Integrand size = 28, antiderivative size = 92 \[ \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {(b d-a e) (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^2 (a+b x)}+\frac {b (d+e x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x)} \]
-1/4*(-a*e+b*d)*(e*x+d)^4*((b*x+a)^2)^(1/2)/e^2/(b*x+a)+1/5*b*(e*x+d)^5*(( b*x+a)^2)^(1/2)/e^2/(b*x+a)
Time = 1.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.97 \[ \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x \sqrt {(a+b x)^2} \left (5 a \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+b x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )\right )}{20 (a+b x)} \]
(x*Sqrt[(a + b*x)^2]*(5*a*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + b* x*(10*d^3 + 20*d^2*e*x + 15*d*e^2*x^2 + 4*e^3*x^3)))/(20*(a + b*x))
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 27, 49, 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.
\(\displaystyle \int \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b (a+b x) (d+e x)^3dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x) (d+e x)^3dx}{a+b x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b (d+e x)^4}{e}+\frac {(a e-b d) (d+e x)^3}{e}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b (d+e x)^5}{5 e^2}-\frac {(d+e x)^4 (b d-a e)}{4 e^2}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/4*((b*d - a*e)*(d + e*x)^4)/e^2 + (b*(d + e*x)^5)/(5*e^2)))/(a + b*x)
3.16.44.3.1 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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.98
method | result | size |
gosper | \(\frac {x \left (4 b \,e^{3} x^{4}+5 x^{3} a \,e^{3}+15 x^{3} b d \,e^{2}+20 a d \,e^{2} x^{2}+20 b \,d^{2} e \,x^{2}+30 x a e \,d^{2}+10 x b \,d^{3}+20 a \,d^{3}\right ) \sqrt {\left (b x +a \right )^{2}}}{20 b x +20 a}\) | \(90\) |
default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (-4 e^{3} x^{3} b^{3}+3 x^{2} a \,b^{2} e^{3}-15 x^{2} b^{3} d \,e^{2}-2 a^{2} b \,e^{3} x +10 x a \,b^{2} d \,e^{2}-20 b^{3} d^{2} e x +a^{3} e^{3}-5 a^{2} b d \,e^{2}+10 a \,b^{2} d^{2} e -10 b^{3} d^{3}\right )}{20 b^{4}}\) | \(121\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b \,e^{3} x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a \,e^{3}+3 b d \,e^{2}\right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a d \,e^{2}+3 b \,d^{2} e \right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a e \,d^{2}+b \,d^{3}\right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a \,d^{3} x}{b x +a}\) | \(153\) |
1/20*x*(4*b*e^3*x^4+5*a*e^3*x^3+15*b*d*e^2*x^3+20*a*d*e^2*x^2+20*b*d^2*e*x ^2+30*a*d^2*e*x+10*b*d^3*x+20*a*d^3)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.75 \[ \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{5} \, b e^{3} x^{5} + a d^{3} x + \frac {1}{4} \, {\left (3 \, b d e^{2} + a e^{3}\right )} x^{4} + {\left (b d^{2} e + a d e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b d^{3} + 3 \, a d^{2} e\right )} x^{2} \]
1/5*b*e^3*x^5 + a*d^3*x + 1/4*(3*b*d*e^2 + a*e^3)*x^4 + (b*d^2*e + a*d*e^2 )*x^3 + 1/2*(b*d^3 + 3*a*d^2*e)*x^2
Time = 1.64 (sec) , antiderivative size = 464, normalized size of antiderivative = 5.04 \[ \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=d^{3} \left (\begin {cases} \left (\frac {a}{2 b} + \frac {x}{2}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {\left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3 a b} & \text {for}\: a b \neq 0 \\x \sqrt {a^{2}} & \text {otherwise} \end {cases}\right ) + 3 d^{2} e \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{2}}{6 b^{2}} + \frac {a x}{6 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \sqrt {a^{2}}}{2} & \text {otherwise} \end {cases}\right ) + 3 d e^{2} \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} x}{12 b^{2}} + \frac {a x^{2}}{12 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{4}}{20 b^{4}} + \frac {a^{3} x}{20 b^{3}} - \frac {a^{2} x^{2}}{20 b^{2}} + \frac {a x^{3}}{20 b} + \frac {x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{6} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {3 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {3 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4} \sqrt {a^{2}}}{4} & \text {otherwise} \end {cases}\right ) \]
d**3*Piecewise(((a/(2*b) + x/2)*sqrt(a**2 + 2*a*b*x + b**2*x**2), Ne(b**2, 0)), ((a**2 + 2*a*b*x)**(3/2)/(3*a*b), Ne(a*b, 0)), (x*sqrt(a**2), True)) + 3*d**2*e*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**2/(6*b**2) + a*x/(6*b) + x**2/3), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a* *2 + 2*a*b*x)**(5/2)/5)/(2*a**2*b**2), Ne(a*b, 0)), (x**2*sqrt(a**2)/2, Tr ue)) + 3*d*e**2*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(a**3/(12*b**3 ) - a**2*x/(12*b**2) + a*x**2/(12*b) + x**3/4), Ne(b**2, 0)), ((a**4*(a**2 + 2*a*b*x)**(3/2)/3 - 2*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x) **(7/2)/7)/(4*a**3*b**3), Ne(a*b, 0)), (x**3*sqrt(a**2)/3, True)) + e**3*P iecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**4/(20*b**4) + a**3*x/(20*b **3) - a**2*x**2/(20*b**2) + a*x**3/(20*b) + x**4/5), Ne(b**2, 0)), ((-a** 6*(a**2 + 2*a*b*x)**(3/2)/3 + 3*a**4*(a**2 + 2*a*b*x)**(5/2)/5 - 3*a**2*(a **2 + 2*a*b*x)**(7/2)/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(8*a**4*b**4), Ne(a*b , 0)), (x**4*sqrt(a**2)/4, True))
Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (66) = 132\).
Time = 0.22 (sec) , antiderivative size = 399, normalized size of antiderivative = 4.34 \[ \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{3} x - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{2} e x}{2 \, b} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d e^{2} x}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} e^{3} x}{2 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e^{3} x^{2}}{5 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{3}}{2 \, b} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d^{2} e}{2 \, b^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} d e^{2}}{2 \, b^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} e^{3}}{2 \, b^{4}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d e^{2} x}{4 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e^{3} x}{20 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{2} e}{b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d e^{2}}{4 \, b^{3}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e^{3}}{20 \, b^{4}} \]
1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*d^3*x - 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 )*a*d^2*e*x/b + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*d*e^2*x/b^2 - 1/2*sq rt(b^2*x^2 + 2*a*b*x + a^2)*a^3*e^3*x/b^3 + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^ (3/2)*e^3*x^2/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d^3/b - 3/2*sqrt(b ^2*x^2 + 2*a*b*x + a^2)*a^2*d^2*e/b^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)* a^3*d*e^2/b^3 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4*e^3/b^4 + 3/4*(b^2*x ^2 + 2*a*b*x + a^2)^(3/2)*d*e^2*x/b^2 - 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/ 2)*a*e^3*x/b^3 + (b^2*x^2 + 2*a*b*x + a^2)^(3/2)*d^2*e/b^2 - 5/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d*e^2/b^3 + 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)* a^2*e^3/b^4
Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (66) = 132\).
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.88 \[ \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{5} \, b e^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, b d e^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a e^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + b d^{2} e x^{3} \mathrm {sgn}\left (b x + a\right ) + a d e^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b d^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a d^{2} e x^{2} \mathrm {sgn}\left (b x + a\right ) + a d^{3} x \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (10 \, a^{2} b^{3} d^{3} - 10 \, a^{3} b^{2} d^{2} e + 5 \, a^{4} b d e^{2} - a^{5} e^{3}\right )} \mathrm {sgn}\left (b x + a\right )}{20 \, b^{4}} \]
1/5*b*e^3*x^5*sgn(b*x + a) + 3/4*b*d*e^2*x^4*sgn(b*x + a) + 1/4*a*e^3*x^4* sgn(b*x + a) + b*d^2*e*x^3*sgn(b*x + a) + a*d*e^2*x^3*sgn(b*x + a) + 1/2*b *d^3*x^2*sgn(b*x + a) + 3/2*a*d^2*e*x^2*sgn(b*x + a) + a*d^3*x*sgn(b*x + a ) + 1/20*(10*a^2*b^3*d^3 - 10*a^3*b^2*d^2*e + 5*a^4*b*d*e^2 - a^5*e^3)*sgn (b*x + a)/b^4
Time = 10.24 (sec) , antiderivative size = 377, normalized size of antiderivative = 4.10 \[ \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=d^3\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {e^3\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}-\frac {a^2\,e^3\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{60\,b^6}-\frac {7\,a\,e^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{60\,b^4}+\frac {d^2\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,b^4}+\frac {3\,d\,e^2\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {5\,a\,d\,e^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{32\,b^5}-\frac {3\,a^2\,d\,e^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2} \]
d^3*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2) + (e^3*x^2*(a^2 + b^2* x^2 + 2*a*b*x)^(3/2))/(5*b^2) - (a^2*e^3*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b ^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(60*b^6) - (7*a*e^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a^3 - 5*a*b^2*x^2 + 3*b*x*(a^2 + b^2*x^2 + 2*a *b*x) - 4*a^2*b*x))/(60*b^4) + (d^2*e*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(8*b^4) + (3*d*e^2*x*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) - (5*a*d*e^2*(8*b^2*(a^2 + b^2*x^2) - 12 *a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(32*b^5) - (3*a^2*d *e^2*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b^2)