Integrand size = 20, antiderivative size = 79 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=-\frac {2 (b d-a e) (B d-A e)}{e^3 \sqrt {d+e x}}-\frac {2 (2 b B d-A b e-a B e) \sqrt {d+e x}}{e^3}+\frac {2 b B (d+e x)^{3/2}}{3 e^3} \] Output:
-2*(-a*e+b*d)*(-A*e+B*d)/e^3/(e*x+d)^(1/2)-2*(-A*b*e-B*a*e+2*B*b*d)*(e*x+d )^(1/2)/e^3+2/3*b*B*(e*x+d)^(3/2)/e^3
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {6 A b e (2 d+e x)+6 a e (2 B d-A e+B e x)+2 b B \left (-8 d^2-4 d e x+e^2 x^2\right )}{3 e^3 \sqrt {d+e x}} \] Input:
Integrate[((a + b*x)*(A + B*x))/(d + e*x)^(3/2),x]
Output:
(6*A*b*e*(2*d + e*x) + 6*a*e*(2*B*d - A*e + B*e*x) + 2*b*B*(-8*d^2 - 4*d*e *x + e^2*x^2))/(3*e^3*Sqrt[d + e*x])
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.
\(\displaystyle \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {a B e+A b e-2 b B d}{e^2 \sqrt {d+e x}}+\frac {(a e-b d) (A e-B d)}{e^2 (d+e x)^{3/2}}+\frac {b B \sqrt {d+e x}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {d+e x} (-a B e-A b e+2 b B d)}{e^3}-\frac {2 (b d-a e) (B d-A e)}{e^3 \sqrt {d+e x}}+\frac {2 b B (d+e x)^{3/2}}{3 e^3}\) |
Input:
Int[((a + b*x)*(A + B*x))/(d + e*x)^(3/2),x]
Output:
(-2*(b*d - a*e)*(B*d - A*e))/(e^3*Sqrt[d + e*x]) - (2*(2*b*B*d - A*b*e - a *B*e)*Sqrt[d + e*x])/e^3 + (2*b*B*(d + e*x)^(3/2))/(3*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 = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.81
method | result | size |
pseudoelliptic | \(\frac {\left (\left (2 b \,x^{2}+6 a x \right ) B -6 A \left (-b x +a \right )\right ) e^{2}+12 d \left (\left (-\frac {2 b x}{3}+a \right ) B +A b \right ) e -16 b B \,d^{2}}{3 \sqrt {e x +d}\, e^{3}}\) | \(64\) |
risch | \(\frac {2 \left (e b B x +3 A b e +3 B a e -5 B b d \right ) \sqrt {e x +d}}{3 e^{3}}-\frac {2 \left (A a \,e^{2}-A b d e -B a d e +b B \,d^{2}\right )}{e^{3} \sqrt {e x +d}}\) | \(72\) |
gosper | \(-\frac {2 \left (-b B \,x^{2} e^{2}-3 A x b \,e^{2}-3 B x a \,e^{2}+4 B x b d e +3 A a \,e^{2}-6 A b d e -6 B a d e +8 b B \,d^{2}\right )}{3 \sqrt {e x +d}\, e^{3}}\) | \(73\) |
trager | \(-\frac {2 \left (-b B \,x^{2} e^{2}-3 A x b \,e^{2}-3 B x a \,e^{2}+4 B x b d e +3 A a \,e^{2}-6 A b d e -6 B a d e +8 b B \,d^{2}\right )}{3 \sqrt {e x +d}\, e^{3}}\) | \(73\) |
orering | \(-\frac {2 \left (-b B \,x^{2} e^{2}-3 A x b \,e^{2}-3 B x a \,e^{2}+4 B x b d e +3 A a \,e^{2}-6 A b d e -6 B a d e +8 b B \,d^{2}\right )}{3 \sqrt {e x +d}\, e^{3}}\) | \(73\) |
derivativedivides | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A b e \sqrt {e x +d}+2 B a e \sqrt {e x +d}-4 B b d \sqrt {e x +d}-\frac {2 \left (A a \,e^{2}-A b d e -B a d e +b B \,d^{2}\right )}{\sqrt {e x +d}}}{e^{3}}\) | \(86\) |
default | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A b e \sqrt {e x +d}+2 B a e \sqrt {e x +d}-4 B b d \sqrt {e x +d}-\frac {2 \left (A a \,e^{2}-A b d e -B a d e +b B \,d^{2}\right )}{\sqrt {e x +d}}}{e^{3}}\) | \(86\) |
Input:
int((b*x+a)*(B*x+A)/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/3*(((2*b*x^2+6*a*x)*B-6*A*(-b*x+a))*e^2+12*d*((-2/3*b*x+a)*B+A*b)*e-16*b *B*d^2)/(e*x+d)^(1/2)/e^3
Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (B b e^{2} x^{2} - 8 \, B b d^{2} - 3 \, A a e^{2} + 6 \, {\left (B a + A b\right )} d e - {\left (4 \, B b d e - 3 \, {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{4} x + d e^{3}\right )}} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(3/2),x, algorithm="fricas")
Output:
2/3*(B*b*e^2*x^2 - 8*B*b*d^2 - 3*A*a*e^2 + 6*(B*a + A*b)*d*e - (4*B*b*d*e - 3*(B*a + A*b)*e^2)*x)*sqrt(e*x + d)/(e^4*x + d*e^3)
Time = 2.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {B b \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {\sqrt {d + e x} \left (A b e + B a e - 2 B b d\right )}{e^{2}} + \frac {\left (- A e + B d\right ) \left (a e - b d\right )}{e^{2} \sqrt {d + e x}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a x + \frac {B b x^{3}}{3} + \frac {x^{2} \left (A b + B a\right )}{2}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)**(3/2),x)
Output:
Piecewise((2*(B*b*(d + e*x)**(3/2)/(3*e**2) + sqrt(d + e*x)*(A*b*e + B*a*e - 2*B*b*d)/e**2 + (-A*e + B*d)*(a*e - b*d)/(e**2*sqrt(d + e*x)))/e, Ne(e, 0)), ((A*a*x + B*b*x**3/3 + x**2*(A*b + B*a)/2)/d**(3/2), True))
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} B b - 3 \, {\left (2 \, B b d - {\left (B a + A b\right )} e\right )} \sqrt {e x + d}}{e^{2}} - \frac {3 \, {\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )}}{\sqrt {e x + d} e^{2}}\right )}}{3 \, e} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(3/2),x, algorithm="maxima")
Output:
2/3*(((e*x + d)^(3/2)*B*b - 3*(2*B*b*d - (B*a + A*b)*e)*sqrt(e*x + d))/e^2 - 3*(B*b*d^2 + A*a*e^2 - (B*a + A*b)*d*e)/(sqrt(e*x + d)*e^2))/e
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=-\frac {2 \, {\left (B b d^{2} - B a d e - A b d e + A a e^{2}\right )}}{\sqrt {e x + d} e^{3}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B b e^{6} - 6 \, \sqrt {e x + d} B b d e^{6} + 3 \, \sqrt {e x + d} B a e^{7} + 3 \, \sqrt {e x + d} A b e^{7}\right )}}{3 \, e^{9}} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(3/2),x, algorithm="giac")
Output:
-2*(B*b*d^2 - B*a*d*e - A*b*d*e + A*a*e^2)/(sqrt(e*x + d)*e^3) + 2/3*((e*x + d)^(3/2)*B*b*e^6 - 6*sqrt(e*x + d)*B*b*d*e^6 + 3*sqrt(e*x + d)*B*a*e^7 + 3*sqrt(e*x + d)*A*b*e^7)/e^9
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\frac {2\,B\,b\,{\left (d+e\,x\right )}^2}{3}-2\,A\,a\,e^2-2\,B\,b\,d^2+2\,A\,b\,e\,\left (d+e\,x\right )+2\,B\,a\,e\,\left (d+e\,x\right )-4\,B\,b\,d\,\left (d+e\,x\right )+2\,A\,b\,d\,e+2\,B\,a\,d\,e}{e^3\,\sqrt {d+e\,x}} \] Input:
int(((A + B*x)*(a + b*x))/(d + e*x)^(3/2),x)
Output:
((2*B*b*(d + e*x)^2)/3 - 2*A*a*e^2 - 2*B*b*d^2 + 2*A*b*e*(d + e*x) + 2*B*a *e*(d + e*x) - 4*B*b*d*(d + e*x) + 2*A*b*d*e + 2*B*a*d*e)/(e^3*(d + e*x)^( 1/2))
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x) (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\frac {2}{3} b^{2} e^{2} x^{2}+4 a b \,e^{2} x -\frac {8}{3} b^{2} d e x -2 a^{2} e^{2}+8 a b d e -\frac {16}{3} b^{2} d^{2}}{\sqrt {e x +d}\, e^{3}} \] Input:
int((b*x+a)*(B*x+A)/(e*x+d)^(3/2),x)
Output:
(2*( - 3*a**2*e**2 + 12*a*b*d*e + 6*a*b*e**2*x - 8*b**2*d**2 - 4*b**2*d*e* x + b**2*e**2*x**2))/(3*sqrt(d + e*x)*e**3)