Integrand size = 22, antiderivative size = 122 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=-\frac {2 (d e-c f)^2 (f g-e h)}{f^4 \sqrt {e+f x}}-\frac {2 (d e-c f) (2 d f g-3 d e h+c f h) \sqrt {e+f x}}{f^4}+\frac {2 d (d f g-3 d e h+2 c f h) (e+f x)^{3/2}}{3 f^4}+\frac {2 d^2 h (e+f x)^{5/2}}{5 f^4} \] Output:
-2*(-c*f+d*e)^2*(-e*h+f*g)/f^4/(f*x+e)^(1/2)-2*(-c*f+d*e)*(c*f*h-3*d*e*h+2 *d*f*g)*(f*x+e)^(1/2)/f^4+2/3*d*(2*c*f*h-3*d*e*h+d*f*g)*(f*x+e)^(3/2)/f^4+ 2/5*d^2*h*(f*x+e)^(5/2)/f^4
Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.06 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {30 c^2 f^2 (-f g+2 e h+f h x)+20 c d f \left (-8 e^2 h+e f (6 g-4 h x)+f^2 x (3 g+h x)\right )+2 d^2 \left (48 e^3 h-8 e^2 f (5 g-3 h x)+f^3 x^2 (5 g+3 h x)-2 e f^2 x (10 g+3 h x)\right )}{15 f^4 \sqrt {e+f x}} \] Input:
Integrate[((c + d*x)^2*(g + h*x))/(e + f*x)^(3/2),x]
Output:
(30*c^2*f^2*(-(f*g) + 2*e*h + f*h*x) + 20*c*d*f*(-8*e^2*h + e*f*(6*g - 4*h *x) + f^2*x*(3*g + h*x)) + 2*d^2*(48*e^3*h - 8*e^2*f*(5*g - 3*h*x) + f^3*x ^2*(5*g + 3*h*x) - 2*e*f^2*x*(10*g + 3*h*x)))/(15*f^4*Sqrt[e + f*x])
Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
| \(\displaystyle \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int \left (\frac {d \sqrt {e+f x} (2 c f h-3 d e h+d f g)}{f^3}+\frac {(c f-d e) (c f h-3 d e h+2 d f g)}{f^3 \sqrt {e+f x}}+\frac {(c f-d e)^2 (f g-e h)}{f^3 (e+f x)^{3/2}}+\frac {d^2 h (e+f x)^{3/2}}{f^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 d (e+f x)^{3/2} (2 c f h-3 d e h+d f g)}{3 f^4}-\frac {2 \sqrt {e+f x} (d e-c f) (c f h-3 d e h+2 d f g)}{f^4}-\frac {2 (d e-c f)^2 (f g-e h)}{f^4 \sqrt {e+f x}}+\frac {2 d^2 h (e+f x)^{5/2}}{5 f^4}\) |
Input:
Int[((c + d*x)^2*(g + h*x))/(e + f*x)^(3/2),x]
Output:
(-2*(d*e - c*f)^2*(f*g - e*h))/(f^4*Sqrt[e + f*x]) - (2*(d*e - c*f)*(2*d*f *g - 3*d*e*h + c*f*h)*Sqrt[e + f*x])/f^4 + (2*d*(d*f*g - 3*d*e*h + 2*c*f*h )*(e + f*x)^(3/2))/(3*f^4) + (2*d^2*h*(e + f*x)^(5/2))/(5*f^4)
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 = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02
| method | result | size |
| pseudoelliptic | \(\frac {\left (\left (6 h \,x^{3}+10 g \,x^{2}\right ) d^{2}+60 x c \left (\frac {h x}{3}+g \right ) d -30 c^{2} \left (-h x +g \right )\right ) f^{3}+60 e \left (\left (-\frac {1}{5} h \,x^{2}-\frac {2}{3} g x \right ) d^{2}+2 c \left (-\frac {2 h x}{3}+g \right ) d +h \,c^{2}\right ) f^{2}-160 \left (\left (-\frac {3 h x}{10}+\frac {g}{2}\right ) d +c h \right ) e^{2} d f +96 d^{2} e^{3} h}{15 \sqrt {f x +e}\, f^{4}}\) | \(125\) |
| risch | \(\frac {2 \left (3 f^{2} d^{2} h \,x^{2}+10 f^{2} c d h x -9 e f x h \,d^{2}+5 f^{2} d^{2} g x +15 c^{2} f^{2} h -50 c d e h f +30 c d \,f^{2} g +33 d^{2} e^{2} h -25 d^{2} e g f \right ) \sqrt {f x +e}}{15 f^{4}}+\frac {2 c^{2} e \,f^{2} h -2 c^{2} g \,f^{3}-4 c d \,e^{2} f h +4 c d e \,f^{2} g +2 d^{2} e^{3} h -2 d^{2} e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(163\) |
| gosper | \(\frac {\frac {2}{5} d^{2} h \,x^{3} f^{3}+\frac {4}{3} c d \,f^{3} h \,x^{2}-\frac {4}{5} d^{2} e \,f^{2} h \,x^{2}+\frac {2}{3} d^{2} f^{3} g \,x^{2}+2 c^{2} f^{3} h x -\frac {16}{3} c d e \,f^{2} h x +4 c d \,f^{3} g x +\frac {16}{5} d^{2} e^{2} f h x -\frac {8}{3} d^{2} e \,f^{2} g x +4 c^{2} e \,f^{2} h -2 c^{2} g \,f^{3}-\frac {32}{3} c d \,e^{2} f h +8 c d e \,f^{2} g +\frac {32}{5} d^{2} e^{3} h -\frac {16}{3} d^{2} e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(169\) |
| trager | \(\frac {\frac {2}{5} d^{2} h \,x^{3} f^{3}+\frac {4}{3} c d \,f^{3} h \,x^{2}-\frac {4}{5} d^{2} e \,f^{2} h \,x^{2}+\frac {2}{3} d^{2} f^{3} g \,x^{2}+2 c^{2} f^{3} h x -\frac {16}{3} c d e \,f^{2} h x +4 c d \,f^{3} g x +\frac {16}{5} d^{2} e^{2} f h x -\frac {8}{3} d^{2} e \,f^{2} g x +4 c^{2} e \,f^{2} h -2 c^{2} g \,f^{3}-\frac {32}{3} c d \,e^{2} f h +8 c d e \,f^{2} g +\frac {32}{5} d^{2} e^{3} h -\frac {16}{3} d^{2} e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(169\) |
| orering | \(\frac {\frac {2}{5} d^{2} h \,x^{3} f^{3}+\frac {4}{3} c d \,f^{3} h \,x^{2}-\frac {4}{5} d^{2} e \,f^{2} h \,x^{2}+\frac {2}{3} d^{2} f^{3} g \,x^{2}+2 c^{2} f^{3} h x -\frac {16}{3} c d e \,f^{2} h x +4 c d \,f^{3} g x +\frac {16}{5} d^{2} e^{2} f h x -\frac {8}{3} d^{2} e \,f^{2} g x +4 c^{2} e \,f^{2} h -2 c^{2} g \,f^{3}-\frac {32}{3} c d \,e^{2} f h +8 c d e \,f^{2} g +\frac {32}{5} d^{2} e^{3} h -\frac {16}{3} d^{2} e^{2} f g}{\sqrt {f x +e}\, f^{4}}\) | \(169\) |
| derivativedivides | \(\frac {\frac {2 h \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+\frac {4 c d f h \left (f x +e \right )^{\frac {3}{2}}}{3}-2 d^{2} e h \left (f x +e \right )^{\frac {3}{2}}+\frac {2 d^{2} f g \left (f x +e \right )^{\frac {3}{2}}}{3}+2 c^{2} f^{2} h \sqrt {f x +e}-8 c d e f h \sqrt {f x +e}+4 c d \,f^{2} g \sqrt {f x +e}+6 d^{2} e^{2} h \sqrt {f x +e}-4 d^{2} e f g \sqrt {f x +e}-\frac {2 \left (-c^{2} e \,f^{2} h +c^{2} g \,f^{3}+2 c d \,e^{2} f h -2 c d e \,f^{2} g -d^{2} e^{3} h +d^{2} e^{2} f g \right )}{\sqrt {f x +e}}}{f^{4}}\) | \(200\) |
| default | \(\frac {\frac {2 h \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+\frac {4 c d f h \left (f x +e \right )^{\frac {3}{2}}}{3}-2 d^{2} e h \left (f x +e \right )^{\frac {3}{2}}+\frac {2 d^{2} f g \left (f x +e \right )^{\frac {3}{2}}}{3}+2 c^{2} f^{2} h \sqrt {f x +e}-8 c d e f h \sqrt {f x +e}+4 c d \,f^{2} g \sqrt {f x +e}+6 d^{2} e^{2} h \sqrt {f x +e}-4 d^{2} e f g \sqrt {f x +e}-\frac {2 \left (-c^{2} e \,f^{2} h +c^{2} g \,f^{3}+2 c d \,e^{2} f h -2 c d e \,f^{2} g -d^{2} e^{3} h +d^{2} e^{2} f g \right )}{\sqrt {f x +e}}}{f^{4}}\) | \(200\) |
Input:
int((d*x+c)^2*(h*x+g)/(f*x+e)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/15*(((6*h*x^3+10*g*x^2)*d^2+60*x*c*(1/3*h*x+g)*d-30*c^2*(-h*x+g))*f^3+60 *e*((-1/5*h*x^2-2/3*g*x)*d^2+2*c*(-2/3*h*x+g)*d+h*c^2)*f^2-160*((-3/10*h*x +1/2*g)*d+c*h)*e^2*d*f+96*d^2*e^3*h)/(f*x+e)^(1/2)/f^4
Time = 0.08 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.48 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, d^{2} f^{3} h x^{3} + {\left (5 \, d^{2} f^{3} g - 2 \, {\left (3 \, d^{2} e f^{2} - 5 \, c d f^{3}\right )} h\right )} x^{2} - 5 \, {\left (8 \, d^{2} e^{2} f - 12 \, c d e f^{2} + 3 \, c^{2} f^{3}\right )} g + 2 \, {\left (24 \, d^{2} e^{3} - 40 \, c d e^{2} f + 15 \, c^{2} e f^{2}\right )} h - {\left (10 \, {\left (2 \, d^{2} e f^{2} - 3 \, c d f^{3}\right )} g - {\left (24 \, d^{2} e^{2} f - 40 \, c d e f^{2} + 15 \, c^{2} f^{3}\right )} h\right )} x\right )} \sqrt {f x + e}}{15 \, {\left (f^{5} x + e f^{4}\right )}} \] Input:
integrate((d*x+c)^2*(h*x+g)/(f*x+e)^(3/2),x, algorithm="fricas")
Output:
2/15*(3*d^2*f^3*h*x^3 + (5*d^2*f^3*g - 2*(3*d^2*e*f^2 - 5*c*d*f^3)*h)*x^2 - 5*(8*d^2*e^2*f - 12*c*d*e*f^2 + 3*c^2*f^3)*g + 2*(24*d^2*e^3 - 40*c*d*e^ 2*f + 15*c^2*e*f^2)*h - (10*(2*d^2*e*f^2 - 3*c*d*f^3)*g - (24*d^2*e^2*f - 40*c*d*e*f^2 + 15*c^2*f^3)*h)*x)*sqrt(f*x + e)/(f^5*x + e*f^4)
Time = 3.64 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.70 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {d^{2} h \left (e + f x\right )^{\frac {5}{2}}}{5 f^{3}} + \frac {\left (e + f x\right )^{\frac {3}{2}} \cdot \left (2 c d f h - 3 d^{2} e h + d^{2} f g\right )}{3 f^{3}} + \frac {\sqrt {e + f x} \left (c^{2} f^{2} h - 4 c d e f h + 2 c d f^{2} g + 3 d^{2} e^{2} h - 2 d^{2} e f g\right )}{f^{3}} + \frac {\left (c f - d e\right )^{2} \left (e h - f g\right )}{f^{3} \sqrt {e + f x}}\right )}{f} & \text {for}\: f \neq 0 \\\frac {c^{2} g x + \frac {d^{2} h x^{4}}{4} + \frac {x^{3} \cdot \left (2 c d h + d^{2} g\right )}{3} + \frac {x^{2} \left (c^{2} h + 2 c d g\right )}{2}}{e^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)**2*(h*x+g)/(f*x+e)**(3/2),x)
Output:
Piecewise((2*(d**2*h*(e + f*x)**(5/2)/(5*f**3) + (e + f*x)**(3/2)*(2*c*d*f *h - 3*d**2*e*h + d**2*f*g)/(3*f**3) + sqrt(e + f*x)*(c**2*f**2*h - 4*c*d* e*f*h + 2*c*d*f**2*g + 3*d**2*e**2*h - 2*d**2*e*f*g)/f**3 + (c*f - d*e)**2 *(e*h - f*g)/(f**3*sqrt(e + f*x)))/f, Ne(f, 0)), ((c**2*g*x + d**2*h*x**4/ 4 + x**3*(2*c*d*h + d**2*g)/3 + x**2*(c**2*h + 2*c*d*g)/2)/e**(3/2), True) )
Time = 0.04 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.42 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (f x + e\right )}^{\frac {5}{2}} d^{2} h + 5 \, {\left (d^{2} f g - {\left (3 \, d^{2} e - 2 \, c d f\right )} h\right )} {\left (f x + e\right )}^{\frac {3}{2}} - 15 \, {\left (2 \, {\left (d^{2} e f - c d f^{2}\right )} g - {\left (3 \, d^{2} e^{2} - 4 \, c d e f + c^{2} f^{2}\right )} h\right )} \sqrt {f x + e}}{f^{3}} - \frac {15 \, {\left ({\left (d^{2} e^{2} f - 2 \, c d e f^{2} + c^{2} f^{3}\right )} g - {\left (d^{2} e^{3} - 2 \, c d e^{2} f + c^{2} e f^{2}\right )} h\right )}}{\sqrt {f x + e} f^{3}}\right )}}{15 \, f} \] Input:
integrate((d*x+c)^2*(h*x+g)/(f*x+e)^(3/2),x, algorithm="maxima")
Output:
2/15*((3*(f*x + e)^(5/2)*d^2*h + 5*(d^2*f*g - (3*d^2*e - 2*c*d*f)*h)*(f*x + e)^(3/2) - 15*(2*(d^2*e*f - c*d*f^2)*g - (3*d^2*e^2 - 4*c*d*e*f + c^2*f^ 2)*h)*sqrt(f*x + e))/f^3 - 15*((d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*g - (d^ 2*e^3 - 2*c*d*e^2*f + c^2*e*f^2)*h)/(sqrt(f*x + e)*f^3))/f
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (110) = 220\).
Time = 0.12 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.81 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=-\frac {2 \, {\left (d^{2} e^{2} f g - 2 \, c d e f^{2} g + c^{2} f^{3} g - d^{2} e^{3} h + 2 \, c d e^{2} f h - c^{2} e f^{2} h\right )}}{\sqrt {f x + e} f^{4}} + \frac {2 \, {\left (5 \, {\left (f x + e\right )}^{\frac {3}{2}} d^{2} f^{17} g - 30 \, \sqrt {f x + e} d^{2} e f^{17} g + 30 \, \sqrt {f x + e} c d f^{18} g + 3 \, {\left (f x + e\right )}^{\frac {5}{2}} d^{2} f^{16} h - 15 \, {\left (f x + e\right )}^{\frac {3}{2}} d^{2} e f^{16} h + 45 \, \sqrt {f x + e} d^{2} e^{2} f^{16} h + 10 \, {\left (f x + e\right )}^{\frac {3}{2}} c d f^{17} h - 60 \, \sqrt {f x + e} c d e f^{17} h + 15 \, \sqrt {f x + e} c^{2} f^{18} h\right )}}{15 \, f^{20}} \] Input:
integrate((d*x+c)^2*(h*x+g)/(f*x+e)^(3/2),x, algorithm="giac")
Output:
-2*(d^2*e^2*f*g - 2*c*d*e*f^2*g + c^2*f^3*g - d^2*e^3*h + 2*c*d*e^2*f*h - c^2*e*f^2*h)/(sqrt(f*x + e)*f^4) + 2/15*(5*(f*x + e)^(3/2)*d^2*f^17*g - 30 *sqrt(f*x + e)*d^2*e*f^17*g + 30*sqrt(f*x + e)*c*d*f^18*g + 3*(f*x + e)^(5 /2)*d^2*f^16*h - 15*(f*x + e)^(3/2)*d^2*e*f^16*h + 45*sqrt(f*x + e)*d^2*e^ 2*f^16*h + 10*(f*x + e)^(3/2)*c*d*f^17*h - 60*sqrt(f*x + e)*c*d*e*f^17*h + 15*sqrt(f*x + e)*c^2*f^18*h)/f^20
Time = 0.06 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.26 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {{\left (e+f\,x\right )}^{3/2}\,\left (2\,d^2\,f\,g-6\,d^2\,e\,h+4\,c\,d\,f\,h\right )}{3\,f^4}-\frac {-2\,h\,c^2\,e\,f^2+2\,g\,c^2\,f^3+4\,h\,c\,d\,e^2\,f-4\,g\,c\,d\,e\,f^2-2\,h\,d^2\,e^3+2\,g\,d^2\,e^2\,f}{f^4\,\sqrt {e+f\,x}}+\frac {2\,\sqrt {e+f\,x}\,\left (c\,f-d\,e\right )\,\left (c\,f\,h-3\,d\,e\,h+2\,d\,f\,g\right )}{f^4}+\frac {2\,d^2\,h\,{\left (e+f\,x\right )}^{5/2}}{5\,f^4} \] Input:
int(((g + h*x)*(c + d*x)^2)/(e + f*x)^(3/2),x)
Output:
((e + f*x)^(3/2)*(2*d^2*f*g - 6*d^2*e*h + 4*c*d*f*h))/(3*f^4) - (2*c^2*f^3 *g - 2*d^2*e^3*h - 2*c^2*e*f^2*h + 2*d^2*e^2*f*g - 4*c*d*e*f^2*g + 4*c*d*e ^2*f*h)/(f^4*(e + f*x)^(1/2)) + (2*(e + f*x)^(1/2)*(c*f - d*e)*(c*f*h - 3* d*e*h + 2*d*f*g))/f^4 + (2*d^2*h*(e + f*x)^(5/2))/(5*f^4)
Time = 0.25 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.39 \[ \int \frac {(c+d x)^2 (g+h x)}{(e+f x)^{3/2}} \, dx=\frac {\frac {2}{5} d^{2} f^{3} h \,x^{3}+\frac {4}{3} c d \,f^{3} h \,x^{2}-\frac {4}{5} d^{2} e \,f^{2} h \,x^{2}+\frac {2}{3} d^{2} f^{3} g \,x^{2}+2 c^{2} f^{3} h x -\frac {16}{3} c d e \,f^{2} h x +4 c d \,f^{3} g x +\frac {16}{5} d^{2} e^{2} f h x -\frac {8}{3} d^{2} e \,f^{2} g x +4 c^{2} e \,f^{2} h -2 c^{2} f^{3} g -\frac {32}{3} c d \,e^{2} f h +8 c d e \,f^{2} g +\frac {32}{5} d^{2} e^{3} h -\frac {16}{3} d^{2} e^{2} f g}{\sqrt {f x +e}\, f^{4}} \] Input:
int((d*x+c)^2*(h*x+g)/(f*x+e)^(3/2),x)
Output:
(2*(30*c**2*e*f**2*h - 15*c**2*f**3*g + 15*c**2*f**3*h*x - 80*c*d*e**2*f*h + 60*c*d*e*f**2*g - 40*c*d*e*f**2*h*x + 30*c*d*f**3*g*x + 10*c*d*f**3*h*x **2 + 48*d**2*e**3*h - 40*d**2*e**2*f*g + 24*d**2*e**2*f*h*x - 20*d**2*e*f **2*g*x - 6*d**2*e*f**2*h*x**2 + 5*d**2*f**3*g*x**2 + 3*d**2*f**3*h*x**3)) /(15*sqrt(e + f*x)*f**4)