3.6.72 \(\int \frac {(d+e x)^4 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [572]

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

[Out]

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

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Rubi [A]
time = 0.07, 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 = {862, 78} \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 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)^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 = 0.03, size = 93, normalized size = 1.15 \begin {gather*} \frac {-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 \left (f^2-g^2 x^2\right )-2 g (e f+2 d g) (d-e x)^2 \log (d-e x)}{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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Maple [A]
time = 0.08, size = 101, normalized size = 1.25

method	result	size
risch	\(-\frac {g^{2} x}{e^{2}}-\frac {\left (-5 d^{2} g^{2}-6 d e f g -e^{2} f^{2}\right ) x +\frac {4 d^{2} g \left (d g +e f \right )}{e}}{e^{2} \left (-e x +d \right )^{2}}-\frac {4 g^{2} \ln \left (-e x +d \right ) d}{e^{3}}-\frac {2 g \ln \left (-e x +d \right ) f}{e^{2}}\)	\(96\)
default	\(-\frac {g^{2} x}{e^{2}}+\frac {d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )^{2}}+\frac {-5 d^{2} g^{2}-6 d e f g -e^{2} f^{2}}{e^{3} \left (-e x +d \right )}-\frac {2 g \left (2 d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(101\)
norman	\(\frac {\left (7 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right ) x^{3}-\frac {d^{4} \left (4 d e \,g^{2}+4 e^{2} f g \right )}{e^{4}}-e^{2} g^{2} x^{5}+\frac {2 d \left (3 g^{2} d^{2} e +4 f g d \,e^{2}+f^{2} e^{3}\right ) x^{2}}{e^{2}}-\frac {d^{2} \left (4 d^{2} g^{2}+2 d e f g -e^{2} f^{2}\right ) x}{e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {2 g \left (2 d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(165\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 102, normalized size = 1.26 \begin {gather*} -g^{2} x e^{\left (-2\right )} - 2 \, {\left (2 \, d g^{2} + f g e\right )} e^{\left (-3\right )} \log \left (x e - d\right ) - \frac {4 \, d^{3} g^{2} + 4 \, d^{2} f g e - {\left (5 \, d^{2} g^{2} e + 6 \, d f g e^{2} + f^{2} e^{3}\right )} x}{x^{2} e^{5} - 2 \, d x e^{4} + 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="maxima")

[Out]

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

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Fricas [A]
time = 3.09, size = 156, normalized size = 1.93 \begin {gather*} -\frac {4 \, d^{3} g^{2} + {\left (g^{2} x^{3} - f^{2} x\right )} e^{3} - 2 \, {\left (d g^{2} x^{2} + 3 \, d f g x\right )} e^{2} - 4 \, {\left (d^{2} g^{2} x - d^{2} f g\right )} e + 2 \, {\left (2 \, d^{3} g^{2} + f g x^{2} e^{3} + 2 \, {\left (d g^{2} x^{2} - d f g x\right )} e^{2} - {\left (4 \, d^{2} g^{2} x - d^{2} f g\right )} e\right )} \log \left (x e - d\right )}{x^{2} e^{5} - 2 \, d x e^{4} + 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]

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

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Sympy [A]
time = 0.44, 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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Giac [A]
time = 2.97, size = 94, normalized size = 1.16 \begin {gather*} -g^{2} x e^{\left (-2\right )} - 2 \, {\left (2 \, d g^{2} + f g e\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) - \frac {{\left (4 \, d^{3} g^{2} + 4 \, d^{2} f g e - {\left (5 \, d^{2} g^{2} e + 6 \, d f g e^{2} + f^{2} e^{3}\right )} x\right )} e^{\left (-3\right )}}{{\left (x e - d\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*(2*d*g^2 + f*g*e)*e^(-3)*log(abs(x*e - d)) - (4*d^3*g^2 + 4*d^2*f*g*e - (5*d^2*g^2*e + 6*d*f
*g*e^2 + f^2*e^3)*x)*e^(-3)/(x*e - d)^2

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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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