3.6.73 \(\int \frac {(d+e x)^3 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [573]

Optimal. Leaf size=61 \[ \frac {(e f+d g)^2}{2 e^3 (d-e x)^2}-\frac {2 g (e f+d g)}{e^3 (d-e x)}-\frac {g^2 \log (d-e x)}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 61, 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, 45} \begin {gather*} -\frac {2 g (d g+e f)}{e^3 (d-e x)}+\frac {(d g+e f)^2}{2 e^3 (d-e x)^2}-\frac {g^2 \log (d-e x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]
time = 0.02, size = 49, normalized size = 0.80 \begin {gather*} \frac {\frac {(e f+d g) (-3 d g+e (f+4 g x))}{(d-e x)^2}-2 g^2 \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)^3,x]

[Out]

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

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Maple [A]
time = 0.08, size = 74, normalized size = 1.21

method	result	size
risch	\(\frac {\frac {2 g \left (d g +e f \right ) x}{e^{2}}-\frac {3 d^{2} g^{2}+2 d e f g -e^{2} f^{2}}{2 e^{3}}}{\left (-e x +d \right )^{2}}-\frac {g^{2} \ln \left (-e x +d \right )}{e^{3}}\)	\(69\)
default	\(-\frac {2 g \left (d g +e f \right )}{e^{3} \left (-e x +d \right )}-\frac {g^{2} \ln \left (-e x +d \right )}{e^{3}}-\frac {-d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{2 e^{3} \left (-e x +d \right )^{2}}\)	\(74\)
norman	\(\frac {\left (2 d \,g^{2}+2 e f g \right ) x^{3}-\frac {d^{2} \left (3 g^{2} d^{2} e +2 f g d \,e^{2}-f^{2} e^{3}\right )}{2 e^{4}}+\frac {\left (5 g^{2} d^{2} e +6 f g d \,e^{2}+f^{2} e^{3}\right ) x^{2}}{2 e^{2}}-\frac {d \left (d^{2} g^{2}-e^{2} f^{2}\right ) x}{e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {g^{2} \ln \left (-e x +d \right )}{e^{3}}\)	\(139\)

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)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 79, normalized size = 1.30 \begin {gather*} -g^{2} e^{\left (-3\right )} \log \left (x e - d\right ) - \frac {3 \, d^{2} g^{2} + 2 \, d f g e - f^{2} e^{2} - 4 \, {\left (d g^{2} e + f g e^{2}\right )} x}{2 \, {\left (x^{2} e^{5} - 2 \, d x e^{4} + d^{2} 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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.22, size = 99, normalized size = 1.62 \begin {gather*} -\frac {3 \, d^{2} g^{2} - {\left (4 \, f g x + f^{2}\right )} e^{2} - 2 \, {\left (2 \, d g^{2} x - d f g\right )} e + 2 \, {\left (g^{2} x^{2} e^{2} - 2 \, d g^{2} x e + d^{2} g^{2}\right )} \log \left (x e - d\right )}{2 \, {\left (x^{2} e^{5} - 2 \, d x e^{4} + d^{2} 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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.26, size = 83, normalized size = 1.36 \begin {gather*} - \frac {3 d^{2} g^{2} + 2 d e f g - e^{2} f^{2} + x \left (- 4 d e g^{2} - 4 e^{2} f g\right )}{2 d^{2} e^{3} - 4 d e^{4} x + 2 e^{5} x^{2}} - \frac {g^{2} \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)**3,x)

[Out]

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

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Giac [A]
time = 1.46, size = 74, normalized size = 1.21 \begin {gather*} -g^{2} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) + \frac {{\left (4 \, {\left (d g^{2} + f g e\right )} x - {\left (3 \, d^{2} g^{2} + 2 \, d f g e - f^{2} e^{2}\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \, {\left (x e - d\right )}^{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)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.07, size = 80, normalized size = 1.31 \begin {gather*} -\frac {\frac {3\,d^2\,g^2+2\,d\,e\,f\,g-e^2\,f^2}{2\,e^3}-\frac {2\,g\,x\,\left (d\,g+e\,f\right )}{e^2}}{d^2-2\,d\,e\,x+e^2\,x^2}-\frac {g^2\,\ln \left (e\,x-d\right )}{e^3} \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)^3,x)

[Out]

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

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