Integrand size = 18, antiderivative size = 73 \[ \int (e x)^m (c+d x) \left (a+b x^2\right ) \, dx=\frac {a c (e x)^{1+m}}{e (1+m)}+\frac {a d (e x)^{2+m}}{e^2 (2+m)}+\frac {b c (e x)^{3+m}}{e^3 (3+m)}+\frac {b d (e x)^{4+m}}{e^4 (4+m)} \] Output:
a*c*(e*x)^(1+m)/e/(1+m)+a*d*(e*x)^(2+m)/e^2/(2+m)+b*c*(e*x)^(3+m)/e^3/(3+m )+b*d*(e*x)^(4+m)/e^4/(4+m)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64 \[ \int (e x)^m (c+d x) \left (a+b x^2\right ) \, dx=x (e x)^m \left (a \left (\frac {c}{1+m}+\frac {d x}{2+m}\right )+b x^2 \left (\frac {c}{3+m}+\frac {d x}{4+m}\right )\right ) \] Input:
Integrate[(e*x)^m*(c + d*x)*(a + b*x^2),x]
Output:
x*(e*x)^m*(a*(c/(1 + m) + (d*x)/(2 + m)) + b*x^2*(c/(3 + m) + (d*x)/(4 + m )))
Time = 0.21 (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.
\(\displaystyle \int \left (a+b x^2\right ) (c+d x) (e x)^m \, dx\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \int \left (a c (e x)^m+\frac {a d (e x)^{m+1}}{e}+\frac {b c (e x)^{m+2}}{e^2}+\frac {b d (e x)^{m+3}}{e^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a c (e x)^{m+1}}{e (m+1)}+\frac {a d (e x)^{m+2}}{e^2 (m+2)}+\frac {b c (e x)^{m+3}}{e^3 (m+3)}+\frac {b d (e x)^{m+4}}{e^4 (m+4)}\) |
Input:
Int[(e*x)^m*(c + d*x)*(a + b*x^2),x]
Output:
(a*c*(e*x)^(1 + m))/(e*(1 + m)) + (a*d*(e*x)^(2 + m))/(e^2*(2 + m)) + (b*c *(e*x)^(3 + m))/(e^3*(3 + m)) + (b*d*(e*x)^(4 + m))/(e^4*(4 + m))
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 | \(\frac {a c x \,{\mathrm e}^{m \ln \left (e x \right )}}{1+m}+\frac {a d \,x^{2} {\mathrm e}^{m \ln \left (e x \right )}}{2+m}+\frac {b c \,x^{3} {\mathrm e}^{m \ln \left (e x \right )}}{3+m}+\frac {b d \,x^{4} {\mathrm e}^{m \ln \left (e x \right )}}{4+m}\) | \(72\) |
gosper | \(\frac {x \left (b d \,m^{3} x^{3}+b c \,m^{3} x^{2}+6 b d \,m^{2} x^{3}+a d \,m^{3} x +7 b c \,m^{2} x^{2}+11 b d m \,x^{3}+a c \,m^{3}+8 a d \,m^{2} x +14 b c m \,x^{2}+6 b d \,x^{3}+9 a c \,m^{2}+19 a d m x +8 b c \,x^{2}+26 a c m +12 a d x +24 a c \right ) \left (e x \right )^{m}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(145\) |
risch | \(\frac {x \left (b d \,m^{3} x^{3}+b c \,m^{3} x^{2}+6 b d \,m^{2} x^{3}+a d \,m^{3} x +7 b c \,m^{2} x^{2}+11 b d m \,x^{3}+a c \,m^{3}+8 a d \,m^{2} x +14 b c m \,x^{2}+6 b d \,x^{3}+9 a c \,m^{2}+19 a d m x +8 b c \,x^{2}+26 a c m +12 a d x +24 a c \right ) \left (e x \right )^{m}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(145\) |
orering | \(\frac {x \left (b d \,m^{3} x^{3}+b c \,m^{3} x^{2}+6 b d \,m^{2} x^{3}+a d \,m^{3} x +7 b c \,m^{2} x^{2}+11 b d m \,x^{3}+a c \,m^{3}+8 a d \,m^{2} x +14 b c m \,x^{2}+6 b d \,x^{3}+9 a c \,m^{2}+19 a d m x +8 b c \,x^{2}+26 a c m +12 a d x +24 a c \right ) \left (e x \right )^{m}}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(145\) |
parallelrisch | \(\frac {x^{4} \left (e x \right )^{m} b d \,m^{3}+6 x^{4} \left (e x \right )^{m} b d \,m^{2}+x^{3} \left (e x \right )^{m} b c \,m^{3}+11 x^{4} \left (e x \right )^{m} b d m +7 x^{3} \left (e x \right )^{m} b c \,m^{2}+x^{2} \left (e x \right )^{m} a d \,m^{3}+6 x^{4} \left (e x \right )^{m} b d +14 x^{3} \left (e x \right )^{m} b c m +8 x^{2} \left (e x \right )^{m} a d \,m^{2}+x \left (e x \right )^{m} a c \,m^{3}+8 x^{3} \left (e x \right )^{m} b c +19 x^{2} \left (e x \right )^{m} a d m +9 x \left (e x \right )^{m} a c \,m^{2}+12 x^{2} \left (e x \right )^{m} a d +26 x \left (e x \right )^{m} a c m +24 x \left (e x \right )^{m} a c}{\left (4+m \right ) \left (3+m \right ) \left (2+m \right ) \left (1+m \right )}\) | \(231\) |
Input:
int((e*x)^m*(d*x+c)*(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
a*c/(1+m)*x*exp(m*ln(e*x))+a*d/(2+m)*x^2*exp(m*ln(e*x))+b*c/(3+m)*x^3*exp( m*ln(e*x))+b*d/(4+m)*x^4*exp(m*ln(e*x))
Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.82 \[ \int (e x)^m (c+d x) \left (a+b x^2\right ) \, dx=\frac {{\left ({\left (b d m^{3} + 6 \, b d m^{2} + 11 \, b d m + 6 \, b d\right )} x^{4} + {\left (b c m^{3} + 7 \, b c m^{2} + 14 \, b c m + 8 \, b c\right )} x^{3} + {\left (a d m^{3} + 8 \, a d m^{2} + 19 \, a d m + 12 \, a d\right )} x^{2} + {\left (a c m^{3} + 9 \, a c m^{2} + 26 \, a c m + 24 \, a c\right )} x\right )} \left (e x\right )^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^2+a),x, algorithm="fricas")
Output:
((b*d*m^3 + 6*b*d*m^2 + 11*b*d*m + 6*b*d)*x^4 + (b*c*m^3 + 7*b*c*m^2 + 14* b*c*m + 8*b*c)*x^3 + (a*d*m^3 + 8*a*d*m^2 + 19*a*d*m + 12*a*d)*x^2 + (a*c* m^3 + 9*a*c*m^2 + 26*a*c*m + 24*a*c)*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 (c+d x) \left (a+b x^2\right ) \, dx=\begin {cases} \frac {- \frac {a c}{3 x^{3}} - \frac {a d}{2 x^{2}} - \frac {b c}{x} + b d \log {\left (x \right )}}{e^{4}} & \text {for}\: m = -4 \\\frac {- \frac {a c}{2 x^{2}} - \frac {a d}{x} + b c \log {\left (x \right )} + b d x}{e^{3}} & \text {for}\: m = -3 \\\frac {- \frac {a c}{x} + a d \log {\left (x \right )} + b c x + \frac {b d x^{2}}{2}}{e^{2}} & \text {for}\: m = -2 \\\frac {a c \log {\left (x \right )} + a d x + \frac {b c x^{2}}{2} + \frac {b d x^{3}}{3}}{e} & \text {for}\: m = -1 \\\frac {a c m^{3} x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 a c m^{2} x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 a c m x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 a c x \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {a d m^{3} x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {8 a d m^{2} x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {19 a d m x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {12 a d x^{2} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {b c m^{3} x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {7 b c m^{2} x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {14 b c m x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {8 b c x^{3} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {b d m^{3} x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 b d m^{2} x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {11 b d m x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {6 b d x^{4} \left (e x\right )^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \] Input:
integrate((e*x)**m*(d*x+c)*(b*x**2+a),x)
Output:
Piecewise(((-a*c/(3*x**3) - a*d/(2*x**2) - b*c/x + b*d*log(x))/e**4, Eq(m, -4)), ((-a*c/(2*x**2) - a*d/x + b*c*log(x) + b*d*x)/e**3, Eq(m, -3)), ((- a*c/x + a*d*log(x) + b*c*x + b*d*x**2/2)/e**2, Eq(m, -2)), ((a*c*log(x) + a*d*x + b*c*x**2/2 + b*d*x**3/3)/e, Eq(m, -1)), (a*c*m**3*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 9*a*c*m**2*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 26*a*c*m*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 5 0*m + 24) + 24*a*c*x*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + a*d *m**3*x**2*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 8*a*d*m**2*x* *2*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 19*a*d*m*x**2*(e*x)** m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 12*a*d*x**2*(e*x)**m/(m**4 + 10 *m**3 + 35*m**2 + 50*m + 24) + b*c*m**3*x**3*(e*x)**m/(m**4 + 10*m**3 + 35 *m**2 + 50*m + 24) + 7*b*c*m**2*x**3*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 14*b*c*m*x**3*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 8*b*c*x**3*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + b*d*m**3*x **4*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 6*b*d*m**2*x**4*(e*x )**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 11*b*d*m*x**4*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 6*b*d*x**4*(e*x)**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24), True))
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int (e x)^m (c+d x) \left (a+b x^2\right ) \, dx=\frac {b d e^{m} x^{4} x^{m}}{m + 4} + \frac {b c e^{m} x^{3} x^{m}}{m + 3} + \frac {a d e^{m} x^{2} x^{m}}{m + 2} + \frac {\left (e x\right )^{m + 1} a c}{e {\left (m + 1\right )}} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^2+a),x, algorithm="maxima")
Output:
b*d*e^m*x^4*x^m/(m + 4) + b*c*e^m*x^3*x^m/(m + 3) + a*d*e^m*x^2*x^m/(m + 2 ) + (e*x)^(m + 1)*a*c/(e*(m + 1))
Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (73) = 146\).
Time = 0.12 (sec) , antiderivative size = 230, normalized size of antiderivative = 3.15 \[ \int (e x)^m (c+d x) \left (a+b x^2\right ) \, dx=\frac {\left (e x\right )^{m} b d m^{3} x^{4} + \left (e x\right )^{m} b c m^{3} x^{3} + 6 \, \left (e x\right )^{m} b d m^{2} x^{4} + \left (e x\right )^{m} a d m^{3} x^{2} + 7 \, \left (e x\right )^{m} b c m^{2} x^{3} + 11 \, \left (e x\right )^{m} b d m x^{4} + \left (e x\right )^{m} a c m^{3} x + 8 \, \left (e x\right )^{m} a d m^{2} x^{2} + 14 \, \left (e x\right )^{m} b c m x^{3} + 6 \, \left (e x\right )^{m} b d x^{4} + 9 \, \left (e x\right )^{m} a c m^{2} x + 19 \, \left (e x\right )^{m} a d m x^{2} + 8 \, \left (e x\right )^{m} b c x^{3} + 26 \, \left (e x\right )^{m} a c m x + 12 \, \left (e x\right )^{m} a d x^{2} + 24 \, \left (e x\right )^{m} a c x}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \] Input:
integrate((e*x)^m*(d*x+c)*(b*x^2+a),x, algorithm="giac")
Output:
((e*x)^m*b*d*m^3*x^4 + (e*x)^m*b*c*m^3*x^3 + 6*(e*x)^m*b*d*m^2*x^4 + (e*x) ^m*a*d*m^3*x^2 + 7*(e*x)^m*b*c*m^2*x^3 + 11*(e*x)^m*b*d*m*x^4 + (e*x)^m*a* c*m^3*x + 8*(e*x)^m*a*d*m^2*x^2 + 14*(e*x)^m*b*c*m*x^3 + 6*(e*x)^m*b*d*x^4 + 9*(e*x)^m*a*c*m^2*x + 19*(e*x)^m*a*d*m*x^2 + 8*(e*x)^m*b*c*x^3 + 26*(e* x)^m*a*c*m*x + 12*(e*x)^m*a*d*x^2 + 24*(e*x)^m*a*c*x)/(m^4 + 10*m^3 + 35*m ^2 + 50*m + 24)
Time = 9.02 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.21 \[ \int (e x)^m (c+d x) \left (a+b x^2\right ) \, dx={\left (e\,x\right )}^m\,\left (\frac {a\,c\,x\,\left (m^3+9\,m^2+26\,m+24\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {b\,c\,x^3\,\left (m^3+7\,m^2+14\,m+8\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {a\,d\,x^2\,\left (m^3+8\,m^2+19\,m+12\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {b\,d\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}\right ) \] Input:
int((e*x)^m*(a + b*x^2)*(c + d*x),x)
Output:
(e*x)^m*((a*c*x*(26*m + 9*m^2 + m^3 + 24))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (b*c*x^3*(14*m + 7*m^2 + m^3 + 8))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (a*d*x^2*(19*m + 8*m^2 + m^3 + 12))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (b*d*x^4*(11*m + 6*m^2 + m^3 + 6))/(50*m + 35*m^2 + 10*m^3 + m^4 + 2 4))
Time = 0.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.99 \[ \int (e x)^m (c+d x) \left (a+b x^2\right ) \, dx=\frac {x^{m} e^{m} x \left (b d \,m^{3} x^{3}+b c \,m^{3} x^{2}+6 b d \,m^{2} x^{3}+a d \,m^{3} x +7 b c \,m^{2} x^{2}+11 b d m \,x^{3}+a c \,m^{3}+8 a d \,m^{2} x +14 b c m \,x^{2}+6 b d \,x^{3}+9 a c \,m^{2}+19 a d m x +8 b c \,x^{2}+26 a c m +12 a d x +24 a c \right )}{m^{4}+10 m^{3}+35 m^{2}+50 m +24} \] Input:
int((e*x)^m*(d*x+c)*(b*x^2+a),x)
Output:
(x**m*e**m*x*(a*c*m**3 + 9*a*c*m**2 + 26*a*c*m + 24*a*c + a*d*m**3*x + 8*a *d*m**2*x + 19*a*d*m*x + 12*a*d*x + b*c*m**3*x**2 + 7*b*c*m**2*x**2 + 14*b *c*m*x**2 + 8*b*c*x**2 + b*d*m**3*x**3 + 6*b*d*m**2*x**3 + 11*b*d*m*x**3 + 6*b*d*x**3))/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24)