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

Optimal. Leaf size=146 \[ \frac {f (d g+e f) \tanh ^{-1}\left (\frac {e x}{d}\right )}{4 d^5 e^2}+\frac {(d g+e f)^2}{16 d^4 e^3 (d-e x)}-\frac {(3 e f-d g) (d g+e f)}{16 d^4 e^3 (d+e x)}-\frac {(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}-\frac {e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2} \]

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Rubi [A]  time = 0.16, antiderivative size = 146, 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} \begin {gather*} -\frac {e^2 f^2-d^2 g^2}{8 d^3 e^3 (d+e x)^2}-\frac {(e f-d g)^2}{12 d^2 e^3 (d+e x)^3}+\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 {f (d g+e f) \tanh ^{-1}\left (\frac {e x}{d}\right )}{4 d^5 e^2} \end {gather*}

Antiderivative was successfully verified.

[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 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)^2 \left (d^2-e^2 x^2\right )^2} \, dx &=\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*}

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Mathematica [A]  time = 0.09, size = 171, normalized size = 1.17 \begin {gather*} \frac {2 d \left (2 d^5 g^2+2 d^4 e g (f+2 g x)+d^3 e^2 f (g x-4 f)+d^2 e^3 f x (f+6 g x)+3 d e^4 f x^2 (2 f+g x)+3 e^5 f^2 x^3\right )+3 e f (e x-d) (d+e x)^3 (d g+e f) \log (d-e x)+3 e f (d-e x) (d+e x)^3 (d g+e f) \log (d+e x)}{24 d^5 e^3 (d-e x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[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)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(f+g x)^2}{(d+e x)^2 \left (d^2-e^2 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.38, size = 337, normalized size = 2.31 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (-(exp(1)*x+d)^-1/exp(1)*g^2*d^2*exp(1)^
6+2*(exp(1)*x+d)^-1/exp(1)*g*d*exp(1)^7*f-(exp(1)*x+d)^-1/exp(1)*exp(1)^8*f^2)/(d^4*exp(1)^8-2*d^4*exp(1)^6*ex
p(2)+d^4*exp(1)^4*exp(2)^2)-(-(-g^2*d^3*exp(1)^6-6*g^2*d^3*exp(1)^4*exp(2)-g^2*d^3*exp(1)^2*exp(2)^2+8*g*d^2*e
xp(1)^5*exp(2)*f+8*g*d^2*exp(1)^3*exp(2)^2*f-d*exp(1)^6*exp(2)*f^2-6*d*exp(1)^4*exp(2)^2*f^2-d*exp(1)^2*exp(2)
^3*f^2)/(exp(1)^2-exp(2))*(exp(1)*x+d)^-1/exp(1)+(-3*g^2*d^2*exp(1)^3*exp(2)-g^2*d^2*exp(1)*exp(2)^2+2*g*d*exp
(1)^4*exp(2)*f+6*g*d*exp(1)^2*exp(2)^2*f-3*exp(1)^3*exp(2)^2*f^2-exp(1)*exp(2)^3*f^2)/(exp(1)^2-exp(2)))/2/d^5
/(exp(2)-exp(1)^2)^2/(-(-(exp(1)*x+d)^-1/exp(1))^2*d^2*exp(1)^4+(-(exp(1)*x+d)^-1/exp(1))^2*d^2*exp(1)^2*exp(2
)-2*(exp(1)*x+d)^-1/exp(1)*d*exp(1)*exp(2)+exp(2))+(g^2*d^2*exp(1)^3+g^2*d^2*exp(1)*exp(2)-g*d*exp(1)^4*f-3*g*
d*exp(1)^2*f*exp(2)+2*exp(1)^3*f^2*exp(2))/(d^5*exp(1)^6-3*d^5*exp(1)^4*exp(2)+3*d^5*exp(1)^2*exp(2)^2-d^5*exp
(2)^3)*ln(abs(-(-(exp(1)*x+d)^-1/exp(1))^2*d^2*exp(1)^4+(-(exp(1)*x+d)^-1/exp(1))^2*d^2*exp(1)^2*exp(2)-2*(exp
(1)*x+d)^-1/exp(1)*d*exp(1)*exp(2)+exp(2)))+(-g^2*d^2*exp(1)^6-6*g^2*d^2*exp(1)^4*exp(2)-g^2*d^2*exp(1)^2*exp(
2)^2+12*g*d*exp(1)^5*f*exp(2)+4*g*d*exp(1)^3*f*exp(2)^2-3*exp(1)^6*f^2*exp(2)-6*exp(1)^4*f^2*exp(2)^2+exp(1)^2
*f^2*exp(2)^3)/2/(2*d^4*exp(1)^6-6*d^4*exp(1)^4*exp(2)+6*d^4*exp(1)^2*exp(2)^2-2*d^4*exp(2)^3)/exp(1)/abs(d)/e
xp(1)^2*ln(abs(2*(exp(1)*x+d)^-1/exp(1)*d^2*exp(1)^4-2*(exp(1)*x+d)^-1/exp(1)*d^2*exp(1)^2*exp(2)+2*d*exp(1)*e
xp(2)-2*exp(1)*abs(d)*exp(1)^2)/abs(2*(exp(1)*x+d)^-1/exp(1)*d^2*exp(1)^4-2*(exp(1)*x+d)^-1/exp(1)*d^2*exp(1)^
2*exp(2)+2*d*exp(1)*exp(2)+2*exp(1)*abs(d)*exp(1)^2))

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maple [A]  time = 0.02, size = 270, normalized size = 1.85 \begin {gather*} \frac {f g}{6 \left (e x +d \right )^{3} d \,e^{2}}-\frac {f^{2}}{12 \left (e x +d \right )^{3} d^{2} e}-\frac {g^{2}}{12 \left (e x +d \right )^{3} e^{3}}+\frac {g^{2}}{8 \left (e x +d \right )^{2} d \,e^{3}}-\frac {f^{2}}{8 \left (e x +d \right )^{2} d^{3} e}-\frac {g^{2}}{16 \left (e x -d \right ) d^{2} e^{3}}+\frac {g^{2}}{16 \left (e x +d \right ) d^{2} e^{3}}-\frac {f g}{8 \left (e x -d \right ) d^{3} e^{2}}-\frac {f g}{8 \left (e x +d \right ) d^{3} e^{2}}-\frac {f^{2}}{16 \left (e x -d \right ) d^{4} e}-\frac {3 f^{2}}{16 \left (e x +d \right ) d^{4} e}-\frac {f g \ln \left (e x -d \right )}{8 d^{4} e^{2}}+\frac {f g \ln \left (e x +d \right )}{8 d^{4} e^{2}}-\frac {f^{2} \ln \left (e x -d \right )}{8 d^{5} e}+\frac {f^{2} \ln \left (e x +d \right )}{8 d^{5} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 197, normalized size = 1.35 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.63, size = 148, normalized size = 1.01 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.36, size = 241, normalized size = 1.65 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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