Integrand size = 25, antiderivative size = 71 \[ \int x^m (A+B x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {a^2 A x^{1+m}}{1+m}+\frac {a (2 A b+a B) x^{2+m}}{2+m}+\frac {b (A b+2 a B) x^{3+m}}{3+m}+\frac {b^2 B x^{4+m}}{4+m} \]
a^2*A*x^(1+m)/(1+m)+a*(2*A*b+B*a)*x^(2+m)/(2+m)+b*(A*b+2*B*a)*x^(3+m)/(3+m )+b^2*B*x^(4+m)/(4+m)
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int x^m (A+B x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {x^{1+m} \left (B (a+b x)^3+(-a B (1+m)+A b (4+m)) \left (\frac {a^2}{1+m}+\frac {2 a b x}{2+m}+\frac {b^2 x^2}{3+m}\right )\right )}{b (4+m)} \]
(x^(1 + m)*(B*(a + b*x)^3 + (-(a*B*(1 + m)) + A*b*(4 + m))*(a^2/(1 + m) + (2*a*b*x)/(2 + m) + (b^2*x^2)/(3 + m))))/(b*(4 + m))
Time = 0.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1184, 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.
| \(\displaystyle \int x^m \left (a^2+2 a b x+b^2 x^2\right ) (A+B x) \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle \frac {\int b^2 x^m (a+b x)^2 (A+B x)dx}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int x^m (a+b x)^2 (A+B x)dx\) |
| \(\Big \downarrow \) 85 |
| \(\displaystyle \int \left (a^2 A x^m+a x^{m+1} (a B+2 A b)+b x^{m+2} (2 a B+A b)+b^2 B x^{m+3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 A x^{m+1}}{m+1}+\frac {a x^{m+2} (a B+2 A b)}{m+2}+\frac {b x^{m+3} (2 a B+A b)}{m+3}+\frac {b^2 B x^{m+4}}{m+4}\) |
(a^2*A*x^(1 + m))/(1 + m) + (a*(2*A*b + a*B)*x^(2 + m))/(2 + m) + (b*(A*b + 2*a*B)*x^(3 + m))/(3 + m) + (b^2*B*x^(4 + m))/(4 + m)
3.9.38.3.1 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[1/c^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}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15
| method | result | size |
| norman | \(\frac {A \,a^{2} x \,{\mathrm e}^{m \ln \left (x \right )}}{1+m}+\frac {B \,b^{2} x^{4} {\mathrm e}^{m \ln \left (x \right )}}{4+m}+\frac {a \left (2 A b +B a \right ) x^{2} {\mathrm e}^{m \ln \left (x \right )}}{2+m}+\frac {b \left (A b +2 B a \right ) x^{3} {\mathrm e}^{m \ln \left (x \right )}}{3+m}\) | \(82\) |
| risch | \(\frac {x^{m} \left (B \,b^{2} m^{3} x^{3}+A \,b^{2} m^{3} x^{2}+2 B a b \,m^{3} x^{2}+6 B \,b^{2} m^{2} x^{3}+2 A a b \,m^{3} x +7 A \,b^{2} m^{2} x^{2}+B \,a^{2} m^{3} x +14 B a b \,m^{2} x^{2}+11 B \,b^{2} m \,x^{3}+A \,a^{2} m^{3}+16 A a b \,m^{2} x +14 A \,b^{2} m \,x^{2}+8 B \,a^{2} m^{2} x +28 B a b m \,x^{2}+6 B \,b^{2} x^{3}+9 A \,a^{2} m^{2}+38 A a b m x +8 A \,b^{2} x^{2}+19 B \,a^{2} m x +16 B a b \,x^{2}+26 A \,a^{2} m +24 a A b x +12 a^{2} B x +24 A \,a^{2}\right ) x}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(245\) |
| gosper | \(\frac {x^{1+m} \left (B \,b^{2} m^{3} x^{3}+A \,b^{2} m^{3} x^{2}+2 B a b \,m^{3} x^{2}+6 B \,b^{2} m^{2} x^{3}+2 A a b \,m^{3} x +7 A \,b^{2} m^{2} x^{2}+B \,a^{2} m^{3} x +14 B a b \,m^{2} x^{2}+11 B \,b^{2} m \,x^{3}+A \,a^{2} m^{3}+16 A a b \,m^{2} x +14 A \,b^{2} m \,x^{2}+8 B \,a^{2} m^{2} x +28 B a b m \,x^{2}+6 B \,b^{2} x^{3}+9 A \,a^{2} m^{2}+38 A a b m x +8 A \,b^{2} x^{2}+19 B \,a^{2} m x +16 B a b \,x^{2}+26 A \,a^{2} m +24 a A b x +12 a^{2} B x +24 A \,a^{2}\right )}{\left (1+m \right ) \left (2+m \right ) \left (3+m \right ) \left (4+m \right )}\) | \(246\) |
| parallelrisch | \(\frac {6 B \,x^{4} x^{m} b^{2} m^{2}+7 A \,x^{3} x^{m} b^{2} m^{2}+11 B \,x^{4} x^{m} b^{2} m +B \,x^{2} x^{m} a^{2} m^{3}+14 A \,x^{3} x^{m} b^{2} m +A x \,x^{m} a^{2} m^{3}+6 B \,x^{4} x^{m} b^{2}+8 A \,x^{3} x^{m} b^{2}+12 B \,x^{2} x^{m} a^{2}+24 A x \,x^{m} a^{2}+28 B \,x^{3} x^{m} a b m +38 A \,x^{2} x^{m} a b m +2 A \,x^{2} x^{m} a b \,m^{3}+14 B \,x^{3} x^{m} a b \,m^{2}+16 A \,x^{2} x^{m} a b \,m^{2}+2 B \,x^{3} x^{m} a b \,m^{3}+B \,x^{4} x^{m} b^{2} m^{3}+A \,x^{3} x^{m} b^{2} m^{3}+8 B \,x^{2} x^{m} a^{2} m^{2}+9 A x \,x^{m} a^{2} m^{2}+16 B \,x^{3} x^{m} a b +19 B \,x^{2} x^{m} a^{2} m +24 A \,x^{2} x^{m} a b +26 A x \,x^{m} a^{2} m}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(333\) |
A*a^2/(1+m)*x*exp(m*ln(x))+B*b^2/(4+m)*x^4*exp(m*ln(x))+a*(2*A*b+B*a)/(2+m )*x^2*exp(m*ln(x))+b*(A*b+2*B*a)/(3+m)*x^3*exp(m*ln(x))
Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (71) = 142\).
Time = 0.30 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.03 \[ \int x^m (A+B x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {{\left ({\left (B b^{2} m^{3} + 6 \, B b^{2} m^{2} + 11 \, B b^{2} m + 6 \, B b^{2}\right )} x^{4} + {\left ({\left (2 \, B a b + A b^{2}\right )} m^{3} + 16 \, B a b + 8 \, A b^{2} + 7 \, {\left (2 \, B a b + A b^{2}\right )} m^{2} + 14 \, {\left (2 \, B a b + A b^{2}\right )} m\right )} x^{3} + {\left ({\left (B a^{2} + 2 \, A a b\right )} m^{3} + 12 \, B a^{2} + 24 \, A a b + 8 \, {\left (B a^{2} + 2 \, A a b\right )} m^{2} + 19 \, {\left (B a^{2} + 2 \, A a b\right )} m\right )} x^{2} + {\left (A a^{2} m^{3} + 9 \, A a^{2} m^{2} + 26 \, A a^{2} m + 24 \, A a^{2}\right )} x\right )} x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \]
((B*b^2*m^3 + 6*B*b^2*m^2 + 11*B*b^2*m + 6*B*b^2)*x^4 + ((2*B*a*b + A*b^2) *m^3 + 16*B*a*b + 8*A*b^2 + 7*(2*B*a*b + A*b^2)*m^2 + 14*(2*B*a*b + A*b^2) *m)*x^3 + ((B*a^2 + 2*A*a*b)*m^3 + 12*B*a^2 + 24*A*a*b + 8*(B*a^2 + 2*A*a* b)*m^2 + 19*(B*a^2 + 2*A*a*b)*m)*x^2 + (A*a^2*m^3 + 9*A*a^2*m^2 + 26*A*a^2 *m + 24*A*a^2)*x)*x^m/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24)
Leaf count of result is larger than twice the leaf count of optimal. 1020 vs. \(2 (63) = 126\).
Time = 0.34 (sec) , antiderivative size = 1020, normalized size of antiderivative = 14.37 \[ \int x^m (A+B x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\begin {cases} - \frac {A a^{2}}{3 x^{3}} - \frac {A a b}{x^{2}} - \frac {A b^{2}}{x} - \frac {B a^{2}}{2 x^{2}} - \frac {2 B a b}{x} + B b^{2} \log {\left (x \right )} & \text {for}\: m = -4 \\- \frac {A a^{2}}{2 x^{2}} - \frac {2 A a b}{x} + A b^{2} \log {\left (x \right )} - \frac {B a^{2}}{x} + 2 B a b \log {\left (x \right )} + B b^{2} x & \text {for}\: m = -3 \\- \frac {A a^{2}}{x} + 2 A a b \log {\left (x \right )} + A b^{2} x + B a^{2} \log {\left (x \right )} + 2 B a b x + \frac {B b^{2} x^{2}}{2} & \text {for}\: m = -2 \\A a^{2} \log {\left (x \right )} + 2 A a b x + \frac {A b^{2} x^{2}}{2} + B a^{2} x + B a b x^{2} + \frac {B b^{2} x^{3}}{3} & \text {for}\: m = -1 \\\frac {A a^{2} m^{3} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 A a^{2} m^{2} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 A a^{2} m x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 A a^{2} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {2 A a b m^{3} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {16 A a b m^{2} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {38 A a b m x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 A a b x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {A b^{2} m^{3} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {7 A b^{2} m^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {14 A b^{2} m x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {8 A b^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {B a^{2} m^{3} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {8 B a^{2} m^{2} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {19 B a^{2} m x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {12 B a^{2} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {2 B a b m^{3} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {14 B a b m^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {28 B a b m x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {16 B a b x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {B b^{2} m^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 B b^{2} m^{2} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {11 B b^{2} m x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 B b^{2} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \]
Piecewise((-A*a**2/(3*x**3) - A*a*b/x**2 - A*b**2/x - B*a**2/(2*x**2) - 2* B*a*b/x + B*b**2*log(x), Eq(m, -4)), (-A*a**2/(2*x**2) - 2*A*a*b/x + A*b** 2*log(x) - B*a**2/x + 2*B*a*b*log(x) + B*b**2*x, Eq(m, -3)), (-A*a**2/x + 2*A*a*b*log(x) + A*b**2*x + B*a**2*log(x) + 2*B*a*b*x + B*b**2*x**2/2, Eq( m, -2)), (A*a**2*log(x) + 2*A*a*b*x + A*b**2*x**2/2 + B*a**2*x + B*a*b*x** 2 + B*b**2*x**3/3, Eq(m, -1)), (A*a**2*m**3*x*x**m/(m**4 + 10*m**3 + 35*m* *2 + 50*m + 24) + 9*A*a**2*m**2*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 26*A*a**2*m*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 24*A*a** 2*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 2*A*a*b*m**3*x**2*x**m/( m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 16*A*a*b*m**2*x**2*x**m/(m**4 + 10 *m**3 + 35*m**2 + 50*m + 24) + 38*A*a*b*m*x**2*x**m/(m**4 + 10*m**3 + 35*m **2 + 50*m + 24) + 24*A*a*b*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 2 4) + A*b**2*m**3*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 7*A*b* *2*m**2*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 14*A*b**2*m*x** 3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 8*A*b**2*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + B*a**2*m**3*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 8*B*a**2*m**2*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 19*B*a**2*m*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 12*B*a**2*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 2*B*a*b*m* *3*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 14*B*a*b*m**2*x**...
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.28 \[ \int x^m (A+B x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {B b^{2} x^{m + 4}}{m + 4} + \frac {2 \, B a b x^{m + 3}}{m + 3} + \frac {A b^{2} x^{m + 3}}{m + 3} + \frac {B a^{2} x^{m + 2}}{m + 2} + \frac {2 \, A a b x^{m + 2}}{m + 2} + \frac {A a^{2} x^{m + 1}}{m + 1} \]
B*b^2*x^(m + 4)/(m + 4) + 2*B*a*b*x^(m + 3)/(m + 3) + A*b^2*x^(m + 3)/(m + 3) + B*a^2*x^(m + 2)/(m + 2) + 2*A*a*b*x^(m + 2)/(m + 2) + A*a^2*x^(m + 1 )/(m + 1)
Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (71) = 142\).
Time = 0.27 (sec) , antiderivative size = 332, normalized size of antiderivative = 4.68 \[ \int x^m (A+B x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {B b^{2} m^{3} x^{4} x^{m} + 2 \, B a b m^{3} x^{3} x^{m} + A b^{2} m^{3} x^{3} x^{m} + 6 \, B b^{2} m^{2} x^{4} x^{m} + B a^{2} m^{3} x^{2} x^{m} + 2 \, A a b m^{3} x^{2} x^{m} + 14 \, B a b m^{2} x^{3} x^{m} + 7 \, A b^{2} m^{2} x^{3} x^{m} + 11 \, B b^{2} m x^{4} x^{m} + A a^{2} m^{3} x x^{m} + 8 \, B a^{2} m^{2} x^{2} x^{m} + 16 \, A a b m^{2} x^{2} x^{m} + 28 \, B a b m x^{3} x^{m} + 14 \, A b^{2} m x^{3} x^{m} + 6 \, B b^{2} x^{4} x^{m} + 9 \, A a^{2} m^{2} x x^{m} + 19 \, B a^{2} m x^{2} x^{m} + 38 \, A a b m x^{2} x^{m} + 16 \, B a b x^{3} x^{m} + 8 \, A b^{2} x^{3} x^{m} + 26 \, A a^{2} m x x^{m} + 12 \, B a^{2} x^{2} x^{m} + 24 \, A a b x^{2} x^{m} + 24 \, A a^{2} x x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \]
(B*b^2*m^3*x^4*x^m + 2*B*a*b*m^3*x^3*x^m + A*b^2*m^3*x^3*x^m + 6*B*b^2*m^2 *x^4*x^m + B*a^2*m^3*x^2*x^m + 2*A*a*b*m^3*x^2*x^m + 14*B*a*b*m^2*x^3*x^m + 7*A*b^2*m^2*x^3*x^m + 11*B*b^2*m*x^4*x^m + A*a^2*m^3*x*x^m + 8*B*a^2*m^2 *x^2*x^m + 16*A*a*b*m^2*x^2*x^m + 28*B*a*b*m*x^3*x^m + 14*A*b^2*m*x^3*x^m + 6*B*b^2*x^4*x^m + 9*A*a^2*m^2*x*x^m + 19*B*a^2*m*x^2*x^m + 38*A*a*b*m*x^ 2*x^m + 16*B*a*b*x^3*x^m + 8*A*b^2*x^3*x^m + 26*A*a^2*m*x*x^m + 12*B*a^2*x ^2*x^m + 24*A*a*b*x^2*x^m + 24*A*a^2*x*x^m)/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24)
Time = 10.16 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.49 \[ \int x^m (A+B x) \left (a^2+2 a b x+b^2 x^2\right ) \, dx=x^m\,\left (\frac {B\,b^2\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {A\,a^2\,x\,\left (m^3+9\,m^2+26\,m+24\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {a\,x^2\,\left (2\,A\,b+B\,a\right )\,\left (m^3+8\,m^2+19\,m+12\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {b\,x^3\,\left (A\,b+2\,B\,a\right )\,\left (m^3+7\,m^2+14\,m+8\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}\right ) \]
x^m*((B*b^2*x^4*(11*m + 6*m^2 + m^3 + 6))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (A*a^2*x*(26*m + 9*m^2 + m^3 + 24))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (a*x^2*(2*A*b + B*a)*(19*m + 8*m^2 + m^3 + 12))/(50*m + 35*m^2 + 10* m^3 + m^4 + 24) + (b*x^3*(A*b + 2*B*a)*(14*m + 7*m^2 + m^3 + 8))/(50*m + 3 5*m^2 + 10*m^3 + m^4 + 24))