Integrand size = 17, antiderivative size = 94 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 (b c-a d)^3}{d^4 \sqrt {c+d x}}+\frac {6 b (b c-a d)^2 \sqrt {c+d x}}{d^4}-\frac {2 b^2 (b c-a d) (c+d x)^{3/2}}{d^4}+\frac {2 b^3 (c+d x)^{5/2}}{5 d^4} \] Output:
2*(-a*d+b*c)^3/d^4/(d*x+c)^(1/2)+6*b*(-a*d+b*c)^2*(d*x+c)^(1/2)/d^4-2*b^2* (-a*d+b*c)*(d*x+c)^(3/2)/d^4+2/5*b^3*(d*x+c)^(5/2)/d^4
Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \left (-5 a^3 d^3+15 a^2 b d^2 (2 c+d x)+5 a b^2 d \left (-8 c^2-4 c d x+d^2 x^2\right )+b^3 \left (16 c^3+8 c^2 d x-2 c d^2 x^2+d^3 x^3\right )\right )}{5 d^4 \sqrt {c+d x}} \] Input:
Integrate[(a + b*x)^3/(c + d*x)^(3/2),x]
Output:
(2*(-5*a^3*d^3 + 15*a^2*b*d^2*(2*c + d*x) + 5*a*b^2*d*(-8*c^2 - 4*c*d*x + d^2*x^2) + b^3*(16*c^3 + 8*c^2*d*x - 2*c*d^2*x^2 + d^3*x^3)))/(5*d^4*Sqrt[ c + d*x])
Time = 0.19 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {53, 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.
\(\displaystyle \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 \sqrt {c+d x} (b c-a d)}{d^3}+\frac {3 b (b c-a d)^2}{d^3 \sqrt {c+d x}}+\frac {(a d-b c)^3}{d^3 (c+d x)^{3/2}}+\frac {b^3 (c+d x)^{3/2}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 b^2 (c+d x)^{3/2} (b c-a d)}{d^4}+\frac {6 b \sqrt {c+d x} (b c-a d)^2}{d^4}+\frac {2 (b c-a d)^3}{d^4 \sqrt {c+d x}}+\frac {2 b^3 (c+d x)^{5/2}}{5 d^4}\) |
Input:
Int[(a + b*x)^3/(c + d*x)^(3/2),x]
Output:
(2*(b*c - a*d)^3)/(d^4*Sqrt[c + d*x]) + (6*b*(b*c - a*d)^2*Sqrt[c + d*x])/ d^4 - (2*b^2*(b*c - a*d)*(c + d*x)^(3/2))/d^4 + (2*b^3*(c + d*x)^(5/2))/(5 *d^4)
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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \left (x^{3} d^{3}-2 c \,d^{2} x^{2}+8 c^{2} d x +16 c^{3}\right ) b^{3}}{5}-16 a \left (-\frac {1}{8} d^{2} x^{2}+\frac {1}{2} c d x +c^{2}\right ) d \,b^{2}+12 a^{2} d^{2} \left (\frac {x d}{2}+c \right ) b -2 a^{3} d^{3}}{\sqrt {x d +c}\, d^{4}}\) | \(94\) |
risch | \(\frac {2 b \left (d^{2} x^{2} b^{2}+5 x a b \,d^{2}-3 x \,b^{2} c d +15 a^{2} d^{2}-25 a b c d +11 b^{2} c^{2}\right ) \sqrt {x d +c}}{5 d^{4}}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{d^{4} \sqrt {x d +c}}\) | \(112\) |
gosper | \(-\frac {2 \left (-d^{3} x^{3} b^{3}-5 x^{2} a \,b^{2} d^{3}+2 x^{2} b^{3} c \,d^{2}-15 x \,a^{2} b \,d^{3}+20 x a \,b^{2} c \,d^{2}-8 x \,b^{3} c^{2} d +5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right )}{5 \sqrt {x d +c}\, d^{4}}\) | \(116\) |
trager | \(-\frac {2 \left (-d^{3} x^{3} b^{3}-5 x^{2} a \,b^{2} d^{3}+2 x^{2} b^{3} c \,d^{2}-15 x \,a^{2} b \,d^{3}+20 x a \,b^{2} c \,d^{2}-8 x \,b^{3} c^{2} d +5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right )}{5 \sqrt {x d +c}\, d^{4}}\) | \(116\) |
orering | \(-\frac {2 \left (-d^{3} x^{3} b^{3}-5 x^{2} a \,b^{2} d^{3}+2 x^{2} b^{3} c \,d^{2}-15 x \,a^{2} b \,d^{3}+20 x a \,b^{2} c \,d^{2}-8 x \,b^{3} c^{2} d +5 a^{3} d^{3}-30 a^{2} b c \,d^{2}+40 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right )}{5 \sqrt {x d +c}\, d^{4}}\) | \(116\) |
derivativedivides | \(\frac {\frac {2 b^{3} \left (x d +c \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} d \left (x d +c \right )^{\frac {3}{2}}-2 b^{3} c \left (x d +c \right )^{\frac {3}{2}}+6 a^{2} b \,d^{2} \sqrt {x d +c}-12 a \,b^{2} c d \sqrt {x d +c}+6 b^{3} c^{2} \sqrt {x d +c}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{\sqrt {x d +c}}}{d^{4}}\) | \(136\) |
default | \(\frac {\frac {2 b^{3} \left (x d +c \right )^{\frac {5}{2}}}{5}+2 a \,b^{2} d \left (x d +c \right )^{\frac {3}{2}}-2 b^{3} c \left (x d +c \right )^{\frac {3}{2}}+6 a^{2} b \,d^{2} \sqrt {x d +c}-12 a \,b^{2} c d \sqrt {x d +c}+6 b^{3} c^{2} \sqrt {x d +c}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{\sqrt {x d +c}}}{d^{4}}\) | \(136\) |
Input:
int((b*x+a)^3/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5*((d^3*x^3-2*c*d^2*x^2+8*c^2*d*x+16*c^3)*b^3-40*a*(-1/8*d^2*x^2+1/2*c*d *x+c^2)*d*b^2+30*a^2*d^2*(1/2*x*d+c)*b-5*a^3*d^3)/(d*x+c)^(1/2)/d^4
Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} d^{3} x^{3} + 16 \, b^{3} c^{3} - 40 \, a b^{2} c^{2} d + 30 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3} - {\left (2 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{2} + {\left (8 \, b^{3} c^{2} d - 20 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}}{5 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:
integrate((b*x+a)^3/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
2/5*(b^3*d^3*x^3 + 16*b^3*c^3 - 40*a*b^2*c^2*d + 30*a^2*b*c*d^2 - 5*a^3*d^ 3 - (2*b^3*c*d^2 - 5*a*b^2*d^3)*x^2 + (8*b^3*c^2*d - 20*a*b^2*c*d^2 + 15*a ^2*b*d^3)*x)*sqrt(d*x + c)/(d^5*x + c*d^4)
Time = 1.87 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (c + d x\right )^{\frac {5}{2}}}{5 d^{3}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \cdot \left (3 a b^{2} d - 3 b^{3} c\right )}{3 d^{3}} + \frac {\sqrt {c + d x} \left (3 a^{2} b d^{2} - 6 a b^{2} c d + 3 b^{3} c^{2}\right )}{d^{3}} - \frac {\left (a d - b c\right )^{3}}{d^{3} \sqrt {c + d x}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{4}}{4 b} & \text {otherwise} \end {cases}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**3/(d*x+c)**(3/2),x)
Output:
Piecewise((2*(b**3*(c + d*x)**(5/2)/(5*d**3) + (c + d*x)**(3/2)*(3*a*b**2* d - 3*b**3*c)/(3*d**3) + sqrt(c + d*x)*(3*a**2*b*d**2 - 6*a*b**2*c*d + 3*b **3*c**2)/d**3 - (a*d - b*c)**3/(d**3*sqrt(c + d*x)))/d, Ne(d, 0)), (Piece wise((a**3*x, Eq(b, 0)), ((a + b*x)**4/(4*b), True))/c**(3/2), True))
Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (d x + c\right )}^{\frac {5}{2}} b^{3} - 5 \, {\left (b^{3} c - a b^{2} d\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 15 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \sqrt {d x + c}}{d^{3}} + \frac {5 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}}{\sqrt {d x + c} d^{3}}\right )}}{5 \, d} \] Input:
integrate((b*x+a)^3/(d*x+c)^(3/2),x, algorithm="maxima")
Output:
2/5*(((d*x + c)^(5/2)*b^3 - 5*(b^3*c - a*b^2*d)*(d*x + c)^(3/2) + 15*(b^3* c^2 - 2*a*b^2*c*d + a^2*b*d^2)*sqrt(d*x + c))/d^3 + 5*(b^3*c^3 - 3*a*b^2*c ^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/(sqrt(d*x + c)*d^3))/d
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}}{\sqrt {d x + c} d^{4}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {5}{2}} b^{3} d^{16} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c d^{16} + 15 \, \sqrt {d x + c} b^{3} c^{2} d^{16} + 5 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} d^{17} - 30 \, \sqrt {d x + c} a b^{2} c d^{17} + 15 \, \sqrt {d x + c} a^{2} b d^{18}\right )}}{5 \, d^{20}} \] Input:
integrate((b*x+a)^3/(d*x+c)^(3/2),x, algorithm="giac")
Output:
2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)/(sqrt(d*x + c)*d^4) + 2/5*((d*x + c)^(5/2)*b^3*d^16 - 5*(d*x + c)^(3/2)*b^3*c*d^16 + 15*sqrt(d *x + c)*b^3*c^2*d^16 + 5*(d*x + c)^(3/2)*a*b^2*d^17 - 30*sqrt(d*x + c)*a*b ^2*c*d^17 + 15*sqrt(d*x + c)*a^2*b*d^18)/d^20
Time = 0.16 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {2\,b^3\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4}-\frac {\left (6\,b^3\,c-6\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,d^4}-\frac {2\,a^3\,d^3-6\,a^2\,b\,c\,d^2+6\,a\,b^2\,c^2\,d-2\,b^3\,c^3}{d^4\,\sqrt {c+d\,x}}+\frac {6\,b\,{\left (a\,d-b\,c\right )}^2\,\sqrt {c+d\,x}}{d^4} \] Input:
int((a + b*x)^3/(c + d*x)^(3/2),x)
Output:
(2*b^3*(c + d*x)^(5/2))/(5*d^4) - ((6*b^3*c - 6*a*b^2*d)*(c + d*x)^(3/2))/ (3*d^4) - (2*a^3*d^3 - 2*b^3*c^3 + 6*a*b^2*c^2*d - 6*a^2*b*c*d^2)/(d^4*(c + d*x)^(1/2)) + (6*b*(a*d - b*c)^2*(c + d*x)^(1/2))/d^4
Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.22 \[ \int \frac {(a+b x)^3}{(c+d x)^{3/2}} \, dx=\frac {\frac {2}{5} b^{3} d^{3} x^{3}+2 a \,b^{2} d^{3} x^{2}-\frac {4}{5} b^{3} c \,d^{2} x^{2}+6 a^{2} b \,d^{3} x -8 a \,b^{2} c \,d^{2} x +\frac {16}{5} b^{3} c^{2} d x -2 a^{3} d^{3}+12 a^{2} b c \,d^{2}-16 a \,b^{2} c^{2} d +\frac {32}{5} b^{3} c^{3}}{\sqrt {d x +c}\, d^{4}} \] Input:
int((b*x+a)^3/(d*x+c)^(3/2),x)
Output:
(2*( - 5*a**3*d**3 + 30*a**2*b*c*d**2 + 15*a**2*b*d**3*x - 40*a*b**2*c**2* d - 20*a*b**2*c*d**2*x + 5*a*b**2*d**3*x**2 + 16*b**3*c**3 + 8*b**3*c**2*d *x - 2*b**3*c*d**2*x**2 + b**3*d**3*x**3))/(5*sqrt(c + d*x)*d**4)