Integrand size = 69, antiderivative size = 57 \[ \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx=-\frac {(c e (2+p)-b f (3+2 p)) \left (d+e x+f x^2\right )^{1+p}}{1+p}+2 c f x \left (d+e x+f x^2\right )^{1+p} \]
Time = 0.74 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.75 \[ \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx=\frac {(-c e (2+p)+b f (3+2 p)+2 c f (1+p) x) (d+x (e+f x))^{1+p}}{1+p} \]
Integrate[(d + e*x + f*x^2)^p*(-2*c*e^2 + 2*c*d*f + 3*b*e*f - c*e^2*p + 2* b*e*f*p + 2*b*f^2*(3 + 2*p)*x + 2*c*f^2*(3 + 2*p)*x^2),x]
Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {2192, 25, 27, 1104}
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.
\(\displaystyle \int \left (d+e x+f x^2\right )^p \left (2 b e f p+3 b e f+2 b f^2 (2 p+3) x+2 c d f-c e^2 p-2 c e^2+2 c f^2 (2 p+3) x^2\right ) \, dx\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {\int -f (2 p+3) (c e (p+2)-b f (2 p+3)) (e+2 f x) \left (f x^2+e x+d\right )^pdx}{f (2 p+3)}+2 c f x \left (d+e x+f x^2\right )^{p+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 c f x \left (d+e x+f x^2\right )^{p+1}-\frac {\int f (2 p+3) (c e (p+2)-b f (2 p+3)) (e+2 f x) \left (f x^2+e x+d\right )^pdx}{f (2 p+3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 c f x \left (d+e x+f x^2\right )^{p+1}-(c e (p+2)-b f (2 p+3)) \int (e+2 f x) \left (f x^2+e x+d\right )^pdx\) |
\(\Big \downarrow \) 1104 |
\(\displaystyle 2 c f x \left (d+e x+f x^2\right )^{p+1}-\frac {(c e (p+2)-b f (2 p+3)) \left (d+e x+f x^2\right )^{p+1}}{p+1}\) |
Int[(d + e*x + f*x^2)^p*(-2*c*e^2 + 2*c*d*f + 3*b*e*f - c*e^2*p + 2*b*e*f* p + 2*b*f^2*(3 + 2*p)*x + 2*c*f^2*(3 + 2*p)*x^2),x]
-(((c*e*(2 + p) - b*f*(3 + 2*p))*(d + e*x + f*x^2)^(1 + p))/(1 + p)) + 2*c *f*x*(d + e*x + f*x^2)^(1 + p)
3.3.77.3.1 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_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol ] :> Simp[d*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Time = 0.66 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89
method | result | size |
gosper | \(\frac {\left (f \,x^{2}+e x +d \right )^{1+p} \left (2 c f x p +2 b f p -c e p +2 c f x +3 b f -2 c e \right )}{1+p}\) | \(51\) |
risch | \(\frac {\left (2 p c \,f^{2} x^{3}+2 b \,f^{2} p \,x^{2}+c e f p \,x^{2}+2 c \,f^{2} x^{3}+2 b e f p x +3 b \,f^{2} x^{2}+2 c d f p x -c \,e^{2} p x +2 b d f p +3 b e f x -c d e p +2 c d f x -2 c \,e^{2} x +3 b d f -2 c d e \right ) \left (f \,x^{2}+e x +d \right )^{p}}{1+p}\) | \(129\) |
norman | \(\frac {d \left (2 b f p -c e p +3 b f -2 c e \right ) {\mathrm e}^{p \ln \left (f \,x^{2}+e x +d \right )}}{1+p}+\frac {\left (2 b e f p +2 c d f p -c \,e^{2} p +3 b e f +2 c d f -2 c \,e^{2}\right ) x \,{\mathrm e}^{p \ln \left (f \,x^{2}+e x +d \right )}}{1+p}+\frac {f \left (2 b f p +c e p +3 b f \right ) x^{2} {\mathrm e}^{p \ln \left (f \,x^{2}+e x +d \right )}}{1+p}+2 c \,f^{2} x^{3} {\mathrm e}^{p \ln \left (f \,x^{2}+e x +d \right )}\) | \(160\) |
parallelrisch | \(\frac {2 x^{3} \left (f \,x^{2}+e x +d \right )^{p} c d \,f^{2} p +2 x^{3} \left (f \,x^{2}+e x +d \right )^{p} c d \,f^{2}+2 x^{2} \left (f \,x^{2}+e x +d \right )^{p} b d \,f^{2} p +x^{2} \left (f \,x^{2}+e x +d \right )^{p} c d e f p +3 x^{2} \left (f \,x^{2}+e x +d \right )^{p} b d \,f^{2}+2 x \left (f \,x^{2}+e x +d \right )^{p} b d e f p +2 x \left (f \,x^{2}+e x +d \right )^{p} c \,d^{2} f p -x \left (f \,x^{2}+e x +d \right )^{p} c d \,e^{2} p +3 x \left (f \,x^{2}+e x +d \right )^{p} b d e f +2 x \left (f \,x^{2}+e x +d \right )^{p} c \,d^{2} f -2 x \left (f \,x^{2}+e x +d \right )^{p} c d \,e^{2}+2 \left (f \,x^{2}+e x +d \right )^{p} b \,d^{2} f p -\left (f \,x^{2}+e x +d \right )^{p} c \,d^{2} e p +3 \left (f \,x^{2}+e x +d \right )^{p} b \,d^{2} f -2 \left (f \,x^{2}+e x +d \right )^{p} c \,d^{2} e}{d \left (1+p \right )}\) | \(321\) |
int((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b*f^2*(3 +2*p)*x+2*c*f^2*(3+2*p)*x^2),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 123 vs. \(2 (56) = 112\).
Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.16 \[ \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx=\frac {{\left (2 \, {\left (c f^{2} p + c f^{2}\right )} x^{3} - 2 \, c d e + 3 \, b d f + {\left (3 \, b f^{2} + {\left (c e f + 2 \, b f^{2}\right )} p\right )} x^{2} - {\left (c d e - 2 \, b d f\right )} p - {\left (2 \, c e^{2} - {\left (2 \, c d + 3 \, b e\right )} f + {\left (c e^{2} - 2 \, {\left (c d + b e\right )} f\right )} p\right )} x\right )} {\left (f x^{2} + e x + d\right )}^{p}}{p + 1} \]
integrate((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b* f^2*(3+2*p)*x+2*c*f^2*(3+2*p)*x^2),x, algorithm="fricas")
(2*(c*f^2*p + c*f^2)*x^3 - 2*c*d*e + 3*b*d*f + (3*b*f^2 + (c*e*f + 2*b*f^2 )*p)*x^2 - (c*d*e - 2*b*d*f)*p - (2*c*e^2 - (2*c*d + 3*b*e)*f + (c*e^2 - 2 *(c*d + b*e)*f)*p)*x)*(f*x^2 + e*x + d)^p/(p + 1)
Leaf count of result is larger than twice the leaf count of optimal. 483 vs. \(2 (51) = 102\).
Time = 58.82 (sec) , antiderivative size = 483, normalized size of antiderivative = 8.47 \[ \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx=\begin {cases} \frac {2 b d f p \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {3 b d f \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {2 b e f p x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {3 b e f x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {2 b f^{2} p x^{2} \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {3 b f^{2} x^{2} \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac {c d e p \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac {2 c d e \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {2 c d f p x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {2 c d f x \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac {c e^{2} p x \left (d + e x + f x^{2}\right )^{p}}{p + 1} - \frac {2 c e^{2} x \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {c e f p x^{2} \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {2 c f^{2} p x^{3} \left (d + e x + f x^{2}\right )^{p}}{p + 1} + \frac {2 c f^{2} x^{3} \left (d + e x + f x^{2}\right )^{p}}{p + 1} & \text {for}\: p \neq -1 \\b f \log {\left (\frac {e}{2 f} + x - \frac {\sqrt {- 4 d f + e^{2}}}{2 f} \right )} + b f \log {\left (\frac {e}{2 f} + x + \frac {\sqrt {- 4 d f + e^{2}}}{2 f} \right )} - c e \log {\left (\frac {e}{2 f} + x - \frac {\sqrt {- 4 d f + e^{2}}}{2 f} \right )} - c e \log {\left (\frac {e}{2 f} + x + \frac {\sqrt {- 4 d f + e^{2}}}{2 f} \right )} + 2 c f x & \text {otherwise} \end {cases} \]
integrate((f*x**2+e*x+d)**p*(-2*c*e**2+2*c*d*f+3*b*e*f-c*e**2*p+2*b*e*f*p+ 2*b*f**2*(3+2*p)*x+2*c*f**2*(3+2*p)*x**2),x)
Piecewise((2*b*d*f*p*(d + e*x + f*x**2)**p/(p + 1) + 3*b*d*f*(d + e*x + f* x**2)**p/(p + 1) + 2*b*e*f*p*x*(d + e*x + f*x**2)**p/(p + 1) + 3*b*e*f*x*( d + e*x + f*x**2)**p/(p + 1) + 2*b*f**2*p*x**2*(d + e*x + f*x**2)**p/(p + 1) + 3*b*f**2*x**2*(d + e*x + f*x**2)**p/(p + 1) - c*d*e*p*(d + e*x + f*x* *2)**p/(p + 1) - 2*c*d*e*(d + e*x + f*x**2)**p/(p + 1) + 2*c*d*f*p*x*(d + e*x + f*x**2)**p/(p + 1) + 2*c*d*f*x*(d + e*x + f*x**2)**p/(p + 1) - c*e** 2*p*x*(d + e*x + f*x**2)**p/(p + 1) - 2*c*e**2*x*(d + e*x + f*x**2)**p/(p + 1) + c*e*f*p*x**2*(d + e*x + f*x**2)**p/(p + 1) + 2*c*f**2*p*x**3*(d + e *x + f*x**2)**p/(p + 1) + 2*c*f**2*x**3*(d + e*x + f*x**2)**p/(p + 1), Ne( p, -1)), (b*f*log(e/(2*f) + x - sqrt(-4*d*f + e**2)/(2*f)) + b*f*log(e/(2* f) + x + sqrt(-4*d*f + e**2)/(2*f)) - c*e*log(e/(2*f) + x - sqrt(-4*d*f + e**2)/(2*f)) - c*e*log(e/(2*f) + x + sqrt(-4*d*f + e**2)/(2*f)) + 2*c*f*x, True))
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.72 \[ \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx=\frac {{\left (2 \, c f^{2} {\left (p + 1\right )} x^{3} + b d f {\left (2 \, p + 3\right )} - c d e {\left (p + 2\right )} + {\left (b f^{2} {\left (2 \, p + 3\right )} + c e f p\right )} x^{2} + {\left (b e f {\left (2 \, p + 3\right )} - {\left (e^{2} {\left (p + 2\right )} - 2 \, d f {\left (p + 1\right )}\right )} c\right )} x\right )} {\left (f x^{2} + e x + d\right )}^{p}}{p + 1} \]
integrate((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b* f^2*(3+2*p)*x+2*c*f^2*(3+2*p)*x^2),x, algorithm="maxima")
(2*c*f^2*(p + 1)*x^3 + b*d*f*(2*p + 3) - c*d*e*(p + 2) + (b*f^2*(2*p + 3) + c*e*f*p)*x^2 + (b*e*f*(2*p + 3) - (e^2*(p + 2) - 2*d*f*(p + 1))*c)*x)*(f *x^2 + e*x + d)^p/(p + 1)
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (56) = 112\).
Time = 0.30 (sec) , antiderivative size = 296, normalized size of antiderivative = 5.19 \[ \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx=\frac {2 \, {\left (f x^{2} + e x + d\right )}^{p} c f^{2} p x^{3} + {\left (f x^{2} + e x + d\right )}^{p} c e f p x^{2} + 2 \, {\left (f x^{2} + e x + d\right )}^{p} b f^{2} p x^{2} + 2 \, {\left (f x^{2} + e x + d\right )}^{p} c f^{2} x^{3} - {\left (f x^{2} + e x + d\right )}^{p} c e^{2} p x + 2 \, {\left (f x^{2} + e x + d\right )}^{p} c d f p x + 2 \, {\left (f x^{2} + e x + d\right )}^{p} b e f p x + 3 \, {\left (f x^{2} + e x + d\right )}^{p} b f^{2} x^{2} - {\left (f x^{2} + e x + d\right )}^{p} c d e p + 2 \, {\left (f x^{2} + e x + d\right )}^{p} b d f p - 2 \, {\left (f x^{2} + e x + d\right )}^{p} c e^{2} x + 2 \, {\left (f x^{2} + e x + d\right )}^{p} c d f x + 3 \, {\left (f x^{2} + e x + d\right )}^{p} b e f x - 2 \, {\left (f x^{2} + e x + d\right )}^{p} c d e + 3 \, {\left (f x^{2} + e x + d\right )}^{p} b d f}{p + 1} \]
integrate((f*x^2+e*x+d)^p*(-2*c*e^2+2*c*d*f+3*b*e*f-c*e^2*p+2*b*e*f*p+2*b* f^2*(3+2*p)*x+2*c*f^2*(3+2*p)*x^2),x, algorithm="giac")
(2*(f*x^2 + e*x + d)^p*c*f^2*p*x^3 + (f*x^2 + e*x + d)^p*c*e*f*p*x^2 + 2*( f*x^2 + e*x + d)^p*b*f^2*p*x^2 + 2*(f*x^2 + e*x + d)^p*c*f^2*x^3 - (f*x^2 + e*x + d)^p*c*e^2*p*x + 2*(f*x^2 + e*x + d)^p*c*d*f*p*x + 2*(f*x^2 + e*x + d)^p*b*e*f*p*x + 3*(f*x^2 + e*x + d)^p*b*f^2*x^2 - (f*x^2 + e*x + d)^p*c *d*e*p + 2*(f*x^2 + e*x + d)^p*b*d*f*p - 2*(f*x^2 + e*x + d)^p*c*e^2*x + 2 *(f*x^2 + e*x + d)^p*c*d*f*x + 3*(f*x^2 + e*x + d)^p*b*e*f*x - 2*(f*x^2 + e*x + d)^p*c*d*e + 3*(f*x^2 + e*x + d)^p*b*d*f)/(p + 1)
Time = 13.57 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.11 \[ \int \left (d+e x+f x^2\right )^p \left (-2 c e^2+2 c d f+3 b e f-c e^2 p+2 b e f p+2 b f^2 (3+2 p) x+2 c f^2 (3+2 p) x^2\right ) \, dx={\left (f\,x^2+e\,x+d\right )}^p\,\left (\frac {x^2\,\left (3\,b\,f^2+2\,b\,f^2\,p+c\,e\,f\,p\right )}{p+1}+2\,c\,f^2\,x^3+\frac {d\,\left (3\,b\,f-2\,c\,e+2\,b\,f\,p-c\,e\,p\right )}{p+1}+\frac {x\,\left (3\,b\,e\,f-2\,c\,e^2+2\,c\,d\,f-c\,e^2\,p+2\,b\,e\,f\,p+2\,c\,d\,f\,p\right )}{p+1}\right ) \]
int((d + e*x + f*x^2)^p*(3*b*e*f - 2*c*e^2 + 2*c*d*f - c*e^2*p + 2*b*f^2*x *(2*p + 3) + 2*c*f^2*x^2*(2*p + 3) + 2*b*e*f*p),x)