Integrand size = 20, antiderivative size = 75 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \left (c d^2-b d e+a e^2\right ) (d+e x)^{5/2}}{5 e^3}-\frac {2 (2 c d-b e) (d+e x)^{7/2}}{7 e^3}+\frac {2 c (d+e x)^{9/2}}{9 e^3} \] Output:
2/5*(a*e^2-b*d*e+c*d^2)*(e*x+d)^(5/2)/e^3-2/7*(-b*e+2*c*d)*(e*x+d)^(7/2)/e ^3+2/9*c*(e*x+d)^(9/2)/e^3
Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 (d+e x)^{5/2} \left (9 e (-2 b d+7 a e+5 b e x)+c \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3} \] Input:
Integrate[(d + e*x)^(3/2)*(a + b*x + c*x^2),x]
Output:
(2*(d + e*x)^(5/2)*(9*e*(-2*b*d + 7*a*e + 5*b*e*x) + c*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3)
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
\(\displaystyle \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \int \left (\frac {(d+e x)^{3/2} \left (a e^2-b d e+c d^2\right )}{e^2}+\frac {(d+e x)^{5/2} (b e-2 c d)}{e^2}+\frac {c (d+e x)^{7/2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (d+e x)^{5/2} \left (a e^2-b d e+c d^2\right )}{5 e^3}-\frac {2 (d+e x)^{7/2} (2 c d-b e)}{7 e^3}+\frac {2 c (d+e x)^{9/2}}{9 e^3}\) |
Input:
Int[(d + e*x)^(3/2)*(a + b*x + c*x^2),x]
Output:
(2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^(5/2))/(5*e^3) - (2*(2*c*d - b*e)*(d + e*x)^(7/2))/(7*e^3) + (2*c*(d + e*x)^(9/2))/(9*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 = 0.61 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (\left (\frac {5}{9} c \,x^{2}+\frac {5}{7} b x +a \right ) e^{2}-\frac {2 d \left (\frac {10 c x}{9}+b \right ) e}{7}+\frac {8 c \,d^{2}}{63}\right )}{5 e^{3}}\) | \(46\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 x^{2} c \,e^{2}+45 x b \,e^{2}-20 c d x e +63 a \,e^{2}-18 b d e +8 c \,d^{2}\right )}{315 e^{3}}\) | \(53\) |
orering | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 x^{2} c \,e^{2}+45 x b \,e^{2}-20 c d x e +63 a \,e^{2}-18 b d e +8 c \,d^{2}\right )}{315 e^{3}}\) | \(53\) |
derivativedivides | \(\frac {\frac {2 c \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (b e -2 c d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\) | \(59\) |
default | \(\frac {\frac {2 c \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (b e -2 c d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\) | \(59\) |
trager | \(\frac {2 \left (35 c \,e^{4} x^{4}+45 b \,e^{4} x^{3}+50 d \,e^{3} c \,x^{3}+63 a \,e^{4} x^{2}+72 b d \,e^{3} x^{2}+3 c \,d^{2} e^{2} x^{2}+126 a d \,e^{3} x +9 d^{2} e^{2} b x -4 c \,d^{3} e x +63 a \,d^{2} e^{2}-18 d^{3} e b +8 c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\) | \(121\) |
risch | \(\frac {2 \left (35 c \,e^{4} x^{4}+45 b \,e^{4} x^{3}+50 d \,e^{3} c \,x^{3}+63 a \,e^{4} x^{2}+72 b d \,e^{3} x^{2}+3 c \,d^{2} e^{2} x^{2}+126 a d \,e^{3} x +9 d^{2} e^{2} b x -4 c \,d^{3} e x +63 a \,d^{2} e^{2}-18 d^{3} e b +8 c \,d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\) | \(121\) |
Input:
int((e*x+d)^(3/2)*(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
2/5*(e*x+d)^(5/2)*((5/9*c*x^2+5/7*b*x+a)*e^2-2/7*d*(10/9*c*x+b)*e+8/63*c*d ^2)/e^3
Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.56 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, c e^{4} x^{4} + 8 \, c d^{4} - 18 \, b d^{3} e + 63 \, a d^{2} e^{2} + 5 \, {\left (10 \, c d e^{3} + 9 \, b e^{4}\right )} x^{3} + 3 \, {\left (c d^{2} e^{2} + 24 \, b d e^{3} + 21 \, a e^{4}\right )} x^{2} - {\left (4 \, c d^{3} e - 9 \, b d^{2} e^{2} - 126 \, a d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \] Input:
integrate((e*x+d)^(3/2)*(c*x^2+b*x+a),x, algorithm="fricas")
Output:
2/315*(35*c*e^4*x^4 + 8*c*d^4 - 18*b*d^3*e + 63*a*d^2*e^2 + 5*(10*c*d*e^3 + 9*b*e^4)*x^3 + 3*(c*d^2*e^2 + 24*b*d*e^3 + 21*a*e^4)*x^2 - (4*c*d^3*e - 9*b*d^2*e^2 - 126*a*d*e^3)*x)*sqrt(e*x + d)/e^3
Time = 0.77 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {c \left (d + e x\right )^{\frac {9}{2}}}{9 e^{2}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (b e - 2 c d\right )}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (a e^{2} - b d e + c d^{2}\right )}{5 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (a x + \frac {b x^{2}}{2} + \frac {c x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x+d)**(3/2)*(c*x**2+b*x+a),x)
Output:
Piecewise((2*(c*(d + e*x)**(9/2)/(9*e**2) + (d + e*x)**(7/2)*(b*e - 2*c*d) /(7*e**2) + (d + e*x)**(5/2)*(a*e**2 - b*d*e + c*d**2)/(5*e**2))/e, Ne(e, 0)), (d**(3/2)*(a*x + b*x**2/2 + c*x**3/3), True))
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} c - 45 \, {\left (2 \, c d - b e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 63 \, {\left (c d^{2} - b d e + a e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{315 \, e^{3}} \] Input:
integrate((e*x+d)^(3/2)*(c*x^2+b*x+a),x, algorithm="maxima")
Output:
2/315*(35*(e*x + d)^(9/2)*c - 45*(2*c*d - b*e)*(e*x + d)^(7/2) + 63*(c*d^2 - b*d*e + a*e^2)*(e*x + d)^(5/2))/e^3
Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (63) = 126\).
Time = 0.40 (sec) , antiderivative size = 345, normalized size of antiderivative = 4.60 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a d^{2} + 210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a d + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} b d^{2}}{e} + 21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} c d^{2}}{e^{2}} + \frac {42 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b d}{e} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} c d}{e^{2}} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b}{e} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} c}{e^{2}}\right )}}{315 \, e} \] Input:
integrate((e*x+d)^(3/2)*(c*x^2+b*x+a),x, algorithm="giac")
Output:
2/315*(315*sqrt(e*x + d)*a*d^2 + 210*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d) *a*d + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*b*d^2/e + 21*(3*(e*x + d) ^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a + 21*(3*(e*x + d)^ (5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*c*d^2/e^2 + 42*(3*(e* x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*b*d/e + 18*(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/e^2 + 9*(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/e + (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 + 3 15*sqrt(e*x + d)*d^4)*c/e^2)/e
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.77 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2\,{\left (d+e\,x\right )}^{5/2}\,\left (35\,c\,{\left (d+e\,x\right )}^2+63\,a\,e^2+63\,c\,d^2+45\,b\,e\,\left (d+e\,x\right )-90\,c\,d\,\left (d+e\,x\right )-63\,b\,d\,e\right )}{315\,e^3} \] Input:
int((d + e*x)^(3/2)*(a + b*x + c*x^2),x)
Output:
(2*(d + e*x)^(5/2)*(35*c*(d + e*x)^2 + 63*a*e^2 + 63*c*d^2 + 45*b*e*(d + e *x) - 90*c*d*(d + e*x) - 63*b*d*e))/(315*e^3)
Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.59 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {2 \sqrt {e x +d}\, \left (35 c \,e^{4} x^{4}+45 b \,e^{4} x^{3}+50 c d \,e^{3} x^{3}+63 a \,e^{4} x^{2}+72 b d \,e^{3} x^{2}+3 c \,d^{2} e^{2} x^{2}+126 a d \,e^{3} x +9 b \,d^{2} e^{2} x -4 c \,d^{3} e x +63 a \,d^{2} e^{2}-18 b \,d^{3} e +8 c \,d^{4}\right )}{315 e^{3}} \] Input:
int((e*x+d)^(3/2)*(c*x^2+b*x+a),x)
Output:
(2*sqrt(d + e*x)*(63*a*d**2*e**2 + 126*a*d*e**3*x + 63*a*e**4*x**2 - 18*b* d**3*e + 9*b*d**2*e**2*x + 72*b*d*e**3*x**2 + 45*b*e**4*x**3 + 8*c*d**4 - 4*c*d**3*e*x + 3*c*d**2*e**2*x**2 + 50*c*d*e**3*x**3 + 35*c*e**4*x**4))/(3 15*e**3)