Integrand size = 17, antiderivative size = 71 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\frac {5 (b c-a d)^2 (c+d x)^{3/5}}{3 d^3}-\frac {5 b (b c-a d) (c+d x)^{8/5}}{4 d^3}+\frac {5 b^2 (c+d x)^{13/5}}{13 d^3} \] Output:
5/3*(-a*d+b*c)^2*(d*x+c)^(3/5)/d^3-5/4*b*(-a*d+b*c)*(d*x+c)^(8/5)/d^3+5/13 *b^2*(d*x+c)^(13/5)/d^3
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\frac {5 (c+d x)^{3/5} \left (52 a^2 d^2+13 a b d (-5 c+3 d x)+b^2 \left (25 c^2-15 c d x+12 d^2 x^2\right )\right )}{156 d^3} \] Input:
Integrate[(a + b*x)^2/(c + d*x)^(2/5),x]
Output:
(5*(c + d*x)^(3/5)*(52*a^2*d^2 + 13*a*b*d*(-5*c + 3*d*x) + b^2*(25*c^2 - 1 5*c*d*x + 12*d^2*x^2)))/(156*d^3)
Time = 0.18 (sec) , antiderivative size = 71, 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.
\(\displaystyle \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {2 b (c+d x)^{3/5} (b c-a d)}{d^2}+\frac {(a d-b c)^2}{d^2 (c+d x)^{2/5}}+\frac {b^2 (c+d x)^{8/5}}{d^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5 b (c+d x)^{8/5} (b c-a d)}{4 d^3}+\frac {5 (c+d x)^{3/5} (b c-a d)^2}{3 d^3}+\frac {5 b^2 (c+d x)^{13/5}}{13 d^3}\) |
Input:
Int[(a + b*x)^2/(c + d*x)^(2/5),x]
Output:
(5*(b*c - a*d)^2*(c + d*x)^(3/5))/(3*d^3) - (5*b*(b*c - a*d)*(c + d*x)^(8/ 5))/(4*d^3) + (5*b^2*(c + d*x)^(13/5))/(13*d^3)
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 = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (\left (\frac {3}{13} b^{2} x^{2}+\frac {3}{4} a b x +a^{2}\right ) d^{2}-\frac {5 c b \left (\frac {3 b x}{13}+a \right ) d}{4}+\frac {25 b^{2} c^{2}}{52}\right )}{3 d^{3}}\) | \(54\) |
derivativedivides | \(\frac {\frac {5 b^{2} \left (x d +c \right )^{\frac {13}{5}}}{13}+\frac {5 \left (a d -b c \right ) b \left (x d +c \right )^{\frac {8}{5}}}{4}+\frac {5 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {3}{5}}}{3}}{d^{3}}\) | \(56\) |
default | \(\frac {\frac {5 b^{2} \left (x d +c \right )^{\frac {13}{5}}}{13}+\frac {5 \left (a d -b c \right ) b \left (x d +c \right )^{\frac {8}{5}}}{4}+\frac {5 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {3}{5}}}{3}}{d^{3}}\) | \(56\) |
gosper | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (12 d^{2} x^{2} b^{2}+39 x a b \,d^{2}-15 x \,b^{2} c d +52 a^{2} d^{2}-65 a b c d +25 b^{2} c^{2}\right )}{156 d^{3}}\) | \(63\) |
trager | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (12 d^{2} x^{2} b^{2}+39 x a b \,d^{2}-15 x \,b^{2} c d +52 a^{2} d^{2}-65 a b c d +25 b^{2} c^{2}\right )}{156 d^{3}}\) | \(63\) |
risch | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (12 d^{2} x^{2} b^{2}+39 x a b \,d^{2}-15 x \,b^{2} c d +52 a^{2} d^{2}-65 a b c d +25 b^{2} c^{2}\right )}{156 d^{3}}\) | \(63\) |
orering | \(\frac {5 \left (x d +c \right )^{\frac {3}{5}} \left (12 d^{2} x^{2} b^{2}+39 x a b \,d^{2}-15 x \,b^{2} c d +52 a^{2} d^{2}-65 a b c d +25 b^{2} c^{2}\right )}{156 d^{3}}\) | \(63\) |
Input:
int((b*x+a)^2/(d*x+c)^(2/5),x,method=_RETURNVERBOSE)
Output:
5/3*(d*x+c)^(3/5)*((3/13*b^2*x^2+3/4*a*b*x+a^2)*d^2-5/4*c*b*(3/13*b*x+a)*d +25/52*b^2*c^2)/d^3
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\frac {5 \, {\left (12 \, b^{2} d^{2} x^{2} + 25 \, b^{2} c^{2} - 65 \, a b c d + 52 \, a^{2} d^{2} - 3 \, {\left (5 \, b^{2} c d - 13 \, a b d^{2}\right )} x\right )} {\left (d x + c\right )}^{\frac {3}{5}}}{156 \, d^{3}} \] Input:
integrate((b*x+a)^2/(d*x+c)^(2/5),x, algorithm="fricas")
Output:
5/156*(12*b^2*d^2*x^2 + 25*b^2*c^2 - 65*a*b*c*d + 52*a^2*d^2 - 3*(5*b^2*c* d - 13*a*b*d^2)*x)*(d*x + c)^(3/5)/d^3
Time = 0.65 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\begin {cases} \frac {5 \left (\frac {b^{2} \left (c + d x\right )^{\frac {13}{5}}}{13 d^{2}} + \frac {\left (c + d x\right )^{\frac {8}{5}} \cdot \left (2 a b d - 2 b^{2} c\right )}{8 d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{5}} \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{3 d^{2}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}}{c^{\frac {2}{5}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2/(d*x+c)**(2/5),x)
Output:
Piecewise((5*(b**2*(c + d*x)**(13/5)/(13*d**2) + (c + d*x)**(8/5)*(2*a*b*d - 2*b**2*c)/(8*d**2) + (c + d*x)**(3/5)*(a**2*d**2 - 2*a*b*c*d + b**2*c** 2)/(3*d**2))/d, Ne(d, 0)), (Piecewise((a**2*x, Eq(b, 0)), ((a + b*x)**3/(3 *b), True))/c**(2/5), True))
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\frac {5 \, {\left (12 \, {\left (d x + c\right )}^{\frac {13}{5}} b^{2} - 39 \, {\left (b^{2} c - a b d\right )} {\left (d x + c\right )}^{\frac {8}{5}} + 52 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {3}{5}}\right )}}{156 \, d^{3}} \] Input:
integrate((b*x+a)^2/(d*x+c)^(2/5),x, algorithm="maxima")
Output:
5/156*(12*(d*x + c)^(13/5)*b^2 - 39*(b^2*c - a*b*d)*(d*x + c)^(8/5) + 52*( b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x + c)^(3/5))/d^3
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\frac {5 \, {\left (52 \, {\left (d x + c\right )}^{\frac {3}{5}} a^{2} + \frac {13 \, {\left (3 \, {\left (d x + c\right )}^{\frac {8}{5}} - 8 \, {\left (d x + c\right )}^{\frac {3}{5}} c\right )} a b}{d} + \frac {{\left (12 \, {\left (d x + c\right )}^{\frac {13}{5}} - 39 \, {\left (d x + c\right )}^{\frac {8}{5}} c + 52 \, {\left (d x + c\right )}^{\frac {3}{5}} c^{2}\right )} b^{2}}{d^{2}}\right )}}{156 \, d} \] Input:
integrate((b*x+a)^2/(d*x+c)^(2/5),x, algorithm="giac")
Output:
5/156*(52*(d*x + c)^(3/5)*a^2 + 13*(3*(d*x + c)^(8/5) - 8*(d*x + c)^(3/5)* c)*a*b/d + (12*(d*x + c)^(13/5) - 39*(d*x + c)^(8/5)*c + 52*(d*x + c)^(3/5 )*c^2)*b^2/d^2)/d
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\frac {5\,{\left (c+d\,x\right )}^{3/5}\,\left (12\,b^2\,{\left (c+d\,x\right )}^2+52\,a^2\,d^2+52\,b^2\,c^2-39\,b^2\,c\,\left (c+d\,x\right )+39\,a\,b\,d\,\left (c+d\,x\right )-104\,a\,b\,c\,d\right )}{156\,d^3} \] Input:
int((a + b*x)^2/(c + d*x)^(2/5),x)
Output:
(5*(c + d*x)^(3/5)*(12*b^2*(c + d*x)^2 + 52*a^2*d^2 + 52*b^2*c^2 - 39*b^2* c*(c + d*x) + 39*a*b*d*(c + d*x) - 104*a*b*c*d))/(156*d^3)
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^2}{(c+d x)^{2/5}} \, dx=\frac {5 \left (d x +c \right )^{\frac {3}{5}} \left (12 b^{2} d^{2} x^{2}+39 a b \,d^{2} x -15 b^{2} c d x +52 a^{2} d^{2}-65 a b c d +25 b^{2} c^{2}\right )}{156 d^{3}} \] Input:
int((b*x+a)^2/(d*x+c)^(2/5),x)
Output:
(5*(c + d*x)**(3/5)*(52*a**2*d**2 - 65*a*b*c*d + 39*a*b*d**2*x + 25*b**2*c **2 - 15*b**2*c*d*x + 12*b**2*d**2*x**2))/(156*d**3)