\(\int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx\) [550]

Optimal result

Integrand size = 29, antiderivative size = 65 \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {2 d g (e f+d g) x}{e^2}-\frac {d (f+g x)^2}{e}-\frac {(f+g x)^3}{3 g}-\frac {2 d (e f+d g)^2 \log (d-e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, 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, 78} \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {2 d (d g+e f)^2 \log (d-e x)}{e^3}-\frac {2 d g x (d g+e f)}{e^2}-\frac {d (f+g x)^2}{e}-\frac {(f+g x)^3}{3 g} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e x \left (6 d^2 g^2+3 d e g (4 f+g x)+e^2 \left (3 f^2+3 f g x+g^2 x^2\right )\right )+6 d (e f+d g)^2 \log (d-e x)}{3 e^3} \]

[In]

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

[Out]

-1/3*(e*x*(6*d^2*g^2 + 3*d*e*g*(4*f + g*x) + e^2*(3*f^2 + 3*f*g*x + g^2*x^2)) + 6*d*(e*f + d*g)^2*Log[d - e*x]
)/e^3

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.35

method	result	size
norman	\(-\frac {g^{2} x^{3}}{3}-\frac {\left (2 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right ) x}{e^{2}}-\frac {g \left (d g +e f \right ) x^{2}}{e}-\frac {2 d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(88\)
default	\(-\frac {\frac {1}{3} g^{2} x^{3} e^{2}+d e \,g^{2} x^{2}+e^{2} f g \,x^{2}+2 d^{2} g^{2} x +4 d e f g x +e^{2} f^{2} x}{e^{2}}-\frac {2 d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(95\)
risch	\(-\frac {g^{2} x^{3}}{3}-\frac {d \,g^{2} x^{2}}{e}-f g \,x^{2}-\frac {2 d^{2} g^{2} x}{e^{2}}-\frac {4 d f g x}{e}-f^{2} x -\frac {2 d^{3} \ln \left (-e x +d \right ) g^{2}}{e^{3}}-\frac {4 d^{2} \ln \left (-e x +d \right ) f g}{e^{2}}-\frac {2 d \ln \left (-e x +d \right ) f^{2}}{e}\)	\(107\)
parallelrisch	\(-\frac {g^{2} x^{3} e^{3}+3 x^{2} d \,e^{2} g^{2}+3 x^{2} e^{3} f g +6 \ln \left (e x -d \right ) d^{3} g^{2}+12 \ln \left (e x -d \right ) d^{2} e f g +6 \ln \left (e x -d \right ) d \,e^{2} f^{2}+6 x \,d^{2} e \,g^{2}+12 x d \,e^{2} f g +3 x \,e^{3} f^{2}}{3 e^{3}}\)	\(116\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.51 \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e^{3} g^{2} x^{3} + 3 \, {\left (e^{3} f g + d e^{2} g^{2}\right )} x^{2} + 3 \, {\left (e^{3} f^{2} + 4 \, d e^{2} f g + 2 \, d^{2} e g^{2}\right )} x + 6 \, {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}\right )} \log \left (e x - d\right )}{3 \, e^{3}} \]

[In]

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

[Out]

-1/3*(e^3*g^2*x^3 + 3*(e^3*f*g + d*e^2*g^2)*x^2 + 3*(e^3*f^2 + 4*d*e^2*f*g + 2*d^2*e*g^2)*x + 6*(d*e^2*f^2 + 2
*d^2*e*f*g + d^3*g^2)*log(e*x - d))/e^3

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=- \frac {2 d \left (d g + e f\right )^{2} \log {\left (- d + e x \right )}}{e^{3}} - \frac {g^{2} x^{3}}{3} - x^{2} \left (\frac {d g^{2}}{e} + f g\right ) - x \left (\frac {2 d^{2} g^{2}}{e^{2}} + \frac {4 d f g}{e} + f^{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.49 \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {e^{2} g^{2} x^{3} + 3 \, {\left (e^{2} f g + d e g^{2}\right )} x^{2} + 3 \, {\left (e^{2} f^{2} + 4 \, d e f g + 2 \, d^{2} g^{2}\right )} x}{3 \, e^{2}} - \frac {2 \, {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.62 \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=-\frac {2 \, {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {e^{3} g^{2} x^{3} + 3 \, e^{3} f g x^{2} + 3 \, d e^{2} g^{2} x^{2} + 3 \, e^{3} f^{2} x + 12 \, d e^{2} f g x + 6 \, d^{2} e g^{2} x}{3 \, e^{3}} \]

[In]

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

[Out]

-2*(d*e^2*f^2 + 2*d^2*e*f*g + d^3*g^2)*log(abs(e*x - d))/e^3 - 1/3*(e^3*g^2*x^3 + 3*e^3*f*g*x^2 + 3*d*e^2*g^2*
x^2 + 3*e^3*f^2*x + 12*d*e^2*f*g*x + 6*d^2*e*g^2*x)/e^3

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.95 \[ \int \frac {(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx=-x^2\,\left (\frac {d\,g^2+2\,e\,f\,g}{2\,e}+\frac {d\,g^2}{2\,e}\right )-x\,\left (\frac {e\,f^2+2\,d\,g\,f}{e}+\frac {d\,\left (\frac {d\,g^2+2\,e\,f\,g}{e}+\frac {d\,g^2}{e}\right )}{e}\right )-\frac {g^2\,x^3}{3}-\frac {\ln \left (e\,x-d\right )\,\left (2\,d^3\,g^2+4\,d^2\,e\,f\,g+2\,d\,e^2\,f^2\right )}{e^3} \]

[In]

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

[Out]

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