Integrand size = 20, antiderivative size = 83 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{5/2}}{5 e^3}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{7/2}}{7 e^3}+\frac {2 b B (d+e x)^{9/2}}{9 e^3} \] Output:
2/5*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(5/2)/e^3-2/7*(-A*b*e-B*a*e+2*B*b*d)*(e* x+d)^(7/2)/e^3+2/9*b*B*(e*x+d)^(9/2)/e^3
Time = 0.06 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\frac {2 (d+e x)^{5/2} \left (9 A b e (-2 d+5 e x)+9 a e (-2 B d+7 A e+5 B e x)+b B \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3} \] Input:
Integrate[(a + b*x)*(A + B*x)*(d + e*x)^(3/2),x]
Output:
(2*(d + e*x)^(5/2)*(9*A*b*e*(-2*d + 5*e*x) + 9*a*e*(-2*B*d + 7*A*e + 5*B*e *x) + b*B*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3)
Time = 0.23 (sec) , antiderivative size = 83, 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 = {86, 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.
\(\displaystyle \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {(d+e x)^{5/2} (a B e+A b e-2 b B d)}{e^2}+\frac {(d+e x)^{3/2} (a e-b d) (A e-B d)}{e^2}+\frac {b B (d+e x)^{7/2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 (d+e x)^{7/2} (-a B e-A b e+2 b B d)}{7 e^3}+\frac {2 (d+e x)^{5/2} (b d-a e) (B d-A e)}{5 e^3}+\frac {2 b B (d+e x)^{9/2}}{9 e^3}\) |
Input:
Int[(a + b*x)*(A + B*x)*(d + e*x)^(3/2),x]
Output:
(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(5/2))/(5*e^3) - (2*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(7/2))/(7*e^3) + (2*b*B*(d + e*x)^(9/2))/(9*e^3)
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Time = 0.65 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (\left (\frac {5 x \left (\frac {7 b x}{9}+a \right ) B}{7}+A \left (\frac {5 b x}{7}+a \right )\right ) e^{2}-\frac {2 \left (\left (\frac {10 b x}{9}+a \right ) B +A b \right ) d e}{7}+\frac {8 b B \,d^{2}}{63}\right )}{5 e^{3}}\) | \(60\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 b B \,x^{2} e^{2}+45 A x b \,e^{2}+45 B x a \,e^{2}-20 B x b d e +63 A a \,e^{2}-18 A b d e -18 B a d e +8 b B \,d^{2}\right )}{315 e^{3}}\) | \(73\) |
derivativedivides | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a e -d b \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a e -d b \right ) \left (A e -B d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\) | \(73\) |
default | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a e -d b \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a e -d b \right ) \left (A e -B d \right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\) | \(73\) |
orering | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 b B \,x^{2} e^{2}+45 A x b \,e^{2}+45 B x a \,e^{2}-20 B x b d e +63 A a \,e^{2}-18 A b d e -18 B a d e +8 b B \,d^{2}\right )}{315 e^{3}}\) | \(73\) |
trager | \(\frac {2 \left (35 B b \,e^{4} x^{4}+45 A b \,e^{4} x^{3}+45 B a \,e^{4} x^{3}+50 b B d \,e^{3} x^{3}+63 A a \,e^{4} x^{2}+72 A b d \,e^{3} x^{2}+72 B a d \,e^{3} x^{2}+3 B b \,d^{2} e^{2} x^{2}+126 A a d \,e^{3} x +9 A b \,d^{2} e^{2} x +9 B a \,d^{2} e^{2} x -4 b B \,d^{3} e x +63 A a \,d^{2} e^{2}-18 A b \,d^{3} e -18 B a \,d^{3} e +8 b B \,d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\) | \(173\) |
risch | \(\frac {2 \left (35 B b \,e^{4} x^{4}+45 A b \,e^{4} x^{3}+45 B a \,e^{4} x^{3}+50 b B d \,e^{3} x^{3}+63 A a \,e^{4} x^{2}+72 A b d \,e^{3} x^{2}+72 B a d \,e^{3} x^{2}+3 B b \,d^{2} e^{2} x^{2}+126 A a d \,e^{3} x +9 A b \,d^{2} e^{2} x +9 B a \,d^{2} e^{2} x -4 b B \,d^{3} e x +63 A a \,d^{2} e^{2}-18 A b \,d^{3} e -18 B a \,d^{3} e +8 b B \,d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\) | \(173\) |
Input:
int((b*x+a)*(B*x+A)*(e*x+d)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5*(e*x+d)^(5/2)*((5/7*x*(7/9*b*x+a)*B+A*(5/7*b*x+a))*e^2-2/7*((10/9*b*x+ a)*B+A*b)*d*e+8/63*b*B*d^2)/e^3
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (71) = 142\).
Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.80 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\frac {2 \, {\left (35 \, B b e^{4} x^{4} + 8 \, B b d^{4} + 63 \, A a d^{2} e^{2} - 18 \, {\left (B a + A b\right )} d^{3} e + 5 \, {\left (10 \, B b d e^{3} + 9 \, {\left (B a + A b\right )} e^{4}\right )} x^{3} + 3 \, {\left (B b d^{2} e^{2} + 21 \, A a e^{4} + 24 \, {\left (B a + A b\right )} d e^{3}\right )} x^{2} - {\left (4 \, B b d^{3} e - 126 \, A a d e^{3} - 9 \, {\left (B a + A b\right )} d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)^(3/2),x, algorithm="fricas")
Output:
2/315*(35*B*b*e^4*x^4 + 8*B*b*d^4 + 63*A*a*d^2*e^2 - 18*(B*a + A*b)*d^3*e + 5*(10*B*b*d*e^3 + 9*(B*a + A*b)*e^4)*x^3 + 3*(B*b*d^2*e^2 + 21*A*a*e^4 + 24*(B*a + A*b)*d*e^3)*x^2 - (4*B*b*d^3*e - 126*A*a*d*e^3 - 9*(B*a + A*b)* d^2*e^2)*x)*sqrt(e*x + d)/e^3
Time = 0.93 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.49 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\begin {cases} \frac {2 \left (\frac {B b \left (d + e x\right )^{\frac {9}{2}}}{9 e^{2}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (A b e + B a e - 2 B b d\right )}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (A a e^{2} - A b d e - B a d e + B b d^{2}\right )}{5 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (A a x + \frac {B b x^{3}}{3} + \frac {x^{2} \left (A b + B a\right )}{2}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)**(3/2),x)
Output:
Piecewise((2*(B*b*(d + e*x)**(9/2)/(9*e**2) + (d + e*x)**(7/2)*(A*b*e + B* a*e - 2*B*b*d)/(7*e**2) + (d + e*x)**(5/2)*(A*a*e**2 - A*b*d*e - B*a*d*e + B*b*d**2)/(5*e**2))/e, Ne(e, 0)), (d**(3/2)*(A*a*x + B*b*x**3/3 + x**2*(A *b + B*a)/2), True))
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} B b - 45 \, {\left (2 \, B b d - {\left (B a + A b\right )} e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 63 \, {\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{315 \, e^{3}} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)^(3/2),x, algorithm="maxima")
Output:
2/315*(35*(e*x + d)^(9/2)*B*b - 45*(2*B*b*d - (B*a + A*b)*e)*(e*x + d)^(7/ 2) + 63*(B*b*d^2 + A*a*e^2 - (B*a + A*b)*d*e)*(e*x + d)^(5/2))/e^3
Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (71) = 142\).
Time = 0.13 (sec) , antiderivative size = 473, normalized size of antiderivative = 5.70 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} A a d^{2} + 210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a d + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a d^{2}}{e} + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A 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 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 )} B b 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 a d}{e} + \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 )} A 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 )} B b 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 a}{e} + \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 )} A 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 )} B b}{e^{2}}\right )}}{315 \, e} \] Input:
integrate((b*x+a)*(B*x+A)*(e*x+d)^(3/2),x, algorithm="giac")
Output:
2/315*(315*sqrt(e*x + d)*A*a*d^2 + 210*((e*x + d)^(3/2) - 3*sqrt(e*x + d)* d)*A*a*d + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a*d^2/e + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*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*a + 21*(3*(e*x + d)^(5/2) - 10*( e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*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*a*d/e + 42*(3*(e*x + d )^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A*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)*B*b*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*a/e + 9*(5*(e*x + d)^(7/2) - 2 1*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*A*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 + 315*sqrt(e*x + d)*d^4)*B*b/e^2)/e
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\frac {2\,{\left (d+e\,x\right )}^{5/2}\,\left (35\,B\,b\,{\left (d+e\,x\right )}^2+63\,A\,a\,e^2+63\,B\,b\,d^2+45\,A\,b\,e\,\left (d+e\,x\right )+45\,B\,a\,e\,\left (d+e\,x\right )-90\,B\,b\,d\,\left (d+e\,x\right )-63\,A\,b\,d\,e-63\,B\,a\,d\,e\right )}{315\,e^3} \] Input:
int((A + B*x)*(a + b*x)*(d + e*x)^(3/2),x)
Output:
(2*(d + e*x)^(5/2)*(35*B*b*(d + e*x)^2 + 63*A*a*e^2 + 63*B*b*d^2 + 45*A*b* e*(d + e*x) + 45*B*a*e*(d + e*x) - 90*B*b*d*(d + e*x) - 63*A*b*d*e - 63*B* a*d*e))/(315*e^3)
Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.67 \[ \int (a+b x) (A+B x) (d+e x)^{3/2} \, dx=\frac {2 \sqrt {e x +d}\, \left (35 b^{2} e^{4} x^{4}+90 a b \,e^{4} x^{3}+50 b^{2} d \,e^{3} x^{3}+63 a^{2} e^{4} x^{2}+144 a b d \,e^{3} x^{2}+3 b^{2} d^{2} e^{2} x^{2}+126 a^{2} d \,e^{3} x +18 a b \,d^{2} e^{2} x -4 b^{2} d^{3} e x +63 a^{2} d^{2} e^{2}-36 a b \,d^{3} e +8 b^{2} d^{4}\right )}{315 e^{3}} \] Input:
int((b*x+a)*(B*x+A)*(e*x+d)^(3/2),x)
Output:
(2*sqrt(d + e*x)*(63*a**2*d**2*e**2 + 126*a**2*d*e**3*x + 63*a**2*e**4*x** 2 - 36*a*b*d**3*e + 18*a*b*d**2*e**2*x + 144*a*b*d*e**3*x**2 + 90*a*b*e**4 *x**3 + 8*b**2*d**4 - 4*b**2*d**3*e*x + 3*b**2*d**2*e**2*x**2 + 50*b**2*d* e**3*x**3 + 35*b**2*e**4*x**4))/(315*e**3)