Integrand size = 17, antiderivative size = 71 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\frac {5 (b c-a d)^2 (c+d x)^{9/5}}{9 d^3}-\frac {5 b (b c-a d) (c+d x)^{14/5}}{7 d^3}+\frac {5 b^2 (c+d x)^{19/5}}{19 d^3} \] Output:
5/9*(-a*d+b*c)^2*(d*x+c)^(9/5)/d^3-5/7*b*(-a*d+b*c)*(d*x+c)^(14/5)/d^3+5/1 9*b^2*(d*x+c)^(19/5)/d^3
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\frac {5 (c+d x)^{9/5} \left (133 a^2 d^2+19 a b d (-5 c+9 d x)+b^2 \left (25 c^2-45 c d x+63 d^2 x^2\right )\right )}{1197 d^3} \] Input:
Integrate[(a + b*x)^2*(c + d*x)^(4/5),x]
Output:
(5*(c + d*x)^(9/5)*(133*a^2*d^2 + 19*a*b*d*(-5*c + 9*d*x) + b^2*(25*c^2 - 45*c*d*x + 63*d^2*x^2)))/(1197*d^3)
Time = 0.20 (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.
\(\displaystyle \int (a+b x)^2 (c+d x)^{4/5} \, dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {2 b (c+d x)^{9/5} (b c-a d)}{d^2}+\frac {(c+d x)^{4/5} (a d-b c)^2}{d^2}+\frac {b^2 (c+d x)^{14/5}}{d^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5 b (c+d x)^{14/5} (b c-a d)}{7 d^3}+\frac {5 (c+d x)^{9/5} (b c-a d)^2}{9 d^3}+\frac {5 b^2 (c+d x)^{19/5}}{19 d^3}\) |
Input:
Int[(a + b*x)^2*(c + d*x)^(4/5),x]
Output:
(5*(b*c - a*d)^2*(c + d*x)^(9/5))/(9*d^3) - (5*b*(b*c - a*d)*(c + d*x)^(14 /5))/(7*d^3) + (5*b^2*(c + d*x)^(19/5))/(19*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 = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(\frac {5 \left (x d +c \right )^{\frac {9}{5}} \left (\left (\frac {9}{19} b^{2} x^{2}+\frac {9}{7} a b x +a^{2}\right ) d^{2}-\frac {5 \left (\frac {9 b x}{19}+a \right ) c b d}{7}+\frac {25 b^{2} c^{2}}{133}\right )}{9 d^{3}}\) | \(54\) |
derivativedivides | \(\frac {\frac {5 b^{2} \left (x d +c \right )^{\frac {19}{5}}}{19}+\frac {5 \left (a d -b c \right ) b \left (x d +c \right )^{\frac {14}{5}}}{7}+\frac {5 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {9}{5}}}{9}}{d^{3}}\) | \(56\) |
default | \(\frac {\frac {5 b^{2} \left (x d +c \right )^{\frac {19}{5}}}{19}+\frac {5 \left (a d -b c \right ) b \left (x d +c \right )^{\frac {14}{5}}}{7}+\frac {5 \left (a d -b c \right )^{2} \left (x d +c \right )^{\frac {9}{5}}}{9}}{d^{3}}\) | \(56\) |
gosper | \(\frac {5 \left (x d +c \right )^{\frac {9}{5}} \left (63 d^{2} x^{2} b^{2}+171 x a b \,d^{2}-45 x \,b^{2} c d +133 a^{2} d^{2}-95 a b c d +25 b^{2} c^{2}\right )}{1197 d^{3}}\) | \(63\) |
orering | \(\frac {5 \left (x d +c \right )^{\frac {9}{5}} \left (63 d^{2} x^{2} b^{2}+171 x a b \,d^{2}-45 x \,b^{2} c d +133 a^{2} d^{2}-95 a b c d +25 b^{2} c^{2}\right )}{1197 d^{3}}\) | \(63\) |
trager | \(\frac {5 \left (63 b^{2} d^{3} x^{3}+171 a b \,d^{3} x^{2}+18 b^{2} c \,d^{2} x^{2}+133 a^{2} d^{3} x +76 a b c \,d^{2} x -20 b^{2} c^{2} d x +133 a^{2} c \,d^{2}-95 a b \,c^{2} d +25 b^{2} c^{3}\right ) \left (x d +c \right )^{\frac {4}{5}}}{1197 d^{3}}\) | \(100\) |
risch | \(\frac {5 \left (63 b^{2} d^{3} x^{3}+171 a b \,d^{3} x^{2}+18 b^{2} c \,d^{2} x^{2}+133 a^{2} d^{3} x +76 a b c \,d^{2} x -20 b^{2} c^{2} d x +133 a^{2} c \,d^{2}-95 a b \,c^{2} d +25 b^{2} c^{3}\right ) \left (x d +c \right )^{\frac {4}{5}}}{1197 d^{3}}\) | \(100\) |
Input:
int((b*x+a)^2*(d*x+c)^(4/5),x,method=_RETURNVERBOSE)
Output:
5/9*(d*x+c)^(9/5)*((9/19*b^2*x^2+9/7*a*b*x+a^2)*d^2-5/7*(9/19*b*x+a)*c*b*d +25/133*b^2*c^2)/d^3
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\frac {5 \, {\left (63 \, b^{2} d^{3} x^{3} + 25 \, b^{2} c^{3} - 95 \, a b c^{2} d + 133 \, a^{2} c d^{2} + 9 \, {\left (2 \, b^{2} c d^{2} + 19 \, a b d^{3}\right )} x^{2} - {\left (20 \, b^{2} c^{2} d - 76 \, a b c d^{2} - 133 \, a^{2} d^{3}\right )} x\right )} {\left (d x + c\right )}^{\frac {4}{5}}}{1197 \, d^{3}} \] Input:
integrate((b*x+a)^2*(d*x+c)^(4/5),x, algorithm="fricas")
Output:
5/1197*(63*b^2*d^3*x^3 + 25*b^2*c^3 - 95*a*b*c^2*d + 133*a^2*c*d^2 + 9*(2* b^2*c*d^2 + 19*a*b*d^3)*x^2 - (20*b^2*c^2*d - 76*a*b*c*d^2 - 133*a^2*d^3)* x)*(d*x + c)^(4/5)/d^3
Time = 0.76 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\begin {cases} \frac {5 \left (\frac {b^{2} \left (c + d x\right )^{\frac {19}{5}}}{19 d^{2}} + \frac {\left (c + d x\right )^{\frac {14}{5}} \cdot \left (2 a b d - 2 b^{2} c\right )}{14 d^{2}} + \frac {\left (c + d x\right )^{\frac {9}{5}} \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{9 d^{2}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {4}{5}} \left (\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)**2*(d*x+c)**(4/5),x)
Output:
Piecewise((5*(b**2*(c + d*x)**(19/5)/(19*d**2) + (c + d*x)**(14/5)*(2*a*b* d - 2*b**2*c)/(14*d**2) + (c + d*x)**(9/5)*(a**2*d**2 - 2*a*b*c*d + b**2*c **2)/(9*d**2))/d, Ne(d, 0)), (c**(4/5)*Piecewise((a**2*x, Eq(b, 0)), ((a + b*x)**3/(3*b), True)), True))
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\frac {5 \, {\left (63 \, {\left (d x + c\right )}^{\frac {19}{5}} b^{2} - 171 \, {\left (b^{2} c - a b d\right )} {\left (d x + c\right )}^{\frac {14}{5}} + 133 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x + c\right )}^{\frac {9}{5}}\right )}}{1197 \, d^{3}} \] Input:
integrate((b*x+a)^2*(d*x+c)^(4/5),x, algorithm="maxima")
Output:
5/1197*(63*(d*x + c)^(19/5)*b^2 - 171*(b^2*c - a*b*d)*(d*x + c)^(14/5) + 1 33*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x + c)^(9/5))/d^3
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (59) = 118\).
Time = 0.13 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.87 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\frac {5 \, {\left (1197 \, {\left (d x + c\right )}^{\frac {4}{5}} a^{2} c + 133 \, {\left (4 \, {\left (d x + c\right )}^{\frac {9}{5}} - 9 \, {\left (d x + c\right )}^{\frac {4}{5}} c\right )} a^{2} + \frac {266 \, {\left (4 \, {\left (d x + c\right )}^{\frac {9}{5}} - 9 \, {\left (d x + c\right )}^{\frac {4}{5}} c\right )} a b c}{d} + \frac {19 \, {\left (18 \, {\left (d x + c\right )}^{\frac {14}{5}} - 56 \, {\left (d x + c\right )}^{\frac {9}{5}} c + 63 \, {\left (d x + c\right )}^{\frac {4}{5}} c^{2}\right )} b^{2} c}{d^{2}} + \frac {38 \, {\left (18 \, {\left (d x + c\right )}^{\frac {14}{5}} - 56 \, {\left (d x + c\right )}^{\frac {9}{5}} c + 63 \, {\left (d x + c\right )}^{\frac {4}{5}} c^{2}\right )} a b}{d} + \frac {3 \, {\left (84 \, {\left (d x + c\right )}^{\frac {19}{5}} - 342 \, {\left (d x + c\right )}^{\frac {14}{5}} c + 532 \, {\left (d x + c\right )}^{\frac {9}{5}} c^{2} - 399 \, {\left (d x + c\right )}^{\frac {4}{5}} c^{3}\right )} b^{2}}{d^{2}}\right )}}{4788 \, d} \] Input:
integrate((b*x+a)^2*(d*x+c)^(4/5),x, algorithm="giac")
Output:
5/4788*(1197*(d*x + c)^(4/5)*a^2*c + 133*(4*(d*x + c)^(9/5) - 9*(d*x + c)^ (4/5)*c)*a^2 + 266*(4*(d*x + c)^(9/5) - 9*(d*x + c)^(4/5)*c)*a*b*c/d + 19* (18*(d*x + c)^(14/5) - 56*(d*x + c)^(9/5)*c + 63*(d*x + c)^(4/5)*c^2)*b^2* c/d^2 + 38*(18*(d*x + c)^(14/5) - 56*(d*x + c)^(9/5)*c + 63*(d*x + c)^(4/5 )*c^2)*a*b/d + 3*(84*(d*x + c)^(19/5) - 342*(d*x + c)^(14/5)*c + 532*(d*x + c)^(9/5)*c^2 - 399*(d*x + c)^(4/5)*c^3)*b^2/d^2)/d
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\frac {5\,{\left (c+d\,x\right )}^{9/5}\,\left (63\,b^2\,{\left (c+d\,x\right )}^2+133\,a^2\,d^2+133\,b^2\,c^2-171\,b^2\,c\,\left (c+d\,x\right )+171\,a\,b\,d\,\left (c+d\,x\right )-266\,a\,b\,c\,d\right )}{1197\,d^3} \] Input:
int((a + b*x)^2*(c + d*x)^(4/5),x)
Output:
(5*(c + d*x)^(9/5)*(63*b^2*(c + d*x)^2 + 133*a^2*d^2 + 133*b^2*c^2 - 171*b ^2*c*(c + d*x) + 171*a*b*d*(c + d*x) - 266*a*b*c*d))/(1197*d^3)
Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int (a+b x)^2 (c+d x)^{4/5} \, dx=\frac {5 \left (d x +c \right )^{\frac {4}{5}} \left (63 b^{2} d^{3} x^{3}+171 a b \,d^{3} x^{2}+18 b^{2} c \,d^{2} x^{2}+133 a^{2} d^{3} x +76 a b c \,d^{2} x -20 b^{2} c^{2} d x +133 a^{2} c \,d^{2}-95 a b \,c^{2} d +25 b^{2} c^{3}\right )}{1197 d^{3}} \] Input:
int((b*x+a)^2*(d*x+c)^(4/5),x)
Output:
(5*(c + d*x)**(4/5)*(133*a**2*c*d**2 + 133*a**2*d**3*x - 95*a*b*c**2*d + 7 6*a*b*c*d**2*x + 171*a*b*d**3*x**2 + 25*b**2*c**3 - 20*b**2*c**2*d*x + 18* b**2*c*d**2*x**2 + 63*b**2*d**3*x**3))/(1197*d**3)