Integrand size = 17, antiderivative size = 100 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=-\frac {5 (b c-a d)^3 (c+d x)^{3/5}}{3 d^4}+\frac {15 b (b c-a d)^2 (c+d x)^{8/5}}{8 d^4}-\frac {15 b^2 (b c-a d) (c+d x)^{13/5}}{13 d^4}+\frac {5 b^3 (c+d x)^{18/5}}{18 d^4} \] Output:
-5/3*(-a*d+b*c)^3*(d*x+c)^(3/5)/d^4+15/8*b*(-a*d+b*c)^2*(d*x+c)^(8/5)/d^4- 15/13*b^2*(-a*d+b*c)*(d*x+c)^(13/5)/d^4+5/18*b^3*(d*x+c)^(18/5)/d^4
Time = 0.06 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=\frac {5 (c+d x)^{3/5} \left (312 a^3 d^3+117 a^2 b d^2 (-5 c+3 d x)+18 a b^2 d \left (25 c^2-15 c d x+12 d^2 x^2\right )+b^3 \left (-125 c^3+75 c^2 d x-60 c d^2 x^2+52 d^3 x^3\right )\right )}{936 d^4} \] Input:
Integrate[(a + b*x)^3/(c + d*x)^(2/5),x]
Output:
(5*(c + d*x)^(3/5)*(312*a^3*d^3 + 117*a^2*b*d^2*(-5*c + 3*d*x) + 18*a*b^2* d*(25*c^2 - 15*c*d*x + 12*d^2*x^2) + b^3*(-125*c^3 + 75*c^2*d*x - 60*c*d^2 *x^2 + 52*d^3*x^3)))/(936*d^4)
Time = 0.20 (sec) , antiderivative size = 100, 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.
\(\displaystyle \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 (c+d x)^{8/5} (b c-a d)}{d^3}+\frac {3 b (c+d x)^{3/5} (b c-a d)^2}{d^3}+\frac {(a d-b c)^3}{d^3 (c+d x)^{2/5}}+\frac {b^3 (c+d x)^{13/5}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {15 b^2 (c+d x)^{13/5} (b c-a d)}{13 d^4}+\frac {15 b (c+d x)^{8/5} (b c-a d)^2}{8 d^4}-\frac {5 (c+d x)^{3/5} (b c-a d)^3}{3 d^4}+\frac {5 b^3 (c+d x)^{18/5}}{18 d^4}\) |
Input:
Int[(a + b*x)^3/(c + d*x)^(2/5),x]
Output:
(-5*(b*c - a*d)^3*(c + d*x)^(3/5))/(3*d^4) + (15*b*(b*c - a*d)^2*(c + d*x) ^(8/5))/(8*d^4) - (15*b^2*(b*c - a*d)*(c + d*x)^(13/5))/(13*d^4) + (5*b^3* (c + d*x)^(18/5))/(18*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 = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\frac {5 b^{3} \left (x d +c \right )^{\frac {18}{5}}}{18}+\frac {15 \left (a d -b c \right ) b^{2} \left (x d +c \right )^{\frac {13}{5}}}{13}+\frac {15 \left (a d -b c \right )^{2} b \left (x d +c \right )^{\frac {8}{5}}}{8}+\frac {5 \left (a d -b c \right )^{3} \left (x d +c \right )^{\frac {3}{5}}}{3}}{d^{4}}\) | \(78\) |
default | \(\frac {\frac {5 b^{3} \left (x d +c \right )^{\frac {18}{5}}}{18}+\frac {15 \left (a d -b c \right ) b^{2} \left (x d +c \right )^{\frac {13}{5}}}{13}+\frac {15 \left (a d -b c \right )^{2} b \left (x d +c \right )^{\frac {8}{5}}}{8}+\frac {5 \left (a d -b c \right )^{3} \left (x d +c \right )^{\frac {3}{5}}}{3}}{d^{4}}\) | \(78\) |
pseudoelliptic | \(\frac {5 \left (\left (\frac {1}{6} b^{3} x^{3}+\frac {9}{13} a \,b^{2} x^{2}+\frac {9}{8} a^{2} b x +a^{3}\right ) d^{3}-\frac {15 c \left (\frac {4}{39} b^{2} x^{2}+\frac {6}{13} a b x +a^{2}\right ) b \,d^{2}}{8}+\frac {75 \left (\frac {b x}{6}+a \right ) c^{2} b^{2} d}{52}-\frac {125 b^{3} c^{3}}{312}\right ) \left (x d +c \right )^{\frac {3}{5}}}{3 d^{4}}\) | \(93\) |
gosper | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (52 d^{3} x^{3} b^{3}+216 x^{2} a \,b^{2} d^{3}-60 x^{2} b^{3} c \,d^{2}+351 x \,a^{2} b \,d^{3}-270 x a \,b^{2} c \,d^{2}+75 x \,b^{3} c^{2} d +312 a^{3} d^{3}-585 a^{2} b c \,d^{2}+450 a \,b^{2} c^{2} d -125 b^{3} c^{3}\right )}{936 d^{4}}\) | \(116\) |
trager | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (52 d^{3} x^{3} b^{3}+216 x^{2} a \,b^{2} d^{3}-60 x^{2} b^{3} c \,d^{2}+351 x \,a^{2} b \,d^{3}-270 x a \,b^{2} c \,d^{2}+75 x \,b^{3} c^{2} d +312 a^{3} d^{3}-585 a^{2} b c \,d^{2}+450 a \,b^{2} c^{2} d -125 b^{3} c^{3}\right )}{936 d^{4}}\) | \(116\) |
risch | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (52 d^{3} x^{3} b^{3}+216 x^{2} a \,b^{2} d^{3}-60 x^{2} b^{3} c \,d^{2}+351 x \,a^{2} b \,d^{3}-270 x a \,b^{2} c \,d^{2}+75 x \,b^{3} c^{2} d +312 a^{3} d^{3}-585 a^{2} b c \,d^{2}+450 a \,b^{2} c^{2} d -125 b^{3} c^{3}\right )}{936 d^{4}}\) | \(116\) |
orering | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (52 d^{3} x^{3} b^{3}+216 x^{2} a \,b^{2} d^{3}-60 x^{2} b^{3} c \,d^{2}+351 x \,a^{2} b \,d^{3}-270 x a \,b^{2} c \,d^{2}+75 x \,b^{3} c^{2} d +312 a^{3} d^{3}-585 a^{2} b c \,d^{2}+450 a \,b^{2} c^{2} d -125 b^{3} c^{3}\right )}{936 d^{4}}\) | \(116\) |
Input:
int((b*x+a)^3/(d*x+c)^(2/5),x,method=_RETURNVERBOSE)
Output:
5/d^4*(1/18*b^3*(d*x+c)^(18/5)+3/13*(a*d-b*c)*b^2*(d*x+c)^(13/5)+3/8*(a*d- b*c)^2*b*(d*x+c)^(8/5)+1/3*(a*d-b*c)^3*(d*x+c)^(3/5))
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=\frac {5 \, {\left (52 \, b^{3} d^{3} x^{3} - 125 \, b^{3} c^{3} + 450 \, a b^{2} c^{2} d - 585 \, a^{2} b c d^{2} + 312 \, a^{3} d^{3} - 12 \, {\left (5 \, b^{3} c d^{2} - 18 \, a b^{2} d^{3}\right )} x^{2} + 3 \, {\left (25 \, b^{3} c^{2} d - 90 \, a b^{2} c d^{2} + 117 \, a^{2} b d^{3}\right )} x\right )} {\left (d x + c\right )}^{\frac {3}{5}}}{936 \, d^{4}} \] Input:
integrate((b*x+a)^3/(d*x+c)^(2/5),x, algorithm="fricas")
Output:
5/936*(52*b^3*d^3*x^3 - 125*b^3*c^3 + 450*a*b^2*c^2*d - 585*a^2*b*c*d^2 + 312*a^3*d^3 - 12*(5*b^3*c*d^2 - 18*a*b^2*d^3)*x^2 + 3*(25*b^3*c^2*d - 90*a *b^2*c*d^2 + 117*a^2*b*d^3)*x)*(d*x + c)^(3/5)/d^4
Time = 0.70 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=\begin {cases} \frac {5 \left (\frac {b^{3} \left (c + d x\right )^{\frac {18}{5}}}{18 d^{3}} + \frac {\left (c + d x\right )^{\frac {13}{5}} \cdot \left (3 a b^{2} d - 3 b^{3} c\right )}{13 d^{3}} + \frac {\left (c + d x\right )^{\frac {8}{5}} \cdot \left (3 a^{2} b d^{2} - 6 a b^{2} c d + 3 b^{3} c^{2}\right )}{8 d^{3}} + \frac {\left (c + d x\right )^{\frac {3}{5}} \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}\right )}{3 d^{3}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{4}}{4 b} & \text {otherwise} \end {cases}}{c^{\frac {2}{5}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**3/(d*x+c)**(2/5),x)
Output:
Piecewise((5*(b**3*(c + d*x)**(18/5)/(18*d**3) + (c + d*x)**(13/5)*(3*a*b* *2*d - 3*b**3*c)/(13*d**3) + (c + d*x)**(8/5)*(3*a**2*b*d**2 - 6*a*b**2*c* d + 3*b**3*c**2)/(8*d**3) + (c + d*x)**(3/5)*(a**3*d**3 - 3*a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3)/(3*d**3))/d, Ne(d, 0)), (Piecewise((a**3*x, Eq(b, 0)), ((a + b*x)**4/(4*b), True))/c**(2/5), True))
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=\frac {5 \, {\left (52 \, {\left (d x + c\right )}^{\frac {18}{5}} b^{3} - 216 \, {\left (b^{3} c - a b^{2} d\right )} {\left (d x + c\right )}^{\frac {13}{5}} + 351 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left (d x + c\right )}^{\frac {8}{5}} - 312 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x + c\right )}^{\frac {3}{5}}\right )}}{936 \, d^{4}} \] Input:
integrate((b*x+a)^3/(d*x+c)^(2/5),x, algorithm="maxima")
Output:
5/936*(52*(d*x + c)^(18/5)*b^3 - 216*(b^3*c - a*b^2*d)*(d*x + c)^(13/5) + 351*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*(d*x + c)^(8/5) - 312*(b^3*c^3 - 3 *a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(d*x + c)^(3/5))/d^4
Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=\frac {5 \, {\left (312 \, {\left (d x + c\right )}^{\frac {3}{5}} a^{3} + \frac {117 \, {\left (3 \, {\left (d x + c\right )}^{\frac {8}{5}} - 8 \, {\left (d x + c\right )}^{\frac {3}{5}} c\right )} a^{2} b}{d} + \frac {18 \, {\left (12 \, {\left (d x + c\right )}^{\frac {13}{5}} - 39 \, {\left (d x + c\right )}^{\frac {8}{5}} c + 52 \, {\left (d x + c\right )}^{\frac {3}{5}} c^{2}\right )} a b^{2}}{d^{2}} + \frac {{\left (52 \, {\left (d x + c\right )}^{\frac {18}{5}} - 216 \, {\left (d x + c\right )}^{\frac {13}{5}} c + 351 \, {\left (d x + c\right )}^{\frac {8}{5}} c^{2} - 312 \, {\left (d x + c\right )}^{\frac {3}{5}} c^{3}\right )} b^{3}}{d^{3}}\right )}}{936 \, d} \] Input:
integrate((b*x+a)^3/(d*x+c)^(2/5),x, algorithm="giac")
Output:
5/936*(312*(d*x + c)^(3/5)*a^3 + 117*(3*(d*x + c)^(8/5) - 8*(d*x + c)^(3/5 )*c)*a^2*b/d + 18*(12*(d*x + c)^(13/5) - 39*(d*x + c)^(8/5)*c + 52*(d*x + c)^(3/5)*c^2)*a*b^2/d^2 + (52*(d*x + c)^(18/5) - 216*(d*x + c)^(13/5)*c + 351*(d*x + c)^(8/5)*c^2 - 312*(d*x + c)^(3/5)*c^3)*b^3/d^3)/d
Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=\frac {5\,b^3\,{\left (c+d\,x\right )}^{18/5}}{18\,d^4}-\frac {\left (15\,b^3\,c-15\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{13/5}}{13\,d^4}+\frac {5\,{\left (a\,d-b\,c\right )}^3\,{\left (c+d\,x\right )}^{3/5}}{3\,d^4}+\frac {15\,b\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{8/5}}{8\,d^4} \] Input:
int((a + b*x)^3/(c + d*x)^(2/5),x)
Output:
(5*b^3*(c + d*x)^(18/5))/(18*d^4) - ((15*b^3*c - 15*a*b^2*d)*(c + d*x)^(13 /5))/(13*d^4) + (5*(a*d - b*c)^3*(c + d*x)^(3/5))/(3*d^4) + (15*b*(a*d - b *c)^2*(c + d*x)^(8/5))/(8*d^4)
Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^3}{(c+d x)^{2/5}} \, dx=\frac {5 \left (d x +c \right )^{\frac {3}{5}} \left (52 b^{3} d^{3} x^{3}+216 a \,b^{2} d^{3} x^{2}-60 b^{3} c \,d^{2} x^{2}+351 a^{2} b \,d^{3} x -270 a \,b^{2} c \,d^{2} x +75 b^{3} c^{2} d x +312 a^{3} d^{3}-585 a^{2} b c \,d^{2}+450 a \,b^{2} c^{2} d -125 b^{3} c^{3}\right )}{936 d^{4}} \] Input:
int((b*x+a)^3/(d*x+c)^(2/5),x)
Output:
(5*(c + d*x)**(3/5)*(312*a**3*d**3 - 585*a**2*b*c*d**2 + 351*a**2*b*d**3*x + 450*a*b**2*c**2*d - 270*a*b**2*c*d**2*x + 216*a*b**2*d**3*x**2 - 125*b* *3*c**3 + 75*b**3*c**2*d*x - 60*b**3*c*d**2*x**2 + 52*b**3*d**3*x**3))/(93 6*d**4)