Integrand size = 29, antiderivative size = 114 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {a A x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {(A b+a B) x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {b B x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)} \] Output:
a*A*x^4*((b*x+a)^2)^(1/2)/(4*b*x+4*a)+(A*b+B*a)*x^5*((b*x+a)^2)^(1/2)/(5*b *x+5*a)+b*B*x^6*((b*x+a)^2)^(1/2)/(6*b*x+6*a)
Time = 1.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.43 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x^4 \sqrt {(a+b x)^2} (3 a (5 A+4 B x)+2 b x (6 A+5 B x))}{60 (a+b x)} \] Input:
Integrate[x^3*(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(x^4*Sqrt[(a + b*x)^2]*(3*a*(5*A + 4*B*x) + 2*b*x*(6*A + 5*B*x)))/(60*(a + b*x))
Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54, 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.
\(\displaystyle \int x^3 \sqrt {a^2+2 a b x+b^2 x^2} (A+B x) \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b x^3 (a+b x) (A+B x)dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^3 (a+b x) (A+B x)dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (b B x^5+(A b+a B) x^4+a A x^3\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {1}{5} x^5 (a B+A b)+\frac {1}{4} a A x^4+\frac {1}{6} b B x^6\right )}{a+b x}\) |
Input:
Int[x^3*(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((a*A*x^4)/4 + ((A*b + a*B)*x^5)/5 + (b*B*x ^6)/6))/(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 = 0.59 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.39
method | result | size |
gosper | \(\frac {x^{4} \left (10 B b \,x^{2}+12 A b x +12 B a x +15 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{60 b x +60 a}\) | \(44\) |
orering | \(\frac {x^{4} \left (10 B b \,x^{2}+12 A b x +12 B a x +15 A a \right ) \sqrt {\left (b x +a \right )^{2}}}{60 b x +60 a}\) | \(44\) |
risch | \(\frac {a A \,x^{4} \sqrt {\left (b x +a \right )^{2}}}{4 b x +4 a}+\frac {\left (A b +B a \right ) x^{5} \sqrt {\left (b x +a \right )^{2}}}{5 b x +5 a}+\frac {b B \,x^{6} \sqrt {\left (b x +a \right )^{2}}}{6 b x +6 a}\) | \(76\) |
default | \(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (-10 b^{4} B \,x^{4}-12 A \,b^{4} x^{3}+8 B a \,b^{3} x^{3}+9 A a \,b^{3} x^{2}-6 B \,a^{2} b^{2} x^{2}-6 A \,a^{2} b^{2} x +4 B \,a^{3} b x +3 A \,a^{3} b -2 a^{4} B \right )}{60 b^{5}}\) | \(101\) |
Input:
int(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/60*x^4*(10*B*b*x^2+12*A*b*x+12*B*a*x+15*A*a)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.24 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{6} \, B b x^{6} + \frac {1}{4} \, A a x^{4} + \frac {1}{5} \, {\left (B a + A b\right )} x^{5} \] Input:
integrate(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="fricas")
Output:
1/6*B*b*x^6 + 1/4*A*a*x^4 + 1/5*(B*a + A*b)*x^5
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (78) = 156\).
Time = 1.90 (sec) , antiderivative size = 357, normalized size of antiderivative = 3.13 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=A \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{4}}{20 b^{4}} + \frac {a^{3} x}{20 b^{3}} - \frac {a^{2} x^{2}}{20 b^{2}} + \frac {a x^{3}}{20 b} + \frac {x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{6} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {3 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {3 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4} \sqrt {a^{2}}}{4} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{5}}{30 b^{5}} - \frac {a^{4} x}{30 b^{4}} + \frac {a^{3} x^{2}}{30 b^{3}} - \frac {a^{2} x^{3}}{30 b^{2}} + \frac {a x^{4}}{30 b} + \frac {x^{5}}{6}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{8} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {4 a^{6} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {6 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} - \frac {4 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {11}{2}}}{11}}{16 a^{5} b^{5}} & \text {for}\: a b \neq 0 \\\frac {x^{5} \sqrt {a^{2}}}{5} & \text {otherwise} \end {cases}\right ) \] Input:
integrate(x**3*(B*x+A)*((b*x+a)**2)**(1/2),x)
Output:
A*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**4/(20*b**4) + a**3*x/(2 0*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)) + B*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(a**5/(30*b**5) - a**4*x/(30*b**4) + a**3*x**2/(30*b**3) - a**2 *x**3/(30*b**2) + a*x**4/(30*b) + x**5/6), Ne(b**2, 0)), ((a**8*(a**2 + 2* a*b*x)**(3/2)/3 - 4*a**6*(a**2 + 2*a*b*x)**(5/2)/5 + 6*a**4*(a**2 + 2*a*b* x)**(7/2)/7 - 4*a**2*(a**2 + 2*a*b*x)**(9/2)/9 + (a**2 + 2*a*b*x)**(11/2)/ 11)/(16*a**5*b**5), Ne(a*b, 0)), (x**5*sqrt(a**2)/5, True))
Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (75) = 150\).
Time = 0.03 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.64 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B x^{3}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{4} x}{2 \, b^{4}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{3} x}{2 \, b^{3}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a x^{2}}{10 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A x^{2}}{5 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{5}}{2 \, b^{5}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{4}}{2 \, b^{4}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} x}{5 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a x}{20 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3}}{15 \, b^{5}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a^{2}}{20 \, b^{4}} \] Input:
integrate(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="maxima")
Output:
1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*x^3/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^4*x/b^4 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^3*x/b^3 - 3/10 *(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a*x^2/b^3 + 1/5*(b^2*x^2 + 2*a*b*x + a^ 2)^(3/2)*A*x^2/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^5/b^5 - 1/2*sqr t(b^2*x^2 + 2*a*b*x + a^2)*A*a^4/b^4 + 2/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2) *B*a^2*x/b^4 - 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*a*x/b^3 - 7/15*(b^2* x^2 + 2*a*b*x + a^2)^(3/2)*B*a^3/b^5 + 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2 )*A*a^2/b^4
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{6} \, B b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B a x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A a x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (2 \, B a^{6} - 3 \, A a^{5} b\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{5}} \] Input:
integrate(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x, algorithm="giac")
Output:
1/6*B*b*x^6*sgn(b*x + a) + 1/5*B*a*x^5*sgn(b*x + a) + 1/5*A*b*x^5*sgn(b*x + a) + 1/4*A*a*x^4*sgn(b*x + a) + 1/60*(2*B*a^6 - 3*A*a^5*b)*sgn(b*x + a)/ b^5
Time = 10.73 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.98 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {A\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}+\frac {B\,x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{6\,b^2}-\frac {7\,A\,a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{60\,b^4}-\frac {B\,a^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{24\,b^5}-\frac {A\,a^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{60\,b^6}-\frac {3\,B\,a\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (4\,b^2\,x^2\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-a^4+9\,a^2\,b^2\,x^2+8\,a^3\,b\,x-7\,a\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )\right )}{40\,b^5} \] Input:
int(x^3*((a + b*x)^2)^(1/2)*(A + B*x),x)
Output:
(A*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b^2) + (B*x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(6*b^2) - (7*A*a*(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) - (B*a^ 2*(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))/(24*b^5) - (A*a^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))/(60*b^6) - (3*B*a*(a^ 2 + b^2*x^2 + 2*a*b*x)^(1/2)*(4*b^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x) - a^4 + 9*a^2*b^2*x^2 + 8*a^3*b*x - 7*a*b*x*(a^2 + b^2*x^2 + 2*a*b*x)))/(40*b^5)
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.21 \[ \int x^3 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x^{4} \left (10 b^{2} x^{2}+24 a b x +15 a^{2}\right )}{60} \] Input:
int(x^3*(B*x+A)*((b*x+a)^2)^(1/2),x)
Output:
(x**4*(15*a**2 + 24*a*b*x + 10*b**2*x**2))/60