Integrand size = 31, antiderivative size = 96 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 (b d-a e)^3}{3 e^4 (d+e x)^{3/2}}-\frac {6 b (b d-a e)^2}{e^4 \sqrt {d+e x}}-\frac {6 b^2 (b d-a e) \sqrt {d+e x}}{e^4}+\frac {2 b^3 (d+e x)^{3/2}}{3 e^4} \]
2/3*(-a*e+b*d)^3/e^4/(e*x+d)^(3/2)+2/3*b^3*(e*x+d)^(3/2)/e^4-6*b*(-a*e+b*d )^2/e^4/(e*x+d)^(1/2)-6*b^2*(-a*e+b*d)*(e*x+d)^(1/2)/e^4
Time = 0.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (a^3 e^3+3 a^2 b e^2 (2 d+3 e x)-3 a b^2 e \left (8 d^2+12 d e x+3 e^2 x^2\right )+b^3 \left (16 d^3+24 d^2 e x+6 d e^2 x^2-e^3 x^3\right )\right )}{3 e^4 (d+e x)^{3/2}} \]
(-2*(a^3*e^3 + 3*a^2*b*e^2*(2*d + 3*e*x) - 3*a*b^2*e*(8*d^2 + 12*d*e*x + 3 *e^2*x^2) + b^3*(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 = 0.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1184, 27, 53, 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.
\(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^3}{(d+e x)^{5/2}}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^3}{(d+e x)^{5/2}}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 (b d-a e)}{e^3 \sqrt {d+e x}}+\frac {3 b (b d-a e)^2}{e^3 (d+e x)^{3/2}}+\frac {(a e-b d)^3}{e^3 (d+e x)^{5/2}}+\frac {b^3 \sqrt {d+e x}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 b^2 \sqrt {d+e x} (b d-a e)}{e^4}-\frac {6 b (b d-a e)^2}{e^4 \sqrt {d+e x}}+\frac {2 (b d-a e)^3}{3 e^4 (d+e x)^{3/2}}+\frac {2 b^3 (d+e x)^{3/2}}{3 e^4}\) |
(2*(b*d - a*e)^3)/(3*e^4*(d + e*x)^(3/2)) - (6*b*(b*d - a*e)^2)/(e^4*Sqrt[ d + e*x]) - (6*b^2*(b*d - a*e)*Sqrt[d + e*x])/e^4 + (2*b^3*(d + e*x)^(3/2) )/(3*e^4)
3.21.48.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (-b x +a \right ) e -2 b d \right ) \left (\left (b^{2} x^{2}+10 a b x +a^{2}\right ) e^{2}+8 b d \left (-b x +a \right ) e -8 b^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(66\) |
risch | \(\frac {2 b^{2} \left (b e x +9 a e -8 b d \right ) \sqrt {e x +d}}{3 e^{4}}-\frac {2 \left (9 b e x +a e +8 b d \right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{3 e^{4} \left (e x +d \right )^{\frac {3}{2}}}\) | \(76\) |
gosper | \(-\frac {2 \left (-b^{3} x^{3} e^{3}-9 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+9 x \,a^{2} b \,e^{3}-36 x a \,b^{2} d \,e^{2}+24 x \,b^{3} d^{2} e +a^{3} e^{3}+6 a^{2} b d \,e^{2}-24 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(115\) |
trager | \(-\frac {2 \left (-b^{3} x^{3} e^{3}-9 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+9 x \,a^{2} b \,e^{3}-36 x a \,b^{2} d \,e^{2}+24 x \,b^{3} d^{2} e +a^{3} e^{3}+6 a^{2} b d \,e^{2}-24 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4}}\) | \(115\) |
derivativedivides | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+6 a \,b^{2} e \sqrt {e x +d}-6 b^{3} d \sqrt {e x +d}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {6 b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(122\) |
default | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {3}{2}}}{3}+6 a \,b^{2} e \sqrt {e x +d}-6 b^{3} d \sqrt {e x +d}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}-\frac {6 b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{\sqrt {e x +d}}}{e^{4}}\) | \(122\) |
-2/3*((-b*x+a)*e-2*b*d)*((b^2*x^2+10*a*b*x+a^2)*e^2+8*b*d*(-b*x+a)*e-8*b^2 *d^2)/(e*x+d)^(3/2)/e^4
Time = 0.47 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]
2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 - 3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a ^2*b*e^3)*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. 461 vs. \(2 (88) = 176\).
Time = 0.34 (sec) , antiderivative size = 461, normalized size of antiderivative = 4.80 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx=\begin {cases} - \frac {2 a^{3} e^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {12 a^{2} b d e^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {18 a^{2} b e^{3} x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {48 a b^{2} d^{2} e}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {72 a b^{2} d e^{2} x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {18 a b^{2} e^{3} x^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {32 b^{3} d^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {48 b^{3} d^{2} e x}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} - \frac {12 b^{3} d e^{2} x^{2}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} + \frac {2 b^{3} e^{3} x^{3}}{3 d e^{4} \sqrt {d + e x} + 3 e^{5} x \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-2*a**3*e**3/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 12*a**2*b*d*e**2/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 18* a**2*b*e**3*x/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 48*a*b** 2*d**2*e/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 72*a*b**2*d*e **2*x/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 18*a*b**2*e**3*x **2/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 32*b**3*d**3/(3*d* e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 48*b**3*d**2*e*x/(3*d*e**4* sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) - 12*b**3*d*e**2*x**2/(3*d*e**4*sq rt(d + e*x) + 3*e**5*x*sqrt(d + e*x)) + 2*b**3*e**3*x**3/(3*d*e**4*sqrt(d + e*x) + 3*e**5*x*sqrt(d + e*x)), Ne(e, 0)), ((a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4/4)/d**(5/2), True))
Time = 0.19 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} b^{3} - 9 \, {\left (b^{3} d - a b^{2} e\right )} \sqrt {e x + d}}{e^{3}} + \frac {b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3} - 9 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{3}}\right )}}{3 \, e} \]
2/3*(((e*x + d)^(3/2)*b^3 - 9*(b^3*d - a*b^2*e)*sqrt(e*x + d))/e^3 + (b^3* d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3 - 9*(b^3*d^2 - 2*a*b^2*d*e + a^2*b*e^2)*(e*x + d))/((e*x + d)^(3/2)*e^3))/e
Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (e x + d\right )} b^{3} d^{2} - b^{3} d^{3} - 18 \, {\left (e x + d\right )} a b^{2} d e + 3 \, a b^{2} d^{2} e + 9 \, {\left (e x + d\right )} a^{2} b e^{2} - 3 \, a^{2} b d e^{2} + a^{3} e^{3}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{3} e^{8} - 9 \, \sqrt {e x + d} b^{3} d e^{8} + 9 \, \sqrt {e x + d} a b^{2} e^{9}\right )}}{3 \, e^{12}} \]
-2/3*(9*(e*x + d)*b^3*d^2 - b^3*d^3 - 18*(e*x + d)*a*b^2*d*e + 3*a*b^2*d^2 *e + 9*(e*x + d)*a^2*b*e^2 - 3*a^2*b*d*e^2 + a^3*e^3)/((e*x + d)^(3/2)*e^4 ) + 2/3*((e*x + d)^(3/2)*b^3*e^8 - 9*sqrt(e*x + d)*b^3*d*e^8 + 9*sqrt(e*x + d)*a*b^2*e^9)/e^12
Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{5/2}} \, dx=\frac {2\,b^3\,{\left (d+e\,x\right )}^3-2\,a^3\,e^3+2\,b^3\,d^3-18\,b^3\,d\,{\left (d+e\,x\right )}^2-18\,b^3\,d^2\,\left (d+e\,x\right )+18\,a\,b^2\,e\,{\left (d+e\,x\right )}^2-18\,a^2\,b\,e^2\,\left (d+e\,x\right )-6\,a\,b^2\,d^2\,e+6\,a^2\,b\,d\,e^2+36\,a\,b^2\,d\,e\,\left (d+e\,x\right )}{3\,e^4\,{\left (d+e\,x\right )}^{3/2}} \]