Integrand size = 27, antiderivative size = 106 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {(A b-a B) x^5}{5 a b (a+b x)^5}-\frac {a^4 B}{4 b^6 (a+b x)^4}+\frac {4 a^3 B}{3 b^6 (a+b x)^3}-\frac {3 a^2 B}{b^6 (a+b x)^2}+\frac {4 a B}{b^6 (a+b x)}+\frac {B \log (a+b x)}{b^6} \] Output:
1/5*(A*b-B*a)*x^5/a/b/(b*x+a)^5-1/4*a^4*B/b^6/(b*x+a)^4+4/3*a^3*B/b^6/(b*x +a)^3-3*a^2*B/b^6/(b*x+a)^2+4*a*B/b^6/(b*x+a)+B*ln(b*x+a)/b^6
Time = 0.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.07 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {137 a^5 B-60 A b^5 x^4+60 a b^4 x^3 (-2 A+5 B x)+60 a^2 b^3 x^2 (-2 A+15 B x)+20 a^3 b^2 x (-3 A+55 B x)+a^4 (-12 A b+625 b B x)+60 B (a+b x)^5 \log (a+b x)}{60 b^6 (a+b x)^5} \] Input:
Integrate[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
Output:
(137*a^5*B - 60*A*b^5*x^4 + 60*a*b^4*x^3*(-2*A + 5*B*x) + 60*a^2*b^3*x^2*( -2*A + 15*B*x) + 20*a^3*b^2*x*(-3*A + 55*B*x) + a^4*(-12*A*b + 625*b*B*x) + 60*B*(a + b*x)^5*Log[a + b*x])/(60*b^6*(a + b*x)^5)
Time = 0.47 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1184, 27, 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.
\(\displaystyle \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^6 \int \frac {x^4 (A+B x)}{b^6 (a+b x)^6}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^4 (A+B x)}{(a+b x)^6}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {B \int \frac {x^4}{(a+b x)^5}dx}{b}+\frac {x^5 (A b-a B)}{5 a b (a+b x)^5}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {B \int \left (\frac {a^4}{b^4 (a+b x)^5}-\frac {4 a^3}{b^4 (a+b x)^4}+\frac {6 a^2}{b^4 (a+b x)^3}-\frac {4 a}{b^4 (a+b x)^2}+\frac {1}{b^4 (a+b x)}\right )dx}{b}+\frac {x^5 (A b-a B)}{5 a b (a+b x)^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B \left (-\frac {a^4}{4 b^5 (a+b x)^4}+\frac {4 a^3}{3 b^5 (a+b x)^3}-\frac {3 a^2}{b^5 (a+b x)^2}+\frac {4 a}{b^5 (a+b x)}+\frac {\log (a+b x)}{b^5}\right )}{b}+\frac {x^5 (A b-a B)}{5 a b (a+b x)^5}\) |
Input:
Int[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
Output:
((A*b - a*B)*x^5)/(5*a*b*(a + b*x)^5) + (B*(-1/4*a^4/(b^5*(a + b*x)^4) + ( 4*a^3)/(3*b^5*(a + b*x)^3) - (3*a^2)/(b^5*(a + b*x)^2) + (4*a)/(b^5*(a + b *x)) + Log[a + b*x]/b^5))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.03 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.05
method | result | size |
norman | \(\frac {-\frac {a^{4} \left (12 A b -137 B a \right )}{60 b^{6}}-\frac {\left (A b -5 B a \right ) x^{4}}{b^{2}}-\frac {a \left (2 A b -15 B a \right ) x^{3}}{b^{3}}-\frac {a^{2} \left (6 A b -55 B a \right ) x^{2}}{3 b^{4}}-\frac {a^{3} \left (12 A b -125 B a \right ) x}{12 b^{5}}}{\left (b x +a \right )^{5}}+\frac {B \ln \left (b x +a \right )}{b^{6}}\) | \(111\) |
default | \(-\frac {a^{4} \left (A b -B a \right )}{5 b^{6} \left (b x +a \right )^{5}}-\frac {2 a^{2} \left (3 A b -5 B a \right )}{3 b^{6} \left (b x +a \right )^{3}}-\frac {A b -5 B a}{b^{6} \left (b x +a \right )}+\frac {B \ln \left (b x +a \right )}{b^{6}}+\frac {a \left (2 A b -5 B a \right )}{b^{6} \left (b x +a \right )^{2}}+\frac {a^{3} \left (4 A b -5 B a \right )}{4 b^{6} \left (b x +a \right )^{4}}\) | \(125\) |
risch | \(\frac {-\frac {a^{4} \left (12 A b -137 B a \right )}{60 b^{6}}-\frac {\left (A b -5 B a \right ) x^{4}}{b^{2}}-\frac {a \left (2 A b -15 B a \right ) x^{3}}{b^{3}}-\frac {a^{2} \left (6 A b -55 B a \right ) x^{2}}{3 b^{4}}-\frac {a^{3} \left (12 A b -125 B a \right ) x}{12 b^{5}}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}+\frac {B \ln \left (b x +a \right )}{b^{6}}\) | \(129\) |
parallelrisch | \(-\frac {-60 B \ln \left (b x +a \right ) x^{5} b^{5}-300 B \ln \left (b x +a \right ) x^{4} a \,b^{4}+60 A \,b^{5} x^{4}-600 B \ln \left (b x +a \right ) x^{3} a^{2} b^{3}-300 B a \,b^{4} x^{4}+120 A a \,b^{4} x^{3}-600 B \ln \left (b x +a \right ) x^{2} a^{3} b^{2}-900 B \,a^{2} b^{3} x^{3}+120 A \,a^{2} b^{3} x^{2}-300 B \ln \left (b x +a \right ) x \,a^{4} b -1100 B \,a^{3} b^{2} x^{2}+60 A \,a^{3} b^{2} x -60 B \ln \left (b x +a \right ) a^{5}-625 B \,a^{4} b x +12 A \,a^{4} b -137 B \,a^{5}}{60 b^{6} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2} \left (b x +a \right )}\) | \(221\) |
Input:
int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
Output:
(-1/60*a^4*(12*A*b-137*B*a)/b^6-(A*b-5*B*a)/b^2*x^4-a*(2*A*b-15*B*a)/b^3*x ^3-1/3*a^2*(6*A*b-55*B*a)/b^4*x^2-1/12*a^3*(12*A*b-125*B*a)/b^5*x)/(b*x+a) ^5+B*ln(b*x+a)/b^6
Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (100) = 200\).
Time = 0.08 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.09 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {137 \, B a^{5} - 12 \, A a^{4} b + 60 \, {\left (5 \, B a b^{4} - A b^{5}\right )} x^{4} + 60 \, {\left (15 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{3} + 20 \, {\left (55 \, B a^{3} b^{2} - 6 \, A a^{2} b^{3}\right )} x^{2} + 5 \, {\left (125 \, B a^{4} b - 12 \, A a^{3} b^{2}\right )} x + 60 \, {\left (B b^{5} x^{5} + 5 \, B a b^{4} x^{4} + 10 \, B a^{2} b^{3} x^{3} + 10 \, B a^{3} b^{2} x^{2} + 5 \, B a^{4} b x + B a^{5}\right )} \log \left (b x + a\right )}{60 \, {\left (b^{11} x^{5} + 5 \, a b^{10} x^{4} + 10 \, a^{2} b^{9} x^{3} + 10 \, a^{3} b^{8} x^{2} + 5 \, a^{4} b^{7} x + a^{5} b^{6}\right )}} \] Input:
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
Output:
1/60*(137*B*a^5 - 12*A*a^4*b + 60*(5*B*a*b^4 - A*b^5)*x^4 + 60*(15*B*a^2*b ^3 - 2*A*a*b^4)*x^3 + 20*(55*B*a^3*b^2 - 6*A*a^2*b^3)*x^2 + 5*(125*B*a^4*b - 12*A*a^3*b^2)*x + 60*(B*b^5*x^5 + 5*B*a*b^4*x^4 + 10*B*a^2*b^3*x^3 + 10 *B*a^3*b^2*x^2 + 5*B*a^4*b*x + B*a^5)*log(b*x + a))/(b^11*x^5 + 5*a*b^10*x ^4 + 10*a^2*b^9*x^3 + 10*a^3*b^8*x^2 + 5*a^4*b^7*x + a^5*b^6)
Time = 1.06 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.62 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {B \log {\left (a + b x \right )}}{b^{6}} + \frac {- 12 A a^{4} b + 137 B a^{5} + x^{4} \left (- 60 A b^{5} + 300 B a b^{4}\right ) + x^{3} \left (- 120 A a b^{4} + 900 B a^{2} b^{3}\right ) + x^{2} \left (- 120 A a^{2} b^{3} + 1100 B a^{3} b^{2}\right ) + x \left (- 60 A a^{3} b^{2} + 625 B a^{4} b\right )}{60 a^{5} b^{6} + 300 a^{4} b^{7} x + 600 a^{3} b^{8} x^{2} + 600 a^{2} b^{9} x^{3} + 300 a b^{10} x^{4} + 60 b^{11} x^{5}} \] Input:
integrate(x**4*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**3,x)
Output:
B*log(a + b*x)/b**6 + (-12*A*a**4*b + 137*B*a**5 + x**4*(-60*A*b**5 + 300* B*a*b**4) + x**3*(-120*A*a*b**4 + 900*B*a**2*b**3) + x**2*(-120*A*a**2*b** 3 + 1100*B*a**3*b**2) + x*(-60*A*a**3*b**2 + 625*B*a**4*b))/(60*a**5*b**6 + 300*a**4*b**7*x + 600*a**3*b**8*x**2 + 600*a**2*b**9*x**3 + 300*a*b**10* x**4 + 60*b**11*x**5)
Time = 0.04 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.60 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {137 \, B a^{5} - 12 \, A a^{4} b + 60 \, {\left (5 \, B a b^{4} - A b^{5}\right )} x^{4} + 60 \, {\left (15 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{3} + 20 \, {\left (55 \, B a^{3} b^{2} - 6 \, A a^{2} b^{3}\right )} x^{2} + 5 \, {\left (125 \, B a^{4} b - 12 \, A a^{3} b^{2}\right )} x}{60 \, {\left (b^{11} x^{5} + 5 \, a b^{10} x^{4} + 10 \, a^{2} b^{9} x^{3} + 10 \, a^{3} b^{8} x^{2} + 5 \, a^{4} b^{7} x + a^{5} b^{6}\right )}} + \frac {B \log \left (b x + a\right )}{b^{6}} \] Input:
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")
Output:
1/60*(137*B*a^5 - 12*A*a^4*b + 60*(5*B*a*b^4 - A*b^5)*x^4 + 60*(15*B*a^2*b ^3 - 2*A*a*b^4)*x^3 + 20*(55*B*a^3*b^2 - 6*A*a^2*b^3)*x^2 + 5*(125*B*a^4*b - 12*A*a^3*b^2)*x)/(b^11*x^5 + 5*a*b^10*x^4 + 10*a^2*b^9*x^3 + 10*a^3*b^8 *x^2 + 5*a^4*b^7*x + a^5*b^6) + B*log(b*x + a)/b^6
Time = 0.20 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.17 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {B \log \left ({\left | b x + a \right |}\right )}{b^{6}} + \frac {60 \, {\left (5 \, B a b^{3} - A b^{4}\right )} x^{4} + 60 \, {\left (15 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} + 20 \, {\left (55 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x^{2} + 5 \, {\left (125 \, B a^{4} - 12 \, A a^{3} b\right )} x + \frac {137 \, B a^{5} - 12 \, A a^{4} b}{b}}{60 \, {\left (b x + a\right )}^{5} b^{5}} \] Input:
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
Output:
B*log(abs(b*x + a))/b^6 + 1/60*(60*(5*B*a*b^3 - A*b^4)*x^4 + 60*(15*B*a^2* b^2 - 2*A*a*b^3)*x^3 + 20*(55*B*a^3*b - 6*A*a^2*b^2)*x^2 + 5*(125*B*a^4 - 12*A*a^3*b)*x + (137*B*a^5 - 12*A*a^4*b)/b)/((b*x + a)^5*b^5)
Time = 10.73 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.52 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {137\,B\,a^5-12\,A\,a^4\,b}{60\,b^6}+\frac {x^3\,\left (15\,B\,a^2-2\,A\,a\,b\right )}{b^3}+\frac {x\,\left (125\,B\,a^4-12\,A\,a^3\,b\right )}{12\,b^5}-\frac {x^4\,\left (A\,b-5\,B\,a\right )}{b^2}+\frac {x^2\,\left (55\,B\,a^3-6\,A\,a^2\,b\right )}{3\,b^4}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5}+\frac {B\,\ln \left (a+b\,x\right )}{b^6} \] Input:
int((x^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)
Output:
((137*B*a^5 - 12*A*a^4*b)/(60*b^6) + (x^3*(15*B*a^2 - 2*A*a*b))/b^3 + (x*( 125*B*a^4 - 12*A*a^3*b))/(12*b^5) - (x^4*(A*b - 5*B*a))/b^2 + (x^2*(55*B*a ^3 - 6*A*a^2*b))/(3*b^4))/(a^5 + b^5*x^5 + 5*a*b^4*x^4 + 10*a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4*b*x) + (B*log(a + b*x))/b^6
Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.39 \[ \int \frac {x^4 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {12 \,\mathrm {log}\left (b x +a \right ) a^{4}+48 \,\mathrm {log}\left (b x +a \right ) a^{3} b x +72 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} x^{2}+48 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} x^{3}+12 \,\mathrm {log}\left (b x +a \right ) b^{4} x^{4}+13 a^{4}+40 a^{3} b x +36 a^{2} b^{2} x^{2}-12 b^{4} x^{4}}{12 b^{5} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x)
Output:
(12*log(a + b*x)*a**4 + 48*log(a + b*x)*a**3*b*x + 72*log(a + b*x)*a**2*b* *2*x**2 + 48*log(a + b*x)*a*b**3*x**3 + 12*log(a + b*x)*b**4*x**4 + 13*a** 4 + 40*a**3*b*x + 36*a**2*b**2*x**2 - 12*b**4*x**4)/(12*b**5*(a**4 + 4*a** 3*b*x + 6*a**2*b**2*x**2 + 4*a*b**3*x**3 + b**4*x**4))