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

Optimal. Leaf size=50 \[ -\frac{g x (d g+e f)}{e^2}-\frac{(d g+e f)^2 \log (d-e x)}{e^3}-\frac{(f+g x)^2}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.0340709, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {799, 43} \[ -\frac{g x (d g+e f)}{e^2}-\frac{(d g+e f)^2 \log (d-e x)}{e^3}-\frac{(f+g x)^2}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 799

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*x)/g)^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]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0192171, size = 43, normalized size = 0.86 \[ -\frac{e g x (2 d g+4 e f+e g x)+2 (d g+e f)^2 \log (d-e x)}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 82, normalized size = 1.6 \begin{align*} -{\frac{{g}^{2}{x}^{2}}{2\,e}}-{\frac{{g}^{2}dx}{{e}^{2}}}-2\,{\frac{fgx}{e}}-{\frac{\ln \left ( ex-d \right ){d}^{2}{g}^{2}}{{e}^{3}}}-2\,{\frac{\ln \left ( ex-d \right ) dfg}{{e}^{2}}}-{\frac{\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)*(g*x+f)^2/(-e^2*x^2+d^2),x)

[Out]

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

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Maxima [A]  time = 0.947378, size = 85, normalized size = 1.7 \begin{align*} -\frac{e g^{2} x^{2} + 2 \,{\left (2 \, e f g + d g^{2}\right )} x}{2 \, e^{2}} - \frac{{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} 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)*(g*x+f)^2/(-e^2*x^2+d^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.78179, size = 136, normalized size = 2.72 \begin{align*} -\frac{e^{2} g^{2} x^{2} + 2 \,{\left (2 \, e^{2} f g + d e g^{2}\right )} x + 2 \,{\left (e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.402451, size = 46, normalized size = 0.92 \begin{align*} - \frac{g^{2} x^{2}}{2 e} - \frac{x \left (d g^{2} + 2 e f g\right )}{e^{2}} - \frac{\left (d g + e f\right )^{2} \log{\left (- d + e x \right )}}{e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.19032, size = 181, normalized size = 3.62 \begin{align*} -\frac{1}{2} \,{\left (d^{2} g^{2} e + 2 \, d f g e^{2} + f^{2} e^{3}\right )} e^{\left (-4\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac{1}{2} \,{\left (g^{2} x^{2} e^{3} + 2 \, d g^{2} x e^{2} + 4 \, f g x e^{3}\right )} e^{\left (-4\right )} - \frac{{\left (d^{3} g^{2} + 2 \, d^{2} f g e + d f^{2} e^{2}\right )} e^{\left (-3\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 )}{2 \,{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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