Integrand size = 19, antiderivative size = 68 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\frac {2 d (c d-b e) (d+e x)^{7/2}}{7 e^3}-\frac {2 (2 c d-b e) (d+e x)^{9/2}}{9 e^3}+\frac {2 c (d+e x)^{11/2}}{11 e^3} \] Output:
2/7*d*(-b*e+c*d)*(e*x+d)^(7/2)/e^3-2/9*(-b*e+2*c*d)*(e*x+d)^(9/2)/e^3+2/11 *c*(e*x+d)^(11/2)/e^3
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\frac {2 (d+e x)^{7/2} \left (11 b e (-2 d+7 e x)+c \left (8 d^2-28 d e x+63 e^2 x^2\right )\right )}{693 e^3} \] Input:
Integrate[(d + e*x)^(5/2)*(b*x + c*x^2),x]
Output:
(2*(d + e*x)^(7/2)*(11*b*e*(-2*d + 7*e*x) + c*(8*d^2 - 28*d*e*x + 63*e^2*x ^2)))/(693*e^3)
Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1140, 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.
\(\displaystyle \int \left (b x+c x^2\right ) (d+e x)^{5/2} \, dx\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \int \left (\frac {(d+e x)^{7/2} (b e-2 c d)}{e^2}+\frac {d (d+e x)^{5/2} (c d-b e)}{e^2}+\frac {c (d+e x)^{9/2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 (d+e x)^{9/2} (2 c d-b e)}{9 e^3}+\frac {2 d (d+e x)^{7/2} (c d-b e)}{7 e^3}+\frac {2 c (d+e x)^{11/2}}{11 e^3}\) |
Input:
Int[(d + e*x)^(5/2)*(b*x + c*x^2),x]
Output:
(2*d*(c*d - b*e)*(d + e*x)^(7/2))/(7*e^3) - (2*(2*c*d - b*e)*(d + e*x)^(9/ 2))/(9*e^3) + (2*c*(d + e*x)^(11/2))/(11*e^3)
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Time = 0.64 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.60
method | result | size |
pseudoelliptic | \(-\frac {4 \left (e x +d \right )^{\frac {7}{2}} \left (-\frac {7 x \left (\frac {9 c x}{11}+b \right ) e^{2}}{2}+d \left (\frac {14 c x}{11}+b \right ) e -\frac {4 c \,d^{2}}{11}\right )}{63 e^{3}}\) | \(41\) |
gosper | \(-\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (-63 x^{2} c \,e^{2}-77 x b \,e^{2}+28 c d x e +22 b d e -8 c \,d^{2}\right )}{693 e^{3}}\) | \(47\) |
derivativedivides | \(\frac {\frac {2 c \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (b e -2 c d \right ) \left (e x +d \right )^{\frac {9}{2}}}{9}-\frac {2 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}}{e^{3}}\) | \(52\) |
default | \(\frac {\frac {2 c \left (e x +d \right )^{\frac {11}{2}}}{11}-\frac {2 \left (-b e +2 c d \right ) \left (e x +d \right )^{\frac {9}{2}}}{9}-\frac {2 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}}{e^{3}}\) | \(53\) |
orering | \(-\frac {2 \left (-63 x^{2} c \,e^{2}-77 x b \,e^{2}+28 c d x e +22 b d e -8 c \,d^{2}\right ) \left (e x +d \right )^{\frac {7}{2}} \left (c \,x^{2}+b x \right )}{693 e^{3} x \left (c x +b \right )}\) | \(66\) |
trager | \(-\frac {2 \left (-63 c \,e^{5} x^{5}-77 b \,e^{5} x^{4}-161 c d \,e^{4} x^{4}-209 b d \,e^{4} x^{3}-113 c \,d^{2} e^{3} x^{3}-165 b \,d^{2} e^{3} x^{2}-3 d^{3} e^{2} c \,x^{2}-11 b \,d^{3} e^{2} x +4 c \,d^{4} e x +22 b \,d^{4} e -8 c \,d^{5}\right ) \sqrt {e x +d}}{693 e^{3}}\) | \(119\) |
risch | \(-\frac {2 \left (-63 c \,e^{5} x^{5}-77 b \,e^{5} x^{4}-161 c d \,e^{4} x^{4}-209 b d \,e^{4} x^{3}-113 c \,d^{2} e^{3} x^{3}-165 b \,d^{2} e^{3} x^{2}-3 d^{3} e^{2} c \,x^{2}-11 b \,d^{3} e^{2} x +4 c \,d^{4} e x +22 b \,d^{4} e -8 c \,d^{5}\right ) \sqrt {e x +d}}{693 e^{3}}\) | \(119\) |
Input:
int((e*x+d)^(5/2)*(c*x^2+b*x),x,method=_RETURNVERBOSE)
Output:
-4/63*(e*x+d)^(7/2)*(-7/2*x*(9/11*c*x+b)*e^2+d*(14/11*c*x+b)*e-4/11*c*d^2) /e^3
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (56) = 112\).
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.74 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\frac {2 \, {\left (63 \, c e^{5} x^{5} + 8 \, c d^{5} - 22 \, b d^{4} e + 7 \, {\left (23 \, c d e^{4} + 11 \, b e^{5}\right )} x^{4} + {\left (113 \, c d^{2} e^{3} + 209 \, b d e^{4}\right )} x^{3} + 3 \, {\left (c d^{3} e^{2} + 55 \, b d^{2} e^{3}\right )} x^{2} - {\left (4 \, c d^{4} e - 11 \, b d^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{693 \, e^{3}} \] Input:
integrate((e*x+d)^(5/2)*(c*x^2+b*x),x, algorithm="fricas")
Output:
2/693*(63*c*e^5*x^5 + 8*c*d^5 - 22*b*d^4*e + 7*(23*c*d*e^4 + 11*b*e^5)*x^4 + (113*c*d^2*e^3 + 209*b*d*e^4)*x^3 + 3*(c*d^3*e^2 + 55*b*d^2*e^3)*x^2 - (4*c*d^4*e - 11*b*d^3*e^2)*x)*sqrt(e*x + d)/e^3
Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (65) = 130\).
Time = 0.35 (sec) , antiderivative size = 245, normalized size of antiderivative = 3.60 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\begin {cases} - \frac {4 b d^{4} \sqrt {d + e x}}{63 e^{2}} + \frac {2 b d^{3} x \sqrt {d + e x}}{63 e} + \frac {10 b d^{2} x^{2} \sqrt {d + e x}}{21} + \frac {38 b d e x^{3} \sqrt {d + e x}}{63} + \frac {2 b e^{2} x^{4} \sqrt {d + e x}}{9} + \frac {16 c d^{5} \sqrt {d + e x}}{693 e^{3}} - \frac {8 c d^{4} x \sqrt {d + e x}}{693 e^{2}} + \frac {2 c d^{3} x^{2} \sqrt {d + e x}}{231 e} + \frac {226 c d^{2} x^{3} \sqrt {d + e x}}{693} + \frac {46 c d e x^{4} \sqrt {d + e x}}{99} + \frac {2 c e^{2} x^{5} \sqrt {d + e x}}{11} & \text {for}\: e \neq 0 \\d^{\frac {5}{2}} \left (\frac {b x^{2}}{2} + \frac {c x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x+d)**(5/2)*(c*x**2+b*x),x)
Output:
Piecewise((-4*b*d**4*sqrt(d + e*x)/(63*e**2) + 2*b*d**3*x*sqrt(d + e*x)/(6 3*e) + 10*b*d**2*x**2*sqrt(d + e*x)/21 + 38*b*d*e*x**3*sqrt(d + e*x)/63 + 2*b*e**2*x**4*sqrt(d + e*x)/9 + 16*c*d**5*sqrt(d + e*x)/(693*e**3) - 8*c*d **4*x*sqrt(d + e*x)/(693*e**2) + 2*c*d**3*x**2*sqrt(d + e*x)/(231*e) + 226 *c*d**2*x**3*sqrt(d + e*x)/693 + 46*c*d*e*x**4*sqrt(d + e*x)/99 + 2*c*e**2 *x**5*sqrt(d + e*x)/11, Ne(e, 0)), (d**(5/2)*(b*x**2/2 + c*x**3/3), True))
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.79 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\frac {2 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} c - 77 \, {\left (2 \, c d - b e\right )} {\left (e x + d\right )}^{\frac {9}{2}} + 99 \, {\left (c d^{2} - b d e\right )} {\left (e x + d\right )}^{\frac {7}{2}}\right )}}{693 \, e^{3}} \] Input:
integrate((e*x+d)^(5/2)*(c*x^2+b*x),x, algorithm="maxima")
Output:
2/693*(63*(e*x + d)^(11/2)*c - 77*(2*c*d - b*e)*(e*x + d)^(9/2) + 99*(c*d^ 2 - b*d*e)*(e*x + d)^(7/2))/e^3
Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (56) = 112\).
Time = 0.12 (sec) , antiderivative size = 418, normalized size of antiderivative = 6.15 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\frac {2 \, {\left (\frac {1155 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} b d^{3}}{e} + \frac {231 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c d^{3}}{e^{2}} + \frac {693 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b d^{2}}{e} + \frac {297 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c d^{2}}{e^{2}} + \frac {297 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b d}{e} + \frac {33 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c d}{e^{2}} + \frac {11 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b}{e} + \frac {5 \, {\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} - 385 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 990 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 1386 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} + 1155 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 693 \, \sqrt {e x + d} d^{5}\right )} c}{e^{2}}\right )}}{3465 \, e} \] Input:
integrate((e*x+d)^(5/2)*(c*x^2+b*x),x, algorithm="giac")
Output:
2/3465*(1155*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*b*d^3/e + 231*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*c*d^3/e^2 + 693*( 3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*b*d^2/e + 297*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*c*d^2/e^2 + 297*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5 /2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*b*d/e + 33*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*c*d/e^2 + 11*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*b/e + 5*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2 )*d + 990*(e*x + d)^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d)^ (3/2)*d^4 - 693*sqrt(e*x + d)*d^5)*c/e^2)/e
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\frac {2\,{\left (d+e\,x\right )}^{7/2}\,\left (63\,c\,{\left (d+e\,x\right )}^2+99\,c\,d^2+77\,b\,e\,\left (d+e\,x\right )-154\,c\,d\,\left (d+e\,x\right )-99\,b\,d\,e\right )}{693\,e^3} \] Input:
int((b*x + c*x^2)*(d + e*x)^(5/2),x)
Output:
(2*(d + e*x)^(7/2)*(63*c*(d + e*x)^2 + 99*c*d^2 + 77*b*e*(d + e*x) - 154*c *d*(d + e*x) - 99*b*d*e))/(693*e^3)
Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.72 \[ \int (d+e x)^{5/2} \left (b x+c x^2\right ) \, dx=\frac {2 \sqrt {e x +d}\, \left (63 c \,e^{5} x^{5}+77 b \,e^{5} x^{4}+161 c d \,e^{4} x^{4}+209 b d \,e^{4} x^{3}+113 c \,d^{2} e^{3} x^{3}+165 b \,d^{2} e^{3} x^{2}+3 c \,d^{3} e^{2} x^{2}+11 b \,d^{3} e^{2} x -4 c \,d^{4} e x -22 b \,d^{4} e +8 c \,d^{5}\right )}{693 e^{3}} \] Input:
int((e*x+d)^(5/2)*(c*x^2+b*x),x)
Output:
(2*sqrt(d + e*x)*( - 22*b*d**4*e + 11*b*d**3*e**2*x + 165*b*d**2*e**3*x**2 + 209*b*d*e**4*x**3 + 77*b*e**5*x**4 + 8*c*d**5 - 4*c*d**4*e*x + 3*c*d**3 *e**2*x**2 + 113*c*d**2*e**3*x**3 + 161*c*d*e**4*x**4 + 63*c*e**5*x**5))/( 693*e**3)