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

Optimal. Leaf size=74 \[ \frac {(e f-d g) (d g+e f) \tanh ^{-1}\left (\frac {e x}{d}\right )}{2 d^3 e^3}+\frac {(f+g x) \left (d^2 g+e^2 f x\right )}{2 d^2 e^2 \left (d^2-e^2 x^2\right )} \]

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Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {723, 208} \begin {gather*} \frac {(f+g x) \left (d^2 g+e^2 f x\right )}{2 d^2 e^2 \left (d^2-e^2 x^2\right )}+\frac {(e f-d g) (d g+e f) \tanh ^{-1}\left (\frac {e x}{d}\right )}{2 d^3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^2/(d^2 - e^2*x^2)^2,x]

[Out]

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

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 723

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[((2*p + 3)*(c*d^2 + a*e^2))/(2*a*c*(p + 1)), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 85, normalized size = 1.15 \begin {gather*} \frac {-2 d^2 f g-d^2 g^2 x-e^2 f^2 x}{2 d^2 e^2 \left (e^2 x^2-d^2\right )}-\frac {\left (d^2 g^2-e^2 f^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{2 d^3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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fricas [B]  time = 0.41, size = 155, normalized size = 2.09 \begin {gather*} -\frac {4 \, d^{3} e f g + 2 \, {\left (d e^{3} f^{2} + d^{3} e g^{2}\right )} x + {\left (d^{2} e^{2} f^{2} - d^{4} g^{2} - {\left (e^{4} f^{2} - d^{2} e^{2} g^{2}\right )} x^{2}\right )} \log \left (e x + d\right ) - {\left (d^{2} e^{2} f^{2} - d^{4} g^{2} - {\left (e^{4} f^{2} - d^{2} e^{2} g^{2}\right )} x^{2}\right )} \log \left (e x - d\right )}{4 \, {\left (d^{3} e^{5} x^{2} - d^{5} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 101, normalized size = 1.36 \begin {gather*} \frac {{\left (d^{2} g^{2} - f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (\frac {{\left | 2 \, x e^{2} - 2 \, {\left | d \right |} e \right |}}{{\left | 2 \, x e^{2} + 2 \, {\left | d \right |} e \right |}}\right )}{4 \, d^{2} {\left | d \right |}} - \frac {{\left (d^{2} g^{2} x + 2 \, d^{2} f g + f^{2} x e^{2}\right )} e^{\left (-2\right )}}{2 \, {\left (x^{2} e^{2} - d^{2}\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 180, normalized size = 2.43 \begin {gather*} -\frac {f g}{2 \left (e x -d \right ) d \,e^{2}}+\frac {f g}{2 \left (e x +d \right ) d \,e^{2}}+\frac {g^{2} \ln \left (e x -d \right )}{4 d \,e^{3}}-\frac {g^{2} \ln \left (e x +d \right )}{4 d \,e^{3}}-\frac {f^{2}}{4 \left (e x -d \right ) d^{2} e}-\frac {f^{2}}{4 \left (e x +d \right ) d^{2} e}-\frac {f^{2} \ln \left (e x -d \right )}{4 d^{3} e}+\frac {f^{2} \ln \left (e x +d \right )}{4 d^{3} e}-\frac {g^{2}}{4 \left (e x -d \right ) e^{3}}-\frac {g^{2}}{4 \left (e x +d \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 111, normalized size = 1.50 \begin {gather*} -\frac {2 \, d^{2} f g + {\left (e^{2} f^{2} + d^{2} g^{2}\right )} x}{2 \, {\left (d^{2} e^{4} x^{2} - d^{4} e^{2}\right )}} + \frac {{\left (e^{2} f^{2} - d^{2} g^{2}\right )} \log \left (e x + d\right )}{4 \, d^{3} e^{3}} - \frac {{\left (e^{2} f^{2} - d^{2} g^{2}\right )} \log \left (e x - d\right )}{4 \, d^{3} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.61, size = 115, normalized size = 1.55 \begin {gather*} \frac {\frac {f\,g}{e^2}+\frac {x\,\left (d^2\,g^2+e^2\,f^2\right )}{2\,d^2\,e^2}}{d^2-e^2\,x^2}-\frac {2\,\mathrm {atanh}\left (\frac {4\,e\,x\,\left (\frac {d^2\,g^2}{4}-\frac {e^2\,f^2}{4}\right )}{d\,\left (d^2\,g^2-e^2\,f^2\right )}\right )\,\left (\frac {d^2\,g^2}{4}-\frac {e^2\,f^2}{4}\right )}{d^3\,e^3} \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,x)

[Out]

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

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sympy [B]  time = 0.71, size = 156, normalized size = 2.11 \begin {gather*} \frac {- 2 d^{2} f g + x \left (- d^{2} g^{2} - e^{2} f^{2}\right )}{- 2 d^{4} e^{2} + 2 d^{2} e^{4} x^{2}} + \frac {\left (d g - e f\right ) \left (d g + e f\right ) \log {\left (- \frac {d \left (d g - e f\right ) \left (d g + e f\right )}{e \left (d^{2} g^{2} - e^{2} f^{2}\right )} + x \right )}}{4 d^{3} e^{3}} - \frac {\left (d g - e f\right ) \left (d g + e f\right ) \log {\left (\frac {d \left (d g - e f\right ) \left (d g + e f\right )}{e \left (d^{2} g^{2} - e^{2} f^{2}\right )} + x \right )}}{4 d^{3} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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