Integrand size = 31, antiderivative size = 94 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 (b d-a e)^3}{5 e^4 (d+e x)^{5/2}}-\frac {2 b (b d-a e)^2}{e^4 (d+e x)^{3/2}}+\frac {6 b^2 (b d-a e)}{e^4 \sqrt {d+e x}}+\frac {2 b^3 \sqrt {d+e x}}{e^4} \]
2/5*(-a*e+b*d)^3/e^4/(e*x+d)^(5/2)-2*b*(-a*e+b*d)^2/e^4/(e*x+d)^(3/2)+6*b^ 2*(-a*e+b*d)/e^4/(e*x+d)^(1/2)+2*b^3*(e*x+d)^(1/2)/e^4
Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (a^3 e^3+a^2 b e^2 (2 d+5 e x)+a b^2 e \left (8 d^2+20 d e x+15 e^2 x^2\right )-b^3 \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )\right )}{5 e^4 (d+e x)^{5/2}} \]
(-2*(a^3*e^3 + a^2*b*e^2*(2*d + 5*e*x) + a*b^2*e*(8*d^2 + 20*d*e*x + 15*e^ 2*x^2) - b^3*(16*d^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3)))/(5*e^4*(d + e*x)^(5/2))
Time = 0.24 (sec) , antiderivative size = 94, 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.
\(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^3}{(d+e x)^{7/2}}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^3}{(d+e x)^{7/2}}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 (b d-a e)}{e^3 (d+e x)^{3/2}}+\frac {3 b (b d-a e)^2}{e^3 (d+e x)^{5/2}}+\frac {(a e-b d)^3}{e^3 (d+e x)^{7/2}}+\frac {b^3}{e^3 \sqrt {d+e x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 b^2 (b d-a e)}{e^4 \sqrt {d+e x}}-\frac {2 b (b d-a e)^2}{e^4 (d+e x)^{3/2}}+\frac {2 (b d-a e)^3}{5 e^4 (d+e x)^{5/2}}+\frac {2 b^3 \sqrt {d+e x}}{e^4}\) |
(2*(b*d - a*e)^3)/(5*e^4*(d + e*x)^(5/2)) - (2*b*(b*d - a*e)^2)/(e^4*(d + e*x)^(3/2)) + (6*b^2*(b*d - a*e))/(e^4*Sqrt[d + e*x]) + (2*b^3*Sqrt[d + e* x])/e^4
3.21.49.3.1 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 = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (-5 x^{3} b^{3}+15 a \,b^{2} x^{2}+5 b \,a^{2} x +a^{3}\right ) e^{3}+2 b d \left (-15 b^{2} x^{2}+10 a b x +a^{2}\right ) e^{2}+8 b^{2} d^{2} \left (-5 b x +a \right ) e -16 b^{3} d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(93\) |
risch | \(\frac {2 b^{3} \sqrt {e x +d}}{e^{4}}-\frac {2 \left (15 b^{2} e^{2} x^{2}+5 a b \,e^{2} x +25 b^{2} d e x +e^{2} a^{2}+3 a b d e +11 b^{2} d^{2}\right ) \left (a e -b d \right )}{5 e^{4} \sqrt {e x +d}\, \left (e^{2} x^{2}+2 d e x +d^{2}\right )}\) | \(104\) |
derivativedivides | \(\frac {2 b^{3} \sqrt {e x +d}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{\left (e x +d \right )^{\frac {3}{2}}}-\frac {6 b^{2} \left (a e -b d \right )}{\sqrt {e x +d}}}{e^{4}}\) | \(114\) |
default | \(\frac {2 b^{3} \sqrt {e x +d}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}{\left (e x +d \right )^{\frac {3}{2}}}-\frac {6 b^{2} \left (a e -b d \right )}{\sqrt {e x +d}}}{e^{4}}\) | \(114\) |
gosper | \(-\frac {2 \left (-5 b^{3} x^{3} e^{3}+15 x^{2} a \,b^{2} e^{3}-30 x^{2} b^{3} d \,e^{2}+5 x \,a^{2} b \,e^{3}+20 x a \,b^{2} d \,e^{2}-40 x \,b^{3} d^{2} e +a^{3} e^{3}+2 a^{2} b d \,e^{2}+8 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(115\) |
trager | \(-\frac {2 \left (-5 b^{3} x^{3} e^{3}+15 x^{2} a \,b^{2} e^{3}-30 x^{2} b^{3} d \,e^{2}+5 x \,a^{2} b \,e^{3}+20 x a \,b^{2} d \,e^{2}-40 x \,b^{3} d^{2} e +a^{3} e^{3}+2 a^{2} b d \,e^{2}+8 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(115\) |
-2/5/(e*x+d)^(5/2)*((-5*b^3*x^3+15*a*b^2*x^2+5*a^2*b*x+a^3)*e^3+2*b*d*(-15 *b^2*x^2+10*a*b*x+a^2)*e^2+8*b^2*d^2*(-5*b*x+a)*e-16*b^3*d^3)/e^4
Time = 0.40 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \, {\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{5 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]
2/5*(5*b^3*e^3*x^3 + 16*b^3*d^3 - 8*a*b^2*d^2*e - 2*a^2*b*d*e^2 - a^3*e^3 + 15*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 + 5*(8*b^3*d^2*e - 4*a*b^2*d*e^2 - a^2* b*e^3)*x)*sqrt(e*x + d)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 665 vs. \(2 (87) = 174\).
Time = 0.48 (sec) , antiderivative size = 665, normalized size of antiderivative = 7.07 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\begin {cases} - \frac {2 a^{3} e^{3}}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} - \frac {4 a^{2} b d e^{2}}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} - \frac {10 a^{2} b e^{3} x}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} - \frac {16 a b^{2} d^{2} e}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} - \frac {40 a b^{2} d e^{2} x}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} - \frac {30 a b^{2} e^{3} x^{2}}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} + \frac {32 b^{3} d^{3}}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} + \frac {80 b^{3} d^{2} e x}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} + \frac {60 b^{3} d e^{2} x^{2}}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} + \frac {10 b^{3} e^{3} x^{3}}{5 d^{2} e^{4} \sqrt {d + e x} + 10 d e^{5} x \sqrt {d + e x} + 5 e^{6} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((-2*a**3*e**3/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) - 4*a**2*b*d*e**2/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) - 10*a**2*b* e**3*x/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x** 2*sqrt(d + e*x)) - 16*a*b**2*d**2*e/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5 *x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) - 40*a*b**2*d*e**2*x/(5*d**2 *e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x )) - 30*a*b**2*e**3*x**2/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) + 32*b**3*d**3/(5*d**2*e**4*sqrt(d + e* x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) + 80*b**3*d**2 *e*x/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2* sqrt(d + e*x)) + 60*b**3*d*e**2*x**2/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e** 5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) + 10*b**3*e**3*x**3/(5*d**2 *e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*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** (7/2), True))
Time = 0.18 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (\frac {5 \, \sqrt {e x + d} b^{3}}{e^{3}} + \frac {b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \, {\left (b^{3} d - a b^{2} e\right )} {\left (e x + d\right )}^{2} - 5 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {5}{2}} e^{3}}\right )}}{5 \, e} \]
2/5*(5*sqrt(e*x + d)*b^3/e^3 + (b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3 + 15*(b^3*d - a*b^2*e)*(e*x + d)^2 - 5*(b^3*d^2 - 2*a*b^2*d*e + a^ 2*b*e^2)*(e*x + d))/((e*x + d)^(5/2)*e^3))/e
Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, \sqrt {e x + d} b^{3}}{e^{4}} + \frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} b^{3} d - 5 \, {\left (e x + d\right )} b^{3} d^{2} + b^{3} d^{3} - 15 \, {\left (e x + d\right )}^{2} a b^{2} e + 10 \, {\left (e x + d\right )} a b^{2} d e - 3 \, a b^{2} d^{2} e - 5 \, {\left (e x + d\right )} a^{2} b e^{2} + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )}}{5 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{4}} \]
2*sqrt(e*x + d)*b^3/e^4 + 2/5*(15*(e*x + d)^2*b^3*d - 5*(e*x + d)*b^3*d^2 + b^3*d^3 - 15*(e*x + d)^2*a*b^2*e + 10*(e*x + d)*a*b^2*d*e - 3*a*b^2*d^2* e - 5*(e*x + d)*a^2*b*e^2 + 3*a^2*b*d*e^2 - a^3*e^3)/((e*x + d)^(5/2)*e^4)
Time = 10.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=-\frac {2\,\left (a^3\,e^3+2\,a^2\,b\,d\,e^2+5\,a^2\,b\,e^3\,x+8\,a\,b^2\,d^2\,e+20\,a\,b^2\,d\,e^2\,x+15\,a\,b^2\,e^3\,x^2-16\,b^3\,d^3-40\,b^3\,d^2\,e\,x-30\,b^3\,d\,e^2\,x^2-5\,b^3\,e^3\,x^3\right )}{5\,e^4\,{\left (d+e\,x\right )}^{5/2}} \]