Integrand size = 20, antiderivative size = 79 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 (b d-a e) (B d-A e)}{3 e^3 (d+e x)^{3/2}}+\frac {2 (2 b B d-A b e-a B e)}{e^3 \sqrt {d+e x}}+\frac {2 b B \sqrt {d+e x}}{e^3} \] Output:
-2/3*(-a*e+b*d)*(-A*e+B*d)/e^3/(e*x+d)^(3/2)+2*(-A*b*e-B*a*e+2*B*b*d)/e^3/ (e*x+d)^(1/2)+2*b*B*(e*x+d)^(1/2)/e^3
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (A b e (2 d+3 e x)+a e (2 B d+A e+3 B e x)-b B \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )}{3 e^3 (d+e x)^{3/2}} \] Input:
Integrate[((a + b*x)*(A + B*x))/(d + e*x)^(5/2),x]
Output:
(-2*(A*b*e*(2*d + 3*e*x) + a*e*(2*B*d + A*e + 3*B*e*x) - b*B*(8*d^2 + 12*d *e*x + 3*e^2*x^2)))/(3*e^3*(d + e*x)^(3/2))
Time = 0.24 (sec) , antiderivative size = 79, 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.
\(\displaystyle \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {a B e+A b e-2 b B d}{e^2 (d+e x)^{3/2}}+\frac {(a e-b d) (A e-B d)}{e^2 (d+e x)^{5/2}}+\frac {b B}{e^2 \sqrt {d+e x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 (-a B e-A b e+2 b B d)}{e^3 \sqrt {d+e x}}-\frac {2 (b d-a e) (B d-A e)}{3 e^3 (d+e x)^{3/2}}+\frac {2 b B \sqrt {d+e x}}{e^3}\) |
Input:
Int[((a + b*x)*(A + B*x))/(d + e*x)^(5/2),x]
Output:
(-2*(b*d - a*e)*(B*d - A*e))/(3*e^3*(d + e*x)^(3/2)) + (2*(2*b*B*d - A*b*e - a*B*e))/(e^3*Sqrt[d + e*x]) + (2*b*B*Sqrt[d + e*x])/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 = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.80
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (\left (-3 b \,x^{2}+3 a x \right ) B +A \left (3 b x +a \right )\right ) e^{2}+2 \left (\left (-6 b x +a \right ) B +A b \right ) d e -8 b B \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3}}\) | \(63\) |
gosper | \(-\frac {2 \left (-3 b B \,x^{2} e^{2}+3 A x b \,e^{2}+3 B x a \,e^{2}-12 B x b d e +A a \,e^{2}+2 A b d e +2 B a d e -8 b B \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3}}\) | \(72\) |
trager | \(-\frac {2 \left (-3 b B \,x^{2} e^{2}+3 A x b \,e^{2}+3 B x a \,e^{2}-12 B x b d e +A a \,e^{2}+2 A b d e +2 B a d e -8 b B \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3}}\) | \(72\) |
orering | \(-\frac {2 \left (-3 b B \,x^{2} e^{2}+3 A x b \,e^{2}+3 B x a \,e^{2}-12 B x b d e +A a \,e^{2}+2 A b d e +2 B a d e -8 b B \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{3}}\) | \(72\) |
derivativedivides | \(\frac {2 b B \sqrt {e x +d}-\frac {2 \left (A b e +B a e -2 B b d \right )}{\sqrt {e x +d}}-\frac {2 \left (A a \,e^{2}-A b d e -B a d e +b B \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\) | \(74\) |
default | \(\frac {2 b B \sqrt {e x +d}-\frac {2 \left (A b e +B a e -2 B b d \right )}{\sqrt {e x +d}}-\frac {2 \left (A a \,e^{2}-A b d e -B a d e +b B \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{3}}\) | \(74\) |
risch | \(\frac {2 b B \sqrt {e x +d}}{e^{3}}-\frac {2 \left (3 A x b \,e^{2}+3 B x a \,e^{2}-6 B x b d e +A a \,e^{2}+2 A b d e +2 B a d e -5 b B \,d^{2}\right )}{3 e^{3} \left (e x +d \right )^{\frac {3}{2}}}\) | \(77\) |
Input:
int((b*x+a)*(B*x+A)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3/(e*x+d)^(3/2)*(((-3*b*x^2+3*a*x)*B+A*(3*b*x+a))*e^2+2*((-6*b*x+a)*B+A *b)*d*e-8*b*B*d^2)/e^3
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} - A a e^{2} - 2 \, {\left (B a + A b\right )} d e + 3 \, {\left (4 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(5/2),x, algorithm="fricas")
Output:
2/3*(3*B*b*e^2*x^2 + 8*B*b*d^2 - A*a*e^2 - 2*(B*a + A*b)*d*e + 3*(4*B*b*d* e - (B*a + A*b)*e^2)*x)*sqrt(e*x + d)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3)
Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (80) = 160\).
Time = 0.37 (sec) , antiderivative size = 355, normalized size of antiderivative = 4.49 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=\begin {cases} - \frac {2 A a e^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} - \frac {4 A b d e}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} - \frac {6 A b e^{2} x}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} - \frac {4 B a d e}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} - \frac {6 B a e^{2} x}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {16 B b d^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {24 B b d e x}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {6 B b e^{2} x^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {A a x + \frac {A b x^{2}}{2} + \frac {B a x^{2}}{2} + \frac {B b x^{3}}{3}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)**(5/2),x)
Output:
Piecewise((-2*A*a*e**2/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) - 4*A*b*d*e/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) - 6*A*b*e**2* x/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) - 4*B*a*d*e/(3*d*e**3* sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) - 6*B*a*e**2*x/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) + 16*B*b*d**2/(3*d*e**3*sqrt(d + e*x) + 3*e **4*x*sqrt(d + e*x)) + 24*B*b*d*e*x/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqr t(d + e*x)) + 6*B*b*e**2*x**2/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)), Ne(e, 0)), ((A*a*x + A*b*x**2/2 + B*a*x**2/2 + B*b*x**3/3)/d**(5/2) , True))
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, \sqrt {e x + d} B b}{e^{2}} - \frac {B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e - 3 \, {\left (2 \, B b d - {\left (B a + A b\right )} e\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{2}}\right )}}{3 \, e} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(5/2),x, algorithm="maxima")
Output:
2/3*(3*sqrt(e*x + d)*B*b/e^2 - (B*b*d^2 + A*a*e^2 - (B*a + A*b)*d*e - 3*(2 *B*b*d - (B*a + A*b)*e)*(e*x + d))/((e*x + d)^(3/2)*e^2))/e
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 \, \sqrt {e x + d} B b}{e^{3}} + \frac {2 \, {\left (6 \, {\left (e x + d\right )} B b d - B b d^{2} - 3 \, {\left (e x + d\right )} B a e - 3 \, {\left (e x + d\right )} A b e + B a d e + A b d e - A a e^{2}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{3}} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(5/2),x, algorithm="giac")
Output:
2*sqrt(e*x + d)*B*b/e^3 + 2/3*(6*(e*x + d)*B*b*d - B*b*d^2 - 3*(e*x + d)*B *a*e - 3*(e*x + d)*A*b*e + B*a*d*e + A*b*d*e - A*a*e^2)/((e*x + d)^(3/2)*e ^3)
Time = 0.99 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2\,A\,a\,e^2-16\,B\,b\,d^2+6\,A\,b\,e^2\,x+6\,B\,a\,e^2\,x-6\,B\,b\,e^2\,x^2+4\,A\,b\,d\,e+4\,B\,a\,d\,e-24\,B\,b\,d\,e\,x}{3\,e^3\,{\left (d+e\,x\right )}^{3/2}} \] Input:
int(((A + B*x)*(a + b*x))/(d + e*x)^(5/2),x)
Output:
-(2*A*a*e^2 - 16*B*b*d^2 + 6*A*b*e^2*x + 6*B*a*e^2*x - 6*B*b*e^2*x^2 + 4*A *b*d*e + 4*B*a*d*e - 24*B*b*d*e*x)/(3*e^3*(d + e*x)^(3/2))
Time = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{5/2}} \, dx=\frac {2 b^{2} e^{2} x^{2}-4 a b \,e^{2} x +8 b^{2} d e x -\frac {2}{3} a^{2} e^{2}-\frac {8}{3} a b d e +\frac {16}{3} b^{2} d^{2}}{\sqrt {e x +d}\, e^{3} \left (e x +d \right )} \] Input:
int((b*x+a)*(B*x+A)/(e*x+d)^(5/2),x)
Output:
(2*( - a**2*e**2 - 4*a*b*d*e - 6*a*b*e**2*x + 8*b**2*d**2 + 12*b**2*d*e*x + 3*b**2*e**2*x**2))/(3*sqrt(d + e*x)*e**3*(d + e*x))