Integrand size = 31, antiderivative size = 100 \[ \int (a+b x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=-\frac {2 (b d-a e)^3 (d+e x)^{3/2}}{3 e^4}+\frac {6 b (b d-a e)^2 (d+e x)^{5/2}}{5 e^4}-\frac {6 b^2 (b d-a e) (d+e x)^{7/2}}{7 e^4}+\frac {2 b^3 (d+e x)^{9/2}}{9 e^4} \]
-2/3*(-a*e+b*d)^3*(e*x+d)^(3/2)/e^4+6/5*b*(-a*e+b*d)^2*(e*x+d)^(5/2)/e^4-6 /7*b^2*(-a*e+b*d)*(e*x+d)^(7/2)/e^4+2/9*b^3*(e*x+d)^(9/2)/e^4
Time = 0.02 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02 \[ \int (a+b x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (d+e x)^{3/2} \left (105 a^3 e^3+63 a^2 b e^2 (-2 d+3 e x)+9 a b^2 e \left (8 d^2-12 d e x+15 e^2 x^2\right )+b^3 \left (-16 d^3+24 d^2 e x-30 d e^2 x^2+35 e^3 x^3\right )\right )}{315 e^4} \]
(2*(d + e*x)^(3/2)*(105*a^3*e^3 + 63*a^2*b*e^2*(-2*d + 3*e*x) + 9*a*b^2*e* (8*d^2 - 12*d*e*x + 15*e^2*x^2) + b^3*(-16*d^3 + 24*d^2*e*x - 30*d*e^2*x^2 + 35*e^3*x^3)))/(315*e^4)
Time = 0.24 (sec) , antiderivative size = 100, 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.
\(\displaystyle \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right ) \sqrt {d+e x} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int b^2 (a+b x)^3 \sqrt {d+e x}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int (a+b x)^3 \sqrt {d+e x}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 (d+e x)^{5/2} (b d-a e)}{e^3}+\frac {3 b (d+e x)^{3/2} (b d-a e)^2}{e^3}+\frac {\sqrt {d+e x} (a e-b d)^3}{e^3}+\frac {b^3 (d+e x)^{7/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 b^2 (d+e x)^{7/2} (b d-a e)}{7 e^4}+\frac {6 b (d+e x)^{5/2} (b d-a e)^2}{5 e^4}-\frac {2 (d+e x)^{3/2} (b d-a e)^3}{3 e^4}+\frac {2 b^3 (d+e x)^{9/2}}{9 e^4}\) |
(-2*(b*d - a*e)^3*(d + e*x)^(3/2))/(3*e^4) + (6*b*(b*d - a*e)^2*(d + e*x)^ (5/2))/(5*e^4) - (6*b^2*(b*d - a*e)*(d + e*x)^(7/2))/(7*e^4) + (2*b^3*(d + e*x)^(9/2))/(9*e^4)
3.21.45.3.1 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 = 0.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.93
method | result | size |
pseudoelliptic | \(\frac {2 \left (\left (\frac {9}{7} a \,b^{2} x^{2}+\frac {9}{5} b \,a^{2} x +\frac {1}{3} x^{3} b^{3}+a^{3}\right ) e^{3}-\frac {6 \left (\frac {5}{21} b^{2} x^{2}+\frac {6}{7} a b x +a^{2}\right ) b d \,e^{2}}{5}+\frac {24 b^{2} \left (\frac {b x}{3}+a \right ) d^{2} e}{35}-\frac {16 b^{3} d^{3}}{105}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3 e^{4}}\) | \(93\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 b^{3} x^{3} e^{3}+135 x^{2} a \,b^{2} e^{3}-30 x^{2} b^{3} d \,e^{2}+189 x \,a^{2} b \,e^{3}-108 x a \,b^{2} d \,e^{2}+24 x \,b^{3} d^{2} e +105 a^{3} e^{3}-126 a^{2} b d \,e^{2}+72 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right )}{315 e^{4}}\) | \(116\) |
derivativedivides | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a e -b d \right ) b^{2}+b \left (2 a b e -2 b^{2} d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a e -b d \right ) \left (2 a b e -2 b^{2} d \right )+b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a e -b d \right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(147\) |
default | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (\left (a e -b d \right ) b^{2}+b \left (2 a b e -2 b^{2} d \right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a e -b d \right ) \left (2 a b e -2 b^{2} d \right )+b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a e -b d \right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}}{e^{4}}\) | \(147\) |
trager | \(\frac {2 \left (35 b^{3} e^{4} x^{4}+135 a \,b^{2} e^{4} x^{3}+5 b^{3} d \,e^{3} x^{3}+189 a^{2} b \,e^{4} x^{2}+27 a \,b^{2} d \,e^{3} x^{2}-6 b^{3} d^{2} e^{2} x^{2}+105 a^{3} e^{4} x +63 b \,a^{2} e^{3} d x -36 b^{2} a \,d^{2} e^{2} x +8 b^{3} d^{3} e x +105 a^{3} e^{3} d -126 a^{2} b \,d^{2} e^{2}+72 a \,b^{2} d^{3} e -16 b^{3} d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(170\) |
risch | \(\frac {2 \left (35 b^{3} e^{4} x^{4}+135 a \,b^{2} e^{4} x^{3}+5 b^{3} d \,e^{3} x^{3}+189 a^{2} b \,e^{4} x^{2}+27 a \,b^{2} d \,e^{3} x^{2}-6 b^{3} d^{2} e^{2} x^{2}+105 a^{3} e^{4} x +63 b \,a^{2} e^{3} d x -36 b^{2} a \,d^{2} e^{2} x +8 b^{3} d^{3} e x +105 a^{3} e^{3} d -126 a^{2} b \,d^{2} e^{2}+72 a \,b^{2} d^{3} e -16 b^{3} d^{4}\right ) \sqrt {e x +d}}{315 e^{4}}\) | \(170\) |
2/3*((9/7*a*b^2*x^2+9/5*b*a^2*x+1/3*x^3*b^3+a^3)*e^3-6/5*(5/21*b^2*x^2+6/7 *a*b*x+a^2)*b*d*e^2+24/35*b^2*(1/3*b*x+a)*d^2*e-16/105*b^3*d^3)*(e*x+d)^(3 /2)/e^4
Time = 0.35 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.64 \[ \int (a+b x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (35 \, b^{3} e^{4} x^{4} - 16 \, b^{3} d^{4} + 72 \, a b^{2} d^{3} e - 126 \, a^{2} b d^{2} e^{2} + 105 \, a^{3} d e^{3} + 5 \, {\left (b^{3} d e^{3} + 27 \, a b^{2} e^{4}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{2} e^{2} - 9 \, a b^{2} d e^{3} - 63 \, a^{2} b e^{4}\right )} x^{2} + {\left (8 \, b^{3} d^{3} e - 36 \, a b^{2} d^{2} e^{2} + 63 \, a^{2} b d e^{3} + 105 \, a^{3} e^{4}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{4}} \]
2/315*(35*b^3*e^4*x^4 - 16*b^3*d^4 + 72*a*b^2*d^3*e - 126*a^2*b*d^2*e^2 + 105*a^3*d*e^3 + 5*(b^3*d*e^3 + 27*a*b^2*e^4)*x^3 - 3*(2*b^3*d^2*e^2 - 9*a* b^2*d*e^3 - 63*a^2*b*e^4)*x^2 + (8*b^3*d^3*e - 36*a*b^2*d^2*e^2 + 63*a^2*b *d*e^3 + 105*a^3*e^4)*x)*sqrt(e*x + d)/e^4
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (92) = 184\).
Time = 1.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.96 \[ \int (a+b x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{3}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (3 a b^{2} e - 3 b^{3} d\right )}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (3 a^{2} b e^{2} - 6 a b^{2} d e + 3 b^{3} d^{2}\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (a^{3} e^{3} - 3 a^{2} b d e^{2} + 3 a b^{2} d^{2} e - b^{3} d^{3}\right )}{3 e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {a^{3} b x + \frac {3 a^{2} b^{2} x^{2}}{2} + a b^{3} x^{3} + \frac {b^{4} x^{4}}{4}}{b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(b**3*(d + e*x)**(9/2)/(9*e**3) + (d + e*x)**(7/2)*(3*a*b**2* e - 3*b**3*d)/(7*e**3) + (d + e*x)**(5/2)*(3*a**2*b*e**2 - 6*a*b**2*d*e + 3*b**3*d**2)/(5*e**3) + (d + e*x)**(3/2)*(a**3*e**3 - 3*a**2*b*d*e**2 + 3* a*b**2*d**2*e - b**3*d**3)/(3*e**3))/e, Ne(e, 0)), (sqrt(d)*Piecewise((a** 3*x, Eq(b, 0)), ((a**3*b*x + 3*a**2*b**2*x**2/2 + a*b**3*x**3 + b**4*x**4/ 4)/b, True)), True))
Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int (a+b x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} b^{3} - 135 \, {\left (b^{3} d - a b^{2} e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 189 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}} - 105 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left (e x + d\right )}^{\frac {3}{2}}\right )}}{315 \, e^{4}} \]
2/315*(35*(e*x + d)^(9/2)*b^3 - 135*(b^3*d - a*b^2*e)*(e*x + d)^(7/2) + 18 9*(b^3*d^2 - 2*a*b^2*d*e + a^2*b*e^2)*(e*x + d)^(5/2) - 105*(b^3*d^3 - 3*a *b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*(e*x + d)^(3/2))/e^4
Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (84) = 168\).
Time = 0.28 (sec) , antiderivative size = 322, normalized size of antiderivative = 3.22 \[ \int (a+b x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{3} d + 105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{3} + \frac {315 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} b d}{e} + \frac {63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a b^{2} d}{e^{2}} + \frac {63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} b}{e} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b^{3} d}{e^{3}} + \frac {27 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a b^{2}}{e^{2}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b^{3}}{e^{3}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*a^3*d + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d) *a^3 + 315*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2*b*d/e + 63*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*b^2*d/e^2 + 63*( 3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a^2*b/e + 9*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35 *sqrt(e*x + d)*d^3)*b^3*d/e^3 + 27*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2) *d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a*b^2/e^2 + (35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d )^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*b^3/e^3)/e
Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int (a+b x) \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2\,b^3\,{\left (d+e\,x\right )}^{9/2}}{9\,e^4}-\frac {\left (6\,b^3\,d-6\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4}+\frac {2\,{\left (a\,e-b\,d\right )}^3\,{\left (d+e\,x\right )}^{3/2}}{3\,e^4}+\frac {6\,b\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4} \]