Integrand size = 17, antiderivative size = 96 \[ \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx=\frac {2 (b c-a d)^3}{3 d^4 (c+d x)^{3/2}}-\frac {6 b (b c-a d)^2}{d^4 \sqrt {c+d x}}-\frac {6 b^2 (b c-a d) \sqrt {c+d x}}{d^4}+\frac {2 b^3 (c+d x)^{3/2}}{3 d^4} \]
2/3*(-a*d+b*c)^3/d^4/(d*x+c)^(3/2)+2/3*b^3*(d*x+c)^(3/2)/d^4-6*b*(-a*d+b*c )^2/d^4/(d*x+c)^(1/2)-6*b^2*(-a*d+b*c)*(d*x+c)^(1/2)/d^4
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (a^3 d^3+3 a^2 b d^2 (2 c+3 d x)-3 a b^2 d \left (8 c^2+12 c d x+3 d^2 x^2\right )+b^3 \left (16 c^3+24 c^2 d x+6 c d^2 x^2-d^3 x^3\right )\right )}{3 d^4 (c+d x)^{3/2}} \]
(-2*(a^3*d^3 + 3*a^2*b*d^2*(2*c + 3*d*x) - 3*a*b^2*d*(8*c^2 + 12*c*d*x + 3 *d^2*x^2) + b^3*(16*c^3 + 24*c^2*d*x + 6*c*d^2*x^2 - d^3*x^3)))/(3*d^4*(c + d*x)^(3/2))
Time = 0.21 (sec) , antiderivative size = 96, 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.
\(\displaystyle \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 (b c-a d)}{d^3 \sqrt {c+d x}}+\frac {3 b (b c-a d)^2}{d^3 (c+d x)^{3/2}}+\frac {(a d-b c)^3}{d^3 (c+d x)^{5/2}}+\frac {b^3 \sqrt {c+d x}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 b^2 \sqrt {c+d x} (b c-a d)}{d^4}-\frac {6 b (b c-a d)^2}{d^4 \sqrt {c+d x}}+\frac {2 (b c-a d)^3}{3 d^4 (c+d x)^{3/2}}+\frac {2 b^3 (c+d x)^{3/2}}{3 d^4}\) |
(2*(b*c - a*d)^3)/(3*d^4*(c + d*x)^(3/2)) - (6*b*(b*c - a*d)^2)/(d^4*Sqrt[ c + d*x]) - (6*b^2*(b*c - a*d)*Sqrt[c + d*x])/d^4 + (2*b^3*(c + d*x)^(3/2) )/(3*d^4)
3.15.36.3.1 Defintions of rubi rules used
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 = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (b^{2} x^{2}+10 a b x +a^{2}\right ) d^{2}+8 b c \left (-b x +a \right ) d -8 b^{2} c^{2}\right ) \left (d \left (-b x +a \right )-2 b c \right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{4}}\) | \(66\) |
risch | \(\frac {2 b^{2} \left (b d x +9 a d -8 b c \right ) \sqrt {d x +c}}{3 d^{4}}-\frac {2 \left (9 b d x +a d +8 b c \right ) \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{3 d^{4} \left (d x +c \right )^{\frac {3}{2}}}\) | \(76\) |
gosper | \(-\frac {2 \left (-d^{3} x^{3} b^{3}-9 x^{2} a \,b^{2} d^{3}+6 x^{2} b^{3} c \,d^{2}+9 x \,a^{2} b \,d^{3}-36 x a \,b^{2} c \,d^{2}+24 x \,b^{3} c^{2} d +a^{3} d^{3}+6 a^{2} b c \,d^{2}-24 a \,b^{2} c^{2} d +16 b^{3} c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{4}}\) | \(115\) |
trager | \(-\frac {2 \left (-d^{3} x^{3} b^{3}-9 x^{2} a \,b^{2} d^{3}+6 x^{2} b^{3} c \,d^{2}+9 x \,a^{2} b \,d^{3}-36 x a \,b^{2} c \,d^{2}+24 x \,b^{3} c^{2} d +a^{3} d^{3}+6 a^{2} b c \,d^{2}-24 a \,b^{2} c^{2} d +16 b^{3} c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{4}}\) | \(115\) |
derivativedivides | \(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {3}{2}}}{3}+6 b^{2} d a \sqrt {d x +c}-6 b^{3} c \sqrt {d x +c}-\frac {6 b \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\sqrt {d x +c}}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{4}}\) | \(122\) |
default | \(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {3}{2}}}{3}+6 b^{2} d a \sqrt {d x +c}-6 b^{3} c \sqrt {d x +c}-\frac {6 b \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\sqrt {d x +c}}-\frac {2 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{4}}\) | \(122\) |
-2/3*((b^2*x^2+10*a*b*x+a^2)*d^2+8*b*c*(-b*x+a)*d-8*b^2*c^2)/(d*x+c)^(3/2) *(d*(-b*x+a)-2*b*c)/d^4
Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (b^{3} d^{3} x^{3} - 16 \, b^{3} c^{3} + 24 \, a b^{2} c^{2} d - 6 \, a^{2} b c d^{2} - a^{3} d^{3} - 3 \, {\left (2 \, b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} c^{2} d - 12 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} \]
2/3*(b^3*d^3*x^3 - 16*b^3*c^3 + 24*a*b^2*c^2*d - 6*a^2*b*c*d^2 - a^3*d^3 - 3*(2*b^3*c*d^2 - 3*a*b^2*d^3)*x^2 - 3*(8*b^3*c^2*d - 12*a*b^2*c*d^2 + 3*a ^2*b*d^3)*x)*sqrt(d*x + c)/(d^6*x^2 + 2*c*d^5*x + c^2*d^4)
Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (88) = 176\).
Time = 0.42 (sec) , antiderivative size = 461, normalized size of antiderivative = 4.80 \[ \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx=\begin {cases} - \frac {2 a^{3} d^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {12 a^{2} b c d^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {18 a^{2} b d^{3} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {48 a b^{2} c^{2} d}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {72 a b^{2} c d^{2} x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {18 a b^{2} d^{3} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {32 b^{3} c^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {48 b^{3} c^{2} d x}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} - \frac {12 b^{3} c d^{2} x^{2}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} + \frac {2 b^{3} d^{3} x^{3}}{3 c d^{4} \sqrt {c + d x} + 3 d^{5} x \sqrt {c + d x}} & \text {for}\: d \neq 0 \\\frac {a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-2*a**3*d**3/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 12*a**2*b*c*d**2/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 18* a**2*b*d**3*x/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 48*a*b** 2*c**2*d/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 72*a*b**2*c*d **2*x/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 18*a*b**2*d**3*x **2/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 32*b**3*c**3/(3*c* d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 48*b**3*c**2*d*x/(3*c*d**4* sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) - 12*b**3*c*d**2*x**2/(3*c*d**4*sq rt(c + d*x) + 3*d**5*x*sqrt(c + d*x)) + 2*b**3*d**3*x**3/(3*c*d**4*sqrt(c + d*x) + 3*d**5*x*sqrt(c + d*x)), Ne(d, 0)), ((a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4/4)/c**(5/2), True))
Time = 0.24 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (d x + c\right )}^{\frac {3}{2}} b^{3} - 9 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {d x + c}}{d^{3}} + \frac {b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3} - 9 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left (d x + c\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{3}}\right )}}{3 \, d} \]
2/3*(((d*x + c)^(3/2)*b^3 - 9*(b^3*c - a*b^2*d)*sqrt(d*x + c))/d^3 + (b^3* c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3 - 9*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*(d*x + c))/((d*x + c)^(3/2)*d^3))/d
Time = 0.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (d x + c\right )} b^{3} c^{2} - b^{3} c^{3} - 18 \, {\left (d x + c\right )} a b^{2} c d + 3 \, a b^{2} c^{2} d + 9 \, {\left (d x + c\right )} a^{2} b d^{2} - 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{4}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{3} d^{8} - 9 \, \sqrt {d x + c} b^{3} c d^{8} + 9 \, \sqrt {d x + c} a b^{2} d^{9}\right )}}{3 \, d^{12}} \]
-2/3*(9*(d*x + c)*b^3*c^2 - b^3*c^3 - 18*(d*x + c)*a*b^2*c*d + 3*a*b^2*c^2 *d + 9*(d*x + c)*a^2*b*d^2 - 3*a^2*b*c*d^2 + a^3*d^3)/((d*x + c)^(3/2)*d^4 ) + 2/3*((d*x + c)^(3/2)*b^3*d^8 - 9*sqrt(d*x + c)*b^3*c*d^8 + 9*sqrt(d*x + c)*a*b^2*d^9)/d^12
Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b x)^3}{(c+d x)^{5/2}} \, dx=\frac {2\,b^3\,{\left (c+d\,x\right )}^3-2\,a^3\,d^3+2\,b^3\,c^3-18\,b^3\,c\,{\left (c+d\,x\right )}^2-18\,b^3\,c^2\,\left (c+d\,x\right )+18\,a\,b^2\,d\,{\left (c+d\,x\right )}^2-18\,a^2\,b\,d^2\,\left (c+d\,x\right )-6\,a\,b^2\,c^2\,d+6\,a^2\,b\,c\,d^2+36\,a\,b^2\,c\,d\,\left (c+d\,x\right )}{3\,d^4\,{\left (c+d\,x\right )}^{3/2}} \]