3.6.62 \(\int \frac {(d+e x)^2 (f+g x)^2}{(d^2-e^2 x^2)^2} \, dx\) [562]

Optimal. Leaf size=50 \[ \frac {g^2 x}{e^2}+\frac {(e f+d g)^2}{e^3 (d-e x)}+\frac {2 g (e f+d g) \log (d-e x)}{e^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 50, 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 {(d g+e f)^2}{e^3 (d-e x)}+\frac {2 g (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)^2*(f + g*x)^2)/(d^2 - e^2*x^2)^2,x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 46, normalized size = 0.92 \begin {gather*} \frac {e g^2 x+\frac {(e f+d g)^2}{d-e x}+2 g (e f+d g) \log (d-e x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 63, normalized size = 1.26

method	result	size
default	\(\frac {g^{2} x}{e^{2}}+\frac {2 g \left (d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{e^{3} \left (-e x +d \right )}\)	\(63\)
risch	\(\frac {g^{2} x}{e^{2}}+\frac {2 g^{2} \ln \left (-e x +d \right ) d}{e^{3}}+\frac {2 g \ln \left (-e x +d \right ) f}{e^{2}}+\frac {d^{2} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {2 d f g}{e^{2} \left (-e x +d \right )}+\frac {f^{2}}{e \left (-e x +d \right )}\)	\(89\)
norman	\(\frac {\frac {d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3}}+\frac {\left (2 d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) x}{e^{2}}-g^{2} x^{3}}{-e^{2} x^{2}+d^{2}}+\frac {2 g \left (d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(g*x+f)^2/(-e^2*x^2+d^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 67, normalized size = 1.34 \begin {gather*} g^{2} x e^{\left (-2\right )} + 2 \, {\left (d g^{2} + f g e\right )} e^{\left (-3\right )} \log \left (x e - d\right ) - \frac {d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}}{x e^{4} - d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.92, size = 96, normalized size = 1.92 \begin {gather*} -\frac {d^{2} g^{2} - {\left (g^{2} x^{2} - f^{2}\right )} e^{2} + {\left (d g^{2} x + 2 \, d f g\right )} e + 2 \, {\left (d^{2} g^{2} - f g x e^{2} - {\left (d g^{2} x - d f g\right )} e\right )} \log \left (x e - d\right )}{x e^{4} - d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.18, size = 61, normalized size = 1.22 \begin {gather*} \frac {- d^{2} g^{2} - 2 d e f g - e^{2} f^{2}}{- d e^{3} + e^{4} x} + \frac {g^{2} x}{e^{2}} + \frac {2 g \left (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)**2*(g*x+f)**2/(-e**2*x**2+d**2)**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.59, size = 68, normalized size = 1.36 \begin {gather*} g^{2} x e^{\left (-2\right )} + 2 \, {\left (d g^{2} + f g e\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) - \frac {{\left (d^{2} g^{2} + 2 \, d f g e + 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)^2*(g*x+f)^2/(-e^2*x^2+d^2)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 2.56, size = 72, normalized size = 1.44 \begin {gather*} \frac {d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {g^2\,x}{e^2}+\frac {\ln \left (e\,x-d\right )\,\left (2\,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)^2)/(d^2 - e^2*x^2)^2,x)

[Out]

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

________________________________________________________________________________________