Integrand size = 22, antiderivative size = 81 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=-\frac {2 A b^3}{3 x^{3/2}}-\frac {2 b^2 (b B+3 A c)}{\sqrt {x}}+6 b c (b B+A c) \sqrt {x}+\frac {2}{3} c^2 (3 b B+A c) x^{3/2}+\frac {2}{5} B c^3 x^{5/2} \] Output:
-2/3*A*b^3/x^(3/2)-2*b^2*(3*A*c+B*b)/x^(1/2)+6*b*c*(A*c+B*b)*x^(1/2)+2/3*c ^2*(A*c+3*B*b)*x^(3/2)+2/5*B*c^3*x^(5/2)
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=\frac {-10 A \left (b^3+9 b^2 c x-9 b c^2 x^2-c^3 x^3\right )+6 B x \left (-5 b^3+15 b^2 c x+5 b c^2 x^2+c^3 x^3\right )}{15 x^{3/2}} \] Input:
Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^(11/2),x]
Output:
(-10*A*(b^3 + 9*b^2*c*x - 9*b*c^2*x^2 - c^3*x^3) + 6*B*x*(-5*b^3 + 15*b^2* c*x + 5*b*c^2*x^2 + c^3*x^3))/(15*x^(3/2))
Time = 0.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {(A+B x) (b+c x)^3}{x^{5/2}}dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \int \left (\frac {A b^3}{x^{5/2}}+\frac {b^2 (3 A c+b B)}{x^{3/2}}+c^2 \sqrt {x} (A c+3 b B)+\frac {3 b c (A c+b B)}{\sqrt {x}}+B c^3 x^{3/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 A b^3}{3 x^{3/2}}-\frac {2 b^2 (3 A c+b B)}{\sqrt {x}}+\frac {2}{3} c^2 x^{3/2} (A c+3 b B)+6 b c \sqrt {x} (A c+b B)+\frac {2}{5} B c^3 x^{5/2}\) |
Input:
Int[((A + B*x)*(b*x + c*x^2)^3)/x^(11/2),x]
Output:
(-2*A*b^3)/(3*x^(3/2)) - (2*b^2*(b*B + 3*A*c))/Sqrt[x] + 6*b*c*(b*B + A*c) *Sqrt[x] + (2*c^2*(3*b*B + A*c)*x^(3/2))/3 + (2*B*c^3*x^(5/2))/5
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 = 0.72 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {2 B \,c^{3} x^{\frac {5}{2}}}{5}+\frac {2 A \,c^{3} x^{\frac {3}{2}}}{3}+2 B b \,c^{2} x^{\frac {3}{2}}+6 A b \,c^{2} \sqrt {x}+6 B \,b^{2} c \sqrt {x}-\frac {2 A \,b^{3}}{3 x^{\frac {3}{2}}}-\frac {2 b^{2} \left (3 A c +B b \right )}{\sqrt {x}}\) | \(75\) |
default | \(\frac {2 B \,c^{3} x^{\frac {5}{2}}}{5}+\frac {2 A \,c^{3} x^{\frac {3}{2}}}{3}+2 B b \,c^{2} x^{\frac {3}{2}}+6 A b \,c^{2} \sqrt {x}+6 B \,b^{2} c \sqrt {x}-\frac {2 A \,b^{3}}{3 x^{\frac {3}{2}}}-\frac {2 b^{2} \left (3 A c +B b \right )}{\sqrt {x}}\) | \(75\) |
gosper | \(-\frac {2 \left (-3 B \,c^{3} x^{4}-5 A \,c^{3} x^{3}-15 x^{3} B b \,c^{2}-45 A b \,c^{2} x^{2}-45 x^{2} B \,b^{2} c +45 A \,b^{2} c x +15 x B \,b^{3}+5 A \,b^{3}\right )}{15 x^{\frac {3}{2}}}\) | \(76\) |
trager | \(-\frac {2 \left (-3 B \,c^{3} x^{4}-5 A \,c^{3} x^{3}-15 x^{3} B b \,c^{2}-45 A b \,c^{2} x^{2}-45 x^{2} B \,b^{2} c +45 A \,b^{2} c x +15 x B \,b^{3}+5 A \,b^{3}\right )}{15 x^{\frac {3}{2}}}\) | \(76\) |
risch | \(-\frac {2 \left (-3 B \,c^{3} x^{4}-5 A \,c^{3} x^{3}-15 x^{3} B b \,c^{2}-45 A b \,c^{2} x^{2}-45 x^{2} B \,b^{2} c +45 A \,b^{2} c x +15 x B \,b^{3}+5 A \,b^{3}\right )}{15 x^{\frac {3}{2}}}\) | \(76\) |
orering | \(-\frac {2 \left (-3 B \,c^{3} x^{4}-5 A \,c^{3} x^{3}-15 x^{3} B b \,c^{2}-45 A b \,c^{2} x^{2}-45 x^{2} B \,b^{2} c +45 A \,b^{2} c x +15 x B \,b^{3}+5 A \,b^{3}\right ) \left (c \,x^{2}+b x \right )^{3}}{15 x^{\frac {9}{2}} \left (c x +b \right )^{3}}\) | \(94\) |
Input:
int((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x,method=_RETURNVERBOSE)
Output:
2/5*B*c^3*x^(5/2)+2/3*A*c^3*x^(3/2)+2*B*b*c^2*x^(3/2)+6*A*b*c^2*x^(1/2)+6* B*b^2*c*x^(1/2)-2/3*A*b^3/x^(3/2)-2*b^2*(3*A*c+B*b)/x^(1/2)
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=\frac {2 \, {\left (3 \, B c^{3} x^{4} - 5 \, A b^{3} + 5 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 45 \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} - 15 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x\right )}}{15 \, x^{\frac {3}{2}}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x, algorithm="fricas")
Output:
2/15*(3*B*c^3*x^4 - 5*A*b^3 + 5*(3*B*b*c^2 + A*c^3)*x^3 + 45*(B*b^2*c + A* b*c^2)*x^2 - 15*(B*b^3 + 3*A*b^2*c)*x)/x^(3/2)
Time = 0.73 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=- \frac {2 A b^{3}}{3 x^{\frac {3}{2}}} - \frac {6 A b^{2} c}{\sqrt {x}} + 6 A b c^{2} \sqrt {x} + \frac {2 A c^{3} x^{\frac {3}{2}}}{3} - \frac {2 B b^{3}}{\sqrt {x}} + 6 B b^{2} c \sqrt {x} + 2 B b c^{2} x^{\frac {3}{2}} + \frac {2 B c^{3} x^{\frac {5}{2}}}{5} \] Input:
integrate((B*x+A)*(c*x**2+b*x)**3/x**(11/2),x)
Output:
-2*A*b**3/(3*x**(3/2)) - 6*A*b**2*c/sqrt(x) + 6*A*b*c**2*sqrt(x) + 2*A*c** 3*x**(3/2)/3 - 2*B*b**3/sqrt(x) + 6*B*b**2*c*sqrt(x) + 2*B*b*c**2*x**(3/2) + 2*B*c**3*x**(5/2)/5
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=\frac {2}{5} \, B c^{3} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {3}{2}} + 6 \, {\left (B b^{2} c + A b c^{2}\right )} \sqrt {x} - \frac {2 \, {\left (A b^{3} + 3 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x, algorithm="maxima")
Output:
2/5*B*c^3*x^(5/2) + 2/3*(3*B*b*c^2 + A*c^3)*x^(3/2) + 6*(B*b^2*c + A*b*c^2 )*sqrt(x) - 2/3*(A*b^3 + 3*(B*b^3 + 3*A*b^2*c)*x)/x^(3/2)
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=\frac {2}{5} \, B c^{3} x^{\frac {5}{2}} + 2 \, B b c^{2} x^{\frac {3}{2}} + \frac {2}{3} \, A c^{3} x^{\frac {3}{2}} + 6 \, B b^{2} c \sqrt {x} + 6 \, A b c^{2} \sqrt {x} - \frac {2 \, {\left (3 \, B b^{3} x + 9 \, A b^{2} c x + A b^{3}\right )}}{3 \, x^{\frac {3}{2}}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x, algorithm="giac")
Output:
2/5*B*c^3*x^(5/2) + 2*B*b*c^2*x^(3/2) + 2/3*A*c^3*x^(3/2) + 6*B*b^2*c*sqrt (x) + 6*A*b*c^2*sqrt(x) - 2/3*(3*B*b^3*x + 9*A*b^2*c*x + A*b^3)/x^(3/2)
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=x^{3/2}\,\left (\frac {2\,A\,c^3}{3}+2\,B\,b\,c^2\right )-\frac {x\,\left (2\,B\,b^3+6\,A\,c\,b^2\right )+\frac {2\,A\,b^3}{3}}{x^{3/2}}+\frac {2\,B\,c^3\,x^{5/2}}{5}+6\,b\,c\,\sqrt {x}\,\left (A\,c+B\,b\right ) \] Input:
int(((b*x + c*x^2)^3*(A + B*x))/x^(11/2),x)
Output:
x^(3/2)*((2*A*c^3)/3 + 2*B*b*c^2) - (x*(2*B*b^3 + 6*A*b^2*c) + (2*A*b^3)/3 )/x^(3/2) + (2*B*c^3*x^(5/2))/5 + 6*b*c*x^(1/2)*(A*c + B*b)
Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^3}{x^{11/2}} \, dx=\frac {\frac {2}{5} b \,c^{3} x^{4}+\frac {2}{3} a \,c^{3} x^{3}+2 b^{2} c^{2} x^{3}+6 a b \,c^{2} x^{2}+6 b^{3} c \,x^{2}-6 a \,b^{2} c x -2 b^{4} x -\frac {2}{3} a \,b^{3}}{\sqrt {x}\, x} \] Input:
int((B*x+A)*(c*x^2+b*x)^3/x^(11/2),x)
Output:
(2*( - 5*a*b**3 - 45*a*b**2*c*x + 45*a*b*c**2*x**2 + 5*a*c**3*x**3 - 15*b* *4*x + 45*b**3*c*x**2 + 15*b**2*c**2*x**3 + 3*b*c**3*x**4))/(15*sqrt(x)*x)