Integrand size = 15, antiderivative size = 70 \[ \int (d+e x)^m \left (a+c x^2\right ) \, dx=\frac {\left (c d^2+a e^2\right ) (d+e x)^{1+m}}{e^3 (1+m)}-\frac {2 c d (d+e x)^{2+m}}{e^3 (2+m)}+\frac {c (d+e x)^{3+m}}{e^3 (3+m)} \]
(a*e^2+c*d^2)*(e*x+d)^(1+m)/e^3/(1+m)-2*c*d*(e*x+d)^(2+m)/e^3/(2+m)+c*(e*x +d)^(3+m)/e^3/(3+m)
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int (d+e x)^m \left (a+c x^2\right ) \, dx=\frac {(d+e x)^{1+m} \left (\frac {c d^2+a e^2}{1+m}-\frac {2 c d (d+e x)}{2+m}+\frac {c (d+e x)^2}{3+m}\right )}{e^3} \]
((d + e*x)^(1 + m)*((c*d^2 + a*e^2)/(1 + m) - (2*c*d*(d + e*x))/(2 + m) + (c*(d + e*x)^2)/(3 + m)))/e^3
Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {476, 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.
\(\displaystyle \int \left (a+c x^2\right ) (d+e x)^m \, dx\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \int \left (\frac {\left (a e^2+c d^2\right ) (d+e x)^m}{e^2}-\frac {2 c d (d+e x)^{m+1}}{e^2}+\frac {c (d+e x)^{m+2}}{e^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a e^2+c d^2\right ) (d+e x)^{m+1}}{e^3 (m+1)}-\frac {2 c d (d+e x)^{m+2}}{e^3 (m+2)}+\frac {c (d+e x)^{m+3}}{e^3 (m+3)}\) |
((c*d^2 + a*e^2)*(d + e*x)^(1 + m))/(e^3*(1 + m)) - (2*c*d*(d + e*x)^(2 + m))/(e^3*(2 + m)) + (c*(d + e*x)^(3 + m))/(e^3*(3 + m))
3.8.23.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Time = 2.60 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.43
method | result | size |
gosper | \(\frac {\left (e x +d \right )^{1+m} \left (c \,e^{2} m^{2} x^{2}+3 c \,e^{2} m \,x^{2}+a \,e^{2} m^{2}-2 c d e m x +2 c \,x^{2} e^{2}+5 a \,e^{2} m -2 x c d e +6 e^{2} a +2 c \,d^{2}\right )}{e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )}\) | \(100\) |
risch | \(\frac {\left (c \,e^{3} m^{2} x^{3}+c d \,e^{2} m^{2} x^{2}+3 c \,e^{3} m \,x^{3}+a \,e^{3} m^{2} x +c d m \,x^{2} e^{2}+2 c \,x^{3} e^{3}+a d \,e^{2} m^{2}+5 a \,e^{3} m x -2 c \,d^{2} e m x +5 a d \,e^{2} m +6 a \,e^{3} x +6 a d \,e^{2}+2 d^{3} c \right ) \left (e x +d \right )^{m}}{\left (2+m \right ) \left (3+m \right ) \left (1+m \right ) e^{3}}\) | \(143\) |
norman | \(\frac {c \,x^{3} {\mathrm e}^{m \ln \left (e x +d \right )}}{3+m}+\frac {d \left (a \,e^{2} m^{2}+5 a \,e^{2} m +6 e^{2} a +2 c \,d^{2}\right ) {\mathrm e}^{m \ln \left (e x +d \right )}}{e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {\left (a \,e^{2} m^{2}+5 a \,e^{2} m -2 c \,d^{2} m +6 e^{2} a \right ) x \,{\mathrm e}^{m \ln \left (e x +d \right )}}{e^{2} \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {c d m \,x^{2} {\mathrm e}^{m \ln \left (e x +d \right )}}{e \left (m^{2}+5 m +6\right )}\) | \(165\) |
parallelrisch | \(\frac {x^{3} \left (e x +d \right )^{m} c \,e^{3} m^{2}+3 x^{3} \left (e x +d \right )^{m} c \,e^{3} m +x^{2} \left (e x +d \right )^{m} c d \,e^{2} m^{2}+2 x^{3} \left (e x +d \right )^{m} c \,e^{3}+x^{2} \left (e x +d \right )^{m} c d \,e^{2} m +x \left (e x +d \right )^{m} a \,e^{3} m^{2}+5 x \left (e x +d \right )^{m} a \,e^{3} m -2 x \left (e x +d \right )^{m} c \,d^{2} e m +\left (e x +d \right )^{m} a d \,e^{2} m^{2}+6 x \left (e x +d \right )^{m} a \,e^{3}+5 \left (e x +d \right )^{m} a d \,e^{2} m +6 \left (e x +d \right )^{m} a d \,e^{2}+2 \left (e x +d \right )^{m} c \,d^{3}}{e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )}\) | \(227\) |
1/e^3*(e*x+d)^(1+m)/(m^3+6*m^2+11*m+6)*(c*e^2*m^2*x^2+3*c*e^2*m*x^2+a*e^2* m^2-2*c*d*e*m*x+2*c*e^2*x^2+5*a*e^2*m-2*c*d*e*x+6*a*e^2+2*c*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (70) = 140\).
Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.13 \[ \int (d+e x)^m \left (a+c x^2\right ) \, dx=\frac {{\left (a d e^{2} m^{2} + 5 \, a d e^{2} m + 2 \, c d^{3} + 6 \, a d e^{2} + {\left (c e^{3} m^{2} + 3 \, c e^{3} m + 2 \, c e^{3}\right )} x^{3} + {\left (c d e^{2} m^{2} + c d e^{2} m\right )} x^{2} + {\left (a e^{3} m^{2} + 6 \, a e^{3} - {\left (2 \, c d^{2} e - 5 \, a e^{3}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{3} m^{3} + 6 \, e^{3} m^{2} + 11 \, e^{3} m + 6 \, e^{3}} \]
(a*d*e^2*m^2 + 5*a*d*e^2*m + 2*c*d^3 + 6*a*d*e^2 + (c*e^3*m^2 + 3*c*e^3*m + 2*c*e^3)*x^3 + (c*d*e^2*m^2 + c*d*e^2*m)*x^2 + (a*e^3*m^2 + 6*a*e^3 - (2 *c*d^2*e - 5*a*e^3)*m)*x)*(e*x + d)^m/(e^3*m^3 + 6*e^3*m^2 + 11*e^3*m + 6* e^3)
Leaf count of result is larger than twice the leaf count of optimal. 952 vs. \(2 (61) = 122\).
Time = 0.58 (sec) , antiderivative size = 952, normalized size of antiderivative = 13.60 \[ \int (d+e x)^m \left (a+c x^2\right ) \, dx=\begin {cases} d^{m} \left (a x + \frac {c x^{3}}{3}\right ) & \text {for}\: e = 0 \\- \frac {a e^{2}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {2 c d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {3 c d^{2}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {4 c d e x \log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {4 c d e x}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {2 c e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} & \text {for}\: m = -3 \\- \frac {a e^{2}}{d e^{3} + e^{4} x} - \frac {2 c d^{2} \log {\left (\frac {d}{e} + x \right )}}{d e^{3} + e^{4} x} - \frac {2 c d^{2}}{d e^{3} + e^{4} x} - \frac {2 c d e x \log {\left (\frac {d}{e} + x \right )}}{d e^{3} + e^{4} x} + \frac {c e^{2} x^{2}}{d e^{3} + e^{4} x} & \text {for}\: m = -2 \\\frac {a \log {\left (\frac {d}{e} + x \right )}}{e} + \frac {c d^{2} \log {\left (\frac {d}{e} + x \right )}}{e^{3}} - \frac {c d x}{e^{2}} + \frac {c x^{2}}{2 e} & \text {for}\: m = -1 \\\frac {a d e^{2} m^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {5 a d e^{2} m \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {6 a d e^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {a e^{3} m^{2} x \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {5 a e^{3} m x \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {6 a e^{3} x \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {2 c d^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} - \frac {2 c d^{2} e m x \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {c d e^{2} m^{2} x^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {c d e^{2} m x^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {c e^{3} m^{2} x^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {3 c e^{3} m x^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {2 c e^{3} x^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} & \text {otherwise} \end {cases} \]
Piecewise((d**m*(a*x + c*x**3/3), Eq(e, 0)), (-a*e**2/(2*d**2*e**3 + 4*d*e **4*x + 2*e**5*x**2) + 2*c*d**2*log(d/e + x)/(2*d**2*e**3 + 4*d*e**4*x + 2 *e**5*x**2) + 3*c*d**2/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 4*c*d*e* x*log(d/e + x)/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 4*c*d*e*x/(2*d** 2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 2*c*e**2*x**2*log(d/e + x)/(2*d**2*e* *3 + 4*d*e**4*x + 2*e**5*x**2), Eq(m, -3)), (-a*e**2/(d*e**3 + e**4*x) - 2 *c*d**2*log(d/e + x)/(d*e**3 + e**4*x) - 2*c*d**2/(d*e**3 + e**4*x) - 2*c* d*e*x*log(d/e + x)/(d*e**3 + e**4*x) + c*e**2*x**2/(d*e**3 + e**4*x), Eq(m , -2)), (a*log(d/e + x)/e + c*d**2*log(d/e + x)/e**3 - c*d*x/e**2 + c*x**2 /(2*e), Eq(m, -1)), (a*d*e**2*m**2*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 5*a*d*e**2*m*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 6*a*d*e**2*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + a*e**3*m**2*x*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 5*a*e**3*m*x*(d + e*x)**m/(e**3*m**3 + 6*e**3*m** 2 + 11*e**3*m + 6*e**3) + 6*a*e**3*x*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 2*c*d**3*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) - 2*c*d**2*e*m*x*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + c*d*e**2*m**2*x**2*(d + e*x)**m/(e**3*m**3 + 6*e* *3*m**2 + 11*e**3*m + 6*e**3) + c*d*e**2*m*x**2*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + c*e**3*m**2*x**3*(d + e*x)**m/(e**3...
Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.27 \[ \int (d+e x)^m \left (a+c x^2\right ) \, dx=\frac {{\left (e x + d\right )}^{m + 1} a}{e {\left (m + 1\right )}} + \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} e^{3} x^{3} + {\left (m^{2} + m\right )} d e^{2} x^{2} - 2 \, d^{2} e m x + 2 \, d^{3}\right )} {\left (e x + d\right )}^{m} c}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} e^{3}} \]
(e*x + d)^(m + 1)*a/(e*(m + 1)) + ((m^2 + 3*m + 2)*e^3*x^3 + (m^2 + m)*d*e ^2*x^2 - 2*d^2*e*m*x + 2*d^3)*(e*x + d)^m*c/((m^3 + 6*m^2 + 11*m + 6)*e^3)
Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (70) = 140\).
Time = 0.30 (sec) , antiderivative size = 237, normalized size of antiderivative = 3.39 \[ \int (d+e x)^m \left (a+c x^2\right ) \, dx=\frac {{\left (e x + d\right )}^{m} c e^{3} m^{2} x^{3} + {\left (e x + d\right )}^{m} c d e^{2} m^{2} x^{2} + 3 \, {\left (e x + d\right )}^{m} c e^{3} m x^{3} + {\left (e x + d\right )}^{m} a e^{3} m^{2} x + {\left (e x + d\right )}^{m} c d e^{2} m x^{2} + 2 \, {\left (e x + d\right )}^{m} c e^{3} x^{3} + {\left (e x + d\right )}^{m} a d e^{2} m^{2} - 2 \, {\left (e x + d\right )}^{m} c d^{2} e m x + 5 \, {\left (e x + d\right )}^{m} a e^{3} m x + 5 \, {\left (e x + d\right )}^{m} a d e^{2} m + 6 \, {\left (e x + d\right )}^{m} a e^{3} x + 2 \, {\left (e x + d\right )}^{m} c d^{3} + 6 \, {\left (e x + d\right )}^{m} a d e^{2}}{e^{3} m^{3} + 6 \, e^{3} m^{2} + 11 \, e^{3} m + 6 \, e^{3}} \]
((e*x + d)^m*c*e^3*m^2*x^3 + (e*x + d)^m*c*d*e^2*m^2*x^2 + 3*(e*x + d)^m*c *e^3*m*x^3 + (e*x + d)^m*a*e^3*m^2*x + (e*x + d)^m*c*d*e^2*m*x^2 + 2*(e*x + d)^m*c*e^3*x^3 + (e*x + d)^m*a*d*e^2*m^2 - 2*(e*x + d)^m*c*d^2*e*m*x + 5 *(e*x + d)^m*a*e^3*m*x + 5*(e*x + d)^m*a*d*e^2*m + 6*(e*x + d)^m*a*e^3*x + 2*(e*x + d)^m*c*d^3 + 6*(e*x + d)^m*a*d*e^2)/(e^3*m^3 + 6*e^3*m^2 + 11*e^ 3*m + 6*e^3)
Time = 9.63 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.33 \[ \int (d+e x)^m \left (a+c x^2\right ) \, dx={\left (d+e\,x\right )}^m\,\left (\frac {c\,x^3\,\left (m^2+3\,m+2\right )}{m^3+6\,m^2+11\,m+6}+\frac {x\,\left (-2\,c\,d^2\,e\,m+a\,e^3\,m^2+5\,a\,e^3\,m+6\,a\,e^3\right )}{e^3\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {d\,\left (2\,c\,d^2+a\,e^2\,m^2+5\,a\,e^2\,m+6\,a\,e^2\right )}{e^3\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {c\,d\,m\,x^2\,\left (m+1\right )}{e\,\left (m^3+6\,m^2+11\,m+6\right )}\right ) \]