Integrand size = 20, antiderivative size = 67 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=a d^2 x+\frac {1}{3} d (c d+2 a f) x^3+\frac {1}{5} f (2 c d+a f) x^5+\frac {1}{7} c f^2 x^7+\frac {b \left (d+f x^2\right )^3}{6 f} \] Output:
a*d^2*x+1/3*d*(2*a*f+c*d)*x^3+1/5*f*(a*f+2*c*d)*x^5+1/7*c*f^2*x^7+1/6*b*(f *x^2+d)^3/f
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.22 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=a d^2 x+\frac {1}{2} b d^2 x^2+\frac {1}{3} d (c d+2 a f) x^3+\frac {1}{2} b d f x^4+\frac {1}{5} f (2 c d+a f) x^5+\frac {1}{6} b f^2 x^6+\frac {1}{7} c f^2 x^7 \] Input:
Integrate[(a + b*x + c*x^2)*(d + f*x^2)^2,x]
Output:
a*d^2*x + (b*d^2*x^2)/2 + (d*(c*d + 2*a*f)*x^3)/3 + (b*d*f*x^4)/2 + (f*(2* c*d + a*f)*x^5)/5 + (b*f^2*x^6)/6 + (c*f^2*x^7)/7
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2017, 290, 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.
\(\displaystyle \int \left (d+f x^2\right )^2 \left (a+b x+c x^2\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (c x^2+a\right ) \left (f x^2+d\right )^2dx+\frac {b \left (d+f x^2\right )^3}{6 f}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \int \left (c f^2 x^6+f (2 c d+a f) x^4+d (c d+2 a f) x^2+a d^2\right )dx+\frac {b \left (d+f x^2\right )^3}{6 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} f x^5 (a f+2 c d)+\frac {1}{3} d x^3 (2 a f+c d)+a d^2 x+\frac {b \left (d+f x^2\right )^3}{6 f}+\frac {1}{7} c f^2 x^7\) |
Input:
Int[(a + b*x + c*x^2)*(d + f*x^2)^2,x]
Output:
a*d^2*x + (d*(c*d + 2*a*f)*x^3)/3 + (f*(2*c*d + a*f)*x^5)/5 + (c*f^2*x^7)/ 7 + (b*(d + f*x^2)^3)/(6*f)
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coeff[Px, x, n - 1] *x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && IGtQ[p , 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] && !MatchQ[Px, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ [{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Coeff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]
Time = 0.78 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {c \,f^{2} x^{7}}{7}+\frac {b \,f^{2} x^{6}}{6}+\frac {\left (a \,f^{2}+2 d f c \right ) x^{5}}{5}+\frac {b d f \,x^{4}}{2}+\frac {\left (2 a d f +c \,d^{2}\right ) x^{3}}{3}+\frac {b \,x^{2} d^{2}}{2}+a \,d^{2} x\) | \(75\) |
norman | \(\frac {c \,f^{2} x^{7}}{7}+\frac {b \,f^{2} x^{6}}{6}+\left (\frac {1}{5} a \,f^{2}+\frac {2}{5} d f c \right ) x^{5}+\frac {b d f \,x^{4}}{2}+\left (\frac {2}{3} a d f +\frac {1}{3} c \,d^{2}\right ) x^{3}+\frac {b \,x^{2} d^{2}}{2}+a \,d^{2} x\) | \(75\) |
gosper | \(\frac {1}{7} c \,f^{2} x^{7}+\frac {1}{6} b \,f^{2} x^{6}+\frac {1}{5} x^{5} a \,f^{2}+\frac {2}{5} x^{5} d f c +\frac {1}{2} b d f \,x^{4}+\frac {2}{3} a d f \,x^{3}+\frac {1}{3} c \,d^{2} x^{3}+\frac {1}{2} b \,x^{2} d^{2}+a \,d^{2} x\) | \(77\) |
risch | \(\frac {1}{7} c \,f^{2} x^{7}+\frac {1}{6} b \,f^{2} x^{6}+\frac {1}{5} x^{5} a \,f^{2}+\frac {2}{5} x^{5} d f c +\frac {1}{2} b d f \,x^{4}+\frac {2}{3} a d f \,x^{3}+\frac {1}{3} c \,d^{2} x^{3}+\frac {1}{2} b \,x^{2} d^{2}+a \,d^{2} x\) | \(77\) |
parallelrisch | \(\frac {1}{7} c \,f^{2} x^{7}+\frac {1}{6} b \,f^{2} x^{6}+\frac {1}{5} x^{5} a \,f^{2}+\frac {2}{5} x^{5} d f c +\frac {1}{2} b d f \,x^{4}+\frac {2}{3} a d f \,x^{3}+\frac {1}{3} c \,d^{2} x^{3}+\frac {1}{2} b \,x^{2} d^{2}+a \,d^{2} x\) | \(77\) |
orering | \(\frac {x \left (30 c \,f^{2} x^{6}+35 b \,f^{2} x^{5}+42 a \,f^{2} x^{4}+84 c d f \,x^{4}+105 b d f \,x^{3}+140 a d f \,x^{2}+70 c \,d^{2} x^{2}+105 b x \,d^{2}+210 a \,d^{2}\right )}{210}\) | \(78\) |
Input:
int((c*x^2+b*x+a)*(f*x^2+d)^2,x,method=_RETURNVERBOSE)
Output:
1/7*c*f^2*x^7+1/6*b*f^2*x^6+1/5*(a*f^2+2*c*d*f)*x^5+1/2*b*d*f*x^4+1/3*(2*a *d*f+c*d^2)*x^3+1/2*b*x^2*d^2+a*d^2*x
Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=\frac {1}{7} \, c f^{2} x^{7} + \frac {1}{6} \, b f^{2} x^{6} + \frac {1}{2} \, b d f x^{4} + \frac {1}{5} \, {\left (2 \, c d f + a f^{2}\right )} x^{5} + \frac {1}{2} \, b d^{2} x^{2} + a d^{2} x + \frac {1}{3} \, {\left (c d^{2} + 2 \, a d f\right )} x^{3} \] Input:
integrate((c*x^2+b*x+a)*(f*x^2+d)^2,x, algorithm="fricas")
Output:
1/7*c*f^2*x^7 + 1/6*b*f^2*x^6 + 1/2*b*d*f*x^4 + 1/5*(2*c*d*f + a*f^2)*x^5 + 1/2*b*d^2*x^2 + a*d^2*x + 1/3*(c*d^2 + 2*a*d*f)*x^3
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=a d^{2} x + \frac {b d^{2} x^{2}}{2} + \frac {b d f x^{4}}{2} + \frac {b f^{2} x^{6}}{6} + \frac {c f^{2} x^{7}}{7} + x^{5} \left (\frac {a f^{2}}{5} + \frac {2 c d f}{5}\right ) + x^{3} \cdot \left (\frac {2 a d f}{3} + \frac {c d^{2}}{3}\right ) \] Input:
integrate((c*x**2+b*x+a)*(f*x**2+d)**2,x)
Output:
a*d**2*x + b*d**2*x**2/2 + b*d*f*x**4/2 + b*f**2*x**6/6 + c*f**2*x**7/7 + x**5*(a*f**2/5 + 2*c*d*f/5) + x**3*(2*a*d*f/3 + c*d**2/3)
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=\frac {1}{7} \, c f^{2} x^{7} + \frac {1}{6} \, b f^{2} x^{6} + \frac {1}{2} \, b d f x^{4} + \frac {1}{5} \, {\left (2 \, c d f + a f^{2}\right )} x^{5} + \frac {1}{2} \, b d^{2} x^{2} + a d^{2} x + \frac {1}{3} \, {\left (c d^{2} + 2 \, a d f\right )} x^{3} \] Input:
integrate((c*x^2+b*x+a)*(f*x^2+d)^2,x, algorithm="maxima")
Output:
1/7*c*f^2*x^7 + 1/6*b*f^2*x^6 + 1/2*b*d*f*x^4 + 1/5*(2*c*d*f + a*f^2)*x^5 + 1/2*b*d^2*x^2 + a*d^2*x + 1/3*(c*d^2 + 2*a*d*f)*x^3
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.13 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=\frac {1}{7} \, c f^{2} x^{7} + \frac {1}{6} \, b f^{2} x^{6} + \frac {2}{5} \, c d f x^{5} + \frac {1}{5} \, a f^{2} x^{5} + \frac {1}{2} \, b d f x^{4} + \frac {1}{3} \, c d^{2} x^{3} + \frac {2}{3} \, a d f x^{3} + \frac {1}{2} \, b d^{2} x^{2} + a d^{2} x \] Input:
integrate((c*x^2+b*x+a)*(f*x^2+d)^2,x, algorithm="giac")
Output:
1/7*c*f^2*x^7 + 1/6*b*f^2*x^6 + 2/5*c*d*f*x^5 + 1/5*a*f^2*x^5 + 1/2*b*d*f* x^4 + 1/3*c*d^2*x^3 + 2/3*a*d*f*x^3 + 1/2*b*d^2*x^2 + a*d^2*x
Time = 15.96 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=x^3\,\left (\frac {c\,d^2}{3}+\frac {2\,a\,f\,d}{3}\right )+x^5\,\left (\frac {a\,f^2}{5}+\frac {2\,c\,d\,f}{5}\right )+\frac {b\,d^2\,x^2}{2}+\frac {b\,f^2\,x^6}{6}+\frac {c\,f^2\,x^7}{7}+a\,d^2\,x+\frac {b\,d\,f\,x^4}{2} \] Input:
int((d + f*x^2)^2*(a + b*x + c*x^2),x)
Output:
x^3*((c*d^2)/3 + (2*a*d*f)/3) + x^5*((a*f^2)/5 + (2*c*d*f)/5) + (b*d^2*x^2 )/2 + (b*f^2*x^6)/6 + (c*f^2*x^7)/7 + a*d^2*x + (b*d*f*x^4)/2
Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \left (a+b x+c x^2\right ) \left (d+f x^2\right )^2 \, dx=\frac {x \left (30 c \,f^{2} x^{6}+35 b \,f^{2} x^{5}+42 a \,f^{2} x^{4}+84 c d f \,x^{4}+105 b d f \,x^{3}+140 a d f \,x^{2}+70 c \,d^{2} x^{2}+105 b \,d^{2} x +210 a \,d^{2}\right )}{210} \] Input:
int((c*x^2+b*x+a)*(f*x^2+d)^2,x)
Output:
(x*(210*a*d**2 + 140*a*d*f*x**2 + 42*a*f**2*x**4 + 105*b*d**2*x + 105*b*d* f*x**3 + 35*b*f**2*x**5 + 70*c*d**2*x**2 + 84*c*d*f*x**4 + 30*c*f**2*x**6) )/210