Integrand size = 20, antiderivative size = 81 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e) (B d-A e) \sqrt {d+e x}}{e^3}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{3/2}}{3 e^3}+\frac {2 b B (d+e x)^{5/2}}{5 e^3} \] Output:
2*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(1/2)/e^3-2/3*(-A*b*e-B*a*e+2*B*b*d)*(e*x+ d)^(3/2)/e^3+2/5*b*B*(e*x+d)^(5/2)/e^3
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (5 A b e (-2 d+e x)+5 a e (-2 B d+3 A e+B e x)+b B \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3} \] Input:
Integrate[((a + b*x)*(A + B*x))/Sqrt[d + e*x],x]
Output:
(2*Sqrt[d + e*x]*(5*A*b*e*(-2*d + e*x) + 5*a*e*(-2*B*d + 3*A*e + B*e*x) + b*B*(8*d^2 - 4*d*e*x + 3*e^2*x^2)))/(15*e^3)
Time = 0.23 (sec) , antiderivative size = 81, 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.
\(\displaystyle \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (\frac {\sqrt {d+e x} (a B e+A b e-2 b B d)}{e^2}+\frac {(a e-b d) (A e-B d)}{e^2 \sqrt {d+e x}}+\frac {b B (d+e x)^{3/2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 (d+e x)^{3/2} (-a B e-A b e+2 b B d)}{3 e^3}+\frac {2 \sqrt {d+e x} (b d-a e) (B d-A e)}{e^3}+\frac {2 b B (d+e x)^{5/2}}{5 e^3}\) |
Input:
Int[((a + b*x)*(A + B*x))/Sqrt[d + e*x],x]
Output:
(2*(b*d - a*e)*(B*d - A*e)*Sqrt[d + e*x])/e^3 - (2*(2*b*B*d - A*b*e - a*B* e)*(d + e*x)^(3/2))/(3*e^3) + (2*b*B*(d + e*x)^(5/2))/(5*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 = 0.62 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {e x +d}\, \left (\left (\frac {\left (\frac {3 b x}{5}+a \right ) x B}{3}+A \left (\frac {b x}{3}+a \right )\right ) e^{2}-\frac {2 \left (\left (\frac {2 b x}{5}+a \right ) B +A b \right ) d e}{3}+\frac {8 b B \,d^{2}}{15}\right )}{e^{3}}\) | \(60\) |
derivativedivides | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a e -d b \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -d b \right ) \left (A e -B d \right ) \sqrt {e x +d}}{e^{3}}\) | \(72\) |
default | \(\frac {\frac {2 b B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a e -d b \right ) B +b \left (A e -B d \right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -d b \right ) \left (A e -B d \right ) \sqrt {e x +d}}{e^{3}}\) | \(72\) |
gosper | \(\frac {2 \sqrt {e x +d}\, \left (3 b B \,x^{2} e^{2}+5 A x b \,e^{2}+5 B x a \,e^{2}-4 B x b d e +15 A a \,e^{2}-10 A b d e -10 B a d e +8 b B \,d^{2}\right )}{15 e^{3}}\) | \(73\) |
trager | \(\frac {2 \sqrt {e x +d}\, \left (3 b B \,x^{2} e^{2}+5 A x b \,e^{2}+5 B x a \,e^{2}-4 B x b d e +15 A a \,e^{2}-10 A b d e -10 B a d e +8 b B \,d^{2}\right )}{15 e^{3}}\) | \(73\) |
risch | \(\frac {2 \sqrt {e x +d}\, \left (3 b B \,x^{2} e^{2}+5 A x b \,e^{2}+5 B x a \,e^{2}-4 B x b d e +15 A a \,e^{2}-10 A b d e -10 B a d e +8 b B \,d^{2}\right )}{15 e^{3}}\) | \(73\) |
orering | \(\frac {2 \sqrt {e x +d}\, \left (3 b B \,x^{2} e^{2}+5 A x b \,e^{2}+5 B x a \,e^{2}-4 B x b d e +15 A a \,e^{2}-10 A b d e -10 B a d e +8 b B \,d^{2}\right )}{15 e^{3}}\) | \(73\) |
Input:
int((b*x+a)*(B*x+A)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)
Output:
2*(e*x+d)^(1/2)*((1/3*(3/5*b*x+a)*x*B+A*(1/3*b*x+a))*e^2-2/3*((2/5*b*x+a)* B+A*b)*d*e+8/15*b*B*d^2)/e^3
Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} + 15 \, A a e^{2} - 10 \, {\left (B a + A b\right )} d e - {\left (4 \, B b d e - 5 \, {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, e^{3}} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(1/2),x, algorithm="fricas")
Output:
2/15*(3*B*b*e^2*x^2 + 8*B*b*d^2 + 15*A*a*e^2 - 10*(B*a + A*b)*d*e - (4*B*b *d*e - 5*(B*a + A*b)*e^2)*x)*sqrt(e*x + d)/e^3
Time = 0.84 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.51 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (\frac {B b \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (A b e + B a e - 2 B b d\right )}{3 e^{2}} + \frac {\sqrt {d + e x} \left (A a e^{2} - A b d e - B a d e + B b d^{2}\right )}{e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {A a x + \frac {B b x^{3}}{3} + \frac {x^{2} \left (A b + B a\right )}{2}}{\sqrt {d}} & \text {otherwise} \end {cases} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)**(1/2),x)
Output:
Piecewise((2*(B*b*(d + e*x)**(5/2)/(5*e**2) + (d + e*x)**(3/2)*(A*b*e + B* a*e - 2*B*b*d)/(3*e**2) + sqrt(d + e*x)*(A*a*e**2 - A*b*d*e - B*a*d*e + B* b*d**2)/e**2)/e, Ne(e, 0)), ((A*a*x + B*b*x**3/3 + x**2*(A*b + B*a)/2)/sqr t(d), True))
Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b - 5 \, {\left (2 \, B b d - {\left (B a + A b\right )} e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )} \sqrt {e x + d}\right )}}{15 \, e^{3}} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(1/2),x, algorithm="maxima")
Output:
2/15*(3*(e*x + d)^(5/2)*B*b - 5*(2*B*b*d - (B*a + A*b)*e)*(e*x + d)^(3/2) + 15*(B*b*d^2 + A*a*e^2 - (B*a + A*b)*d*e)*sqrt(e*x + d))/e^3
Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {e x + d} A a + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a}{e} + \frac {5 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b}{e^{2}}\right )}}{15 \, e} \] Input:
integrate((b*x+a)*(B*x+A)/(e*x+d)^(1/2),x, algorithm="giac")
Output:
2/15*(15*sqrt(e*x + d)*A*a + 5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*B*a/e + 5*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*b/e + (3*(e*x + d)^(5/2) - 10 *(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b/e^2)/e
Time = 0.97 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2\,\sqrt {d+e\,x}\,\left (3\,B\,b\,{\left (d+e\,x\right )}^2+15\,A\,a\,e^2+15\,B\,b\,d^2+5\,A\,b\,e\,\left (d+e\,x\right )+5\,B\,a\,e\,\left (d+e\,x\right )-10\,B\,b\,d\,\left (d+e\,x\right )-15\,A\,b\,d\,e-15\,B\,a\,d\,e\right )}{15\,e^3} \] Input:
int(((A + B*x)*(a + b*x))/(d + e*x)^(1/2),x)
Output:
(2*(d + e*x)^(1/2)*(3*B*b*(d + e*x)^2 + 15*A*a*e^2 + 15*B*b*d^2 + 5*A*b*e* (d + e*x) + 5*B*a*e*(d + e*x) - 10*B*b*d*(d + e*x) - 15*A*b*d*e - 15*B*a*d *e))/(15*e^3)
Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b x) (A+B x)}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {e x +d}\, \left (3 b^{2} e^{2} x^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right )}{15 e^{3}} \] Input:
int((b*x+a)*(B*x+A)/(e*x+d)^(1/2),x)
Output:
(2*sqrt(d + e*x)*(15*a**2*e**2 - 20*a*b*d*e + 10*a*b*e**2*x + 8*b**2*d**2 - 4*b**2*d*e*x + 3*b**2*e**2*x**2))/(15*e**3)