Integrand size = 24, antiderivative size = 122 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 d (B d-A e) (c d-b e)}{3 e^4 (d+e x)^{3/2}}-\frac {2 (B d (3 c d-2 b e)-A e (2 c d-b e))}{e^4 \sqrt {d+e x}}-\frac {2 (3 B c d-b B e-A c e) \sqrt {d+e x}}{e^4}+\frac {2 B c (d+e x)^{3/2}}{3 e^4} \] Output:
2/3*d*(-A*e+B*d)*(-b*e+c*d)/e^4/(e*x+d)^(3/2)-2*(B*d*(-2*b*e+3*c*d)-A*e*(- b*e+2*c*d))/e^4/(e*x+d)^(1/2)-2*(-A*c*e-B*b*e+3*B*c*d)*(e*x+d)^(1/2)/e^4+2 /3*B*c*(e*x+d)^(3/2)/e^4
Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 \left (A e \left (-b e (2 d+3 e x)+c \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )+B \left (b e \left (8 d^2+12 d e x+3 e^2 x^2\right )+c \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )\right )\right )}{3 e^4 (d+e x)^{3/2}} \] Input:
Integrate[((A + B*x)*(b*x + c*x^2))/(d + e*x)^(5/2),x]
Output:
(2*(A*e*(-(b*e*(2*d + 3*e*x)) + c*(8*d^2 + 12*d*e*x + 3*e^2*x^2)) + B*(b*e *(8*d^2 + 12*d*e*x + 3*e^2*x^2) + c*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e^3*x^3))))/(3*e^4*(d + e*x)^(3/2))
Time = 0.45 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1195, 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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \int \left (\frac {A c e+b B e-3 B c d}{e^3 \sqrt {d+e x}}+\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^3 (d+e x)^{3/2}}-\frac {d (B d-A e) (c d-b e)}{e^3 (d+e x)^{5/2}}+\frac {B c \sqrt {d+e x}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {d+e x} (-A c e-b B e+3 B c d)}{e^4}-\frac {2 (B d (3 c d-2 b e)-A e (2 c d-b e))}{e^4 \sqrt {d+e x}}+\frac {2 d (B d-A e) (c d-b e)}{3 e^4 (d+e x)^{3/2}}+\frac {2 B c (d+e x)^{3/2}}{3 e^4}\) |
Input:
Int[((A + B*x)*(b*x + c*x^2))/(d + e*x)^(5/2),x]
Output:
(2*d*(B*d - A*e)*(c*d - b*e))/(3*e^4*(d + e*x)^(3/2)) - (2*(B*d*(3*c*d - 2 *b*e) - A*e*(2*c*d - b*e)))/(e^4*Sqrt[d + e*x]) - (2*(3*B*c*d - b*B*e - A* c*e)*Sqrt[d + e*x])/e^4 + (2*B*c*(d + e*x)^(3/2))/(3*e^4)
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Time = 0.76 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79
method | result | size |
pseudoelliptic | \(\frac {\frac {2 \left (B c \,x^{3}+3 \left (A c +B b \right ) x^{2}-3 A b x \right ) e^{3}}{3}-\frac {4 \left (3 B c \,x^{2}+6 \left (-A c -B b \right ) x +A b \right ) d \,e^{2}}{3}+\frac {16 d^{2} \left (-3 B c x +A c +B b \right ) e}{3}-\frac {32 B c \,d^{3}}{3}}{\left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(96\) |
risch | \(\frac {2 \left (B c x e +3 A c e +3 B b e -8 B c d \right ) \sqrt {e x +d}}{3 e^{4}}-\frac {2 \left (3 A x b \,e^{3}-6 A x c d \,e^{2}-6 B x b d \,e^{2}+9 B x c \,d^{2} e +2 A b d \,e^{2}-5 A c \,d^{2} e -5 B b \,d^{2} e +8 B c \,d^{3}\right )}{3 e^{4} \left (e x +d \right )^{\frac {3}{2}}}\) | \(114\) |
gosper | \(-\frac {2 \left (-B c \,x^{3} e^{3}-3 A c \,e^{3} x^{2}-3 B b \,e^{3} x^{2}+6 B c d \,e^{2} x^{2}+3 A x b \,e^{3}-12 A x c d \,e^{2}-12 B x b d \,e^{2}+24 B x c \,d^{2} e +2 A b d \,e^{2}-8 A c \,d^{2} e -8 B b \,d^{2} e +16 B c \,d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(121\) |
trager | \(-\frac {2 \left (-B c \,x^{3} e^{3}-3 A c \,e^{3} x^{2}-3 B b \,e^{3} x^{2}+6 B c d \,e^{2} x^{2}+3 A x b \,e^{3}-12 A x c d \,e^{2}-12 B x b d \,e^{2}+24 B x c \,d^{2} e +2 A b d \,e^{2}-8 A c \,d^{2} e -8 B b \,d^{2} e +16 B c \,d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(121\) |
derivativedivides | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A c e \sqrt {e x +d}+2 B b e \sqrt {e x +d}-6 B c d \sqrt {e x +d}-\frac {2 \left (A b \,e^{2}-2 A c d e -2 B b d e +3 B c \,d^{2}\right )}{\sqrt {e x +d}}+\frac {2 d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{4}}\) | \(122\) |
default | \(\frac {\frac {2 B c \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A c e \sqrt {e x +d}+2 B b e \sqrt {e x +d}-6 B c d \sqrt {e x +d}-\frac {2 \left (A b \,e^{2}-2 A c d e -2 B b d e +3 B c \,d^{2}\right )}{\sqrt {e x +d}}+\frac {2 d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{4}}\) | \(122\) |
orering | \(-\frac {2 \left (-B c \,x^{3} e^{3}-3 A c \,e^{3} x^{2}-3 B b \,e^{3} x^{2}+6 B c d \,e^{2} x^{2}+3 A x b \,e^{3}-12 A x c d \,e^{2}-12 B x b d \,e^{2}+24 B x c \,d^{2} e +2 A b d \,e^{2}-8 A c \,d^{2} e -8 B b \,d^{2} e +16 B c \,d^{3}\right ) \left (c \,x^{2}+b x \right )}{3 e^{4} x \left (c x +b \right ) \left (e x +d \right )^{\frac {3}{2}}}\) | \(140\) |
Input:
int((B*x+A)*(c*x^2+b*x)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
Output:
2/3*((B*c*x^3+3*(A*c+B*b)*x^2-3*A*b*x)*e^3-2*(3*B*c*x^2+6*(-A*c-B*b)*x+A*b )*d*e^2+8*d^2*(-3*B*c*x+A*c+B*b)*e-16*B*c*d^3)/(e*x+d)^(3/2)/e^4
Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (B c e^{3} x^{3} - 16 \, B c d^{3} - 2 \, A b d e^{2} + 8 \, {\left (B b + A c\right )} d^{2} e - 3 \, {\left (2 \, B c d e^{2} - {\left (B b + A c\right )} e^{3}\right )} x^{2} - 3 \, {\left (8 \, B c d^{2} e + A b e^{3} - 4 \, {\left (B b + A c\right )} d e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^(5/2),x, algorithm="fricas")
Output:
2/3*(B*c*e^3*x^3 - 16*B*c*d^3 - 2*A*b*d*e^2 + 8*(B*b + A*c)*d^2*e - 3*(2*B *c*d*e^2 - (B*b + A*c)*e^3)*x^2 - 3*(8*B*c*d^2*e + A*b*e^3 - 4*(B*b + A*c) *d*e^2)*x)*sqrt(e*x + d)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 539 vs. \(2 (121) = 242\).
Time = 0.48 (sec) , antiderivative size = 539, normalized size of antiderivative = 4.42 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=\begin {cases} - \frac {4 A b d e^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {6 A b e^{3} x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {16 A c d^{2} e}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {24 A c d e^{2} x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {6 A c e^{3} x^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {16 B b d^{2} e}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {24 B b d e^{2} x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {6 B b e^{3} x^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {32 B c d^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {48 B c d^{2} e x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {12 B c d e^{2} x^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {2 B c e^{3} x^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {\frac {A b x^{2}}{2} + \frac {A c x^{3}}{3} + \frac {B b x^{3}}{3} + \frac {B c x^{4}}{4}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((B*x+A)*(c*x**2+b*x)/(e*x+d)**(5/2),x)
Output:
Piecewise((-4*A*b*d*e**2/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 6*A*b*e**3*x/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 16*A*c *d**2*e/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 24*A*c*d*e**2* x/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 6*A*c*e**3*x**2/(3*d *e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 16*B*b*d**2*e/(3*d*e**4*sq rt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 24*B*b*d*e**2*x/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 6*B*b*e**3*x**2/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 32*B*c*d**3/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x *sqrt(d + e*x)) - 48*B*c*d**2*e*x/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt( d + e*x)) - 12*B*c*d*e**2*x**2/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 2*B*c*e**3*x**3/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) , Ne(e, 0)), ((A*b*x**2/2 + A*c*x**3/3 + B*b*x**3/3 + B*c*x**4/4)/d**(5/2) , True))
Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} B c - 3 \, {\left (3 \, B c d - {\left (B b + A c\right )} e\right )} \sqrt {e x + d}}{e^{3}} + \frac {B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e - 3 \, {\left (3 \, B c d^{2} + A b e^{2} - 2 \, {\left (B b + A c\right )} d e\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{3}}\right )}}{3 \, e} \] Input:
integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^(5/2),x, algorithm="maxima")
Output:
2/3*(((e*x + d)^(3/2)*B*c - 3*(3*B*c*d - (B*b + A*c)*e)*sqrt(e*x + d))/e^3 + (B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e - 3*(3*B*c*d^2 + A*b*e^2 - 2*( B*b + A*c)*d*e)*(e*x + d))/((e*x + d)^(3/2)*e^3))/e
Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (e x + d\right )} B c d^{2} - B c d^{3} - 6 \, {\left (e x + d\right )} B b d e - 6 \, {\left (e x + d\right )} A c d e + B b d^{2} e + A c d^{2} e + 3 \, {\left (e x + d\right )} A b e^{2} - A b d e^{2}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B c e^{8} - 9 \, \sqrt {e x + d} B c d e^{8} + 3 \, \sqrt {e x + d} B b e^{9} + 3 \, \sqrt {e x + d} A c e^{9}\right )}}{3 \, e^{12}} \] Input:
integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^(5/2),x, algorithm="giac")
Output:
-2/3*(9*(e*x + d)*B*c*d^2 - B*c*d^3 - 6*(e*x + d)*B*b*d*e - 6*(e*x + d)*A* c*d*e + B*b*d^2*e + A*c*d^2*e + 3*(e*x + d)*A*b*e^2 - A*b*d*e^2)/((e*x + d )^(3/2)*e^4) + 2/3*((e*x + d)^(3/2)*B*c*e^8 - 9*sqrt(e*x + d)*B*c*d*e^8 + 3*sqrt(e*x + d)*B*b*e^9 + 3*sqrt(e*x + d)*A*c*e^9)/e^12
Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2\,B\,c\,{\left (d+e\,x\right )}^3+2\,B\,c\,d^3+2\,A\,b\,d\,e^2-2\,A\,c\,d^2\,e-2\,B\,b\,d^2\,e-6\,A\,b\,e^2\,\left (d+e\,x\right )+6\,A\,c\,e\,{\left (d+e\,x\right )}^2+6\,B\,b\,e\,{\left (d+e\,x\right )}^2-18\,B\,c\,d\,{\left (d+e\,x\right )}^2-18\,B\,c\,d^2\,\left (d+e\,x\right )+12\,A\,c\,d\,e\,\left (d+e\,x\right )+12\,B\,b\,d\,e\,\left (d+e\,x\right )}{3\,e^4\,{\left (d+e\,x\right )}^{3/2}} \] Input:
int(((b*x + c*x^2)*(A + B*x))/(d + e*x)^(5/2),x)
Output:
(2*B*c*(d + e*x)^3 + 2*B*c*d^3 + 2*A*b*d*e^2 - 2*A*c*d^2*e - 2*B*b*d^2*e - 6*A*b*e^2*(d + e*x) + 6*A*c*e*(d + e*x)^2 + 6*B*b*e*(d + e*x)^2 - 18*B*c* d*(d + e*x)^2 - 18*B*c*d^2*(d + e*x) + 12*A*c*d*e*(d + e*x) + 12*B*b*d*e*( d + e*x))/(3*e^4*(d + e*x)^(3/2))
Time = 0.21 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {\frac {2}{3} b c \,e^{3} x^{3}+2 a c \,e^{3} x^{2}+2 b^{2} e^{3} x^{2}-4 b c d \,e^{2} x^{2}-2 a b \,e^{3} x +8 a c d \,e^{2} x +8 b^{2} d \,e^{2} x -16 b c \,d^{2} e x -\frac {4}{3} a b d \,e^{2}+\frac {16}{3} a c \,d^{2} e +\frac {16}{3} b^{2} d^{2} e -\frac {32}{3} b c \,d^{3}}{\sqrt {e x +d}\, e^{4} \left (e x +d \right )} \] Input:
int((B*x+A)*(c*x^2+b*x)/(e*x+d)^(5/2),x)
Output:
(2*( - 2*a*b*d*e**2 - 3*a*b*e**3*x + 8*a*c*d**2*e + 12*a*c*d*e**2*x + 3*a* c*e**3*x**2 + 8*b**2*d**2*e + 12*b**2*d*e**2*x + 3*b**2*e**3*x**2 - 16*b*c *d**3 - 24*b*c*d**2*e*x - 6*b*c*d*e**2*x**2 + b*c*e**3*x**3))/(3*sqrt(d + e*x)*e**4*(d + e*x))