Integrand size = 29, antiderivative size = 210 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {a^3 A x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^2 (3 A b+a B) x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {3 a b (A b+a B) x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {b^2 (A b+3 a B) x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac {b^3 B x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]
1/5*a^3*A*x^5*((b*x+a)^2)^(1/2)/(b*x+a)+1/6*a^2*(3*A*b+B*a)*x^6*((b*x+a)^2 )^(1/2)/(b*x+a)+3/7*a*b*(A*b+B*a)*x^7*((b*x+a)^2)^(1/2)/(b*x+a)+1/8*b^2*(A *b+3*B*a)*x^8*((b*x+a)^2)^(1/2)/(b*x+a)+1/9*b^3*B*x^9*((b*x+a)^2)^(1/2)/(b *x+a)
Time = 1.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.41 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^5 \sqrt {(a+b x)^2} \left (84 a^3 (6 A+5 B x)+180 a^2 b x (7 A+6 B x)+135 a b^2 x^2 (8 A+7 B x)+35 b^3 x^3 (9 A+8 B x)\right )}{2520 (a+b x)} \]
(x^5*Sqrt[(a + b*x)^2]*(84*a^3*(6*A + 5*B*x) + 180*a^2*b*x*(7*A + 6*B*x) + 135*a*b^2*x^2*(8*A + 7*B*x) + 35*b^3*x^3*(9*A + 8*B*x)))/(2520*(a + b*x))
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 27, 85, 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.
\(\displaystyle \int x^4 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 x^4 (a+b x)^3 (A+B x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^4 (a+b x)^3 (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^3 B x^8+b^2 (A b+3 a B) x^7+3 a b (A b+a B) x^6+a^2 (3 A b+a B) x^5+a^3 A x^4\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{5} a^3 A x^5+\frac {1}{6} a^2 x^6 (a B+3 A b)+\frac {1}{8} b^2 x^8 (3 a B+A b)+\frac {3}{7} a b x^7 (a B+A b)+\frac {1}{9} b^3 B x^9\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((a^3*A*x^5)/5 + (a^2*(3*A*b + a*B)*x^6)/6 + (3*a*b*(A*b + a*B)*x^7)/7 + (b^2*(A*b + 3*a*B)*x^8)/8 + (b^3*B*x^9)/9))/ (a + b*x)
3.7.66.3.1 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[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
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 = 0.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(\frac {x^{5} \left (280 x^{4} B \,b^{3}+315 A \,b^{3} x^{3}+945 B a \,b^{2} x^{3}+1080 A a \,b^{2} x^{2}+1080 B \,a^{2} b \,x^{2}+1260 A \,a^{2} b x +420 a^{3} B x +504 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 \left (b x +a \right )^{3}}\) | \(92\) |
default | \(\frac {x^{5} \left (280 x^{4} B \,b^{3}+315 A \,b^{3} x^{3}+945 B a \,b^{2} x^{3}+1080 A a \,b^{2} x^{2}+1080 B \,a^{2} b \,x^{2}+1260 A \,a^{2} b x +420 a^{3} B x +504 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 \left (b x +a \right )^{3}}\) | \(92\) |
risch | \(\frac {b^{3} B \,x^{9} \sqrt {\left (b x +a \right )^{2}}}{9 b x +9 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{3}+3 B a \,b^{2}\right ) x^{8}}{8 b x +8 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A a \,b^{2}+3 B b \,a^{2}\right ) x^{7}}{7 b x +7 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{6}}{6 b x +6 a}+\frac {a^{3} A \,x^{5} \sqrt {\left (b x +a \right )^{2}}}{5 b x +5 a}\) | \(156\) |
1/2520*x^5*(280*B*b^3*x^4+315*A*b^3*x^3+945*B*a*b^2*x^3+1080*A*a*b^2*x^2+1 080*B*a^2*b*x^2+1260*A*a^2*b*x+420*B*a^3*x+504*A*a^3)*((b*x+a)^2)^(3/2)/(b *x+a)^3
Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{9} \, B b^{3} x^{9} + \frac {1}{5} \, A a^{3} x^{5} + \frac {1}{8} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{8} + \frac {3}{7} \, {\left (B a^{2} b + A a b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{6} \]
1/9*B*b^3*x^9 + 1/5*A*a^3*x^5 + 1/8*(3*B*a*b^2 + A*b^3)*x^8 + 3/7*(B*a^2*b + A*a*b^2)*x^7 + 1/6*(B*a^3 + 3*A*a^2*b)*x^6
Leaf count of result is larger than twice the leaf count of optimal. 5448 vs. \(2 (155) = 310\).
Time = 0.93 (sec) , antiderivative size = 5448, normalized size of antiderivative = 25.94 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\text {Too large to display} \]
Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*b**2*x**8/9 + x**7*(A*b**4 + 19*B*a*b**3/9)/(8*b**2) + x**6*(4*A*a*b**3 + 46*B*a**2*b**2/9 - 15*a*(A* b**4 + 19*B*a*b**3/9)/(8*b))/(7*b**2) + x**5*(6*A*a**2*b**2 + 4*B*a**3*b - 7*a**2*(A*b**4 + 19*B*a*b**3/9)/(8*b**2) - 13*a*(4*A*a*b**3 + 46*B*a**2*b **2/9 - 15*a*(A*b**4 + 19*B*a*b**3/9)/(8*b))/(7*b))/(6*b**2) + x**4*(4*A*a **3*b + B*a**4 - 6*a**2*(4*A*a*b**3 + 46*B*a**2*b**2/9 - 15*a*(A*b**4 + 19 *B*a*b**3/9)/(8*b))/(7*b**2) - 11*a*(6*A*a**2*b**2 + 4*B*a**3*b - 7*a**2*( A*b**4 + 19*B*a*b**3/9)/(8*b**2) - 13*a*(4*A*a*b**3 + 46*B*a**2*b**2/9 - 1 5*a*(A*b**4 + 19*B*a*b**3/9)/(8*b))/(7*b))/(6*b))/(5*b**2) + x**3*(A*a**4 - 5*a**2*(6*A*a**2*b**2 + 4*B*a**3*b - 7*a**2*(A*b**4 + 19*B*a*b**3/9)/(8* b**2) - 13*a*(4*A*a*b**3 + 46*B*a**2*b**2/9 - 15*a*(A*b**4 + 19*B*a*b**3/9 )/(8*b))/(7*b))/(6*b**2) - 9*a*(4*A*a**3*b + B*a**4 - 6*a**2*(4*A*a*b**3 + 46*B*a**2*b**2/9 - 15*a*(A*b**4 + 19*B*a*b**3/9)/(8*b))/(7*b**2) - 11*a*( 6*A*a**2*b**2 + 4*B*a**3*b - 7*a**2*(A*b**4 + 19*B*a*b**3/9)/(8*b**2) - 13 *a*(4*A*a*b**3 + 46*B*a**2*b**2/9 - 15*a*(A*b**4 + 19*B*a*b**3/9)/(8*b))/( 7*b))/(6*b))/(5*b))/(4*b**2) + x**2*(-4*a**2*(4*A*a**3*b + B*a**4 - 6*a**2 *(4*A*a*b**3 + 46*B*a**2*b**2/9 - 15*a*(A*b**4 + 19*B*a*b**3/9)/(8*b))/(7* b**2) - 11*a*(6*A*a**2*b**2 + 4*B*a**3*b - 7*a**2*(A*b**4 + 19*B*a*b**3/9) /(8*b**2) - 13*a*(4*A*a*b**3 + 46*B*a**2*b**2/9 - 15*a*(A*b**4 + 19*B*a*b* *3/9)/(8*b))/(7*b))/(6*b))/(5*b**2) - 7*a*(A*a**4 - 5*a**2*(6*A*a**2*b*...
Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (145) = 290\).
Time = 0.21 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.72 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x^{4}}{9 \, b^{2}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a x^{3}}{72 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A x^{3}}{8 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{5} x}{4 \, b^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{4} x}{4 \, b^{4}} + \frac {37 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{2} x^{2}}{168 \, b^{4}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a x^{2}}{56 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{6}}{4 \, b^{6}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{5}}{4 \, b^{5}} - \frac {121 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{3} x}{504 \, b^{5}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{2} x}{56 \, b^{4}} + \frac {125 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{4}}{504 \, b^{6}} - \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{3}}{280 \, b^{5}} \]
1/9*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*x^4/b^2 - 13/72*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a*x^3/b^3 + 1/8*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*x^3/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^5*x/b^5 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^4*x/b^4 + 37/168*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^2*x^2 /b^4 - 11/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a*x^2/b^3 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^6/b^6 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^5 /b^5 - 121/504*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^3*x/b^5 + 13/56*(b^2*x^ 2 + 2*a*b*x + a^2)^(5/2)*A*a^2*x/b^4 + 125/504*(b^2*x^2 + 2*a*b*x + a^2)^( 5/2)*B*a^4/b^6 - 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a^3/b^5
Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.71 \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{9} \, B b^{3} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{8} \, B a b^{2} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{8} \, A b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, B a^{2} b x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, A a b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, B a^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{2} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A a^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (5 \, B a^{9} - 9 \, A a^{8} b\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, b^{6}} \]
1/9*B*b^3*x^9*sgn(b*x + a) + 3/8*B*a*b^2*x^8*sgn(b*x + a) + 1/8*A*b^3*x^8* sgn(b*x + a) + 3/7*B*a^2*b*x^7*sgn(b*x + a) + 3/7*A*a*b^2*x^7*sgn(b*x + a) + 1/6*B*a^3*x^6*sgn(b*x + a) + 1/2*A*a^2*b*x^6*sgn(b*x + a) + 1/5*A*a^3*x ^5*sgn(b*x + a) - 1/2520*(5*B*a^9 - 9*A*a^8*b)*sgn(b*x + a)/b^6
Timed out. \[ \int x^4 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^4\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]