3.6.61 \(\int \frac {(d+e x)^3 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\) [561]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {862, 78} \begin {gather*} \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} \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 78

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 862

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/e)*x)^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*}

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Mathematica [A]
time = 0.04, size = 83, normalized size = 1.06 \begin {gather*} \frac {2 e g (2 e f+3 d g) x+e^2 g^2 x^2+\frac {4 d (e f+d g)^2}{d-e x}+2 \left (e^2 f^2+6 d e f g+5 d^2 g^2\right ) \log (d-e x)}{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)

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Maple [A]
time = 0.09, size = 93, normalized size = 1.19

method	result	size
default	\(\frac {g \left (\frac {1}{2} e g \,x^{2}+3 d g x +2 e f x \right )}{e^{2}}+\frac {\left (5 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {2 d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}\)	\(93\)
risch	\(\frac {g^{2} x^{2}}{2 e}+\frac {3 g^{2} d x}{e^{2}}+\frac {2 g f x}{e}+\frac {5 \ln \left (-e x +d \right ) d^{2} g^{2}}{e^{3}}+\frac {6 \ln \left (-e x +d \right ) d f g}{e^{2}}+\frac {\ln \left (-e x +d \right ) f^{2}}{e}+\frac {2 d^{3} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {4 d^{2} f g}{e^{2} \left (-e x +d \right )}+\frac {2 d \,f^{2}}{e \left (-e x +d \right )}\)	\(132\)
norman	\(\frac {\frac {d \left (5 d^{2} g^{2}+6 d e f g +2 e^{2} f^{2}\right ) x}{e^{2}}+\frac {d^{2} \left (5 d^{2} g^{2}+8 d e f g +4 e^{2} f^{2}\right )}{2 e^{3}}-\frac {e \,g^{2} x^{4}}{2}-g \left (3 d g +2 e f \right ) x^{3}}{-e^{2} x^{2}+d^{2}}+\frac {\left (5 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(135\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 103, normalized size = 1.32 \begin {gather*} {\left (5 \, d^{2} g^{2} + 6 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right ) + \frac {1}{2} \, {\left (g^{2} x^{2} e + 2 \, {\left (3 \, d g^{2} + 2 \, f g e\right )} x\right )} e^{\left (-2\right )} - \frac {2 \, {\left (d^{3} g^{2} + 2 \, d^{2} f g e + d f^{2} e^{2}\right )}}{x e^{4} - d 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]

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

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Fricas [A]
time = 1.77, size = 155, normalized size = 1.99 \begin {gather*} -\frac {4 \, d^{3} g^{2} - {\left (g^{2} x^{3} + 4 \, f g x^{2}\right )} e^{3} - {\left (5 \, d g^{2} x^{2} - 4 \, d f g x - 4 \, d f^{2}\right )} e^{2} + 2 \, {\left (3 \, d^{2} g^{2} x + 4 \, d^{2} f g\right )} e + 2 \, {\left (5 \, d^{3} g^{2} - f^{2} x e^{3} - {\left (6 \, d f g x - d f^{2}\right )} e^{2} - {\left (5 \, d^{2} g^{2} x - 6 \, d^{2} f g\right )} e\right )} \log \left (x e - d\right )}{2 \, {\left (x e^{4} - 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*(4*d^3*g^2 - (g^2*x^3 + 4*f*g*x^2)*e^3 - (5*d*g^2*x^2 - 4*d*f*g*x - 4*d*f^2)*e^2 + 2*(3*d^2*g^2*x + 4*d^2
*f*g)*e + 2*(5*d^3*g^2 - f^2*x*e^3 - (6*d*f*g*x - d*f^2)*e^2 - (5*d^2*g^2*x - 6*d^2*f*g)*e)*log(x*e - d))/(x*e
^4 - d*e^3)

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Sympy [A]
time = 0.28, 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

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Giac [A]
time = 2.00, size = 104, normalized size = 1.33 \begin {gather*} {\left (5 \, d^{2} g^{2} + 6 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) + \frac {1}{2} \, {\left (g^{2} x^{2} e^{3} + 6 \, d g^{2} x e^{2} + 4 \, f g x e^{3}\right )} e^{\left (-4\right )} - \frac {2 \, {\left (d^{3} g^{2} + 2 \, d^{2} f g e + d f^{2} e^{2}\right )} e^{\left (-3\right )}}{x e - d} \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]

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

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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))

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