\(\int \frac {(d+e x) (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\) [563]

Optimal result

Integrand size = 27, antiderivative size = 86 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {(e f+d g)^2}{2 d e^3 (d-e x)}-\frac {(e f-3 d g) (e f+d g) \log (d-e x)}{4 d^2 e^3}+\frac {(e f-d g)^2 \log (d+e x)}{4 d^2 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {813, 90} \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {(e f-d g)^2 \log (d+e x)}{4 d^2 e^3}-\frac {(e f-3 d g) (d g+e f) \log (d-e x)}{4 d^2 e^3}+\frac {(d g+e f)^2}{2 d e^3 (d-e x)} \]

[In]

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

[Out]

(e*f + d*g)^2/(2*d*e^3*(d - e*x)) - ((e*f - 3*d*g)*(e*f + d*g)*Log[d - e*x])/(4*d^2*e^3) + ((e*f - d*g)^2*Log[
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 813

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^m*(
f + g*x)^(p + 1)*(a/f + (c/g)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m}, x] && EqQ[c*f^2 + a*g^2, 0] && (Integer
Q[p] || (GtQ[a, 0] && GtQ[f, 0] && EqQ[p, -1]))

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {2 d (e f+d g)^2+\left (-e^2 f^2+2 d e f g+3 d^2 g^2\right ) (d-e x) \log (d-e x)+(e f-d g)^2 (d-e x) \log (d+e x)}{4 d^2 e^3 (d-e x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{2 d \,e^{3} \left (-e x +d \right )}+\frac {\left (3 d^{2} g^{2}+2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{4 d^{2} e^{3}}+\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{4 d^{2} e^{3}}\)	\(112\)
norman	\(\frac {-\frac {-d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{2 e^{3}}+\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) x}{2 d \,e^{2}}}{-e^{2} x^{2}+d^{2}}+\frac {\left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) \ln \left (e x +d \right )}{4 d^{2} e^{3}}+\frac {\left (3 d^{2} g^{2}+2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{4 d^{2} e^{3}}\)	\(149\)
risch	\(\frac {d \,g^{2}}{2 e^{3} \left (-e x +d \right )}+\frac {f g}{e^{2} \left (-e x +d \right )}+\frac {f^{2}}{2 d e \left (-e x +d \right )}+\frac {\ln \left (-e x -d \right ) g^{2}}{4 e^{3}}-\frac {\ln \left (-e x -d \right ) f g}{2 d \,e^{2}}+\frac {\ln \left (-e x -d \right ) f^{2}}{4 d^{2} e}+\frac {3 \ln \left (e x -d \right ) g^{2}}{4 e^{3}}+\frac {\ln \left (e x -d \right ) f g}{2 d \,e^{2}}-\frac {\ln \left (e x -d \right ) f^{2}}{4 d^{2} e}\)	\(161\)
parallelrisch	\(\frac {3 \ln \left (e x -d \right ) x \,d^{2} e \,g^{2}+2 \ln \left (e x -d \right ) x d \,e^{2} f g -\ln \left (e x -d \right ) x \,e^{3} f^{2}+\ln \left (e x +d \right ) x \,d^{2} e \,g^{2}-2 \ln \left (e x +d \right ) x d \,e^{2} f g +\ln \left (e x +d \right ) x \,e^{3} f^{2}-3 \ln \left (e x -d \right ) d^{3} g^{2}-2 \ln \left (e x -d \right ) d^{2} e f g +\ln \left (e x -d \right ) d \,e^{2} f^{2}-\ln \left (e x +d \right ) d^{3} g^{2}+2 \ln \left (e x +d \right ) d^{2} e f g -\ln \left (e x +d \right ) d \,e^{2} f^{2}-2 d^{3} g^{2}-4 d^{2} e f g -2 d \,e^{2} f^{2}}{4 d^{2} e^{3} \left (e x -d \right )}\)	\(231\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (81) = 162\).

Time = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.95 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=-\frac {2 \, d e^{2} f^{2} + 4 \, d^{2} e f g + 2 \, d^{3} g^{2} + {\left (d e^{2} f^{2} - 2 \, d^{2} e f g + d^{3} g^{2} - {\left (e^{3} f^{2} - 2 \, d e^{2} f g + d^{2} e g^{2}\right )} x\right )} \log \left (e x + d\right ) - {\left (d e^{2} f^{2} - 2 \, d^{2} e f g - 3 \, d^{3} g^{2} - {\left (e^{3} f^{2} - 2 \, d e^{2} f g - 3 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{4 \, {\left (d^{2} e^{4} x - d^{3} e^{3}\right )}} \]

[In]

integrate((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2)^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 + 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*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^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 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {- d^{2} g^{2} - 2 d e f g - e^{2} f^{2}}{- 2 d^{2} e^{3} + 2 d e^{4} x} + \frac {\left (d g - e f\right )^{2} \log {\left (x + \frac {2 d^{3} g^{2} - d \left (d g - e f\right )^{2}}{d^{2} e g^{2} + 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} + \frac {\left (d g + e f\right ) \left (3 d g - e f\right ) \log {\left (x + \frac {2 d^{3} g^{2} - d \left (d g + e f\right ) \left (3 d g - e f\right )}{d^{2} e g^{2} + 2 d e^{2} f g - e^{3} f^{2}} \right )}}{4 d^{2} e^{3}} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.33 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=-\frac {e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}}{2 \, {\left (d e^{4} x - d^{2} e^{3}\right )}} + \frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x + d\right )}{4 \, d^{2} e^{3}} - \frac {{\left (e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2}\right )} \log \left (e x - d\right )}{4 \, d^{2} e^{3}} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.36 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {{\left (e^{2} f^{2} - 2 \, d e f g + d^{2} g^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{4 \, d^{2} e^{3}} - \frac {{\left (e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{4 \, d^{2} e^{3}} - \frac {d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}}{2 \, {\left (e x - d\right )} d^{2} e^{3}} \]

[In]

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

[Out]

1/4*(e^2*f^2 - 2*d*e*f*g + d^2*g^2)*log(abs(e*x + d))/(d^2*e^3) - 1/4*(e^2*f^2 - 2*d*e*f*g - 3*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 = 11.88 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.29 \[ \int \frac {(d+e x) (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2}{2\,d\,e^3\,\left (d-e\,x\right )}+\frac {\ln \left (d+e\,x\right )\,\left (d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2\right )}{4\,d^2\,e^3}+\frac {\ln \left (d-e\,x\right )\,\left (3\,d^2\,g^2+2\,d\,e\,f\,g-e^2\,f^2\right )}{4\,d^2\,e^3} \]

[In]

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

[Out]

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