \[ - \frac {- 4 d^{6} g^{2} - 16 d^{5} e f g + 8 d^{4} e^{2} f^{2} + x^{4} \left (- 3 d^{2} e^{4} g^{2} + 6 d e^{5} f g + 15 e^{6} f^{2}\right ) + x^{3} \left (- 3 d^{3} e^{3} g^{2} + 6 d^{2} e^{4} f g + 15 d e^{5} f^{2}\right ) + x^{2} \left (5 d^{4} e^{2} g^{2} - 10 d^{3} e^{3} f g - 25 d^{2} e^{4} f^{2}\right ) + x \left (- 7 d^{5} e g^{2} - 10 d^{4} e^{2} f g - 25 d^{3} e^{3} f^{2}\right )}{48 d^{10} e^{3} + 48 d^{9} e^{4} x - 96 d^{8} e^{5} x^{2} - 96 d^{7} e^{6} x^{3} + 48 d^{6} e^{7} x^{4} + 48 d^{5} e^{8} x^{5}} + \frac {\left (d^{2} g^{2} - 2 d e f g - 5 e^{2} f^{2}\right ) \log {\left (- \frac {d}{e} + x \right )}}{32 d^{6} e^{3}} - \frac {\left (d^{2} g^{2} - 2 d e f g - 5 e^{2} f^{2}\right ) \log {\left (\frac {d}{e} + x \right )}}{32 d^{6} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-4*d**6*g**2 - 16*d**5*e*f*g + 8*d**4*e**2*f**2 + x**4*(-3*d**2*e**4*g**2 + 6*d*e**5*f*g + 15*e**6*f**2) + x
**3*(-3*d**3*e**3*g**2 + 6*d**2*e**4*f*g + 15*d*e**5*f**2) + x**2*(5*d**4*e**2*g**2 - 10*d**3*e**3*f*g - 25*d*
*2*e**4*f**2) + x*(-7*d**5*e*g**2 - 10*d**4*e**2*f*g - 25*d**3*e**3*f**2))/(48*d**10*e**3 + 48*d**9*e**4*x - 9
6*d**8*e**5*x**2 - 96*d**7*e**6*x**3 + 48*d**6*e**7*x**4 + 48*d**5*e**8*x**5) + (d**2*g**2 - 2*d*e*f*g - 5*e**
2*f**2)*log(-d/e + x)/(32*d**6*e**3) - (d**2*g**2 - 2*d*e*f*g - 5*e**2*f**2)*log(d/e + x)/(32*d**6*e**3)

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