Integrand size = 27, antiderivative size = 66 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=4 a^3 A b x+3 a^2 A b^2 x^2+\frac {4}{3} a A b^3 x^3+\frac {1}{4} A b^4 x^4+\frac {B (a+b x)^5}{5 b}+a^4 A \log (x) \] Output:
4*a^3*A*b*x+3*a^2*A*b^2*x^2+4/3*a*A*b^3*x^3+1/4*A*b^4*x^4+1/5*B*(b*x+a)^5/ b+a^4*A*ln(x)
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=a^4 B x+2 a^3 b x (2 A+B x)+a^2 b^2 x^2 (3 A+2 B x)+\frac {1}{3} a b^3 x^3 (4 A+3 B x)+\frac {1}{20} b^4 x^4 (5 A+4 B x)+a^4 A \log (x) \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x,x]
Output:
a^4*B*x + 2*a^3*b*x*(2*A + B*x) + a^2*b^2*x^2*(3*A + 2*B*x) + (a*b^3*x^3*( 4*A + 3*B*x))/3 + (b^4*x^4*(5*A + 4*B*x))/20 + a^4*A*Log[x]
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1184, 27, 90, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2 (A+B x)}{x} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^4 (a+b x)^4 (A+B x)}{x}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^4 (A+B x)}{x}dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle A \int \frac {(a+b x)^4}{x}dx+\frac {B (a+b x)^5}{5 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle A \int \left (\frac {a^4}{x}+4 b a^3+6 b^2 x a^2+4 b^3 x^2 a+b^4 x^3\right )dx+\frac {B (a+b x)^5}{5 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle A \left (a^4 \log (x)+4 a^3 b x+3 a^2 b^2 x^2+\frac {4}{3} a b^3 x^3+\frac {b^4 x^4}{4}\right )+\frac {B (a+b x)^5}{5 b}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2)/x,x]
Output:
(B*(a + b*x)^5)/(5*b) + A*(4*a^3*b*x + 3*a^2*b^2*x^2 + (4*a*b^3*x^3)/3 + ( b^4*x^4)/4 + a^4*Log[x])
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*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 0.87 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.39
method | result | size |
norman | \(\left (\frac {1}{4} A \,b^{4}+B a \,b^{3}\right ) x^{4}+\left (\frac {4}{3} A a \,b^{3}+2 B \,a^{2} b^{2}\right ) x^{3}+\left (3 a^{2} A \,b^{2}+2 B \,a^{3} b \right ) x^{2}+\left (4 A \,a^{3} b +a^{4} B \right ) x +\frac {b^{4} B \,x^{5}}{5}+a^{4} A \ln \left (x \right )\) | \(92\) |
default | \(\frac {b^{4} B \,x^{5}}{5}+\frac {A \,b^{4} x^{4}}{4}+B a \,b^{3} x^{4}+\frac {4 A a \,b^{3} x^{3}}{3}+2 B \,a^{2} b^{2} x^{3}+3 A \,a^{2} b^{2} x^{2}+2 B \,a^{3} b \,x^{2}+4 A \,a^{3} b x +a^{4} B x +a^{4} A \ln \left (x \right )\) | \(94\) |
risch | \(\frac {b^{4} B \,x^{5}}{5}+\frac {A \,b^{4} x^{4}}{4}+B a \,b^{3} x^{4}+\frac {4 A a \,b^{3} x^{3}}{3}+2 B \,a^{2} b^{2} x^{3}+3 A \,a^{2} b^{2} x^{2}+2 B \,a^{3} b \,x^{2}+4 A \,a^{3} b x +a^{4} B x +a^{4} A \ln \left (x \right )\) | \(94\) |
parallelrisch | \(\frac {b^{4} B \,x^{5}}{5}+\frac {A \,b^{4} x^{4}}{4}+B a \,b^{3} x^{4}+\frac {4 A a \,b^{3} x^{3}}{3}+2 B \,a^{2} b^{2} x^{3}+3 A \,a^{2} b^{2} x^{2}+2 B \,a^{3} b \,x^{2}+4 A \,a^{3} b x +a^{4} B x +a^{4} A \ln \left (x \right )\) | \(94\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x,x,method=_RETURNVERBOSE)
Output:
(1/4*A*b^4+B*a*b^3)*x^4+(4/3*A*a*b^3+2*B*a^2*b^2)*x^3+(3*A*a^2*b^2+2*B*a^3 *b)*x^2+(4*A*a^3*b+B*a^4)*x+1/5*b^4*B*x^5+a^4*A*ln(x)
Time = 0.07 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=\frac {1}{5} \, B b^{4} x^{5} + A a^{4} \log \left (x\right ) + \frac {1}{4} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + \frac {2}{3} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} + 4 \, A a^{3} b\right )} x \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x,x, algorithm="fricas")
Output:
1/5*B*b^4*x^5 + A*a^4*log(x) + 1/4*(4*B*a*b^3 + A*b^4)*x^4 + 2/3*(3*B*a^2* b^2 + 2*A*a*b^3)*x^3 + (2*B*a^3*b + 3*A*a^2*b^2)*x^2 + (B*a^4 + 4*A*a^3*b) *x
Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.44 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=A a^{4} \log {\left (x \right )} + \frac {B b^{4} x^{5}}{5} + x^{4} \left (\frac {A b^{4}}{4} + B a b^{3}\right ) + x^{3} \cdot \left (\frac {4 A a b^{3}}{3} + 2 B a^{2} b^{2}\right ) + x^{2} \cdot \left (3 A a^{2} b^{2} + 2 B a^{3} b\right ) + x \left (4 A a^{3} b + B a^{4}\right ) \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**2/x,x)
Output:
A*a**4*log(x) + B*b**4*x**5/5 + x**4*(A*b**4/4 + B*a*b**3) + x**3*(4*A*a*b **3/3 + 2*B*a**2*b**2) + x**2*(3*A*a**2*b**2 + 2*B*a**3*b) + x*(4*A*a**3*b + B*a**4)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=\frac {1}{5} \, B b^{4} x^{5} + A a^{4} \log \left (x\right ) + \frac {1}{4} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + \frac {2}{3} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} + {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + {\left (B a^{4} + 4 \, A a^{3} b\right )} x \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x,x, algorithm="maxima")
Output:
1/5*B*b^4*x^5 + A*a^4*log(x) + 1/4*(4*B*a*b^3 + A*b^4)*x^4 + 2/3*(3*B*a^2* b^2 + 2*A*a*b^3)*x^3 + (2*B*a^3*b + 3*A*a^2*b^2)*x^2 + (B*a^4 + 4*A*a^3*b) *x
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=\frac {1}{5} \, B b^{4} x^{5} + B a b^{3} x^{4} + \frac {1}{4} \, A b^{4} x^{4} + 2 \, B a^{2} b^{2} x^{3} + \frac {4}{3} \, A a b^{3} x^{3} + 2 \, B a^{3} b x^{2} + 3 \, A a^{2} b^{2} x^{2} + B a^{4} x + 4 \, A a^{3} b x + A a^{4} \log \left ({\left | x \right |}\right ) \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x,x, algorithm="giac")
Output:
1/5*B*b^4*x^5 + B*a*b^3*x^4 + 1/4*A*b^4*x^4 + 2*B*a^2*b^2*x^3 + 4/3*A*a*b^ 3*x^3 + 2*B*a^3*b*x^2 + 3*A*a^2*b^2*x^2 + B*a^4*x + 4*A*a^3*b*x + A*a^4*lo g(abs(x))
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=x\,\left (B\,a^4+4\,A\,b\,a^3\right )+x^4\,\left (\frac {A\,b^4}{4}+B\,a\,b^3\right )+\frac {B\,b^4\,x^5}{5}+A\,a^4\,\ln \left (x\right )+a^2\,b\,x^2\,\left (3\,A\,b+2\,B\,a\right )+\frac {2\,a\,b^2\,x^3\,\left (2\,A\,b+3\,B\,a\right )}{3} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2)/x,x)
Output:
x*(B*a^4 + 4*A*a^3*b) + x^4*((A*b^4)/4 + B*a*b^3) + (B*b^4*x^5)/5 + A*a^4* log(x) + a^2*b*x^2*(3*A*b + 2*B*a) + (2*a*b^2*x^3*(2*A*b + 3*B*a))/3
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{x} \, dx=\mathrm {log}\left (x \right ) a^{5}+5 a^{4} b x +5 a^{3} b^{2} x^{2}+\frac {10 a^{2} b^{3} x^{3}}{3}+\frac {5 a \,b^{4} x^{4}}{4}+\frac {b^{5} x^{5}}{5} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2/x,x)
Output:
(60*log(x)*a**5 + 300*a**4*b*x + 300*a**3*b**2*x**2 + 200*a**2*b**3*x**3 + 75*a*b**4*x**4 + 12*b**5*x**5)/60