Integrand size = 29, antiderivative size = 149 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (e f+d g) (3 e f+7 d g)}{e^3 (d-e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3} \]
-(18*d^2*g^2+12*d*e*f*g+e^2*f^2)*x/e^2-g*(3*d*g+e*f)*x^2/e-1/3*g^2*x^3+4*d ^3*(d*g+e*f)^2/e^3/(-e*x+d)^2-4*d^2*(d*g+e*f)*(7*d*g+3*e*f)/e^3/(-e*x+d)-2 *d*(19*d^2*g^2+18*d*e*f*g+3*e^2*f^2)*ln(-e*x+d)/e^3
Time = 0.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}+\frac {4 d^2 \left (3 e^2 f^2+10 d e f g+7 d^2 g^2\right )}{e^3 (-d+e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3} \]
-(((e^2*f^2 + 12*d*e*f*g + 18*d^2*g^2)*x)/e^2) - (g*(e*f + 3*d*g)*x^2)/e - (g^2*x^3)/3 + (4*d^3*(e*f + d*g)^2)/(e^3*(d - e*x)^2) + (4*d^2*(3*e^2*f^2 + 10*d*e*f*g + 7*d^2*g^2))/(e^3*(-d + e*x)) - (2*d*(3*e^2*f^2 + 18*d*e*f* g + 19*d^2*g^2)*Log[d - e*x])/e^3
Time = 0.39 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {639, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 639 |
| \(\displaystyle \int \frac {(d+e x)^3 (f+g x)^2}{(d-e x)^3}dx\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \int \left (-\frac {8 d^3 (d g+e f)^2}{e^2 (e x-d)^3}-\frac {2 d \left (19 d^2 g^2+18 d e f g+3 e^2 f^2\right )}{e^2 (e x-d)}+\frac {-18 d^2 g^2-12 d e f g-e^2 f^2}{e^2}+\frac {4 d^2 (-7 d g-3 e f) (d g+e f)}{e^2 (d-e x)^2}-\frac {2 g x (3 d g+e f)}{e}-g^2 x^2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 d^3 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (d g+e f) (7 d g+3 e f)}{e^3 (d-e x)}-\frac {x \left (18 d^2 g^2+12 d e f g+e^2 f^2\right )}{e^2}-\frac {2 d \left (19 d^2 g^2+18 d e f g+3 e^2 f^2\right ) \log (d-e x)}{e^3}-\frac {g x^2 (3 d g+e f)}{e}-\frac {g^2 x^3}{3}\) |
-(((e^2*f^2 + 12*d*e*f*g + 18*d^2*g^2)*x)/e^2) - (g*(e*f + 3*d*g)*x^2)/e - (g^2*x^3)/3 + (4*d^3*(e*f + d*g)^2)/(e^3*(d - e*x)^2) - (4*d^2*(e*f + d*g )*(3*e*f + 7*d*g))/(e^3*(d - e*x)) - (2*d*(3*e^2*f^2 + 18*d*e*f*g + 19*d^2 *g^2)*Log[d - e*x])/e^3
3.6.70.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(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]))
Int[((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^ 2)^(p_.), x_Symbol] :> Int[(c + d*x)^(m + p)*(e + f*x)^n*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (I ntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[m]))
Time = 0.44 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.17
| method | result | size |
| default | \(-\frac {\frac {1}{3} g^{2} x^{3} e^{2}+3 d e \,g^{2} x^{2}+e^{2} f g \,x^{2}+18 d^{2} g^{2} x +12 d e f g x +e^{2} f^{2} x}{e^{2}}-\frac {2 d \left (19 d^{2} g^{2}+18 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}-\frac {4 d^{2} \left (7 d^{2} g^{2}+10 d e f g +3 e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}+\frac {4 d^{3} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )^{2}}\) | \(174\) |
| risch | \(-\frac {g^{2} x^{3}}{3}-\frac {3 d \,g^{2} x^{2}}{e}-f g \,x^{2}-\frac {18 d^{2} g^{2} x}{e^{2}}-\frac {12 d f g x}{e}-f^{2} x -\frac {\left (-28 d^{4} g^{2}-40 f g e \,d^{3}-12 d^{2} e^{2} f^{2}\right ) x +\frac {8 d^{3} \left (3 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right )}{e}}{e^{2} \left (-e x +d \right )^{2}}-\frac {38 d^{3} \ln \left (-e x +d \right ) g^{2}}{e^{3}}-\frac {36 d^{2} \ln \left (-e x +d \right ) f g}{e^{2}}-\frac {6 d \ln \left (-e x +d \right ) f^{2}}{e}\) | \(181\) |
| norman | \(\frac {\left (\frac {191}{3} d^{4} g^{2}+64 f g e \,d^{3}+14 d^{2} e^{2} f^{2}\right ) x^{3}+\left (-\frac {52}{3} d^{2} g^{2} e^{2}-12 d f g \,e^{3}-f^{2} e^{4}\right ) x^{5}+\frac {d^{2} \left (41 g^{2} e \,d^{3}+51 e^{2} f g \,d^{2}+16 e^{3} f^{2} d \right ) x^{2}}{e^{2}}-\frac {d^{4} \left (30 g^{2} e \,d^{3}+34 e^{2} f g \,d^{2}+8 e^{3} f^{2} d \right )}{e^{4}}-\frac {g^{2} e^{4} x^{7}}{3}-\frac {d^{4} \left (38 d^{2} g^{2}+36 d e f g +5 e^{2} f^{2}\right ) x}{e^{2}}-e^{3} g \left (3 d g +e f \right ) x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {2 d \left (19 d^{2} g^{2}+18 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\) | \(254\) |
| parallelrisch | \(-\frac {g^{2} e^{5} x^{5}+7 x^{4} d \,e^{4} g^{2}+3 x^{4} e^{5} f g +114 \ln \left (e x -d \right ) x^{2} d^{3} e^{2} g^{2}+108 \ln \left (e x -d \right ) x^{2} d^{2} e^{3} f g +18 \ln \left (e x -d \right ) x^{2} d \,e^{4} f^{2}+37 x^{3} d^{2} e^{3} g^{2}+30 x^{3} d \,e^{4} f g +3 x^{3} e^{5} f^{2}-228 \ln \left (e x -d \right ) x \,d^{4} e \,g^{2}-216 \ln \left (e x -d \right ) x \,d^{3} e^{2} f g -36 \ln \left (e x -d \right ) x \,d^{2} e^{3} f^{2}+114 \ln \left (e x -d \right ) d^{5} g^{2}+108 \ln \left (e x -d \right ) d^{4} e f g +18 \ln \left (e x -d \right ) d^{3} e^{2} f^{2}-228 x \,d^{4} e \,g^{2}-222 x \,d^{3} e^{2} f g -45 x \,d^{2} e^{3} f^{2}+171 g^{2} d^{5}+165 f g \,d^{4} e +30 f^{2} d^{3} e^{2}}{3 e^{3} \left (e x -d \right )^{2}}\) | \(315\) |
-1/e^2*(1/3*g^2*x^3*e^2+3*d*e*g^2*x^2+e^2*f*g*x^2+18*d^2*g^2*x+12*d*e*f*g* x+e^2*f^2*x)-2*d*(19*d^2*g^2+18*d*e*f*g+3*e^2*f^2)*ln(-e*x+d)/e^3-4*d^2/e^ 3*(7*d^2*g^2+10*d*e*f*g+3*e^2*f^2)/(-e*x+d)+4*d^3*(d^2*g^2+2*d*e*f*g+e^2*f ^2)/e^3/(-e*x+d)^2
Time = 0.36 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.97 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {e^{5} g^{2} x^{5} + 24 \, d^{3} e^{2} f^{2} + 96 \, d^{4} e f g + 72 \, d^{5} g^{2} + {\left (3 \, e^{5} f g + 7 \, d e^{4} g^{2}\right )} x^{4} + {\left (3 \, e^{5} f^{2} + 30 \, d e^{4} f g + 37 \, d^{2} e^{3} g^{2}\right )} x^{3} - 3 \, {\left (2 \, d e^{4} f^{2} + 23 \, d^{2} e^{3} f g + 33 \, d^{3} e^{2} g^{2}\right )} x^{2} - 3 \, {\left (11 \, d^{2} e^{3} f^{2} + 28 \, d^{3} e^{2} f g + 10 \, d^{4} e g^{2}\right )} x + 6 \, {\left (3 \, d^{3} e^{2} f^{2} + 18 \, d^{4} e f g + 19 \, d^{5} g^{2} + {\left (3 \, d e^{4} f^{2} + 18 \, d^{2} e^{3} f g + 19 \, d^{3} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (3 \, d^{2} e^{3} f^{2} + 18 \, d^{3} e^{2} f g + 19 \, d^{4} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{3 \, {\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]
-1/3*(e^5*g^2*x^5 + 24*d^3*e^2*f^2 + 96*d^4*e*f*g + 72*d^5*g^2 + (3*e^5*f* g + 7*d*e^4*g^2)*x^4 + (3*e^5*f^2 + 30*d*e^4*f*g + 37*d^2*e^3*g^2)*x^3 - 3 *(2*d*e^4*f^2 + 23*d^2*e^3*f*g + 33*d^3*e^2*g^2)*x^2 - 3*(11*d^2*e^3*f^2 + 28*d^3*e^2*f*g + 10*d^4*e*g^2)*x + 6*(3*d^3*e^2*f^2 + 18*d^4*e*f*g + 19*d ^5*g^2 + (3*d*e^4*f^2 + 18*d^2*e^3*f*g + 19*d^3*e^2*g^2)*x^2 - 2*(3*d^2*e^ 3*f^2 + 18*d^3*e^2*f*g + 19*d^4*e*g^2)*x)*log(e*x - d))/(e^5*x^2 - 2*d*e^4 *x + d^2*e^3)
Time = 0.60 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.19 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=- \frac {2 d \left (19 d^{2} g^{2} + 18 d e f g + 3 e^{2} f^{2}\right ) \log {\left (- d + e x \right )}}{e^{3}} - \frac {g^{2} x^{3}}{3} - x^{2} \cdot \left (\frac {3 d g^{2}}{e} + f g\right ) - x \left (\frac {18 d^{2} g^{2}}{e^{2}} + \frac {12 d f g}{e} + f^{2}\right ) - \frac {24 d^{5} g^{2} + 32 d^{4} e f g + 8 d^{3} e^{2} f^{2} + x \left (- 28 d^{4} e g^{2} - 40 d^{3} e^{2} f g - 12 d^{2} e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} \]
-2*d*(19*d**2*g**2 + 18*d*e*f*g + 3*e**2*f**2)*log(-d + e*x)/e**3 - g**2*x **3/3 - x**2*(3*d*g**2/e + f*g) - x*(18*d**2*g**2/e**2 + 12*d*f*g/e + f**2 ) - (24*d**5*g**2 + 32*d**4*e*f*g + 8*d**3*e**2*f**2 + x*(-28*d**4*e*g**2 - 40*d**3*e**2*f*g - 12*d**2*e**3*f**2))/(d**2*e**3 - 2*d*e**4*x + e**5*x* *2)
Time = 0.19 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {4 \, {\left (2 \, d^{3} e^{2} f^{2} + 8 \, d^{4} e f g + 6 \, d^{5} g^{2} - {\left (3 \, d^{2} e^{3} f^{2} + 10 \, d^{3} e^{2} f g + 7 \, d^{4} e g^{2}\right )} x\right )}}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} - \frac {e^{2} g^{2} x^{3} + 3 \, {\left (e^{2} f g + 3 \, d e g^{2}\right )} x^{2} + 3 \, {\left (e^{2} f^{2} + 12 \, d e f g + 18 \, d^{2} g^{2}\right )} x}{3 \, e^{2}} - \frac {2 \, {\left (3 \, d e^{2} f^{2} + 18 \, d^{2} e f g + 19 \, d^{3} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]
-4*(2*d^3*e^2*f^2 + 8*d^4*e*f*g + 6*d^5*g^2 - (3*d^2*e^3*f^2 + 10*d^3*e^2* f*g + 7*d^4*e*g^2)*x)/(e^5*x^2 - 2*d*e^4*x + d^2*e^3) - 1/3*(e^2*g^2*x^3 + 3*(e^2*f*g + 3*d*e*g^2)*x^2 + 3*(e^2*f^2 + 12*d*e*f*g + 18*d^2*g^2)*x)/e^ 2 - 2*(3*d*e^2*f^2 + 18*d^2*e*f*g + 19*d^3*g^2)*log(e*x - d)/e^3
Time = 0.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.24 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {2 \, {\left (3 \, d e^{2} f^{2} + 18 \, d^{2} e f g + 19 \, d^{3} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {4 \, {\left (2 \, d^{3} e^{2} f^{2} + 8 \, d^{4} e f g + 6 \, d^{5} g^{2} - {\left (3 \, d^{2} e^{3} f^{2} + 10 \, d^{3} e^{2} f g + 7 \, d^{4} e g^{2}\right )} x\right )}}{{\left (e x - d\right )}^{2} e^{3}} - \frac {e^{9} g^{2} x^{3} + 3 \, e^{9} f g x^{2} + 9 \, d e^{8} g^{2} x^{2} + 3 \, e^{9} f^{2} x + 36 \, d e^{8} f g x + 54 \, d^{2} e^{7} g^{2} x}{3 \, e^{9}} \]
-2*(3*d*e^2*f^2 + 18*d^2*e*f*g + 19*d^3*g^2)*log(abs(e*x - d))/e^3 - 4*(2* d^3*e^2*f^2 + 8*d^4*e*f*g + 6*d^5*g^2 - (3*d^2*e^3*f^2 + 10*d^3*e^2*f*g + 7*d^4*e*g^2)*x)/((e*x - d)^2*e^3) - 1/3*(e^9*g^2*x^3 + 3*e^9*f*g*x^2 + 9*d *e^8*g^2*x^2 + 3*e^9*f^2*x + 36*d*e^8*f*g*x + 54*d^2*e^7*g^2*x)/e^9
Time = 0.11 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.61 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {x\,\left (28\,d^4\,g^2+40\,d^3\,e\,f\,g+12\,d^2\,e^2\,f^2\right )-\frac {8\,\left (3\,d^5\,g^2+4\,d^4\,e\,f\,g+d^3\,e^2\,f^2\right )}{e}}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-x\,\left (\frac {3\,d^2\,e\,g^2+6\,d\,e^2\,f\,g+e^3\,f^2}{e^3}+\frac {3\,d\,\left (\frac {g\,\left (3\,d\,g+2\,e\,f\right )}{e}+\frac {3\,d\,g^2}{e}\right )}{e}-\frac {3\,d^2\,g^2}{e^2}\right )-x^2\,\left (\frac {g\,\left (3\,d\,g+2\,e\,f\right )}{2\,e}+\frac {3\,d\,g^2}{2\,e}\right )-\frac {g^2\,x^3}{3}-\frac {\ln \left (e\,x-d\right )\,\left (38\,d^3\,g^2+36\,d^2\,e\,f\,g+6\,d\,e^2\,f^2\right )}{e^3} \]
(x*(28*d^4*g^2 + 12*d^2*e^2*f^2 + 40*d^3*e*f*g) - (8*(3*d^5*g^2 + d^3*e^2* f^2 + 4*d^4*e*f*g))/e)/(d^2*e^2 + e^4*x^2 - 2*d*e^3*x) - x*((e^3*f^2 + 3*d ^2*e*g^2 + 6*d*e^2*f*g)/e^3 + (3*d*((g*(3*d*g + 2*e*f))/e + (3*d*g^2)/e))/ e - (3*d^2*g^2)/e^2) - x^2*((g*(3*d*g + 2*e*f))/(2*e) + (3*d*g^2)/(2*e)) - (g^2*x^3)/3 - (log(e*x - d)*(38*d^3*g^2 + 6*d*e^2*f^2 + 36*d^2*e*f*g))/e^ 3