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

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Rubi [A]  time = 0.08, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 31

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

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 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])

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.91, size = 76, normalized size = 1.23 \[ -\frac {2 \, d e g^{2} x - {\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right ) + {\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, d e^{3}} \]

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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giac [A]  time = 0.15, size = 81, normalized size = 1.31 \[ -g^{2} x e^{\left (-2\right )} - f g e^{\left (-2\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac {{\left (d^{2} g^{2} + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (\frac {{\left | 2 \, x e^{2} - 2 \, {\left | d \right |} e \right |}}{{\left | 2 \, x e^{2} + 2 \, {\left | d \right |} e \right |}}\right )}{2 \, {\left | d \right |}} \]

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)

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maple [A]  time = 0.01, size = 107, normalized size = 1.73 \[ -\frac {d \,g^{2} \ln \left (e x -d \right )}{2 e^{3}}+\frac {d \,g^{2} \ln \left (e x +d \right )}{2 e^{3}}-\frac {f^{2} \ln \left (e x -d \right )}{2 d e}+\frac {f^{2} \ln \left (e x +d \right )}{2 d e}-\frac {f g \ln \left (e x -d \right )}{e^{2}}-\frac {f g \ln \left (e x +d \right )}{e^{2}}-\frac {g^{2} x}{e^{2}} \]

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/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+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 = 0.44, size = 82, normalized size = 1.32 \[ -\frac {g^{2} x}{e^{2}} + \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{2 \, d e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, d e^{3}} \]

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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mupad [B]  time = 0.15, size = 81, normalized size = 1.31 \[ \frac {\ln \left (d+e\,x\right )\,\left (d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2\right )}{2\,d\,e^3}-\frac {g^2\,x}{e^2}-\frac {\ln \left (d-e\,x\right )\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{2\,d\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)^2/(d^2 - e^2*x^2),x)

[Out]

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

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sympy [B]  time = 0.64, size = 112, normalized size = 1.81 \[ - \frac {g^{2} x}{e^{2}} + \frac {\left (d g - e f\right )^{2} \log {\left (x + \frac {2 d^{2} f g + \frac {d \left (d g - e f\right )^{2}}{e}}{d^{2} g^{2} + e^{2} f^{2}} \right )}}{2 d e^{3}} - \frac {\left (d g + e f\right )^{2} \log {\left (x + \frac {2 d^{2} f g - \frac {d \left (d g + e f\right )^{2}}{e}}{d^{2} g^{2} + e^{2} f^{2}} \right )}}{2 d e^{3}} \]

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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