\(\int \frac {(d+e x)^5 (f+g x)^2}{(d^2-e^2 x^2)^3} \, dx\) [571]

Optimal result

Integrand size = 29, antiderivative size = 118 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {g (2 e f+5 d g) x}{e^2}-\frac {g^2 x^2}{2 e}+\frac {2 d^2 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d (e f+d g) (e f+3 d g)}{e^3 (d-e x)}-\frac {\left (e^2 f^2+10 d e f g+13 d^2 g^2\right ) \log (d-e x)}{e^3} \]

[Out]

-g*(5*d*g+2*e*f)*x/e^2-1/2*g^2*x^2/e+2*d^2*(d*g+e*f)^2/e^3/(-e*x+d)^2-4*d*(d*g+e*f)*(3*d*g+e*f)/e^3/(-e*x+d)-(
13*d^2*g^2+10*d*e*f*g+e^2*f^2)*ln(-e*x+d)/e^3

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {862, 90} \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {2 d^2 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {\left (13 d^2 g^2+10 d e f g+e^2 f^2\right ) \log (d-e x)}{e^3}-\frac {4 d (3 d g+e f) (d g+e f)}{e^3 (d-e x)}-\frac {g x (5 d g+2 e f)}{e^2}-\frac {g^2 x^2}{2 e} \]

[In]

Int[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-((g*(2*e*f + 5*d*g)*x)/e^2) - (g^2*x^2)/(2*e) + (2*d^2*(e*f + d*g)^2)/(e^3*(d - e*x)^2) - (4*d*(e*f + d*g)*(e
*f + 3*d*g))/(e^3*(d - e*x)) - ((e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*Log[d - e*x])/e^3

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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*} \text {integral}& = \int \frac {(d+e x)^2 (f+g x)^2}{(d-e x)^3} \, dx \\ & = \int \left (-\frac {g (2 e f+5 d g)}{e^2}-\frac {g^2 x}{e}+\frac {4 d (-e f-3 d g) (e f+d g)}{e^2 (d-e x)^2}-\frac {4 d^2 (e f+d g)^2}{e^2 (-d+e x)^3}+\frac {-e^2 f^2-10 d e f g-13 d^2 g^2}{e^2 (-d+e x)}\right ) \, dx \\ & = -\frac {g (2 e f+5 d g) x}{e^2}-\frac {g^2 x^2}{2 e}+\frac {2 d^2 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d (e f+d g) (e f+3 d g)}{e^3 (d-e x)}-\frac {\left (e^2 f^2+10 d e f g+13 d^2 g^2\right ) \log (d-e x)}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {2 e g (2 e f+5 d g) x+e^2 g^2 x^2-\frac {4 d^2 (e f+d g)^2}{(d-e x)^2}+\frac {8 d \left (e^2 f^2+4 d e f g+3 d^2 g^2\right )}{d-e x}+2 \left (e^2 f^2+10 d e f g+13 d^2 g^2\right ) \log (d-e x)}{2 e^3} \]

[In]

Integrate[((d + e*x)^5*(f + g*x)^2)/(d^2 - e^2*x^2)^3,x]

[Out]

-1/2*(2*e*g*(2*e*f + 5*d*g)*x + e^2*g^2*x^2 - (4*d^2*(e*f + d*g)^2)/(d - e*x)^2 + (8*d*(e^2*f^2 + 4*d*e*f*g +
3*d^2*g^2))/(d - e*x) + 2*(e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*Log[d - e*x])/e^3

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.13

method	result	size
default	\(-\frac {g \left (\frac {1}{2} e g \,x^{2}+5 d g x +2 e f x \right )}{e^{2}}+\frac {\left (-13 d^{2} g^{2}-10 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}-\frac {4 d \left (3 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}+\frac {2 d^{2} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )^{2}}\)	\(133\)
risch	\(-\frac {g^{2} x^{2}}{2 e}-\frac {5 g^{2} d x}{e^{2}}-\frac {2 g f x}{e}+\frac {\left (12 d^{3} g^{2}+16 d^{2} e f g +4 d \,e^{2} f^{2}\right ) x -\frac {2 d^{2} \left (5 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right )}{e}}{e^{2} \left (-e x +d \right )^{2}}-\frac {13 \ln \left (-e x +d \right ) d^{2} g^{2}}{e^{3}}-\frac {10 \ln \left (-e x +d \right ) d f g}{e^{2}}-\frac {\ln \left (-e x +d \right ) f^{2}}{e}\)	\(150\)
norman	\(\frac {\left (22 d^{3} g^{2}+20 d^{2} e f g +4 d \,e^{2} f^{2}\right ) x^{3}-\frac {d^{4} \left (11 d^{2} g^{2} e +12 d f g \,e^{2}+2 f^{2} e^{3}\right )}{e^{4}}-\frac {e^{3} g^{2} x^{6}}{2}+\frac {d^{2} \left (31 d^{2} g^{2} e +40 d f g \,e^{2}+12 f^{2} e^{3}\right ) x^{2}}{2 e^{2}}-e^{2} g \left (5 d g +2 e f \right ) x^{5}-\frac {d^{4} g \left (13 d g +10 e f \right ) x}{e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {\left (13 d^{2} g^{2}+10 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(201\)
parallelrisch	\(-\frac {g^{2} e^{4} x^{4}+26 \ln \left (e x -d \right ) x^{2} d^{2} e^{2} g^{2}+20 \ln \left (e x -d \right ) x^{2} d \,e^{3} f g +2 \ln \left (e x -d \right ) x^{2} e^{4} f^{2}+8 x^{3} d \,e^{3} g^{2}+4 x^{3} e^{4} f g -52 \ln \left (e x -d \right ) x \,d^{3} e \,g^{2}-40 \ln \left (e x -d \right ) x \,d^{2} e^{2} f g -4 \ln \left (e x -d \right ) x d \,e^{3} f^{2}+26 \ln \left (e x -d \right ) d^{4} g^{2}+20 \ln \left (e x -d \right ) d^{3} e f g +2 \ln \left (e x -d \right ) d^{2} e^{2} f^{2}-52 x \,d^{3} e \,g^{2}-44 x \,d^{2} e^{2} f g -8 x d \,e^{3} f^{2}+39 d^{4} g^{2}+32 f g e \,d^{3}+4 d^{2} e^{2} f^{2}}{2 e^{3} \left (e x -d \right )^{2}}\)	\(272\)

[In]

int((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x,method=_RETURNVERBOSE)

[Out]

-g/e^2*(1/2*e*g*x^2+5*d*g*x+2*e*f*x)+(-13*d^2*g^2-10*d*e*f*g-e^2*f^2)/e^3*ln(-e*x+d)-4*d/e^3*(3*d^2*g^2+4*d*e*
f*g+e^2*f^2)/(-e*x+d)+2*d^2*(d^2*g^2+2*d*e*f*g+e^2*f^2)/e^3/(-e*x+d)^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (118) = 236\).

Time = 0.36 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.04 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {e^{4} g^{2} x^{4} + 4 \, d^{2} e^{2} f^{2} + 24 \, d^{3} e f g + 20 \, d^{4} g^{2} + 4 \, {\left (e^{4} f g + 2 \, d e^{3} g^{2}\right )} x^{3} - {\left (8 \, d e^{3} f g + 19 \, d^{2} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (4 \, d e^{3} f^{2} + 14 \, d^{2} e^{2} f g + 7 \, d^{3} e g^{2}\right )} x + 2 \, {\left (d^{2} e^{2} f^{2} + 10 \, d^{3} e f g + 13 \, d^{4} g^{2} + {\left (e^{4} f^{2} + 10 \, d e^{3} f g + 13 \, d^{2} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (d e^{3} f^{2} + 10 \, d^{2} e^{2} f g + 13 \, d^{3} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{2 \, {\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="fricas")

[Out]

-1/2*(e^4*g^2*x^4 + 4*d^2*e^2*f^2 + 24*d^3*e*f*g + 20*d^4*g^2 + 4*(e^4*f*g + 2*d*e^3*g^2)*x^3 - (8*d*e^3*f*g +
19*d^2*e^2*g^2)*x^2 - 2*(4*d*e^3*f^2 + 14*d^2*e^2*f*g + 7*d^3*e*g^2)*x + 2*(d^2*e^2*f^2 + 10*d^3*e*f*g + 13*d
^4*g^2 + (e^4*f^2 + 10*d*e^3*f*g + 13*d^2*e^2*g^2)*x^2 - 2*(d*e^3*f^2 + 10*d^2*e^2*f*g + 13*d^3*e*g^2)*x)*log(
e*x - d))/(e^5*x^2 - 2*d*e^4*x + d^2*e^3)

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.28 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=- x \left (\frac {5 d g^{2}}{e^{2}} + \frac {2 f g}{e}\right ) - \frac {10 d^{4} g^{2} + 12 d^{3} e f g + 2 d^{2} e^{2} f^{2} + x \left (- 12 d^{3} e g^{2} - 16 d^{2} e^{2} f g - 4 d e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} - \frac {g^{2} x^{2}}{2 e} - \frac {\left (13 d^{2} g^{2} + 10 d e f g + e^{2} f^{2}\right ) \log {\left (- d + e x \right )}}{e^{3}} \]

[In]

integrate((e*x+d)**5*(g*x+f)**2/(-e**2*x**2+d**2)**3,x)

[Out]

-x*(5*d*g**2/e**2 + 2*f*g/e) - (10*d**4*g**2 + 12*d**3*e*f*g + 2*d**2*e**2*f**2 + x*(-12*d**3*e*g**2 - 16*d**2
*e**2*f*g - 4*d*e**3*f**2))/(d**2*e**3 - 2*d*e**4*x + e**5*x**2) - g**2*x**2/(2*e) - (13*d**2*g**2 + 10*d*e*f*
g + e**2*f**2)*log(-d + e*x)/e**3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {2 \, {\left (d^{2} e^{2} f^{2} + 6 \, d^{3} e f g + 5 \, d^{4} g^{2} - 2 \, {\left (d e^{3} f^{2} + 4 \, d^{2} e^{2} f g + 3 \, d^{3} e g^{2}\right )} x\right )}}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} - \frac {e g^{2} x^{2} + 2 \, {\left (2 \, e f g + 5 \, d g^{2}\right )} x}{2 \, e^{2}} - \frac {{\left (e^{2} f^{2} + 10 \, d e f g + 13 \, d^{2} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="maxima")

[Out]

-2*(d^2*e^2*f^2 + 6*d^3*e*f*g + 5*d^4*g^2 - 2*(d*e^3*f^2 + 4*d^2*e^2*f*g + 3*d^3*e*g^2)*x)/(e^5*x^2 - 2*d*e^4*
x + d^2*e^3) - 1/2*(e*g^2*x^2 + 2*(2*e*f*g + 5*d*g^2)*x)/e^2 - (e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*log(e*x - d
)/e^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.21 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {{\left (e^{2} f^{2} + 10 \, d e f g + 13 \, d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {2 \, {\left (d^{2} e^{2} f^{2} + 6 \, d^{3} e f g + 5 \, d^{4} g^{2} - 2 \, {\left (d e^{3} f^{2} + 4 \, d^{2} e^{2} f g + 3 \, d^{3} e g^{2}\right )} x\right )}}{{\left (e x - d\right )}^{2} e^{3}} - \frac {e^{5} g^{2} x^{2} + 4 \, e^{5} f g x + 10 \, d e^{4} g^{2} x}{2 \, e^{6}} \]

[In]

integrate((e*x+d)^5*(g*x+f)^2/(-e^2*x^2+d^2)^3,x, algorithm="giac")

[Out]

-(e^2*f^2 + 10*d*e*f*g + 13*d^2*g^2)*log(abs(e*x - d))/e^3 - 2*(d^2*e^2*f^2 + 6*d^3*e*f*g + 5*d^4*g^2 - 2*(d*e
^3*f^2 + 4*d^2*e^2*f*g + 3*d^3*e*g^2)*x)/((e*x - d)^2*e^3) - 1/2*(e^5*g^2*x^2 + 4*e^5*f*g*x + 10*d*e^4*g^2*x)/
e^6

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {(d+e x)^5 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {\frac {2\,\left (5\,d^4\,g^2+6\,d^3\,e\,f\,g+d^2\,e^2\,f^2\right )}{e}-x\,\left (12\,d^3\,g^2+16\,d^2\,e\,f\,g+4\,d\,e^2\,f^2\right )}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-x\,\left (\frac {2\,g\,\left (d\,g+e\,f\right )}{e^2}+\frac {3\,d\,g^2}{e^2}\right )-\frac {\ln \left (e\,x-d\right )\,\left (13\,d^2\,g^2+10\,d\,e\,f\,g+e^2\,f^2\right )}{e^3}-\frac {g^2\,x^2}{2\,e} \]

[In]

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

[Out]

- ((2*(5*d^4*g^2 + d^2*e^2*f^2 + 6*d^3*e*f*g))/e - x*(12*d^3*g^2 + 4*d*e^2*f^2 + 16*d^2*e*f*g))/(d^2*e^2 + e^4
*x^2 - 2*d*e^3*x) - x*((2*g*(d*g + e*f))/e^2 + (3*d*g^2)/e^2) - (log(e*x - d)*(13*d^2*g^2 + e^2*f^2 + 10*d*e*f
*g))/e^3 - (g^2*x^2)/(2*e)