Integrand size = 29, antiderivative size = 210 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {a^3 A x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {a^2 (3 A b+a B) x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a b (A b+a B) x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {b^2 (A b+3 a B) x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {b^3 B x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)} \] Output:
a^3*A*x^4*((b*x+a)^2)^(1/2)/(4*b*x+4*a)+a^2*(3*A*b+B*a)*x^5*((b*x+a)^2)^(1 /2)/(5*b*x+5*a)+a*b*(A*b+B*a)*x^6*((b*x+a)^2)^(1/2)/(2*b*x+2*a)+b^2*(A*b+3 *B*a)*x^7*((b*x+a)^2)^(1/2)/(7*b*x+7*a)+b^3*B*x^8*((b*x+a)^2)^(1/2)/(8*b*x +8*a)
Time = 1.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.41 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^4 \sqrt {(a+b x)^2} \left (14 a^3 (5 A+4 B x)+28 a^2 b x (6 A+5 B x)+20 a b^2 x^2 (7 A+6 B x)+5 b^3 x^3 (8 A+7 B x)\right )}{280 (a+b x)} \] Input:
Integrate[x^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
(x^4*Sqrt[(a + b*x)^2]*(14*a^3*(5*A + 4*B*x) + 28*a^2*b*x*(6*A + 5*B*x) + 20*a*b^2*x^2*(7*A + 6*B*x) + 5*b^3*x^3*(8*A + 7*B*x)))/(280*(a + b*x))
Time = 0.44 (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.
\(\displaystyle \int x^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 x^3 (a+b x)^3 (A+B x)dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^3 (a+b x)^3 (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b^3 B x^7+b^2 (A b+3 a B) x^6+3 a b (A b+a B) x^5+a^2 (3 A b+a B) x^4+a^3 A x^3\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{4} a^3 A x^4+\frac {1}{5} a^2 x^5 (a B+3 A b)+\frac {1}{7} b^2 x^7 (3 a B+A b)+\frac {1}{2} a b x^6 (a B+A b)+\frac {1}{8} b^3 B x^8\right )}{a+b x}\) |
Input:
Int[x^3*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((a^3*A*x^4)/4 + (a^2*(3*A*b + a*B)*x^5)/5 + (a*b*(A*b + a*B)*x^6)/2 + (b^2*(A*b + 3*a*B)*x^7)/7 + (b^3*B*x^8)/8))/(a + b*x)
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 = 1.01 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.44
method | result | size |
gosper | \(\frac {x^{4} \left (35 B \,b^{3} x^{4}+40 A \,b^{3} x^{3}+120 B a \,b^{2} x^{3}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+168 A \,a^{2} b x +56 B \,a^{3} x +70 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 \left (b x +a \right )^{3}}\) | \(92\) |
default | \(\frac {x^{4} \left (35 B \,b^{3} x^{4}+40 A \,b^{3} x^{3}+120 B a \,b^{2} x^{3}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+168 A \,a^{2} b x +56 B \,a^{3} x +70 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 \left (b x +a \right )^{3}}\) | \(92\) |
orering | \(\frac {x^{4} \left (35 B \,b^{3} x^{4}+40 A \,b^{3} x^{3}+120 B a \,b^{2} x^{3}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+168 A \,a^{2} b x +56 B \,a^{3} x +70 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{280 \left (b x +a \right )^{3}}\) | \(101\) |
risch | \(\frac {b^{3} B \,x^{8} \sqrt {\left (b x +a \right )^{2}}}{8 b x +8 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (A \,b^{3}+3 B a \,b^{2}\right ) x^{7}}{7 b x +7 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A a \,b^{2}+3 B \,a^{2} b \right ) x^{6}}{6 b x +6 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A \,a^{2} b +B \,a^{3}\right ) x^{5}}{5 b x +5 a}+\frac {a^{3} A \,x^{4} \sqrt {\left (b x +a \right )^{2}}}{4 b x +4 a}\) | \(156\) |
Input:
int(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/280*x^4*(35*B*b^3*x^4+40*A*b^3*x^3+120*B*a*b^2*x^3+140*A*a*b^2*x^2+140*B *a^2*b*x^2+168*A*a^2*b*x+56*B*a^3*x+70*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3
Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{8} \, B b^{3} x^{8} + \frac {1}{4} \, A a^{3} x^{4} + \frac {1}{7} \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, {\left (B a^{2} b + A a b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{5} \] Input:
integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
Output:
1/8*B*b^3*x^8 + 1/4*A*a^3*x^4 + 1/7*(3*B*a*b^2 + A*b^3)*x^7 + 1/2*(B*a^2*b + A*a*b^2)*x^6 + 1/5*(B*a^3 + 3*A*a^2*b)*x^5
Leaf count of result is larger than twice the leaf count of optimal. 3364 vs. \(2 (153) = 306\).
Time = 0.87 (sec) , antiderivative size = 3364, normalized size of antiderivative = 16.02 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\text {Too large to display} \] Input:
integrate(x**3*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
Output:
Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*b**2*x**7/8 + x**6*(A*b**4 + 17*B*a*b**3/8)/(7*b**2) + x**5*(4*A*a*b**3 + 41*B*a**2*b**2/8 - 13*a*(A* b**4 + 17*B*a*b**3/8)/(7*b))/(6*b**2) + x**4*(6*A*a**2*b**2 + 4*B*a**3*b - 6*a**2*(A*b**4 + 17*B*a*b**3/8)/(7*b**2) - 11*a*(4*A*a*b**3 + 41*B*a**2*b **2/8 - 13*a*(A*b**4 + 17*B*a*b**3/8)/(7*b))/(6*b))/(5*b**2) + x**3*(4*A*a **3*b + B*a**4 - 5*a**2*(4*A*a*b**3 + 41*B*a**2*b**2/8 - 13*a*(A*b**4 + 17 *B*a*b**3/8)/(7*b))/(6*b**2) - 9*a*(6*A*a**2*b**2 + 4*B*a**3*b - 6*a**2*(A *b**4 + 17*B*a*b**3/8)/(7*b**2) - 11*a*(4*A*a*b**3 + 41*B*a**2*b**2/8 - 13 *a*(A*b**4 + 17*B*a*b**3/8)/(7*b))/(6*b))/(5*b))/(4*b**2) + x**2*(A*a**4 - 4*a**2*(6*A*a**2*b**2 + 4*B*a**3*b - 6*a**2*(A*b**4 + 17*B*a*b**3/8)/(7*b **2) - 11*a*(4*A*a*b**3 + 41*B*a**2*b**2/8 - 13*a*(A*b**4 + 17*B*a*b**3/8) /(7*b))/(6*b))/(5*b**2) - 7*a*(4*A*a**3*b + B*a**4 - 5*a**2*(4*A*a*b**3 + 41*B*a**2*b**2/8 - 13*a*(A*b**4 + 17*B*a*b**3/8)/(7*b))/(6*b**2) - 9*a*(6* A*a**2*b**2 + 4*B*a**3*b - 6*a**2*(A*b**4 + 17*B*a*b**3/8)/(7*b**2) - 11*a *(4*A*a*b**3 + 41*B*a**2*b**2/8 - 13*a*(A*b**4 + 17*B*a*b**3/8)/(7*b))/(6* b))/(5*b))/(4*b))/(3*b**2) + x*(-3*a**2*(4*A*a**3*b + B*a**4 - 5*a**2*(4*A *a*b**3 + 41*B*a**2*b**2/8 - 13*a*(A*b**4 + 17*B*a*b**3/8)/(7*b))/(6*b**2) - 9*a*(6*A*a**2*b**2 + 4*B*a**3*b - 6*a**2*(A*b**4 + 17*B*a*b**3/8)/(7*b* *2) - 11*a*(4*A*a*b**3 + 41*B*a**2*b**2/8 - 13*a*(A*b**4 + 17*B*a*b**3/8)/ (7*b))/(6*b))/(5*b))/(4*b**2) - 5*a*(A*a**4 - 4*a**2*(6*A*a**2*b**2 + 4...
Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (145) = 290\).
Time = 0.04 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.43 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B x^{3}}{8 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{4} x}{4 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{3} x}{4 \, b^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a x^{2}}{56 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A x^{2}}{7 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{5}}{4 \, b^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{4}}{4 \, b^{4}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{2} x}{56 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a x}{14 \, b^{3}} - \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a^{3}}{280 \, b^{5}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A a^{2}}{70 \, b^{4}} \] Input:
integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
Output:
1/8*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*x^3/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a ^2)^(3/2)*B*a^4*x/b^4 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^3*x/b^3 - 11/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a*x^2/b^3 + 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*x^2/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^5/b^5 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a^4/b^4 + 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^2*x/b^4 - 3/14*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*a*x/b^3 - 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^3/b^5 + 17/70*(b^2*x^2 + 2*a*b *x + a^2)^(5/2)*A*a^2/b^4
Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{8} \, B b^{3} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, B a b^{2} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{7} \, A b^{3} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a^{2} b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a b^{2} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B a^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, A a^{2} b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A a^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (B a^{8} - 2 \, A a^{7} b\right )} \mathrm {sgn}\left (b x + a\right )}{280 \, b^{5}} \] Input:
integrate(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
Output:
1/8*B*b^3*x^8*sgn(b*x + a) + 3/7*B*a*b^2*x^7*sgn(b*x + a) + 1/7*A*b^3*x^7* sgn(b*x + a) + 1/2*B*a^2*b*x^6*sgn(b*x + a) + 1/2*A*a*b^2*x^6*sgn(b*x + a) + 1/5*B*a^3*x^5*sgn(b*x + a) + 3/5*A*a^2*b*x^5*sgn(b*x + a) + 1/4*A*a^3*x ^4*sgn(b*x + a) + 1/280*(B*a^8 - 2*A*a^7*b)*sgn(b*x + a)/b^5
Timed out. \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^3\,\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \] Input:
int(x^3*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
Output:
int(x^3*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int x^3 (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^{4} \left (35 b^{4} x^{4}+160 a \,b^{3} x^{3}+280 a^{2} b^{2} x^{2}+224 a^{3} b x +70 a^{4}\right )}{280} \] Input:
int(x^3*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
Output:
(x**4*(70*a**4 + 224*a**3*b*x + 280*a**2*b**2*x**2 + 160*a*b**3*x**3 + 35* b**4*x**4))/280