Integrand size = 22, antiderivative size = 126 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {2 (b d-a e)^2 (B d-A e)}{7 e^4 (d+e x)^{7/2}}-\frac {2 (b d-a e) (3 b B d-2 A b e-a B e)}{5 e^4 (d+e x)^{5/2}}+\frac {2 b (3 b B d-A b e-2 a B e)}{3 e^4 (d+e x)^{3/2}}-\frac {2 b^2 B}{e^4 \sqrt {d+e x}} \] Output:
2/7*(-a*e+b*d)^2*(-A*e+B*d)/e^4/(e*x+d)^(7/2)-2/5*(-a*e+b*d)*(-2*A*b*e-B*a *e+3*B*b*d)/e^4/(e*x+d)^(5/2)+2/3*b*(-A*b*e-2*B*a*e+3*B*b*d)/e^4/(e*x+d)^( 3/2)-2*b^2*B/e^4/(e*x+d)^(1/2)
Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2 \left (3 a^2 e^2 (2 B d+5 A e+7 B e x)+2 a b e \left (3 A e (2 d+7 e x)+B \left (8 d^2+28 d e x+35 e^2 x^2\right )\right )+b^2 \left (A e \left (8 d^2+28 d e x+35 e^2 x^2\right )+3 B \left (16 d^3+56 d^2 e x+70 d e^2 x^2+35 e^3 x^3\right )\right )\right )}{105 e^4 (d+e x)^{7/2}} \] Input:
Integrate[((a + b*x)^2*(A + B*x))/(d + e*x)^(9/2),x]
Output:
(-2*(3*a^2*e^2*(2*B*d + 5*A*e + 7*B*e*x) + 2*a*b*e*(3*A*e*(2*d + 7*e*x) + B*(8*d^2 + 28*d*e*x + 35*e^2*x^2)) + b^2*(A*e*(8*d^2 + 28*d*e*x + 35*e^2*x ^2) + 3*B*(16*d^3 + 56*d^2*e*x + 70*d*e^2*x^2 + 35*e^3*x^3))))/(105*e^4*(d + e*x)^(7/2))
Time = 0.27 (sec) , antiderivative size = 126, 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.
| \(\displaystyle \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int \left (\frac {b (2 a B e+A b e-3 b B d)}{e^3 (d+e x)^{5/2}}+\frac {(a e-b d) (a B e+2 A b e-3 b B d)}{e^3 (d+e x)^{7/2}}+\frac {(a e-b d)^2 (A e-B d)}{e^3 (d+e x)^{9/2}}+\frac {b^2 B}{e^3 (d+e x)^{3/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b (-2 a B e-A b e+3 b B d)}{3 e^4 (d+e x)^{3/2}}-\frac {2 (b d-a e) (-a B e-2 A b e+3 b B d)}{5 e^4 (d+e x)^{5/2}}+\frac {2 (b d-a e)^2 (B d-A e)}{7 e^4 (d+e x)^{7/2}}-\frac {2 b^2 B}{e^4 \sqrt {d+e x}}\) |
Input:
Int[((a + b*x)^2*(A + B*x))/(d + e*x)^(9/2),x]
Output:
(2*(b*d - a*e)^2*(B*d - A*e))/(7*e^4*(d + e*x)^(7/2)) - (2*(b*d - a*e)*(3* b*B*d - 2*A*b*e - a*B*e))/(5*e^4*(d + e*x)^(5/2)) + (2*b*(3*b*B*d - A*b*e - 2*a*B*e))/(3*e^4*(d + e*x)^(3/2)) - (2*b^2*B)/(e^4*Sqrt[d + e*x])
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 = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94
| method | result | size |
| pseudoelliptic | \(\frac {\left (-70 x^{2} \left (3 B x +A \right ) b^{2}-84 a x \left (\frac {5 B x}{3}+A \right ) b -30 a^{2} \left (\frac {7 B x}{5}+A \right )\right ) e^{3}-24 \left (\frac {7 x \left (\frac {15 B x}{2}+A \right ) b^{2}}{3}+a \left (\frac {14 B x}{3}+A \right ) b +\frac {a^{2} B}{2}\right ) d \,e^{2}-16 b \left (\left (21 B x +A \right ) b +2 B a \right ) d^{2} e -96 b^{2} B \,d^{3}}{105 \left (e x +d \right )^{\frac {7}{2}} e^{4}}\) | \(118\) |
| derivativedivides | \(\frac {-\frac {2 b^{2} B}{\sqrt {e x +d}}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{7 \left (e x +d \right )^{\frac {7}{2}}}-\frac {2 b \left (A b e +2 B a e -3 B b d \right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B a b d e +3 b^{2} B \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}}{e^{4}}\) | \(159\) |
| default | \(\frac {-\frac {2 b^{2} B}{\sqrt {e x +d}}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{7 \left (e x +d \right )^{\frac {7}{2}}}-\frac {2 b \left (A b e +2 B a e -3 B b d \right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B a b d e +3 b^{2} B \,d^{2}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}}{e^{4}}\) | \(159\) |
| gosper | \(-\frac {2 \left (105 b^{2} B \,x^{3} e^{3}+35 A \,x^{2} b^{2} e^{3}+70 B \,x^{2} a b \,e^{3}+210 B \,x^{2} b^{2} d \,e^{2}+42 A x a b \,e^{3}+28 A x \,b^{2} d \,e^{2}+21 B x \,a^{2} e^{3}+56 B x a b d \,e^{2}+168 B x \,b^{2} d^{2} e +15 a^{2} A \,e^{3}+12 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +6 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e +48 b^{2} B \,d^{3}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} e^{4}}\) | \(169\) |
| trager | \(-\frac {2 \left (105 b^{2} B \,x^{3} e^{3}+35 A \,x^{2} b^{2} e^{3}+70 B \,x^{2} a b \,e^{3}+210 B \,x^{2} b^{2} d \,e^{2}+42 A x a b \,e^{3}+28 A x \,b^{2} d \,e^{2}+21 B x \,a^{2} e^{3}+56 B x a b d \,e^{2}+168 B x \,b^{2} d^{2} e +15 a^{2} A \,e^{3}+12 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +6 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e +48 b^{2} B \,d^{3}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} e^{4}}\) | \(169\) |
| orering | \(-\frac {2 \left (105 b^{2} B \,x^{3} e^{3}+35 A \,x^{2} b^{2} e^{3}+70 B \,x^{2} a b \,e^{3}+210 B \,x^{2} b^{2} d \,e^{2}+42 A x a b \,e^{3}+28 A x \,b^{2} d \,e^{2}+21 B x \,a^{2} e^{3}+56 B x a b d \,e^{2}+168 B x \,b^{2} d^{2} e +15 a^{2} A \,e^{3}+12 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +6 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e +48 b^{2} B \,d^{3}\right )}{105 \left (e x +d \right )^{\frac {7}{2}} e^{4}}\) | \(169\) |
Input:
int((b*x+a)^2*(B*x+A)/(e*x+d)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/105*((-70*x^2*(3*B*x+A)*b^2-84*a*x*(5/3*B*x+A)*b-30*a^2*(7/5*B*x+A))*e^3 -24*(7/3*x*(15/2*B*x+A)*b^2+a*(14/3*B*x+A)*b+1/2*a^2*B)*d*e^2-16*b*((21*B* x+A)*b+2*B*a)*d^2*e-96*b^2*B*d^3)/(e*x+d)^(7/2)/e^4
Time = 0.10 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (105 \, B b^{2} e^{3} x^{3} + 48 \, B b^{2} d^{3} + 15 \, A a^{2} e^{3} + 8 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 6 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 35 \, {\left (6 \, B b^{2} d e^{2} + {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 7 \, {\left (24 \, B b^{2} d^{2} e + 4 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^(9/2),x, algorithm="fricas")
Output:
-2/105*(105*B*b^2*e^3*x^3 + 48*B*b^2*d^3 + 15*A*a^2*e^3 + 8*(2*B*a*b + A*b ^2)*d^2*e + 6*(B*a^2 + 2*A*a*b)*d*e^2 + 35*(6*B*b^2*d*e^2 + (2*B*a*b + A*b ^2)*e^3)*x^2 + 7*(24*B*b^2*d^2*e + 4*(2*B*a*b + A*b^2)*d*e^2 + 3*(B*a^2 + 2*A*a*b)*e^3)*x)*sqrt(e*x + d)/(e^8*x^4 + 4*d*e^7*x^3 + 6*d^2*e^6*x^2 + 4* d^3*e^5*x + d^4*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 1323 vs. \(2 (128) = 256\).
Time = 0.76 (sec) , antiderivative size = 1323, normalized size of antiderivative = 10.50 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx =\text {Too large to display} \] Input:
integrate((b*x+a)**2*(B*x+A)/(e*x+d)**(9/2),x)
Output:
Piecewise((-30*A*a**2*e**3/(105*d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x* sqrt(d + e*x) + 315*d*e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x )) - 24*A*a*b*d*e**2/(105*d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 315*d*e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 8 4*A*a*b*e**3*x/(105*d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x ) + 315*d*e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 16*A*b* *2*d**2*e/(105*d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 3 15*d*e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 56*A*b**2*d* e**2*x/(105*d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 315* d*e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 70*A*b**2*e**3* x**2/(105*d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 315*d* e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 12*B*a**2*d*e**2/ (105*d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 315*d*e**6* x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 42*B*a**2*e**3*x/(105* d**3*e**4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 315*d*e**6*x**2* sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 32*B*a*b*d**2*e/(105*d**3*e **4*sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 315*d*e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 112*B*a*b*d*e**2*x/(105*d**3*e**4 *sqrt(d + e*x) + 315*d**2*e**5*x*sqrt(d + e*x) + 315*d*e**6*x**2*sqrt(d + e*x) + 105*e**7*x**3*sqrt(d + e*x)) - 140*B*a*b*e**3*x**2/(105*d**3*e**...
Time = 0.03 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.24 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (105 \, {\left (e x + d\right )}^{3} B b^{2} - 15 \, B b^{2} d^{3} + 15 \, A a^{2} e^{3} + 15 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - 15 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} - 35 \, {\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} {\left (e x + d\right )}^{2} + 21 \, {\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} {\left (e x + d\right )}\right )}}{105 \, {\left (e x + d\right )}^{\frac {7}{2}} e^{4}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^(9/2),x, algorithm="maxima")
Output:
-2/105*(105*(e*x + d)^3*B*b^2 - 15*B*b^2*d^3 + 15*A*a^2*e^3 + 15*(2*B*a*b + A*b^2)*d^2*e - 15*(B*a^2 + 2*A*a*b)*d*e^2 - 35*(3*B*b^2*d - (2*B*a*b + A *b^2)*e)*(e*x + d)^2 + 21*(3*B*b^2*d^2 - 2*(2*B*a*b + A*b^2)*d*e + (B*a^2 + 2*A*a*b)*e^2)*(e*x + d))/((e*x + d)^(7/2)*e^4)
Time = 0.14 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (105 \, {\left (e x + d\right )}^{3} B b^{2} - 105 \, {\left (e x + d\right )}^{2} B b^{2} d + 63 \, {\left (e x + d\right )} B b^{2} d^{2} - 15 \, B b^{2} d^{3} + 70 \, {\left (e x + d\right )}^{2} B a b e + 35 \, {\left (e x + d\right )}^{2} A b^{2} e - 84 \, {\left (e x + d\right )} B a b d e - 42 \, {\left (e x + d\right )} A b^{2} d e + 30 \, B a b d^{2} e + 15 \, A b^{2} d^{2} e + 21 \, {\left (e x + d\right )} B a^{2} e^{2} + 42 \, {\left (e x + d\right )} A a b e^{2} - 15 \, B a^{2} d e^{2} - 30 \, A a b d e^{2} + 15 \, A a^{2} e^{3}\right )}}{105 \, {\left (e x + d\right )}^{\frac {7}{2}} e^{4}} \] Input:
integrate((b*x+a)^2*(B*x+A)/(e*x+d)^(9/2),x, algorithm="giac")
Output:
-2/105*(105*(e*x + d)^3*B*b^2 - 105*(e*x + d)^2*B*b^2*d + 63*(e*x + d)*B*b ^2*d^2 - 15*B*b^2*d^3 + 70*(e*x + d)^2*B*a*b*e + 35*(e*x + d)^2*A*b^2*e - 84*(e*x + d)*B*a*b*d*e - 42*(e*x + d)*A*b^2*d*e + 30*B*a*b*d^2*e + 15*A*b^ 2*d^2*e + 21*(e*x + d)*B*a^2*e^2 + 42*(e*x + d)*A*a*b*e^2 - 15*B*a^2*d*e^2 - 30*A*a*b*d*e^2 + 15*A*a^2*e^3)/((e*x + d)^(7/2)*e^4)
Time = 0.06 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx=-\frac {2\,A\,a^2}{7\,e\,{\left (d+e\,x\right )}^{7/2}}-\frac {2\,b\,\left (A\,b+2\,B\,a\right )\,\left (50\,d^2-42\,d\,\left (d+e\,x\right )+35\,e^2\,x^2+70\,d\,e\,x\right )}{105\,e^3\,{\left (d+e\,x\right )}^{7/2}}-\frac {2\,B\,b^2\,\left (21\,d^2\,\left (d+e\,x\right )-5\,d^3+35\,e^3\,x^3+70\,d\,e^2\,x^2+35\,d^2\,e\,x\right )}{35\,e^4\,{\left (d+e\,x\right )}^{7/2}}-\frac {2\,a\,\left (2\,d+7\,e\,x\right )\,\left (2\,A\,b+B\,a\right )}{35\,e^2\,{\left (d+e\,x\right )}^{7/2}} \] Input:
int(((A + B*x)*(a + b*x)^2)/(d + e*x)^(9/2),x)
Output:
- (2*A*a^2)/(7*e*(d + e*x)^(7/2)) - (2*b*(A*b + 2*B*a)*(50*d^2 - 42*d*(d + e*x) + 35*e^2*x^2 + 70*d*e*x))/(105*e^3*(d + e*x)^(7/2)) - (2*B*b^2*(21*d ^2*(d + e*x) - 5*d^3 + 35*e^3*x^3 + 70*d*e^2*x^2 + 35*d^2*e*x))/(35*e^4*(d + e*x)^(7/2)) - (2*a*(2*d + 7*e*x)*(2*A*b + B*a))/(35*e^2*(d + e*x)^(7/2) )
Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.15 \[ \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{9/2}} \, dx=\frac {-2 b^{3} e^{3} x^{3}-2 a \,b^{2} e^{3} x^{2}-4 b^{3} d \,e^{2} x^{2}-\frac {6}{5} a^{2} b \,e^{3} x -\frac {8}{5} a \,b^{2} d \,e^{2} x -\frac {16}{5} b^{3} d^{2} e x -\frac {2}{7} a^{3} e^{3}-\frac {12}{35} a^{2} b d \,e^{2}-\frac {16}{35} a \,b^{2} d^{2} e -\frac {32}{35} b^{3} d^{3}}{\sqrt {e x +d}\, e^{4} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )} \] Input:
int((b*x+a)^2*(B*x+A)/(e*x+d)^(9/2),x)
Output:
(2*( - 5*a**3*e**3 - 6*a**2*b*d*e**2 - 21*a**2*b*e**3*x - 8*a*b**2*d**2*e - 28*a*b**2*d*e**2*x - 35*a*b**2*e**3*x**2 - 16*b**3*d**3 - 56*b**3*d**2*e *x - 70*b**3*d*e**2*x**2 - 35*b**3*e**3*x**3))/(35*sqrt(d + e*x)*e**4*(d** 3 + 3*d**2*e*x + 3*d*e**2*x**2 + e**3*x**3))