Integrand size = 20, antiderivative size = 101 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=-\frac {(B d-A e) (a+b x)^3}{3 e (b d-a e) (d+e x)^3}-\frac {B (b d-a e)^2}{2 e^4 (d+e x)^2}+\frac {2 b B (b d-a e)}{e^4 (d+e x)}+\frac {b^2 B \log (d+e x)}{e^4} \] Output:
-1/3*(-A*e+B*d)*(b*x+a)^3/e/(-a*e+b*d)/(e*x+d)^3-1/2*B*(-a*e+b*d)^2/e^4/(e *x+d)^2+2*b*B*(-a*e+b*d)/e^4/(e*x+d)+b^2*B*ln(e*x+d)/e^4
Time = 0.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=\frac {-a^2 e^2 (2 A e+B (d+3 e x))-2 a b e \left (A e (d+3 e x)+2 B \left (d^2+3 d e x+3 e^2 x^2\right )\right )+b^2 \left (-2 A e \left (d^2+3 d e x+3 e^2 x^2\right )+B d \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )+6 b^2 B (d+e x)^3 \log (d+e x)}{6 e^4 (d+e x)^3} \] Input:
Integrate[((a + b*x)^2*(A + B*x))/(d + e*x)^4,x]
Output:
(-(a^2*e^2*(2*A*e + B*(d + 3*e*x))) - 2*a*b*e*(A*e*(d + 3*e*x) + 2*B*(d^2 + 3*d*e*x + 3*e^2*x^2)) + b^2*(-2*A*e*(d^2 + 3*d*e*x + 3*e^2*x^2) + B*d*(1 1*d^2 + 27*d*e*x + 18*e^2*x^2)) + 6*b^2*B*(d + e*x)^3*Log[d + e*x])/(6*e^4 *(d + e*x)^3)
Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {87, 49, 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.
\(\displaystyle \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {B \int \frac {(a+b x)^2}{(d+e x)^3}dx}{e}-\frac {(a+b x)^3 (B d-A e)}{3 e (d+e x)^3 (b d-a e)}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {B \int \left (\frac {b^2}{e^2 (d+e x)}-\frac {2 (b d-a e) b}{e^2 (d+e x)^2}+\frac {(a e-b d)^2}{e^2 (d+e x)^3}\right )dx}{e}-\frac {(a+b x)^3 (B d-A e)}{3 e (d+e x)^3 (b d-a e)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B \left (\frac {2 b (b d-a e)}{e^3 (d+e x)}-\frac {(b d-a e)^2}{2 e^3 (d+e x)^2}+\frac {b^2 \log (d+e x)}{e^3}\right )}{e}-\frac {(a+b x)^3 (B d-A e)}{3 e (d+e x)^3 (b d-a e)}\) |
Input:
Int[((a + b*x)^2*(A + B*x))/(d + e*x)^4,x]
Output:
-1/3*((B*d - A*e)*(a + b*x)^3)/(e*(b*d - a*e)*(d + e*x)^3) + (B*(-1/2*(b*d - a*e)^2/(e^3*(d + e*x)^2) + (2*b*(b*d - a*e))/(e^3*(d + e*x)) + (b^2*Log [d + e*x])/e^3))/e
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Time = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.55
method | result | size |
risch | \(\frac {-\frac {b \left (A b e +2 B a e -3 B b d \right ) x^{2}}{e^{2}}-\frac {\left (2 A a b \,e^{2}+2 A \,b^{2} d e +B \,a^{2} e^{2}+4 B a b d e -9 b^{2} B \,d^{2}\right ) x}{2 e^{3}}-\frac {2 a^{2} A \,e^{3}+2 A a b d \,e^{2}+2 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}+4 B a b \,d^{2} e -11 b^{2} B \,d^{3}}{6 e^{4}}}{\left (e x +d \right )^{3}}+\frac {b^{2} B \ln \left (e x +d \right )}{e^{4}}\) | \(157\) |
norman | \(\frac {-\frac {2 a^{2} A \,e^{3}+2 A a b d \,e^{2}+2 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}+4 B a b \,d^{2} e -11 b^{2} B \,d^{3}}{6 e^{4}}-\frac {\left (A \,b^{2} e +2 B a b e -3 b^{2} B d \right ) x^{2}}{e^{2}}-\frac {\left (2 A a b \,e^{2}+2 A \,b^{2} d e +B \,a^{2} e^{2}+4 B a b d e -9 b^{2} B \,d^{2}\right ) x}{2 e^{3}}}{\left (e x +d \right )^{3}}+\frac {b^{2} B \ln \left (e x +d \right )}{e^{4}}\) | \(161\) |
default | \(-\frac {b \left (A b e +2 B a e -3 B b d \right )}{e^{4} \left (e x +d \right )}-\frac {2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B a b d e +3 b^{2} B \,d^{2}}{2 e^{4} \left (e x +d \right )^{2}}+\frac {b^{2} B \ln \left (e x +d \right )}{e^{4}}-\frac {a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}}{3 e^{4} \left (e x +d \right )^{3}}\) | \(164\) |
parallelrisch | \(-\frac {-6 B \ln \left (e x +d \right ) x^{3} b^{2} e^{3}-18 B \ln \left (e x +d \right ) x^{2} b^{2} d \,e^{2}+6 A \,x^{2} b^{2} e^{3}-18 B \ln \left (e x +d \right ) x \,b^{2} d^{2} e +12 B \,x^{2} a b \,e^{3}-18 B \,x^{2} b^{2} d \,e^{2}+6 A x a b \,e^{3}+6 A x \,b^{2} d \,e^{2}-6 B \ln \left (e x +d \right ) b^{2} d^{3}+3 B x \,a^{2} e^{3}+12 B x a b d \,e^{2}-27 B x \,b^{2} d^{2} e +2 a^{2} A \,e^{3}+2 A a b d \,e^{2}+2 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}+4 B a b \,d^{2} e -11 b^{2} B \,d^{3}}{6 e^{4} \left (e x +d \right )^{3}}\) | \(225\) |
Input:
int((b*x+a)^2*(B*x+A)/(e*x+d)^4,x,method=_RETURNVERBOSE)
Output:
(-b*(A*b*e+2*B*a*e-3*B*b*d)/e^2*x^2-1/2*(2*A*a*b*e^2+2*A*b^2*d*e+B*a^2*e^2 +4*B*a*b*d*e-9*B*b^2*d^2)/e^3*x-1/6*(2*A*a^2*e^3+2*A*a*b*d*e^2+2*A*b^2*d^2 *e+B*a^2*d*e^2+4*B*a*b*d^2*e-11*B*b^2*d^3)/e^4)/(e*x+d)^3+b^2*B*ln(e*x+d)/ e^4
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (97) = 194\).
Time = 0.07 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.19 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=\frac {11 \, B b^{2} d^{3} - 2 \, A a^{2} e^{3} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 6 \, {\left (3 \, B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 3 \, {\left (9 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} - {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x + 6 \, {\left (B b^{2} e^{3} x^{3} + 3 \, B b^{2} d e^{2} x^{2} + 3 \, B b^{2} d^{2} e x + B b^{2} d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^4,x, algorithm="fricas")
Output:
1/6*(11*B*b^2*d^3 - 2*A*a^2*e^3 - 2*(2*B*a*b + A*b^2)*d^2*e - (B*a^2 + 2*A *a*b)*d*e^2 + 6*(3*B*b^2*d*e^2 - (2*B*a*b + A*b^2)*e^3)*x^2 + 3*(9*B*b^2*d ^2*e - 2*(2*B*a*b + A*b^2)*d*e^2 - (B*a^2 + 2*A*a*b)*e^3)*x + 6*(B*b^2*e^3 *x^3 + 3*B*b^2*d*e^2*x^2 + 3*B*b^2*d^2*e*x + B*b^2*d^3)*log(e*x + d))/(e^7 *x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (88) = 176\).
Time = 2.36 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.09 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=\frac {B b^{2} \log {\left (d + e x \right )}}{e^{4}} + \frac {- 2 A a^{2} e^{3} - 2 A a b d e^{2} - 2 A b^{2} d^{2} e - B a^{2} d e^{2} - 4 B a b d^{2} e + 11 B b^{2} d^{3} + x^{2} \left (- 6 A b^{2} e^{3} - 12 B a b e^{3} + 18 B b^{2} d e^{2}\right ) + x \left (- 6 A a b e^{3} - 6 A b^{2} d e^{2} - 3 B a^{2} e^{3} - 12 B a b d e^{2} + 27 B b^{2} d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \] Input:
integrate((b*x+a)**2*(B*x+A)/(e*x+d)**4,x)
Output:
B*b**2*log(d + e*x)/e**4 + (-2*A*a**2*e**3 - 2*A*a*b*d*e**2 - 2*A*b**2*d** 2*e - B*a**2*d*e**2 - 4*B*a*b*d**2*e + 11*B*b**2*d**3 + x**2*(-6*A*b**2*e* *3 - 12*B*a*b*e**3 + 18*B*b**2*d*e**2) + x*(-6*A*a*b*e**3 - 6*A*b**2*d*e** 2 - 3*B*a**2*e**3 - 12*B*a*b*d*e**2 + 27*B*b**2*d**2*e))/(6*d**3*e**4 + 18 *d**2*e**5*x + 18*d*e**6*x**2 + 6*e**7*x**3)
Time = 0.04 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.82 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=\frac {11 \, B b^{2} d^{3} - 2 \, A a^{2} e^{3} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 6 \, {\left (3 \, B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 3 \, {\left (9 \, B b^{2} d^{2} e - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} - {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac {B b^{2} \log \left (e x + d\right )}{e^{4}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^4,x, algorithm="maxima")
Output:
1/6*(11*B*b^2*d^3 - 2*A*a^2*e^3 - 2*(2*B*a*b + A*b^2)*d^2*e - (B*a^2 + 2*A *a*b)*d*e^2 + 6*(3*B*b^2*d*e^2 - (2*B*a*b + A*b^2)*e^3)*x^2 + 3*(9*B*b^2*d ^2*e - 2*(2*B*a*b + A*b^2)*d*e^2 - (B*a^2 + 2*A*a*b)*e^3)*x)/(e^7*x^3 + 3* d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) + B*b^2*log(e*x + d)/e^4
Time = 0.13 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=\frac {B b^{2} \log \left ({\left | e x + d \right |}\right )}{e^{4}} + \frac {6 \, {\left (3 \, B b^{2} d e - 2 \, B a b e^{2} - A b^{2} e^{2}\right )} x^{2} + 3 \, {\left (9 \, B b^{2} d^{2} - 4 \, B a b d e - 2 \, A b^{2} d e - B a^{2} e^{2} - 2 \, A a b e^{2}\right )} x + \frac {11 \, B b^{2} d^{3} - 4 \, B a b d^{2} e - 2 \, A b^{2} d^{2} e - B a^{2} d e^{2} - 2 \, A a b d e^{2} - 2 \, A a^{2} e^{3}}{e}}{6 \, {\left (e x + d\right )}^{3} e^{3}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^4,x, algorithm="giac")
Output:
B*b^2*log(abs(e*x + d))/e^4 + 1/6*(6*(3*B*b^2*d*e - 2*B*a*b*e^2 - A*b^2*e^ 2)*x^2 + 3*(9*B*b^2*d^2 - 4*B*a*b*d*e - 2*A*b^2*d*e - B*a^2*e^2 - 2*A*a*b* e^2)*x + (11*B*b^2*d^3 - 4*B*a*b*d^2*e - 2*A*b^2*d^2*e - B*a^2*d*e^2 - 2*A *a*b*d*e^2 - 2*A*a^2*e^3)/e)/((e*x + d)^3*e^3)
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.76 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=\frac {B\,b^2\,\ln \left (d+e\,x\right )}{e^4}-\frac {\frac {B\,a^2\,d\,e^2+2\,A\,a^2\,e^3+4\,B\,a\,b\,d^2\,e+2\,A\,a\,b\,d\,e^2-11\,B\,b^2\,d^3+2\,A\,b^2\,d^2\,e}{6\,e^4}+\frac {x\,\left (B\,a^2\,e^2+4\,B\,a\,b\,d\,e+2\,A\,a\,b\,e^2-9\,B\,b^2\,d^2+2\,A\,b^2\,d\,e\right )}{2\,e^3}+\frac {b\,x^2\,\left (A\,b\,e+2\,B\,a\,e-3\,B\,b\,d\right )}{e^2}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \] Input:
int(((A + B*x)*(a + b*x)^2)/(d + e*x)^4,x)
Output:
(B*b^2*log(d + e*x))/e^4 - ((2*A*a^2*e^3 - 11*B*b^2*d^3 + 2*A*b^2*d^2*e + B*a^2*d*e^2 + 2*A*a*b*d*e^2 + 4*B*a*b*d^2*e)/(6*e^4) + (x*(B*a^2*e^2 - 9*B *b^2*d^2 + 2*A*a*b*e^2 + 2*A*b^2*d*e + 4*B*a*b*d*e))/(2*e^3) + (b*x^2*(A*b *e + 2*B*a*e - 3*B*b*d))/e^2)/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x)
Time = 0.16 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.78 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^4} \, dx=\frac {6 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{4}+18 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{3} e x +18 \,\mathrm {log}\left (e x +d \right ) b^{3} d^{2} e^{2} x^{2}+6 \,\mathrm {log}\left (e x +d \right ) b^{3} d \,e^{3} x^{3}-2 a^{3} d \,e^{3}-3 a^{2} b \,d^{2} e^{2}-9 a^{2} b d \,e^{3} x +6 a \,b^{2} e^{4} x^{3}+5 b^{3} d^{4}+9 b^{3} d^{3} e x -6 b^{3} d \,e^{3} x^{3}}{6 d \,e^{4} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )} \] Input:
int((b*x+a)^2*(B*x+A)/(e*x+d)^4,x)
Output:
(6*log(d + e*x)*b**3*d**4 + 18*log(d + e*x)*b**3*d**3*e*x + 18*log(d + e*x )*b**3*d**2*e**2*x**2 + 6*log(d + e*x)*b**3*d*e**3*x**3 - 2*a**3*d*e**3 - 3*a**2*b*d**2*e**2 - 9*a**2*b*d*e**3*x + 6*a*b**2*e**4*x**3 + 5*b**3*d**4 + 9*b**3*d**3*e*x - 6*b**3*d*e**3*x**3)/(6*d*e**4*(d**3 + 3*d**2*e*x + 3*d *e**2*x**2 + e**3*x**3))