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

Optimal. Leaf size=65 \[ -\frac{2 d g x (d g+e f)}{e^2}-\frac{2 d (d g+e f)^2 \log (d-e x)}{e^3}-\frac{d (f+g x)^2}{e}-\frac{(f+g x)^3}{3 g} \]

[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

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Rubi [A]  time = 0.0600605, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {848, 77} \[ -\frac{2 d g x (d g+e f)}{e^2}-\frac{2 d (d g+e f)^2 \log (d-e x)}{e^3}-\frac{d (f+g x)^2}{e}-\frac{(f+g x)^3}{3 g} \]

Antiderivative was successfully verified.

[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 848

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*x)/e)^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]))

Rule 77

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]))))

Rubi steps

\begin{align*} \int \frac{(d+e x)^2 (f+g x)^2}{d^2-e^2 x^2} \, dx &=\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]  time = 0.0348435, size = 73, normalized size = 1.12 \[ -\frac{e x \left (6 d^2 g^2+3 d e g (4 f+g x)+e^2 \left (3 f^2+3 f g x+g^2 x^2\right )\right )+6 d (d g+e f)^2 \log (d-e x)}{3 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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])/(3
*e^3)

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Maple [A]  time = 0.045, size = 110, normalized size = 1.7 \begin{align*} -{\frac{{g}^{2}{x}^{3}}{3}}-{\frac{d{x}^{2}{g}^{2}}{e}}-{x}^{2}fg-2\,{\frac{{d}^{2}{g}^{2}x}{{e}^{2}}}-4\,{\frac{dfgx}{e}}-x{f}^{2}-2\,{\frac{{d}^{3}\ln \left ( ex-d \right ){g}^{2}}{{e}^{3}}}-4\,{\frac{{d}^{2}\ln \left ( ex-d \right ) fg}{{e}^{2}}}-2\,{\frac{d\ln \left ( ex-d \right ){f}^{2}}{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.954936, size = 131, normalized size = 2.02 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.80452, size = 204, normalized size = 3.14 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.540506, size = 75, normalized size = 1.15 \begin{align*} - \frac{2 d \left (d g + e f\right )^{2} \log{\left (- d + e x \right )}}{e^{3}} - \frac{g^{2} x^{3}}{3} - \frac{x^{2} \left (d g^{2} + e f g\right )}{e} - \frac{x \left (2 d^{2} g^{2} + 4 d e f g + e^{2} f^{2}\right )}{e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Giac [B]  time = 1.15648, size = 232, normalized size = 3.57 \begin{align*} -{\left (d^{3} g^{2} e + 2 \, d^{2} f g e^{2} + d f^{2} e^{3}\right )} e^{\left (-4\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac{1}{3} \,{\left (g^{2} x^{3} e^{6} + 3 \, d g^{2} x^{2} e^{5} + 6 \, d^{2} g^{2} x e^{4} + 3 \, f g x^{2} e^{6} + 12 \, d f g x e^{5} + 3 \, f^{2} x e^{6}\right )} e^{\left (-6\right )} - \frac{{\left (d^{4} g^{2} e^{2} + 2 \, d^{3} f g e^{3} + d^{2} f^{2} e^{4}\right )} e^{\left (-5\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 )}{{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d^3*g^2*e + 2*d^2*f*g*e^2 + d*f^2*e^3)*e^(-4)*log(abs(x^2*e^2 - d^2)) - 1/3*(g^2*x^3*e^6 + 3*d*g^2*x^2*e^5 +
6*d^2*g^2*x*e^4 + 3*f*g*x^2*e^6 + 12*d*f*g*x*e^5 + 3*f^2*x*e^6)*e^(-6) - (d^4*g^2*e^2 + 2*d^3*f*g*e^3 + d^2*f
^2*e^4)*e^(-5)*log(abs(2*x*e^2 - 2*abs(d)*e)/abs(2*x*e^2 + 2*abs(d)*e))/abs(d)