\(\int \frac {(f+g x)^2}{(d+e x) (d^2-e^2 x^2)} \, dx\) [553]

Optimal result

Integrand size = 29, antiderivative size = 86 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=-\frac {(e f-d g)^2}{2 d e^3 (d+e x)}-\frac {(e f+d g)^2 \log (d-e x)}{4 d^2 e^3}+\frac {(e f-d g) (e f+3 d g) \log (d+e x)}{4 d^2 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {862, 90} \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=\frac {(3 d g+e f) (e f-d g) \log (d+e x)}{4 d^2 e^3}-\frac {(d g+e f)^2 \log (d-e x)}{4 d^2 e^3}-\frac {(e f-d g)^2}{2 d e^3 (d+e x)} \]

[In]

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

[Out]

-1/2*(e*f - d*g)^2/(d*e^3*(d + e*x)) - ((e*f + d*g)^2*Log[d - e*x])/(4*d^2*e^3) + ((e*f - d*g)*(e*f + 3*d*g)*L
og[d + e*x])/(4*d^2*e^3)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rubi steps \begin{align*} \text {integral}& = \int \frac {(f+g x)^2}{(d-e x) (d+e x)^2} \, dx \\ & = \int \left (\frac {(e f+d g)^2}{4 d^2 e^2 (d-e x)}+\frac {(-e f+d g)^2}{2 d e^2 (d+e x)^2}+\frac {(e f-d g) (e f+3 d g)}{4 d^2 e^2 (d+e x)}\right ) \, dx \\ & = -\frac {(e f-d g)^2}{2 d e^3 (d+e x)}-\frac {(e f+d g)^2 \log (d-e x)}{4 d^2 e^3}+\frac {(e f-d g) (e f+3 d g) \log (d+e x)}{4 d^2 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=\frac {-(e f+d g)^2 (d+e x) \log (d-e x)+(e f-d g) (2 d (-e f+d g)+(e f+3 d g) (d+e x) \log (d+e x))}{4 d^2 e^3 (d+e x)} \]

[In]

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

[Out]

(-((e*f + d*g)^2*(d + e*x)*Log[d - e*x]) + (e*f - d*g)*(2*d*(-(e*f) + d*g) + (e*f + 3*d*g)*(d + e*x)*Log[d + e
*x]))/(4*d^2*e^3*(d + e*x))

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {\left (-d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{4 d^{2} e^{3}}+\frac {\left (-3 d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{4 d^{2} e^{3}}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{2 e^{3} d \left (e x +d \right )}\)	\(112\)
norman	\(\frac {-d^{2} g^{2}+2 d e f g -e^{2} f^{2}}{2 d \,e^{3} \left (e x +d \right )}-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{4 d^{2} e^{3}}-\frac {\left (3 d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (e x +d \right )}{4 d^{2} e^{3}}\)	\(113\)
risch	\(-\frac {d \,g^{2}}{2 e^{3} \left (e x +d \right )}+\frac {f g}{e^{2} \left (e x +d \right )}-\frac {f^{2}}{2 e d \left (e x +d \right )}-\frac {\ln \left (e x -d \right ) g^{2}}{4 e^{3}}-\frac {\ln \left (e x -d \right ) f g}{2 d \,e^{2}}-\frac {\ln \left (e x -d \right ) f^{2}}{4 d^{2} e}-\frac {3 \ln \left (-e x -d \right ) g^{2}}{4 e^{3}}+\frac {\ln \left (-e x -d \right ) f g}{2 d \,e^{2}}+\frac {\ln \left (-e x -d \right ) f^{2}}{4 d^{2} e}\)	\(158\)
parallelrisch	\(-\frac {\ln \left (e x -d \right ) x \,d^{2} e \,g^{2}+2 \ln \left (e x -d \right ) x d \,e^{2} f g +\ln \left (e x -d \right ) x \,e^{3} f^{2}+3 \ln \left (e x +d \right ) x \,d^{2} e \,g^{2}-2 \ln \left (e x +d \right ) x d \,e^{2} f g -\ln \left (e x +d \right ) x \,e^{3} f^{2}+\ln \left (e x -d \right ) d^{3} g^{2}+2 \ln \left (e x -d \right ) d^{2} e f g +\ln \left (e x -d \right ) d \,e^{2} f^{2}+3 \ln \left (e x +d \right ) d^{3} g^{2}-2 \ln \left (e x +d \right ) d^{2} e f g -\ln \left (e x +d \right ) d \,e^{2} f^{2}+2 d^{3} g^{2}-4 d^{2} e f g +2 d \,e^{2} f^{2}}{4 d^{2} e^{3} \left (e x +d \right )}\)	\(228\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (80) = 160\).

Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.92 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=-\frac {2 \, d e^{2} f^{2} - 4 \, d^{2} e f g + 2 \, d^{3} g^{2} - {\left (d e^{2} f^{2} + 2 \, d^{2} e f g - 3 \, d^{3} g^{2} + {\left (e^{3} f^{2} + 2 \, d e^{2} f g - 3 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x + d\right ) + {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2} + {\left (e^{3} f^{2} + 2 \, d e^{2} f g + d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{4 \, {\left (d^{2} e^{4} x + d^{3} e^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (75) = 150\).

Time = 0.47 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.12 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=- \frac {d^{2} g^{2} - 2 d e f g + e^{2} f^{2}}{2 d^{2} e^{3} + 2 d e^{4} x} - \frac {\left (d g - e f\right ) \left (3 d g + e f\right ) \log {\left (x + \frac {- 2 d^{3} g^{2} + d \left (d g - e f\right ) \left (3 d g + e f\right )}{d^{2} e g^{2} - 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} - \frac {\left (d g + e f\right )^{2} \log {\left (x + \frac {- 2 d^{3} g^{2} + d \left (d g + e f\right )^{2}}{d^{2} e g^{2} - 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.31 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=-\frac {e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}}{2 \, {\left (d e^{4} x + d^{2} e^{3}\right )}} + \frac {{\left (e^{2} f^{2} + 2 \, d e f g - 3 \, d^{2} g^{2}\right )} \log \left (e x + d\right )}{4 \, d^{2} e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{4 \, d^{2} e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.34 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=\frac {{\left (e^{2} f^{2} + 2 \, d e f g - 3 \, d^{2} g^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{4 \, d^{2} e^{3}} - \frac {{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{4 \, d^{2} e^{3}} - \frac {d e^{2} f^{2} - 2 \, d^{2} e f g + d^{3} g^{2}}{2 \, {\left (e x + d\right )} d^{2} e^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.02 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.27 \[ \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )} \, dx=\frac {\ln \left (d+e\,x\right )\,\left (-3\,d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^2\,e^3}-\frac {\ln \left (d-e\,x\right )\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^2\,e^3}-\frac {d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2}{2\,d\,e^3\,\left (d+e\,x\right )} \]

[In]

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

[Out]

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