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

Optimal. Leaf size=78 \[ \frac {2 d (d g+e f)^2}{e^3 (d-e x)}+\frac {(5 d g+e f) (d g+e f) \log (d-e x)}{e^3}+\frac {g x (3 d g+2 e f)}{e^2}+\frac {g^2 x^2}{2 e} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 78, 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 {2 d (d g+e f)^2}{e^3 (d-e x)}+\frac {g x (3 d g+2 e f)}{e^2}+\frac {(5 d g+e f) (d g+e f) \log (d-e x)}{e^3}+\frac {g^2 x^2}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 83, normalized size = 1.06 \begin {gather*} \frac {2 \left (5 d^2 g^2+6 d e f g+e^2 f^2\right ) \log (d-e x)+\frac {4 d (d g+e f)^2}{d-e x}+2 e g x (3 d g+2 e f)+e^2 g^2 x^2}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*g*(2*e*f + 3*d*g)*x + e^2*g^2*x^2 + (4*d*(e*f + d*g)^2)/(d - e*x) + 2*(e^2*f^2 + 6*d*e*f*g + 5*d^2*g^2)*L
og[d - e*x])/(2*e^3)

________________________________________________________________________________________

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

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.39, size = 157, normalized size = 2.01 \begin {gather*} \frac {e^{3} g^{2} x^{3} - 4 \, d e^{2} f^{2} - 8 \, d^{2} e f g - 4 \, d^{3} g^{2} + {\left (4 \, e^{3} f g + 5 \, d e^{2} g^{2}\right )} x^{2} - 2 \, {\left (2 \, d e^{2} f g + 3 \, d^{2} e g^{2}\right )} x - 2 \, {\left (d e^{2} f^{2} + 6 \, d^{2} e f g + 5 \, d^{3} g^{2} - {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 5 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{2 \, {\left (e^{4} x - d e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^3*g^2*x^3 - 4*d*e^2*f^2 - 8*d^2*e*f*g - 4*d^3*g^2 + (4*e^3*f*g + 5*d*e^2*g^2)*x^2 - 2*(2*d*e^2*f*g + 3*
d^2*e*g^2)*x - 2*(d*e^2*f^2 + 6*d^2*e*f*g + 5*d^3*g^2 - (e^3*f^2 + 6*d*e^2*f*g + 5*d^2*e*g^2)*x)*log(e*x - d))
/(e^4*x - d*e^3)

________________________________________________________________________________________

giac [B]  time = 0.18, size = 212, normalized size = 2.72 \begin {gather*} \frac {1}{2} \, {\left (5 \, d^{2} g^{2} e^{3} + 6 \, d f g e^{4} + f^{2} e^{5}\right )} e^{\left (-6\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) + \frac {1}{2} \, {\left (g^{2} x^{2} e^{7} + 6 \, d g^{2} x e^{6} + 4 \, f g x e^{7}\right )} e^{\left (-8\right )} + \frac {{\left (5 \, d^{3} g^{2} e^{2} + 6 \, d^{2} f g e^{3} + d 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 )}{2 \, {\left | d \right |}} - \frac {2 \, {\left (d^{4} g^{2} e^{3} + 2 \, d^{3} f g e^{4} + d^{2} f^{2} e^{5} + {\left (d^{3} g^{2} e^{4} + 2 \, d^{2} f g e^{5} + d f^{2} e^{6}\right )} x\right )} e^{\left (-6\right )}}{x^{2} e^{2} - d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 138, normalized size = 1.77 \begin {gather*} \frac {g^{2} x^{2}}{2 e}-\frac {2 d^{3} g^{2}}{\left (e x -d \right ) e^{3}}-\frac {4 d^{2} f g}{\left (e x -d \right ) e^{2}}+\frac {5 d^{2} g^{2} \ln \left (e x -d \right )}{e^{3}}-\frac {2 d \,f^{2}}{\left (e x -d \right ) e}+\frac {6 d f g \ln \left (e x -d \right )}{e^{2}}+\frac {3 d \,g^{2} x}{e^{2}}+\frac {f^{2} \ln \left (e x -d \right )}{e}+\frac {2 f g x}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.44, size = 104, normalized size = 1.33 \begin {gather*} -\frac {2 \, {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac {e g^{2} x^{2} + 2 \, {\left (2 \, e f g + 3 \, d g^{2}\right )} x}{2 \, e^{2}} + \frac {{\left (e^{2} f^{2} + 6 \, d e f g + 5 \, d^{2} 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)^3*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.53, size = 116, normalized size = 1.49 \begin {gather*} x\,\left (\frac {d\,g^2+2\,e\,f\,g}{e^2}+\frac {2\,d\,g^2}{e^2}\right )+\frac {\ln \left (e\,x-d\right )\,\left (5\,d^2\,g^2+6\,d\,e\,f\,g+e^2\,f^2\right )}{e^3}+\frac {g^2\,x^2}{2\,e}+\frac {2\,\left (d^3\,g^2+2\,d^2\,e\,f\,g+d\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.59, size = 94, normalized size = 1.21 \begin {gather*} x \left (\frac {3 d g^{2}}{e^{2}} + \frac {2 f g}{e}\right ) + \frac {- 2 d^{3} g^{2} - 4 d^{2} e f g - 2 d e^{2} f^{2}}{- d e^{3} + e^{4} x} + \frac {g^{2} x^{2}}{2 e} + \frac {\left (d g + e f\right ) \left (5 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)**3*(g*x+f)**2/(-e**2*x**2+d**2)**2,x)

[Out]

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

________________________________________________________________________________________