3.571 \(\int \frac{(d+e x)^5 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\)

Optimal. Leaf size=118 \[ -\frac{\left (13 d^2 g^2+10 d e f g+e^2 f^2\right ) \log (d-e x)}{e^3}+\frac{2 d^2 (d g+e f)^2}{e^3 (d-e x)^2}-\frac{4 d (3 d g+e f) (d g+e f)}{e^3 (d-e x)}-\frac{g x (5 d g+2 e f)}{e^2}-\frac{g^2 x^2}{2 e} \]

[Out]

-((g*(2*e*f + 5*d*g)*x)/e^2) - (g^2*x^2)/(2*e) + (2*d^2*(e*f + d*g)^2)/(e^3*(d - e*x)^2) - (4*d*(e*f + d*g)*(e
*f + 3*d*g))/(e^3*(d - e*x)) - ((e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*Log[d - e*x])/e^3

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Rubi [A]  time = 0.143006, antiderivative size = 118, 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, 88} \[ -\frac{\left (13 d^2 g^2+10 d e f g+e^2 f^2\right ) \log (d-e x)}{e^3}+\frac{2 d^2 (d g+e f)^2}{e^3 (d-e x)^2}-\frac{4 d (3 d g+e f) (d g+e f)}{e^3 (d-e x)}-\frac{g x (5 d g+2 e f)}{e^2}-\frac{g^2 x^2}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-((g*(2*e*f + 5*d*g)*x)/e^2) - (g^2*x^2)/(2*e) + (2*d^2*(e*f + d*g)^2)/(e^3*(d - e*x)^2) - (4*d*(e*f + d*g)*(e
*f + 3*d*g))/(e^3*(d - e*x)) - ((e^2*f^2 + 10*d*e*f*g + 13*d^2*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 88

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]))

Rubi steps

\begin{align*} \int \frac{(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx &=\int \frac{(d+e x)^2 (f+g x)^2}{(d-e x)^3} \, dx\\ &=\int \left (-\frac{g (2 e f+5 d g)}{e^2}-\frac{g^2 x}{e}+\frac{4 d (-e f-3 d g) (e f+d g)}{e^2 (d-e x)^2}-\frac{4 d^2 (e f+d g)^2}{e^2 (-d+e x)^3}+\frac{-e^2 f^2-10 d e f g-13 d^2 g^2}{e^2 (-d+e x)}\right ) \, dx\\ &=-\frac{g (2 e f+5 d g) x}{e^2}-\frac{g^2 x^2}{2 e}+\frac{2 d^2 (e f+d g)^2}{e^3 (d-e x)^2}-\frac{4 d (e f+d g) (e f+3 d g)}{e^3 (d-e x)}-\frac{\left (e^2 f^2+10 d e f g+13 d^2 g^2\right ) \log (d-e x)}{e^3}\\ \end{align*}

Mathematica [A]  time = 0.0928828, size = 118, normalized size = 1. \[ -\frac{\frac{8 d \left (3 d^2 g^2+4 d e f g+e^2 f^2\right )}{d-e x}+2 \left (13 d^2 g^2+10 d e f g+e^2 f^2\right ) \log (d-e x)-\frac{4 d^2 (d g+e f)^2}{(d-e x)^2}+2 e g x (5 d g+2 e f)+e^2 g^2 x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-(2*e*g*(2*e*f + 5*d*g)*x + e^2*g^2*x^2 - (4*d^2*(e*f + d*g)^2)/(d - e*x)^2 + (8*d*(e^2*f^2 + 4*d*e*f*g + 3*d^
2*g^2))/(d - e*x) + 2*(e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*Log[d - e*x])/(2*e^3)

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Maple [A]  time = 0.051, size = 198, normalized size = 1.7 \begin{align*} -{\frac{{g}^{2}{x}^{2}}{2\,e}}-5\,{\frac{d{g}^{2}x}{{e}^{2}}}-2\,{\frac{fgx}{e}}-13\,{\frac{\ln \left ( ex-d \right ){d}^{2}{g}^{2}}{{e}^{3}}}-10\,{\frac{\ln \left ( ex-d \right ) dfg}{{e}^{2}}}-{\frac{\ln \left ( ex-d \right ){f}^{2}}{e}}+2\,{\frac{{d}^{4}{g}^{2}}{{e}^{3} \left ( ex-d \right ) ^{2}}}+4\,{\frac{{d}^{3}fg}{{e}^{2} \left ( ex-d \right ) ^{2}}}+2\,{\frac{{d}^{2}{f}^{2}}{e \left ( ex-d \right ) ^{2}}}+12\,{\frac{{d}^{3}{g}^{2}}{{e}^{3} \left ( ex-d \right ) }}+16\,{\frac{{d}^{2}fg}{{e}^{2} \left ( ex-d \right ) }}+4\,{\frac{d{f}^{2}}{e \left ( ex-d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x)

[Out]

-1/2*g^2*x^2/e-5*g^2/e^2*d*x-2*g/e*f*x-13/e^3*ln(e*x-d)*d^2*g^2-10/e^2*ln(e*x-d)*d*f*g-1/e*ln(e*x-d)*f^2+2*d^4
/e^3/(e*x-d)^2*g^2+4*d^3/e^2/(e*x-d)^2*f*g+2*d^2/e/(e*x-d)^2*f^2+12*d^3/e^3/(e*x-d)*g^2+16*d^2/e^2/(e*x-d)*f*g
+4*d/e/(e*x-d)*f^2

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Maxima [A]  time = 0.975768, size = 201, normalized size = 1.7 \begin{align*} -\frac{2 \,{\left (d^{2} e^{2} f^{2} + 6 \, d^{3} e f g + 5 \, d^{4} g^{2} - 2 \,{\left (d e^{3} f^{2} + 4 \, d^{2} e^{2} f g + 3 \, d^{3} e g^{2}\right )} x\right )}}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} - \frac{e g^{2} x^{2} + 2 \,{\left (2 \, e f g + 5 \, d g^{2}\right )} x}{2 \, e^{2}} - \frac{{\left (e^{2} f^{2} + 10 \, d e f g + 13 \, 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)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

-2*(d^2*e^2*f^2 + 6*d^3*e*f*g + 5*d^4*g^2 - 2*(d*e^3*f^2 + 4*d^2*e^2*f*g + 3*d^3*e*g^2)*x)/(e^5*x^2 - 2*d*e^4*
x + d^2*e^3) - 1/2*(e*g^2*x^2 + 2*(2*e*f*g + 5*d*g^2)*x)/e^2 - (e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*log(e*x - d
)/e^3

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Fricas [B]  time = 1.73834, size = 504, normalized size = 4.27 \begin{align*} -\frac{e^{4} g^{2} x^{4} + 4 \, d^{2} e^{2} f^{2} + 24 \, d^{3} e f g + 20 \, d^{4} g^{2} + 4 \,{\left (e^{4} f g + 2 \, d e^{3} g^{2}\right )} x^{3} -{\left (8 \, d e^{3} f g + 19 \, d^{2} e^{2} g^{2}\right )} x^{2} - 2 \,{\left (4 \, d e^{3} f^{2} + 14 \, d^{2} e^{2} f g + 7 \, d^{3} e g^{2}\right )} x + 2 \,{\left (d^{2} e^{2} f^{2} + 10 \, d^{3} e f g + 13 \, d^{4} g^{2} +{\left (e^{4} f^{2} + 10 \, d e^{3} f g + 13 \, d^{2} e^{2} g^{2}\right )} x^{2} - 2 \,{\left (d e^{3} f^{2} + 10 \, d^{2} e^{2} f g + 13 \, d^{3} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{2 \,{\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

-1/2*(e^4*g^2*x^4 + 4*d^2*e^2*f^2 + 24*d^3*e*f*g + 20*d^4*g^2 + 4*(e^4*f*g + 2*d*e^3*g^2)*x^3 - (8*d*e^3*f*g +
19*d^2*e^2*g^2)*x^2 - 2*(4*d*e^3*f^2 + 14*d^2*e^2*f*g + 7*d^3*e*g^2)*x + 2*(d^2*e^2*f^2 + 10*d^3*e*f*g + 13*d
^4*g^2 + (e^4*f^2 + 10*d*e^3*f*g + 13*d^2*e^2*g^2)*x^2 - 2*(d*e^3*f^2 + 10*d^2*e^2*f*g + 13*d^3*e*g^2)*x)*log(
e*x - d))/(e^5*x^2 - 2*d*e^4*x + d^2*e^3)

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Sympy [A]  time = 1.34291, size = 148, normalized size = 1.25 \begin{align*} \frac{- 10 d^{4} g^{2} - 12 d^{3} e f g - 2 d^{2} e^{2} f^{2} + x \left (12 d^{3} e g^{2} + 16 d^{2} e^{2} f g + 4 d e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} - \frac{g^{2} x^{2}}{2 e} - \frac{x \left (5 d g^{2} + 2 e f g\right )}{e^{2}} - \frac{\left (13 d^{2} g^{2} + 10 d e f g + e^{2} f^{2}\right ) \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)**5*(g*x+f)**2/(-e**2*x**2+d**2)**3,x)

[Out]

(-10*d**4*g**2 - 12*d**3*e*f*g - 2*d**2*e**2*f**2 + x*(12*d**3*e*g**2 + 16*d**2*e**2*f*g + 4*d*e**3*f**2))/(d*
*2*e**3 - 2*d*e**4*x + e**5*x**2) - g**2*x**2/(2*e) - x*(5*d*g**2 + 2*e*f*g)/e**2 - (13*d**2*g**2 + 10*d*e*f*g
+ e**2*f**2)*log(-d + e*x)/e**3

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Giac [B]  time = 1.14559, size = 369, normalized size = 3.13 \begin{align*} -\frac{1}{2} \,{\left (13 \, d^{2} g^{2} e^{5} + 10 \, d f g e^{6} + f^{2} e^{7}\right )} e^{\left (-8\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac{1}{2} \,{\left (g^{2} x^{2} e^{11} + 10 \, d g^{2} x e^{10} + 4 \, f g x e^{11}\right )} e^{\left (-12\right )} - \frac{{\left (13 \, d^{3} g^{2} e^{4} + 10 \, d^{2} f g e^{5} + d f^{2} e^{6}\right )} e^{\left (-7\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 |}} - \frac{2 \,{\left (5 \, d^{6} g^{2} e^{5} + 6 \, d^{5} f g e^{6} + d^{4} f^{2} e^{7} - 2 \,{\left (3 \, d^{3} g^{2} e^{8} + 4 \, d^{2} f g e^{9} + d f^{2} e^{10}\right )} x^{3} -{\left (7 \, d^{4} g^{2} e^{7} + 10 \, d^{3} f g e^{8} + 3 \, d^{2} f^{2} e^{9}\right )} x^{2} + 4 \,{\left (d^{5} g^{2} e^{6} + d^{4} f g e^{7}\right )} x\right )} e^{\left (-8\right )}}{{\left (x^{2} e^{2} - d^{2}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="giac")

[Out]

-1/2*(13*d^2*g^2*e^5 + 10*d*f*g*e^6 + f^2*e^7)*e^(-8)*log(abs(x^2*e^2 - d^2)) - 1/2*(g^2*x^2*e^11 + 10*d*g^2*x
*e^10 + 4*f*g*x*e^11)*e^(-12) - 1/2*(13*d^3*g^2*e^4 + 10*d^2*f*g*e^5 + d*f^2*e^6)*e^(-7)*log(abs(2*x*e^2 - 2*a
bs(d)*e)/abs(2*x*e^2 + 2*abs(d)*e))/abs(d) - 2*(5*d^6*g^2*e^5 + 6*d^5*f*g*e^6 + d^4*f^2*e^7 - 2*(3*d^3*g^2*e^8
+ 4*d^2*f*g*e^9 + d*f^2*e^10)*x^3 - (7*d^4*g^2*e^7 + 10*d^3*f*g*e^8 + 3*d^2*f^2*e^9)*x^2 + 4*(d^5*g^2*e^6 + d
^4*f*g*e^7)*x)*e^(-8)/(x^2*e^2 - d^2)^2