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

Optimal. Leaf size=62 \[ -\frac{(d g+e f)^2 \log (d-e x)}{2 d e^3}+\frac{(e f-d g)^2 \log (d+e x)}{2 d e^3}-\frac{g^2 x}{e^2} \]

[Out]

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

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Rubi [A]  time = 0.0800188, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {702, 633, 31} \[ -\frac{(d g+e f)^2 \log (d-e x)}{2 d e^3}+\frac{(e f-d g)^2 \log (d+e x)}{2 d e^3}-\frac{g^2 x}{e^2} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^2/(d^2 - e^2*x^2),x]

[Out]

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

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0221898, size = 55, normalized size = 0.89 \[ \frac{\left (d^2 g^2+e^2 f^2\right ) \tanh ^{-1}\left (\frac{e x}{d}\right )-d e g \left (f \log \left (d^2-e^2 x^2\right )+g x\right )}{d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.049, size = 107, normalized size = 1.7 \begin{align*} -{\frac{{g}^{2}x}{{e}^{2}}}-{\frac{d\ln \left ( ex-d \right ){g}^{2}}{2\,{e}^{3}}}-{\frac{\ln \left ( ex-d \right ) fg}{{e}^{2}}}-{\frac{\ln \left ( ex-d \right ){f}^{2}}{2\,de}}+{\frac{d\ln \left ( ex+d \right ){g}^{2}}{2\,{e}^{3}}}-{\frac{\ln \left ( ex+d \right ) fg}{{e}^{2}}}+{\frac{\ln \left ( ex+d \right ){f}^{2}}{2\,de}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.992196, size = 111, normalized size = 1.79 \begin{align*} -\frac{g^{2} x}{e^{2}} + \frac{{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{2 \, d e^{3}} - \frac{{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.74218, size = 165, normalized size = 2.66 \begin{align*} -\frac{2 \, d e g^{2} x -{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right ) +{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.733054, size = 112, normalized size = 1.81 \begin{align*} - \frac{g^{2} x}{e^{2}} + \frac{\left (d g - e f\right )^{2} \log{\left (x + \frac{2 d^{2} f g + \frac{d \left (d g - e f\right )^{2}}{e}}{d^{2} g^{2} + e^{2} f^{2}} \right )}}{2 d e^{3}} - \frac{\left (d g + e f\right )^{2} \log{\left (x + \frac{2 d^{2} f g - \frac{d \left (d g + e f\right )^{2}}{e}}{d^{2} g^{2} + e^{2} f^{2}} \right )}}{2 d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12473, size = 109, normalized size = 1.76 \begin{align*} -g^{2} x e^{\left (-2\right )} - f g e^{\left (-2\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \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 )}{2 \,{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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