Integrand size = 31, antiderivative size = 96 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx=-\frac {2 (b d-a e)^3 \sqrt {d+e x}}{e^4}+\frac {2 b (b d-a e)^2 (d+e x)^{3/2}}{e^4}-\frac {6 b^2 (b d-a e) (d+e x)^{5/2}}{5 e^4}+\frac {2 b^3 (d+e x)^{7/2}}{7 e^4} \]
2*b*(-a*e+b*d)^2*(e*x+d)^(3/2)/e^4-6/5*b^2*(-a*e+b*d)*(e*x+d)^(5/2)/e^4+2/ 7*b^3*(e*x+d)^(7/2)/e^4-2*(-a*e+b*d)^3*(e*x+d)^(1/2)/e^4
Time = 0.02 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (35 a^3 e^3+35 a^2 b e^2 (-2 d+e x)+7 a b^2 e \left (8 d^2-4 d e x+3 e^2 x^2\right )+b^3 \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )\right )}{35 e^4} \]
(2*Sqrt[d + e*x]*(35*a^3*e^3 + 35*a^2*b*e^2*(-2*d + e*x) + 7*a*b^2*e*(8*d^ 2 - 4*d*e*x + 3*e^2*x^2) + b^3*(-16*d^3 + 8*d^2*e*x - 6*d*e^2*x^2 + 5*e^3* x^3)))/(35*e^4)
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.
\(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle \frac {\int \frac {b^2 (a+b x)^3}{\sqrt {d+e x}}dx}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(a+b x)^3}{\sqrt {d+e x}}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \int \left (-\frac {3 b^2 (d+e x)^{3/2} (b d-a e)}{e^3}+\frac {3 b \sqrt {d+e x} (b d-a e)^2}{e^3}+\frac {(a e-b d)^3}{e^3 \sqrt {d+e x}}+\frac {b^3 (d+e x)^{5/2}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 b^2 (d+e x)^{5/2} (b d-a e)}{5 e^4}+\frac {2 b (d+e x)^{3/2} (b d-a e)^2}{e^4}-\frac {2 \sqrt {d+e x} (b d-a e)^3}{e^4}+\frac {2 b^3 (d+e x)^{7/2}}{7 e^4}\) |
(-2*(b*d - a*e)^3*Sqrt[d + e*x])/e^4 + (2*b*(b*d - a*e)^2*(d + e*x)^(3/2)) /e^4 - (6*b^2*(b*d - a*e)*(d + e*x)^(5/2))/(5*e^4) + (2*b^3*(d + e*x)^(7/2 ))/(7*e^4)
3.21.46.3.1 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 = 92, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {e x +d}\, \left (\left (\frac {3}{5} a \,b^{2} x^{2}+b \,a^{2} x +\frac {1}{7} x^{3} b^{3}+a^{3}\right ) e^{3}-2 \left (\frac {3}{35} b^{2} x^{2}+\frac {2}{5} a b x +a^{2}\right ) b d \,e^{2}+\frac {8 \left (\frac {b x}{7}+a \right ) b^{2} d^{2} e}{5}-\frac {16 b^{3} d^{3}}{35}\right )}{e^{4}}\) | \(92\) |
gosper | \(\frac {2 \left (5 b^{3} x^{3} e^{3}+21 x^{2} a \,b^{2} e^{3}-6 x^{2} b^{3} d \,e^{2}+35 x \,a^{2} b \,e^{3}-28 x a \,b^{2} d \,e^{2}+8 x \,b^{3} d^{2} e +35 a^{3} e^{3}-70 a^{2} b d \,e^{2}+56 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \sqrt {e x +d}}{35 e^{4}}\) | \(116\) |
trager | \(\frac {2 \left (5 b^{3} x^{3} e^{3}+21 x^{2} a \,b^{2} e^{3}-6 x^{2} b^{3} d \,e^{2}+35 x \,a^{2} b \,e^{3}-28 x a \,b^{2} d \,e^{2}+8 x \,b^{3} d^{2} e +35 a^{3} e^{3}-70 a^{2} b d \,e^{2}+56 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \sqrt {e x +d}}{35 e^{4}}\) | \(116\) |
risch | \(\frac {2 \left (5 b^{3} x^{3} e^{3}+21 x^{2} a \,b^{2} e^{3}-6 x^{2} b^{3} d \,e^{2}+35 x \,a^{2} b \,e^{3}-28 x a \,b^{2} d \,e^{2}+8 x \,b^{3} d^{2} e +35 a^{3} e^{3}-70 a^{2} b d \,e^{2}+56 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \sqrt {e x +d}}{35 e^{4}}\) | \(116\) |
derivativedivides | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a e -b d \right ) b^{2}+b \left (2 a b e -2 b^{2} d \right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a e -b d \right ) \left (2 a b e -2 b^{2} d \right )+b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -b d \right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {e x +d}}{e^{4}}\) | \(146\) |
default | \(\frac {\frac {2 b^{3} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a e -b d \right ) b^{2}+b \left (2 a b e -2 b^{2} d \right )\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a e -b d \right ) \left (2 a b e -2 b^{2} d \right )+b \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -b d \right ) \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {e x +d}}{e^{4}}\) | \(146\) |
2*(e*x+d)^(1/2)*((3/5*a*b^2*x^2+b*a^2*x+1/7*x^3*b^3+a^3)*e^3-2*(3/35*b^2*x ^2+2/5*a*b*x+a^2)*b*d*e^2+8/5*(1/7*b*x+a)*b^2*d^2*e-16/35*b^3*d^3)/e^4
Time = 0.37 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 56 \, a b^{2} d^{2} e - 70 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + {\left (8 \, b^{3} d^{2} e - 28 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{35 \, e^{4}} \]
2/35*(5*b^3*e^3*x^3 - 16*b^3*d^3 + 56*a*b^2*d^2*e - 70*a^2*b*d*e^2 + 35*a^ 3*e^3 - 3*(2*b^3*d*e^2 - 7*a*b^2*e^3)*x^2 + (8*b^3*d^2*e - 28*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d)/e^4
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (88) = 176\).
Time = 1.00 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.02 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (3 a b^{2} e - 3 b^{3} d\right )}{5 e^{3}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (3 a^{2} b e^{2} - 6 a b^{2} d e + 3 b^{3} d^{2}\right )}{3 e^{3}} + \frac {\sqrt {d + e x} \left (a^{3} e^{3} - 3 a^{2} b d e^{2} + 3 a b^{2} d^{2} e - b^{3} d^{3}\right )}{e^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {a^{3} b x + \frac {3 a^{2} b^{2} x^{2}}{2} + a b^{3} x^{3} + \frac {b^{4} x^{4}}{4}}{b} & \text {otherwise} \end {cases}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
Piecewise((2*(b**3*(d + e*x)**(7/2)/(7*e**3) + (d + e*x)**(5/2)*(3*a*b**2* e - 3*b**3*d)/(5*e**3) + (d + e*x)**(3/2)*(3*a**2*b*e**2 - 6*a*b**2*d*e + 3*b**3*d**2)/(3*e**3) + sqrt(d + e*x)*(a**3*e**3 - 3*a**2*b*d*e**2 + 3*a*b **2*d**2*e - b**3*d**3)/e**3)/e, Ne(e, 0)), (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))/sq rt(d), True))
Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{3} - 21 \, {\left (b^{3} d - a b^{2} e\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}} - 35 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {e x + d}\right )}}{35 \, e^{4}} \]
2/35*(5*(e*x + d)^(7/2)*b^3 - 21*(b^3*d - a*b^2*e)*(e*x + d)^(5/2) + 35*(b ^3*d^2 - 2*a*b^2*d*e + a^2*b*e^2)*(e*x + d)^(3/2) - 35*(b^3*d^3 - 3*a*b^2* d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(e*x + d))/e^4
Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (35 \, \sqrt {e x + d} a^{3} + \frac {35 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} b}{e} + \frac {7 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a b^{2}}{e^{2}} + \frac {{\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b^{3}}{e^{3}}\right )}}{35 \, e} \]
2/35*(35*sqrt(e*x + d)*a^3 + 35*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2* b/e + 7*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)* a*b^2/e^2 + (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/e^3)/e
Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )}{\sqrt {d+e x}} \, dx=\frac {2\,b^3\,{\left (d+e\,x\right )}^{7/2}}{7\,e^4}-\frac {\left (6\,b^3\,d-6\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}+\frac {2\,{\left (a\,e-b\,d\right )}^3\,\sqrt {d+e\,x}}{e^4}+\frac {2\,b\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{e^4} \]