3.6.52 \(\int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx\) [552]

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

[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 = 0.05, 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 = {716, 647, 31} \begin {gather*} -\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} \end {gather*}

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 647

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 716

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 \begin {gather*} \frac {\left (e^2 f^2+d^2 g^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )-d e g \left (g x+f \log \left (d^2-e^2 x^2\right )\right )}{d e^3} \end {gather*}

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 = 0.09, size = 84, normalized size = 1.35

method	result	size
norman	\(-\frac {g^{2} x}{e^{2}}+\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{2 e^{3} d}-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{2 d \,e^{3}}\)	\(82\)
default	\(-\frac {g^{2} x}{e^{2}}+\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{2 e^{3} d}+\frac {\left (-d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{2 d \,e^{3}}\)	\(84\)
risch	\(-\frac {g^{2} x}{e^{2}}-\frac {d \ln \left (e x -d \right ) g^{2}}{2 e^{3}}-\frac {\ln \left (e x -d \right ) f g}{e^{2}}-\frac {\ln \left (e x -d \right ) f^{2}}{2 e d}+\frac {d \ln \left (-e x -d \right ) g^{2}}{2 e^{3}}-\frac {\ln \left (-e x -d \right ) f g}{e^{2}}+\frac {\ln \left (-e x -d \right ) f^{2}}{2 e d}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 81, normalized size = 1.31 \begin {gather*} -g^{2} x e^{\left (-2\right )} + \frac {{\left (d^{2} g^{2} - 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e + d\right )}{2 \, d} - \frac {{\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right )}{2 \, d} \end {gather*}

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

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Fricas [A]
time = 1.80, size = 87, normalized size = 1.40 \begin {gather*} -\frac {{\left (2 \, d g^{2} x e^{2} + 2 \, d f g e^{2} \log \left (x^{2} e^{2} - d^{2}\right ) - {\left (d^{2} g^{2} + f^{2} e^{2}\right )} e \log \left (\frac {x^{2} e^{2} + 2 \, d x e + d^{2}}{x^{2} e^{2} - d^{2}}\right )\right )} e^{\left (-4\right )}}{2 \, d} \end {gather*}

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (51) = 102\).
time = 0.29, size = 112, normalized size = 1.81 \begin {gather*} - \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 {gather*}

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 = 0.92, size = 81, normalized size = 1.31 \begin {gather*} -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 {gather*}

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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Mupad [B]
time = 0.15, size = 81, normalized size = 1.31 \begin {gather*} \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} \end {gather*}

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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