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

Optimal result

Integrand size = 22, antiderivative size = 62 \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {g^2 x}{e^2}-\frac {(e f+d g)^2 \log (d-e x)}{2 d e^3}+\frac {(e f-d g)^2 \log (d+e x)}{2 d e^3} \]

[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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {716, 647, 31} \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {(d g+e f)^2 \log (d-e x)}{2 d e^3}+\frac {(e f-d g)^2 \log (d+e x)}{2 d e^3}-\frac {g^2 x}{e^2} \]

[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*} \text {integral}& = \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] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89 \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=\frac {\left (e^2 f^2+d^2 g^2\right ) \text {arctanh}\left (\frac {e x}{d}\right )-d e g \left (g x+f \log \left (d^2-e^2 x^2\right )\right )}{d e^3} \]

[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)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.32

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	\(-\frac {g^{2} x}{e^{2}}+\frac {\left (-d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{2 d \,e^{3}}+\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{2 e^{3} d}\)	\(84\)
parallelrisch	\(-\frac {\ln \left (e x -d \right ) d^{2} g^{2}+2 \ln \left (e x -d \right ) d e f g +\ln \left (e x -d \right ) e^{2} f^{2}-\ln \left (e x +d \right ) d^{2} g^{2}+2 \ln \left (e x +d \right ) d e f g -\ln \left (e x +d \right ) e^{2} f^{2}+2 x d e \,g^{2}}{2 d \,e^{3}}\)	\(102\)
risch	\(-\frac {g^{2} x}{e^{2}}+\frac {d \ln \left (-e x -d \right ) g^{2}}{2 e^{3}}-\frac {\ln \left (-e x -d \right ) f g}{e^{2}}+\frac {\ln \left (-e x -d \right ) f^{2}}{2 e d}-\frac {d \ln \left (e x -d \right ) g^{2}}{2 e^{3}}-\frac {\ln \left (e x -d \right ) f g}{e^{2}}-\frac {\ln \left (e x -d \right ) f^{2}}{2 e d}\)	\(116\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.23 \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {2 \, d e g^{2} x - {\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right ) + {\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, d e^{3}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (51) = 102\).

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.81 \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=- \frac {g^{2} x}{e^{2}} + \frac {\left (d g - e f\right )^{2} \log {\left (x + \frac {2 d^{2} f g + \frac {d \left (d g - e f\right )^{2}}{e}}{d^{2} g^{2} + e^{2} f^{2}} \right )}}{2 d e^{3}} - \frac {\left (d g + e f\right )^{2} \log {\left (x + \frac {2 d^{2} f g - \frac {d \left (d g + e f\right )^{2}}{e}}{d^{2} g^{2} + e^{2} f^{2}} \right )}}{2 d e^{3}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.32 \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {g^{2} x}{e^{2}} + \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{2 \, d e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, d e^{3}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.35 \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {g^{2} x}{e^{2}} + \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{2 \, d e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{2 \, d e^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.31 \[ \int \frac {(f+g x)^2}{d^2-e^2 x^2} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2\right )}{2\,d\,e^3}-\frac {g^2\,x}{e^2}-\frac {\ln \left (d-e\,x\right )\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{2\,d\,e^3} \]

[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)