Integrand size = 16, antiderivative size = 67 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=-\frac {2 a (A b-a B) (a+b x)^{5/2}}{5 b^3}+\frac {2 (A b-2 a B) (a+b x)^{7/2}}{7 b^3}+\frac {2 B (a+b x)^{9/2}}{9 b^3} \] Output:
-2/5*a*(A*b-B*a)*(b*x+a)^(5/2)/b^3+2/7*(A*b-2*B*a)*(b*x+a)^(7/2)/b^3+2/9*B *(b*x+a)^(9/2)/b^3
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 (a+b x)^{5/2} \left (8 a^2 B+5 b^2 x (9 A+7 B x)-2 a b (9 A+10 B x)\right )}{315 b^3} \] Input:
Integrate[x*(a + b*x)^(3/2)*(A + B*x),x]
Output:
(2*(a + b*x)^(5/2)*(8*a^2*B + 5*b^2*x*(9*A + 7*B*x) - 2*a*b*(9*A + 10*B*x) ))/(315*b^3)
Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.
\(\displaystyle \int x (a+b x)^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {(a+b x)^{5/2} (A b-2 a B)}{b^2}+\frac {a (a+b x)^{3/2} (a B-A b)}{b^2}+\frac {B (a+b x)^{7/2}}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (a+b x)^{7/2} (A b-2 a B)}{7 b^3}-\frac {2 a (a+b x)^{5/2} (A b-a B)}{5 b^3}+\frac {2 B (a+b x)^{9/2}}{9 b^3}\) |
Input:
Int[x*(a + b*x)^(3/2)*(A + B*x),x]
Output:
(-2*a*(A*b - a*B)*(a + b*x)^(5/2))/(5*b^3) + (2*(A*b - 2*a*B)*(a + b*x)^(7 /2))/(7*b^3) + (2*B*(a + b*x)^(9/2))/(9*b^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 = 0.36 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(-\frac {4 \left (-\frac {5 \left (\frac {7 B x}{9}+A \right ) x \,b^{2}}{2}+a \left (\frac {10 B x}{9}+A \right ) b -\frac {4 a^{2} B}{9}\right ) \left (b x +a \right )^{\frac {5}{2}}}{35 b^{3}}\) | \(41\) |
gosper | \(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-35 b^{2} B \,x^{2}-45 A \,b^{2} x +20 B a b x +18 a b A -8 a^{2} B \right )}{315 b^{3}}\) | \(47\) |
orering | \(-\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (-35 b^{2} B \,x^{2}-45 A \,b^{2} x +20 B a b x +18 a b A -8 a^{2} B \right )}{315 b^{3}}\) | \(47\) |
derivativedivides | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{3}}\) | \(52\) |
default | \(\frac {\frac {2 B \left (b x +a \right )^{\frac {9}{2}}}{9}+\frac {2 \left (A b -2 B a \right ) \left (b x +a \right )^{\frac {7}{2}}}{7}-\frac {2 a \left (A b -B a \right ) \left (b x +a \right )^{\frac {5}{2}}}{5}}{b^{3}}\) | \(52\) |
trager | \(-\frac {2 \left (-35 B \,x^{4} b^{4}-45 A \,x^{3} b^{4}-50 B \,x^{3} a \,b^{3}-72 A \,x^{2} a \,b^{3}-3 B \,x^{2} a^{2} b^{2}-9 A x \,a^{2} b^{2}+4 B x \,a^{3} b +18 A \,a^{3} b -8 B \,a^{4}\right ) \sqrt {b x +a}}{315 b^{3}}\) | \(95\) |
risch | \(-\frac {2 \left (-35 B \,x^{4} b^{4}-45 A \,x^{3} b^{4}-50 B \,x^{3} a \,b^{3}-72 A \,x^{2} a \,b^{3}-3 B \,x^{2} a^{2} b^{2}-9 A x \,a^{2} b^{2}+4 B x \,a^{3} b +18 A \,a^{3} b -8 B \,a^{4}\right ) \sqrt {b x +a}}{315 b^{3}}\) | \(95\) |
Input:
int(x*(b*x+a)^(3/2)*(B*x+A),x,method=_RETURNVERBOSE)
Output:
-4/35*(-5/2*(7/9*B*x+A)*x*b^2+a*(10/9*B*x+A)*b-4/9*a^2*B)*(b*x+a)^(5/2)/b^ 3
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.42 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (35 \, B b^{4} x^{4} + 8 \, B a^{4} - 18 \, A a^{3} b + 5 \, {\left (10 \, B a b^{3} + 9 \, A b^{4}\right )} x^{3} + 3 \, {\left (B a^{2} b^{2} + 24 \, A a b^{3}\right )} x^{2} - {\left (4 \, B a^{3} b - 9 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x + a}}{315 \, b^{3}} \] Input:
integrate(x*(b*x+a)^(3/2)*(B*x+A),x, algorithm="fricas")
Output:
2/315*(35*B*b^4*x^4 + 8*B*a^4 - 18*A*a^3*b + 5*(10*B*a*b^3 + 9*A*b^4)*x^3 + 3*(B*a^2*b^2 + 24*A*a*b^3)*x^2 - (4*B*a^3*b - 9*A*a^2*b^2)*x)*sqrt(b*x + a)/b^3
Time = 0.70 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.22 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\begin {cases} \frac {2 \left (\frac {B \left (a + b x\right )^{\frac {9}{2}}}{9 b} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (A b - 2 B a\right )}{7 b} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- A a b + B a^{2}\right )}{5 b}\right )}{b^{2}} & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{2}}{2} + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x*(b*x+a)**(3/2)*(B*x+A),x)
Output:
Piecewise((2*(B*(a + b*x)**(9/2)/(9*b) + (a + b*x)**(7/2)*(A*b - 2*B*a)/(7 *b) + (a + b*x)**(5/2)*(-A*a*b + B*a**2)/(5*b))/b**2, Ne(b, 0)), (a**(3/2) *(A*x**2/2 + B*x**3/3), True))
Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} B - 45 \, {\left (2 \, B a - A b\right )} {\left (b x + a\right )}^{\frac {7}{2}} + 63 \, {\left (B a^{2} - A a b\right )} {\left (b x + a\right )}^{\frac {5}{2}}\right )}}{315 \, b^{3}} \] Input:
integrate(x*(b*x+a)^(3/2)*(B*x+A),x, algorithm="maxima")
Output:
2/315*(35*(b*x + a)^(9/2)*B - 45*(2*B*a - A*b)*(b*x + a)^(7/2) + 63*(B*a^2 - A*a*b)*(b*x + a)^(5/2))/b^3
Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (56) = 112\).
Time = 0.12 (sec) , antiderivative size = 275, normalized size of antiderivative = 4.10 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \, {\left (\frac {105 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} A a^{2}}{b} + \frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} B a^{2}}{b^{2}} + \frac {42 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} A a}{b} + \frac {18 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} B a}{b^{2}} + \frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} A}{b} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} B}{b^{2}}\right )}}{315 \, b} \] Input:
integrate(x*(b*x+a)^(3/2)*(B*x+A),x, algorithm="giac")
Output:
2/315*(105*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*A*a^2/b + 21*(3*(b*x + a) ^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*B*a^2/b^2 + 42*(3*(b *x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*A*a/b + 18*(5 *(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 - 35*sqrt (b*x + a)*a^3)*B*a/b^2 + 9*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35* (b*x + a)^(3/2)*a^2 - 35*sqrt(b*x + a)*a^3)*A/b + (35*(b*x + a)^(9/2) - 18 0*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*B/b^2)/b
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2\,{\left (a+b\,x\right )}^{5/2}\,\left (63\,B\,a^2+35\,B\,{\left (a+b\,x\right )}^2-63\,A\,a\,b+45\,A\,b\,\left (a+b\,x\right )-90\,B\,a\,\left (a+b\,x\right )\right )}{315\,b^3} \] Input:
int(x*(A + B*x)*(a + b*x)^(3/2),x)
Output:
(2*(a + b*x)^(5/2)*(63*B*a^2 + 35*B*(a + b*x)^2 - 63*A*a*b + 45*A*b*(a + b *x) - 90*B*a*(a + b*x)))/(315*b^3)
Time = 0.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int x (a+b x)^{3/2} (A+B x) \, dx=\frac {2 \sqrt {b x +a}\, \left (7 b^{4} x^{4}+19 a \,b^{3} x^{3}+15 a^{2} b^{2} x^{2}+a^{3} b x -2 a^{4}\right )}{63 b^{2}} \] Input:
int(x*(b*x+a)^(3/2)*(B*x+A),x)
Output:
(2*sqrt(a + b*x)*( - 2*a**4 + a**3*b*x + 15*a**2*b**2*x**2 + 19*a*b**3*x** 3 + 7*b**4*x**4))/(63*b**2)