Integrand size = 20, antiderivative size = 75 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=\frac {1}{3} A b^3 x^3+\frac {1}{4} b^2 (b B+3 A c) x^4+\frac {3}{5} b c (b B+A c) x^5+\frac {1}{6} c^2 (3 b B+A c) x^6+\frac {1}{7} B c^3 x^7 \] Output:
1/3*A*b^3*x^3+1/4*b^2*(3*A*c+B*b)*x^4+3/5*b*c*(A*c+B*b)*x^5+1/6*c^2*(A*c+3 *B*b)*x^6+1/7*B*c^3*x^7
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=\frac {1}{3} A b^3 x^3+\frac {1}{4} b^2 (b B+3 A c) x^4+\frac {3}{5} b c (b B+A c) x^5+\frac {1}{6} c^2 (3 b B+A c) x^6+\frac {1}{7} B c^3 x^7 \] Input:
Integrate[((A + B*x)*(b*x + c*x^2)^3)/x,x]
Output:
(A*b^3*x^3)/3 + (b^2*(b*B + 3*A*c)*x^4)/4 + (3*b*c*(b*B + A*c)*x^5)/5 + (c ^2*(3*b*B + A*c)*x^6)/6 + (B*c^3*x^7)/7
Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {9, 85, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^2 (A+B x) (b+c x)^3dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (A b^3 x^2+b^2 x^3 (3 A c+b B)+c^2 x^5 (A c+3 b B)+3 b c x^4 (A c+b B)+B c^3 x^6\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} A b^3 x^3+\frac {1}{4} b^2 x^4 (3 A c+b B)+\frac {1}{6} c^2 x^6 (A c+3 b B)+\frac {3}{5} b c x^5 (A c+b B)+\frac {1}{7} B c^3 x^7\) |
Input:
Int[((A + B*x)*(b*x + c*x^2)^3)/x,x]
Output:
(A*b^3*x^3)/3 + (b^2*(b*B + 3*A*c)*x^4)/4 + (3*b*c*(b*B + A*c)*x^5)/5 + (c ^2*(3*b*B + A*c)*x^6)/6 + (B*c^3*x^7)/7
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Time = 0.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\frac {B \,c^{3} x^{7}}{7}+\left (\frac {1}{6} A \,c^{3}+\frac {1}{2} B b \,c^{2}\right ) x^{6}+\left (\frac {3}{5} A b \,c^{2}+\frac {3}{5} B \,b^{2} c \right ) x^{5}+\left (\frac {3}{4} A \,b^{2} c +\frac {1}{4} B \,b^{3}\right ) x^{4}+\frac {A \,b^{3} x^{3}}{3}\) | \(75\) |
gosper | \(\frac {x^{3} \left (60 B \,c^{3} x^{4}+70 A \,c^{3} x^{3}+210 x^{3} B b \,c^{2}+252 A b \,c^{2} x^{2}+252 x^{2} B \,b^{2} c +315 A \,b^{2} c x +105 x B \,b^{3}+140 A \,b^{3}\right )}{420}\) | \(76\) |
default | \(\frac {B \,c^{3} x^{7}}{7}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{6}}{6}+\frac {\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{5}}{5}+\frac {\left (3 A \,b^{2} c +B \,b^{3}\right ) x^{4}}{4}+\frac {A \,b^{3} x^{3}}{3}\) | \(76\) |
risch | \(\frac {1}{7} B \,c^{3} x^{7}+\frac {1}{6} x^{6} A \,c^{3}+\frac {1}{2} x^{6} B b \,c^{2}+\frac {3}{5} x^{5} A b \,c^{2}+\frac {3}{5} x^{5} B \,b^{2} c +\frac {3}{4} x^{4} A \,b^{2} c +\frac {1}{4} B \,b^{3} x^{4}+\frac {1}{3} A \,b^{3} x^{3}\) | \(78\) |
parallelrisch | \(\frac {1}{7} B \,c^{3} x^{7}+\frac {1}{6} x^{6} A \,c^{3}+\frac {1}{2} x^{6} B b \,c^{2}+\frac {3}{5} x^{5} A b \,c^{2}+\frac {3}{5} x^{5} B \,b^{2} c +\frac {3}{4} x^{4} A \,b^{2} c +\frac {1}{4} B \,b^{3} x^{4}+\frac {1}{3} A \,b^{3} x^{3}\) | \(78\) |
orering | \(\frac {\left (60 B \,c^{3} x^{4}+70 A \,c^{3} x^{3}+210 x^{3} B b \,c^{2}+252 A b \,c^{2} x^{2}+252 x^{2} B \,b^{2} c +315 A \,b^{2} c x +105 x B \,b^{3}+140 A \,b^{3}\right ) \left (c \,x^{2}+b x \right )^{3}}{420 \left (c x +b \right )^{3}}\) | \(91\) |
Input:
int((B*x+A)*(c*x^2+b*x)^3/x,x,method=_RETURNVERBOSE)
Output:
1/7*B*c^3*x^7+(1/6*A*c^3+1/2*B*b*c^2)*x^6+(3/5*A*b*c^2+3/5*B*b^2*c)*x^5+(3 /4*A*b^2*c+1/4*B*b^3)*x^4+1/3*A*b^3*x^3
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=\frac {1}{7} \, B c^{3} x^{7} + \frac {1}{3} \, A b^{3} x^{3} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{5} \, {\left (B b^{2} c + A b c^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{4} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^3/x,x, algorithm="fricas")
Output:
1/7*B*c^3*x^7 + 1/3*A*b^3*x^3 + 1/6*(3*B*b*c^2 + A*c^3)*x^6 + 3/5*(B*b^2*c + A*b*c^2)*x^5 + 1/4*(B*b^3 + 3*A*b^2*c)*x^4
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=\frac {A b^{3} x^{3}}{3} + \frac {B c^{3} x^{7}}{7} + x^{6} \left (\frac {A c^{3}}{6} + \frac {B b c^{2}}{2}\right ) + x^{5} \cdot \left (\frac {3 A b c^{2}}{5} + \frac {3 B b^{2} c}{5}\right ) + x^{4} \cdot \left (\frac {3 A b^{2} c}{4} + \frac {B b^{3}}{4}\right ) \] Input:
integrate((B*x+A)*(c*x**2+b*x)**3/x,x)
Output:
A*b**3*x**3/3 + B*c**3*x**7/7 + x**6*(A*c**3/6 + B*b*c**2/2) + x**5*(3*A*b *c**2/5 + 3*B*b**2*c/5) + x**4*(3*A*b**2*c/4 + B*b**3/4)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=\frac {1}{7} \, B c^{3} x^{7} + \frac {1}{3} \, A b^{3} x^{3} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{5} \, {\left (B b^{2} c + A b c^{2}\right )} x^{5} + \frac {1}{4} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{4} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^3/x,x, algorithm="maxima")
Output:
1/7*B*c^3*x^7 + 1/3*A*b^3*x^3 + 1/6*(3*B*b*c^2 + A*c^3)*x^6 + 3/5*(B*b^2*c + A*b*c^2)*x^5 + 1/4*(B*b^3 + 3*A*b^2*c)*x^4
Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=\frac {1}{7} \, B c^{3} x^{7} + \frac {1}{2} \, B b c^{2} x^{6} + \frac {1}{6} \, A c^{3} x^{6} + \frac {3}{5} \, B b^{2} c x^{5} + \frac {3}{5} \, A b c^{2} x^{5} + \frac {1}{4} \, B b^{3} x^{4} + \frac {3}{4} \, A b^{2} c x^{4} + \frac {1}{3} \, A b^{3} x^{3} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^3/x,x, algorithm="giac")
Output:
1/7*B*c^3*x^7 + 1/2*B*b*c^2*x^6 + 1/6*A*c^3*x^6 + 3/5*B*b^2*c*x^5 + 3/5*A* b*c^2*x^5 + 1/4*B*b^3*x^4 + 3/4*A*b^2*c*x^4 + 1/3*A*b^3*x^3
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=x^4\,\left (\frac {B\,b^3}{4}+\frac {3\,A\,c\,b^2}{4}\right )+x^6\,\left (\frac {A\,c^3}{6}+\frac {B\,b\,c^2}{2}\right )+\frac {A\,b^3\,x^3}{3}+\frac {B\,c^3\,x^7}{7}+\frac {3\,b\,c\,x^5\,\left (A\,c+B\,b\right )}{5} \] Input:
int(((b*x + c*x^2)^3*(A + B*x))/x,x)
Output:
x^4*((B*b^3)/4 + (3*A*b^2*c)/4) + x^6*((A*c^3)/6 + (B*b*c^2)/2) + (A*b^3*x ^3)/3 + (B*c^3*x^7)/7 + (3*b*c*x^5*(A*c + B*b))/5
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x} \, dx=\frac {x^{3} \left (60 b \,c^{3} x^{4}+70 a \,c^{3} x^{3}+210 b^{2} c^{2} x^{3}+252 a b \,c^{2} x^{2}+252 b^{3} c \,x^{2}+315 a \,b^{2} c x +105 b^{4} x +140 a \,b^{3}\right )}{420} \] Input:
int((B*x+A)*(c*x^2+b*x)^3/x,x)
Output:
(x**3*(140*a*b**3 + 315*a*b**2*c*x + 252*a*b*c**2*x**2 + 70*a*c**3*x**3 + 105*b**4*x + 252*b**3*c*x**2 + 210*b**2*c**2*x**3 + 60*b*c**3*x**4))/420