3.4.74 \(\int \frac {(d+e x)^4 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\)

Optimal. Leaf size=81 \[ -\frac {(d g+e f) (5 d g+e f)}{e^3 (d-e x)}+\frac {d (d g+e f)^2}{e^3 (d-e x)^2}-\frac {2 g (2 d g+e f) \log (d-e x)}{e^3}-\frac {g^2 x}{e^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 81, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {(d g+e f) (5 d g+e f)}{e^3 (d-e x)}+\frac {d (d g+e f)^2}{e^3 (d-e x)^2}-\frac {2 g (2 d g+e f) \log (d-e x)}{e^3}-\frac {g^2 x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))))

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]))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 93, normalized size = 1.15 \begin {gather*} \frac {-4 d^3 g^2+4 d^2 e g (g x-f)+2 d e^2 g x (3 f+g x)-2 g (d-e x)^2 (2 d g+e f) \log (d-e x)+e^3 x \left (f^2-g^2 x^2\right )}{e^3 (d-e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 159, normalized size = 1.96 \begin {gather*} -\frac {e^{3} g^{2} x^{3} - 2 \, d e^{2} g^{2} x^{2} + 4 \, d^{2} e f g + 4 \, d^{3} g^{2} - {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 4 \, d^{2} e g^{2}\right )} x + 2 \, {\left (d^{2} e f g + 2 \, d^{3} g^{2} + {\left (e^{3} f g + 2 \, d e^{2} g^{2}\right )} x^{2} - 2 \, {\left (d e^{2} f g + 2 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

-(e^3*g^2*x^3 - 2*d*e^2*g^2*x^2 + 4*d^2*e*f*g + 4*d^3*g^2 - (e^3*f^2 + 6*d*e^2*f*g + 4*d^2*e*g^2)*x + 2*(d^2*e
*f*g + 2*d^3*g^2 + (e^3*f*g + 2*d*e^2*g^2)*x^2 - 2*(d*e^2*f*g + 2*d^2*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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giac [B]  time = 0.21, size = 227, normalized size = 2.80 \begin {gather*} -g^{2} x e^{\left (-2\right )} - {\left (2 \, d g^{2} e^{3} + f g e^{4}\right )} e^{\left (-6\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac {{\left (2 \, d^{2} g^{2} e^{4} + d f g e^{5}\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 )}{{\left | d \right |}} - \frac {{\left (4 \, d^{5} g^{2} e^{3} + 4 \, d^{4} f g e^{4} - {\left (5 \, d^{2} g^{2} e^{6} + 6 \, d f g e^{7} + f^{2} e^{8}\right )} x^{3} - 2 \, {\left (3 \, d^{3} g^{2} e^{5} + 4 \, d^{2} f g e^{6} + d f^{2} e^{7}\right )} x^{2} + {\left (3 \, d^{4} g^{2} e^{4} + 2 \, d^{3} f g e^{5} - d^{2} f^{2} e^{6}\right )} x\right )} e^{\left (-6\right )}}{{\left (x^{2} e^{2} - d^{2}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="giac")

[Out]

-g^2*x*e^(-2) - (2*d*g^2*e^3 + f*g*e^4)*e^(-6)*log(abs(x^2*e^2 - d^2)) - (2*d^2*g^2*e^4 + d*f*g*e^5)*e^(-7)*lo
g(abs(2*x*e^2 - 2*abs(d)*e)/abs(2*x*e^2 + 2*abs(d)*e))/abs(d) - (4*d^5*g^2*e^3 + 4*d^4*f*g*e^4 - (5*d^2*g^2*e^
6 + 6*d*f*g*e^7 + f^2*e^8)*x^3 - 2*(3*d^3*g^2*e^5 + 4*d^2*f*g*e^6 + d*f^2*e^7)*x^2 + (3*d^4*g^2*e^4 + 2*d^3*f*
g*e^5 - d^2*f^2*e^6)*x)*e^(-6)/(x^2*e^2 - d^2)^2

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maple [A]  time = 0.01, size = 151, normalized size = 1.86 \begin {gather*} \frac {d^{3} g^{2}}{\left (e x -d \right )^{2} e^{3}}+\frac {2 d^{2} f g}{\left (e x -d \right )^{2} e^{2}}+\frac {d \,f^{2}}{\left (e x -d \right )^{2} e}+\frac {5 d^{2} g^{2}}{\left (e x -d \right ) e^{3}}+\frac {6 d f g}{\left (e x -d \right ) e^{2}}-\frac {4 d \,g^{2} \ln \left (e x -d \right )}{e^{3}}+\frac {f^{2}}{\left (e x -d \right ) e}-\frac {2 f g \ln \left (e x -d \right )}{e^{2}}-\frac {g^{2} x}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 105, normalized size = 1.30 \begin {gather*} -\frac {g^{2} x}{e^{2}} - \frac {4 \, d^{2} e f g + 4 \, d^{3} g^{2} - {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 5 \, d^{2} e g^{2}\right )} x}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} - \frac {2 \, {\left (e f g + 2 \, d g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.60, size = 107, normalized size = 1.32 \begin {gather*} -\frac {\frac {4\,\left (d^3\,g^2+e\,f\,d^2\,g\right )}{e}-x\,\left (5\,d^2\,g^2+6\,d\,e\,f\,g+e^2\,f^2\right )}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-\frac {g^2\,x}{e^2}-\frac {\ln \left (e\,x-d\right )\,\left (4\,d\,g^2+2\,e\,f\,g\right )}{e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.87, size = 102, normalized size = 1.26 \begin {gather*} - \frac {4 d^{3} g^{2} + 4 d^{2} e f g + x \left (- 5 d^{2} e g^{2} - 6 d e^{2} f g - e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} - \frac {g^{2} x}{e^{2}} - \frac {2 g \left (2 d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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