Integrand size = 31, antiderivative size = 135 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(A b-a B) (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^3}+\frac {(b B d+A b e-2 a B e) (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^3}+\frac {B e (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3} \]
1/4*(A*b-B*a)*(-a*e+b*d)*(b*x+a)^3*((b*x+a)^2)^(1/2)/b^3+1/5*(A*b*e-2*B*a* e+B*b*d)*(b*x+a)^4*((b*x+a)^2)^(1/2)/b^3+1/6*B*e*(b*x+a)^5*((b*x+a)^2)^(1/ 2)/b^3
Time = 1.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x \sqrt {(a+b x)^2} \left (10 a^3 (3 A (2 d+e x)+B x (3 d+2 e x))+3 a b^2 x^2 (5 A (4 d+3 e x)+3 B x (5 d+4 e x))+15 a^2 b x (B x (4 d+3 e x)+A (6 d+4 e x))+b^3 x^3 (3 A (5 d+4 e x)+2 B x (6 d+5 e x))\right )}{60 (a+b x)} \]
(x*Sqrt[(a + b*x)^2]*(10*a^3*(3*A*(2*d + e*x) + B*x*(3*d + 2*e*x)) + 3*a*b ^2*x^2*(5*A*(4*d + 3*e*x) + 3*B*x*(5*d + 4*e*x)) + 15*a^2*b*x*(B*x*(4*d + 3*e*x) + A*(6*d + 4*e*x)) + b^3*x^3*(3*A*(5*d + 4*e*x) + 2*B*x*(6*d + 5*e* x))))/(60*(a + b*x))
Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1187, 27, 86, 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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) (d+e x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 (a+b x)^3 (A+B x) (d+e x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^3 (A+B x) (d+e x)dx}{a+b x}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B e (a+b x)^5}{b^2}+\frac {(b B d+A b e-2 a B e) (a+b x)^4}{b^2}+\frac {(A b-a B) (b d-a e) (a+b x)^3}{b^2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {(a+b x)^5 (-2 a B e+A b e+b B d)}{5 b^3}+\frac {(a+b x)^4 (A b-a B) (b d-a e)}{4 b^3}+\frac {B e (a+b x)^6}{6 b^3}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(((A*b - a*B)*(b*d - a*e)*(a + b*x)^4)/(4*b ^3) + ((b*B*d + A*b*e - 2*a*B*e)*(a + b*x)^5)/(5*b^3) + (B*e*(a + b*x)^6)/ (6*b^3)))/(a + b*x)
3.18.24.3.1 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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[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, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.33
method | result | size |
gosper | \(\frac {x \left (10 x^{5} b^{3} B e +12 x^{4} A \,b^{3} e +36 x^{4} B e \,b^{2} a +12 x^{4} B \,b^{3} d +45 x^{3} A a \,b^{2} e +15 x^{3} A d \,b^{3}+45 x^{3} B e b \,a^{2}+45 x^{3} B a \,b^{2} d +60 x^{2} A \,a^{2} b e +60 x^{2} d A \,b^{2} a +20 x^{2} B e \,a^{3}+60 x^{2} B \,a^{2} b d +30 x A \,a^{3} e +90 x d A b \,a^{2}+30 x B \,a^{3} d +60 d A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (b x +a \right )^{3}}\) | \(180\) |
default | \(\frac {x \left (10 x^{5} b^{3} B e +12 x^{4} A \,b^{3} e +36 x^{4} B e \,b^{2} a +12 x^{4} B \,b^{3} d +45 x^{3} A a \,b^{2} e +15 x^{3} A d \,b^{3}+45 x^{3} B e b \,a^{2}+45 x^{3} B a \,b^{2} d +60 x^{2} A \,a^{2} b e +60 x^{2} d A \,b^{2} a +20 x^{2} B e \,a^{3}+60 x^{2} B \,a^{2} b d +30 x A \,a^{3} e +90 x d A b \,a^{2}+30 x B \,a^{3} d +60 d A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (b x +a \right )^{3}}\) | \(180\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, x^{6} b^{3} B e}{6 b x +6 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b^{3} \left (A e +B d \right )+3 B e \,b^{2} a \right ) x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A d \,b^{3}+3 \left (A e +B d \right ) b^{2} a +3 B e b \,a^{2}\right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 d A \,b^{2} a +3 \left (A e +B d \right ) b \,a^{2}+B e \,a^{3}\right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 d A b \,a^{2}+\left (A e +B d \right ) a^{3}\right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} A d x}{b x +a}\) | \(231\) |
1/60*x*(10*B*b^3*e*x^5+12*A*b^3*e*x^4+36*B*a*b^2*e*x^4+12*B*b^3*d*x^4+45*A *a*b^2*e*x^3+15*A*b^3*d*x^3+45*B*a^2*b*e*x^3+45*B*a*b^2*d*x^3+60*A*a^2*b*e *x^2+60*A*a*b^2*d*x^2+20*B*a^3*e*x^2+60*B*a^2*b*d*x^2+30*A*a^3*e*x+90*A*a^ 2*b*d*x+30*B*a^3*d*x+60*A*a^3*d)*((b*x+a)^2)^(3/2)/(b*x+a)^3
Time = 0.40 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.08 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{6} \, B b^{3} e x^{6} + A a^{3} d x + \frac {1}{5} \, {\left (B b^{3} d + {\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} x^{5} + \frac {1}{4} \, {\left ({\left (3 \, B a b^{2} + A b^{3}\right )} d + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e\right )} x^{4} + \frac {1}{3} \, {\left (3 \, {\left (B a^{2} b + A a b^{2}\right )} d + {\left (B a^{3} + 3 \, A a^{2} b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a^{3} e + {\left (B a^{3} + 3 \, A a^{2} b\right )} d\right )} x^{2} \]
1/6*B*b^3*e*x^6 + A*a^3*d*x + 1/5*(B*b^3*d + (3*B*a*b^2 + A*b^3)*e)*x^5 + 1/4*((3*B*a*b^2 + A*b^3)*d + 3*(B*a^2*b + A*a*b^2)*e)*x^4 + 1/3*(3*(B*a^2* b + A*a*b^2)*d + (B*a^3 + 3*A*a^2*b)*e)*x^3 + 1/2*(A*a^3*e + (B*a^3 + 3*A* a^2*b)*d)*x^2
Leaf count of result is larger than twice the leaf count of optimal. 2076 vs. \(2 (104) = 208\).
Time = 1.45 (sec) , antiderivative size = 2076, normalized size of antiderivative = 15.38 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\text {Too large to display} \]
Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*b**2*e*x**5/6 + x**4*(A*b** 4*e + 13*B*a*b**3*e/6 + B*b**4*d)/(5*b**2) + x**3*(4*A*a*b**3*e + A*b**4*d + 31*B*a**2*b**2*e/6 + 4*B*a*b**3*d - 9*a*(A*b**4*e + 13*B*a*b**3*e/6 + B *b**4*d)/(5*b))/(4*b**2) + x**2*(6*A*a**2*b**2*e + 4*A*a*b**3*d + 4*B*a**3 *b*e + 6*B*a**2*b**2*d - 4*a**2*(A*b**4*e + 13*B*a*b**3*e/6 + B*b**4*d)/(5 *b**2) - 7*a*(4*A*a*b**3*e + A*b**4*d + 31*B*a**2*b**2*e/6 + 4*B*a*b**3*d - 9*a*(A*b**4*e + 13*B*a*b**3*e/6 + B*b**4*d)/(5*b))/(4*b))/(3*b**2) + x*( 4*A*a**3*b*e + 6*A*a**2*b**2*d + B*a**4*e + 4*B*a**3*b*d - 3*a**2*(4*A*a*b **3*e + A*b**4*d + 31*B*a**2*b**2*e/6 + 4*B*a*b**3*d - 9*a*(A*b**4*e + 13* B*a*b**3*e/6 + B*b**4*d)/(5*b))/(4*b**2) - 5*a*(6*A*a**2*b**2*e + 4*A*a*b* *3*d + 4*B*a**3*b*e + 6*B*a**2*b**2*d - 4*a**2*(A*b**4*e + 13*B*a*b**3*e/6 + B*b**4*d)/(5*b**2) - 7*a*(4*A*a*b**3*e + A*b**4*d + 31*B*a**2*b**2*e/6 + 4*B*a*b**3*d - 9*a*(A*b**4*e + 13*B*a*b**3*e/6 + B*b**4*d)/(5*b))/(4*b)) /(3*b))/(2*b**2) + (A*a**4*e + 4*A*a**3*b*d + B*a**4*d - 2*a**2*(6*A*a**2* b**2*e + 4*A*a*b**3*d + 4*B*a**3*b*e + 6*B*a**2*b**2*d - 4*a**2*(A*b**4*e + 13*B*a*b**3*e/6 + B*b**4*d)/(5*b**2) - 7*a*(4*A*a*b**3*e + A*b**4*d + 31 *B*a**2*b**2*e/6 + 4*B*a*b**3*d - 9*a*(A*b**4*e + 13*B*a*b**3*e/6 + B*b**4 *d)/(5*b))/(4*b))/(3*b**2) - 3*a*(4*A*a**3*b*e + 6*A*a**2*b**2*d + B*a**4* e + 4*B*a**3*b*d - 3*a**2*(4*A*a*b**3*e + A*b**4*d + 31*B*a**2*b**2*e/6 + 4*B*a*b**3*d - 9*a*(A*b**4*e + 13*B*a*b**3*e/6 + B*b**4*d)/(5*b))/(4*b*...
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (96) = 192\).
Time = 0.20 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.88 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A d x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} e x}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a d}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3} e}{4 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (B d + A e\right )} a x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B e x}{6 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (B d + A e\right )} a^{2}}{4 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a e}{30 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (B d + A e\right )}}{5 \, b^{2}} \]
1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*d*x + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^ (3/2)*B*a^2*e*x/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a*d/b + 1/4*(b ^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^3*e/b^3 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^ (3/2)*(B*d + A*e)*a*x/b + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*e*x/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(B*d + A*e)*a^2/b^2 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a*e/b^3 + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*(B*d + A*e)/b^2
Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (96) = 192\).
Time = 0.30 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.25 \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{6} \, B b^{3} e x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B b^{3} d x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, B a b^{2} e x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b^{3} e x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a b^{2} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a^{2} b e x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, A a b^{2} e x^{4} \mathrm {sgn}\left (b x + a\right ) + B a^{2} b d x^{3} \mathrm {sgn}\left (b x + a\right ) + A a b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a^{3} e x^{3} \mathrm {sgn}\left (b x + a\right ) + A a^{2} b e x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a^{3} d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, A a^{2} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{3} e x^{2} \mathrm {sgn}\left (b x + a\right ) + A a^{3} d x \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (3 \, B a^{5} b d - 15 \, A a^{4} b^{2} d - B a^{6} e + 3 \, A a^{5} b e\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{3}} \]
1/6*B*b^3*e*x^6*sgn(b*x + a) + 1/5*B*b^3*d*x^5*sgn(b*x + a) + 3/5*B*a*b^2* e*x^5*sgn(b*x + a) + 1/5*A*b^3*e*x^5*sgn(b*x + a) + 3/4*B*a*b^2*d*x^4*sgn( b*x + a) + 1/4*A*b^3*d*x^4*sgn(b*x + a) + 3/4*B*a^2*b*e*x^4*sgn(b*x + a) + 3/4*A*a*b^2*e*x^4*sgn(b*x + a) + B*a^2*b*d*x^3*sgn(b*x + a) + A*a*b^2*d*x ^3*sgn(b*x + a) + 1/3*B*a^3*e*x^3*sgn(b*x + a) + A*a^2*b*e*x^3*sgn(b*x + a ) + 1/2*B*a^3*d*x^2*sgn(b*x + a) + 3/2*A*a^2*b*d*x^2*sgn(b*x + a) + 1/2*A* a^3*e*x^2*sgn(b*x + a) + A*a^3*d*x*sgn(b*x + a) - 1/60*(3*B*a^5*b*d - 15*A *a^4*b^2*d - B*a^6*e + 3*A*a^5*b*e)*sgn(b*x + a)/b^3
Timed out. \[ \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int \left (A+B\,x\right )\,\left (d+e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]