Integrand size = 18, antiderivative size = 73 \[ \int (e x)^m (A+B x) \left (a+c x^2\right ) \, dx=\frac {a A (e x)^{1+m}}{e (1+m)}+\frac {a B (e x)^{2+m}}{e^2 (2+m)}+\frac {A c (e x)^{3+m}}{e^3 (3+m)}+\frac {B c (e x)^{4+m}}{e^4 (4+m)} \]
a*A*(e*x)^(1+m)/e/(1+m)+a*B*(e*x)^(2+m)/e^2/(2+m)+A*c*(e*x)^(3+m)/e^3/(3+m )+B*c*(e*x)^(4+m)/e^4/(4+m)
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64 \[ \int (e x)^m (A+B x) \left (a+c x^2\right ) \, dx=x (e x)^m \left (a \left (\frac {A}{1+m}+\frac {B x}{2+m}\right )+c x^2 \left (\frac {A}{3+m}+\frac {B x}{4+m}\right )\right ) \]
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {522, 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.
\(\displaystyle \int \left (a+c x^2\right ) (A+B x) (e x)^m \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (a A (e x)^m+\frac {a B (e x)^{m+1}}{e}+\frac {A c (e x)^{m+2}}{e^2}+\frac {B c (e x)^{m+3}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a A (e x)^{m+1}}{e (m+1)}+\frac {a B (e x)^{m+2}}{e^2 (m+2)}+\frac {A c (e x)^{m+3}}{e^3 (m+3)}+\frac {B c (e x)^{m+4}}{e^4 (m+4)}\) |
(a*A*(e*x)^(1 + m))/(e*(1 + m)) + (a*B*(e*x)^(2 + m))/(e^2*(2 + m)) + (A*c *(e*x)^(3 + m))/(e^3*(3 + m)) + (B*c*(e*x)^(4 + m))/(e^4*(4 + m))
3.5.88.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99
method | result | size |
norman | \(\frac {A c \,x^{3} {\mathrm e}^{m \ln \left (e x \right )}}{3+m}+\frac {B a \,x^{2} {\mathrm e}^{m \ln \left (e x \right )}}{2+m}+\frac {B c \,x^{4} {\mathrm e}^{m \ln \left (e x \right )}}{4+m}+\frac {a A x \,{\mathrm e}^{m \ln \left (e x \right )}}{1+m}\) | \(72\) |
gosper | \(\frac {x \left (B c \,m^{3} x^{3}+A c \,m^{3} x^{2}+6 B c \,m^{2} x^{3}+7 A c \,m^{2} x^{2}+B a \,m^{3} x +11 B c m \,x^{3}+A a \,m^{3}+14 A c m \,x^{2}+8 B a \,m^{2} x +6 B c \,x^{3}+9 A a \,m^{2}+8 A c \,x^{2}+19 B a m x +26 A a m +12 a B x +24 a A \right ) \left (e x \right )^{m}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(145\) |
risch | \(\frac {x \left (B c \,m^{3} x^{3}+A c \,m^{3} x^{2}+6 B c \,m^{2} x^{3}+7 A c \,m^{2} x^{2}+B a \,m^{3} x +11 B c m \,x^{3}+A a \,m^{3}+14 A c m \,x^{2}+8 B a \,m^{2} x +6 B c \,x^{3}+9 A a \,m^{2}+8 A c \,x^{2}+19 B a m x +26 A a m +12 a B x +24 a A \right ) \left (e x \right )^{m}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(145\) |
parallelrisch | \(\frac {B \,x^{4} \left (e x \right )^{m} c \,m^{3}+A \,x^{3} \left (e x \right )^{m} c \,m^{3}+6 B \,x^{4} \left (e x \right )^{m} c \,m^{2}+7 A \,x^{3} \left (e x \right )^{m} c \,m^{2}+11 B \,x^{4} \left (e x \right )^{m} c m +B \,x^{2} \left (e x \right )^{m} a \,m^{3}+14 A \,x^{3} \left (e x \right )^{m} c m +A x \left (e x \right )^{m} a \,m^{3}+6 B \,x^{4} \left (e x \right )^{m} c +8 B \,x^{2} \left (e x \right )^{m} a \,m^{2}+8 A \,x^{3} \left (e x \right )^{m} c +9 A x \left (e x \right )^{m} a \,m^{2}+19 B \,x^{2} \left (e x \right )^{m} a m +26 A x \left (e x \right )^{m} a m +12 B \,x^{2} \left (e x \right )^{m} a +24 A x \left (e x \right )^{m} a}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(231\) |
A*c/(3+m)*x^3*exp(m*ln(e*x))+B*a/(2+m)*x^2*exp(m*ln(e*x))+B*c/(4+m)*x^4*ex p(m*ln(e*x))+a*A/(1+m)*x*exp(m*ln(e*x))
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.82 \[ \int (e x)^m (A+B x) \left (a+c x^2\right ) \, dx=\frac {{\left ({\left (B c m^{3} + 6 \, B c m^{2} + 11 \, B c m + 6 \, B c\right )} x^{4} + {\left (A c m^{3} + 7 \, A c m^{2} + 14 \, A c m + 8 \, A c\right )} x^{3} + {\left (B a m^{3} + 8 \, B a m^{2} + 19 \, B a m + 12 \, B a\right )} x^{2} + {\left (A a m^{3} + 9 \, A a m^{2} + 26 \, A a m + 24 \, A a\right )} x\right )} \left (e x\right )^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \]
((B*c*m^3 + 6*B*c*m^2 + 11*B*c*m + 6*B*c)*x^4 + (A*c*m^3 + 7*A*c*m^2 + 14* A*c*m + 8*A*c)*x^3 + (B*a*m^3 + 8*B*a*m^2 + 19*B*a*m + 12*B*a)*x^2 + (A*a* m^3 + 9*A*a*m^2 + 26*A*a*m + 24*A*a)*x)*(e*x)^m/(m^4 + 10*m^3 + 35*m^2 + 5 0*m + 24)
Leaf count of result is larger than twice the leaf count of optimal. 658 vs. \(2 (65) = 130\).
Time = 0.38 (sec) , antiderivative size = 658, normalized size of antiderivative = 9.01 \[ \int (e x)^m (A+B x) \left (a+c x^2\right ) \, dx=\begin {cases} \frac {- \frac {A a}{3 x^{3}} - \frac {A c}{x} - \frac {B a}{2 x^{2}} + B c \log {\left (x \right )}}{e^{4}} & \text {for}\: m = -4 \\\frac {- \frac {A a}{2 x^{2}} + A c \log {\left (x \right )} - \frac {B a}{x} + B c x}{e^{3}} & \text {for}\: m = -3 \\\frac {- \frac {A a}{x} + A c x + B a \log {\left (x \right )} + \frac {B c x^{2}}{2}}{e^{2}} & \text {for}\: m = -2 \\\frac {A a \log {\left (x \right )} + \frac {A c x^{2}}{2} + B a x + \frac {B c x^{3}}{3}}{e} & \text {for}\: m = -1 \\\frac {A a m^{3} x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 A a m^{2} x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 A a m x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 A a x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {A c m^{3} x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {7 A c m^{2} x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {14 A c m x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {8 A c x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {B a m^{3} x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {8 B a m^{2} x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {19 B a m x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {12 B a x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {B c m^{3} x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 B c m^{2} x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {11 B c m x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 B c x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \]
Piecewise(((-A*a/(3*x**3) - A*c/x - B*a/(2*x**2) + B*c*log(x))/e**4, Eq(m, -4)), ((-A*a/(2*x**2) + A*c*log(x) - B*a/x + B*c*x)/e**3, Eq(m, -3)), ((- A*a/x + A*c*x + B*a*log(x) + B*c*x**2/2)/e**2, Eq(m, -2)), ((A*a*log(x) + A*c*x**2/2 + B*a*x + B*c*x**3/3)/e, Eq(m, -1)), (A*a*m**3*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 9*A*a*m**2*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 26*A*a*m*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 5 0*m + 24) + 24*A*a*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + A*c *m**3*x**3*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 7*A*c*m**2*x* *3*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 14*A*c*m*x**3*(e*x)** m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 8*A*c*x**3*(e*x)**m/(m**4 + 10* m**3 + 35*m**2 + 50*m + 24) + B*a*m**3*x**2*(e*x)**m/(m**4 + 10*m**3 + 35* m**2 + 50*m + 24) + 8*B*a*m**2*x**2*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 5 0*m + 24) + 19*B*a*m*x**2*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 12*B*a*x**2*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + B*c*m**3*x **4*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 6*B*c*m**2*x**4*(e*x )**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 11*B*c*m*x**4*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 6*B*c*x**4*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24), True))
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int (e x)^m (A+B x) \left (a+c x^2\right ) \, dx=\frac {B c e^{m} x^{4} x^{m}}{m + 4} + \frac {A c e^{m} x^{3} x^{m}}{m + 3} + \frac {B a e^{m} x^{2} x^{m}}{m + 2} + \frac {\left (e x\right )^{m + 1} A a}{e {\left (m + 1\right )}} \]
B*c*e^m*x^4*x^m/(m + 4) + A*c*e^m*x^3*x^m/(m + 3) + B*a*e^m*x^2*x^m/(m + 2 ) + (e*x)^(m + 1)*A*a/(e*(m + 1))
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (73) = 146\).
Time = 0.28 (sec) , antiderivative size = 230, normalized size of antiderivative = 3.15 \[ \int (e x)^m (A+B x) \left (a+c x^2\right ) \, dx=\frac {\left (e x\right )^{m} B c m^{3} x^{4} + \left (e x\right )^{m} A c m^{3} x^{3} + 6 \, \left (e x\right )^{m} B c m^{2} x^{4} + \left (e x\right )^{m} B a m^{3} x^{2} + 7 \, \left (e x\right )^{m} A c m^{2} x^{3} + 11 \, \left (e x\right )^{m} B c m x^{4} + \left (e x\right )^{m} A a m^{3} x + 8 \, \left (e x\right )^{m} B a m^{2} x^{2} + 14 \, \left (e x\right )^{m} A c m x^{3} + 6 \, \left (e x\right )^{m} B c x^{4} + 9 \, \left (e x\right )^{m} A a m^{2} x + 19 \, \left (e x\right )^{m} B a m x^{2} + 8 \, \left (e x\right )^{m} A c x^{3} + 26 \, \left (e x\right )^{m} A a m x + 12 \, \left (e x\right )^{m} B a x^{2} + 24 \, \left (e x\right )^{m} A a x}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \]
((e*x)^m*B*c*m^3*x^4 + (e*x)^m*A*c*m^3*x^3 + 6*(e*x)^m*B*c*m^2*x^4 + (e*x) ^m*B*a*m^3*x^2 + 7*(e*x)^m*A*c*m^2*x^3 + 11*(e*x)^m*B*c*m*x^4 + (e*x)^m*A* a*m^3*x + 8*(e*x)^m*B*a*m^2*x^2 + 14*(e*x)^m*A*c*m*x^3 + 6*(e*x)^m*B*c*x^4 + 9*(e*x)^m*A*a*m^2*x + 19*(e*x)^m*B*a*m*x^2 + 8*(e*x)^m*A*c*x^3 + 26*(e* x)^m*A*a*m*x + 12*(e*x)^m*B*a*x^2 + 24*(e*x)^m*A*a*x)/(m^4 + 10*m^3 + 35*m ^2 + 50*m + 24)
Time = 10.01 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.21 \[ \int (e x)^m (A+B x) \left (a+c x^2\right ) \, dx={\left (e\,x\right )}^m\,\left (\frac {A\,a\,x\,\left (m^3+9\,m^2+26\,m+24\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {B\,a\,x^2\,\left (m^3+8\,m^2+19\,m+12\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {A\,c\,x^3\,\left (m^3+7\,m^2+14\,m+8\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {B\,c\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}\right ) \]
(e*x)^m*((A*a*x*(26*m + 9*m^2 + m^3 + 24))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (B*a*x^2*(19*m + 8*m^2 + m^3 + 12))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (A*c*x^3*(14*m + 7*m^2 + m^3 + 8))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (B*c*x^4*(11*m + 6*m^2 + m^3 + 6))/(50*m + 35*m^2 + 10*m^3 + m^4 + 2 4))