\(\int (d+e x) (a+b x^2+c x^4)^2 \, dx\) [6]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 112 \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=a^2 d x+\frac {1}{2} a^2 e x^2+\frac {2}{3} a b d x^3+\frac {1}{2} a b e x^4+\frac {1}{5} \left (b^2+2 a c\right ) d x^5+\frac {1}{6} \left (b^2+2 a c\right ) e x^6+\frac {2}{7} b c d x^7+\frac {1}{4} b c e x^8+\frac {1}{9} c^2 d x^9+\frac {1}{10} c^2 e x^{10} \]

[Out]

a^2*d*x+1/2*a^2*e*x^2+2/3*a*b*d*x^3+1/2*a*b*e*x^4+1/5*(2*a*c+b^2)*d*x^5+1/6*(2*a*c+b^2)*e*x^6+2/7*b*c*d*x^7+1/
4*b*c*e*x^8+1/9*c^2*d*x^9+1/10*c^2*e*x^10

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1685} \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=a^2 d x+\frac {1}{2} a^2 e x^2+\frac {1}{5} d x^5 \left (2 a c+b^2\right )+\frac {1}{6} e x^6 \left (2 a c+b^2\right )+\frac {2}{3} a b d x^3+\frac {1}{2} a b e x^4+\frac {2}{7} b c d x^7+\frac {1}{4} b c e x^8+\frac {1}{9} c^2 d x^9+\frac {1}{10} c^2 e x^{10} \]

[In]

Int[(d + e*x)*(a + b*x^2 + c*x^4)^2,x]

[Out]

a^2*d*x + (a^2*e*x^2)/2 + (2*a*b*d*x^3)/3 + (a*b*e*x^4)/2 + ((b^2 + 2*a*c)*d*x^5)/5 + ((b^2 + 2*a*c)*e*x^6)/6
+ (2*b*c*d*x^7)/7 + (b*c*e*x^8)/4 + (c^2*d*x^9)/9 + (c^2*e*x^10)/10

Rule 1685

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2 + c*x^4)^
p, x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 d+a^2 e x+2 a b d x^2+2 a b e x^3+\left (b^2+2 a c\right ) d x^4+\left (b^2+2 a c\right ) e x^5+2 b c d x^6+2 b c e x^7+c^2 d x^8+c^2 e x^9\right ) \, dx \\ & = a^2 d x+\frac {1}{2} a^2 e x^2+\frac {2}{3} a b d x^3+\frac {1}{2} a b e x^4+\frac {1}{5} \left (b^2+2 a c\right ) d x^5+\frac {1}{6} \left (b^2+2 a c\right ) e x^6+\frac {2}{7} b c d x^7+\frac {1}{4} b c e x^8+\frac {1}{9} c^2 d x^9+\frac {1}{10} c^2 e x^{10} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=\frac {630 a^2 x (2 d+e x)+42 b^2 x^5 (6 d+5 e x)+45 b c x^7 (8 d+7 e x)+14 c^2 x^9 (10 d+9 e x)+42 a \left (5 b x^3 (4 d+3 e x)+2 c x^5 (6 d+5 e x)\right )}{1260} \]

[In]

Integrate[(d + e*x)*(a + b*x^2 + c*x^4)^2,x]

[Out]

(630*a^2*x*(2*d + e*x) + 42*b^2*x^5*(6*d + 5*e*x) + 45*b*c*x^7*(8*d + 7*e*x) + 14*c^2*x^9*(10*d + 9*e*x) + 42*
a*(5*b*x^3*(4*d + 3*e*x) + 2*c*x^5*(6*d + 5*e*x)))/1260

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.85

method result size
default \(a^{2} d x +\frac {a^{2} e \,x^{2}}{2}+\frac {2 x^{3} d a b}{3}+\frac {a b e \,x^{4}}{2}+\frac {\left (2 a c +b^{2}\right ) d \,x^{5}}{5}+\frac {\left (2 a c +b^{2}\right ) e \,x^{6}}{6}+\frac {2 x^{7} b c d}{7}+\frac {b c e \,x^{8}}{4}+\frac {c^{2} d \,x^{9}}{9}+\frac {c^{2} e \,x^{10}}{10}\) \(95\)
norman \(\frac {c^{2} e \,x^{10}}{10}+\frac {c^{2} d \,x^{9}}{9}+\frac {b c e \,x^{8}}{4}+\frac {2 x^{7} b c d}{7}+\left (\frac {1}{3} a c e +\frac {1}{6} b^{2} e \right ) x^{6}+\left (\frac {2}{5} a c d +\frac {1}{5} b^{2} d \right ) x^{5}+\frac {a b e \,x^{4}}{2}+\frac {2 x^{3} d a b}{3}+\frac {a^{2} e \,x^{2}}{2}+a^{2} d x\) \(99\)
gosper \(\frac {1}{10} c^{2} e \,x^{10}+\frac {1}{9} c^{2} d \,x^{9}+\frac {1}{4} b c e \,x^{8}+\frac {2}{7} x^{7} b c d +\frac {1}{3} x^{6} a c e +\frac {1}{6} x^{6} b^{2} e +\frac {2}{5} a c d \,x^{5}+\frac {1}{5} x^{5} b^{2} d +\frac {1}{2} a b e \,x^{4}+\frac {2}{3} x^{3} d a b +\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\) \(101\)
risch \(\frac {1}{10} c^{2} e \,x^{10}+\frac {1}{9} c^{2} d \,x^{9}+\frac {1}{4} b c e \,x^{8}+\frac {2}{7} x^{7} b c d +\frac {1}{3} x^{6} a c e +\frac {1}{6} x^{6} b^{2} e +\frac {2}{5} a c d \,x^{5}+\frac {1}{5} x^{5} b^{2} d +\frac {1}{2} a b e \,x^{4}+\frac {2}{3} x^{3} d a b +\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\) \(101\)
parallelrisch \(\frac {1}{10} c^{2} e \,x^{10}+\frac {1}{9} c^{2} d \,x^{9}+\frac {1}{4} b c e \,x^{8}+\frac {2}{7} x^{7} b c d +\frac {1}{3} x^{6} a c e +\frac {1}{6} x^{6} b^{2} e +\frac {2}{5} a c d \,x^{5}+\frac {1}{5} x^{5} b^{2} d +\frac {1}{2} a b e \,x^{4}+\frac {2}{3} x^{3} d a b +\frac {1}{2} a^{2} e \,x^{2}+a^{2} d x\) \(101\)

[In]

int((e*x+d)*(c*x^4+b*x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

a^2*d*x+1/2*a^2*e*x^2+2/3*x^3*d*a*b+1/2*a*b*e*x^4+1/5*(2*a*c+b^2)*d*x^5+1/6*(2*a*c+b^2)*e*x^6+2/7*x^7*b*c*d+1/
4*b*c*e*x^8+1/9*c^2*d*x^9+1/10*c^2*e*x^10

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=\frac {1}{10} \, c^{2} e x^{10} + \frac {1}{9} \, c^{2} d x^{9} + \frac {1}{4} \, b c e x^{8} + \frac {2}{7} \, b c d x^{7} + \frac {1}{6} \, {\left (b^{2} + 2 \, a c\right )} e x^{6} + \frac {1}{2} \, a b e x^{4} + \frac {1}{5} \, {\left (b^{2} + 2 \, a c\right )} d x^{5} + \frac {2}{3} \, a b d x^{3} + \frac {1}{2} \, a^{2} e x^{2} + a^{2} d x \]

[In]

integrate((e*x+d)*(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

1/10*c^2*e*x^10 + 1/9*c^2*d*x^9 + 1/4*b*c*e*x^8 + 2/7*b*c*d*x^7 + 1/6*(b^2 + 2*a*c)*e*x^6 + 1/2*a*b*e*x^4 + 1/
5*(b^2 + 2*a*c)*d*x^5 + 2/3*a*b*d*x^3 + 1/2*a^2*e*x^2 + a^2*d*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04 \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=a^{2} d x + \frac {a^{2} e x^{2}}{2} + \frac {2 a b d x^{3}}{3} + \frac {a b e x^{4}}{2} + \frac {2 b c d x^{7}}{7} + \frac {b c e x^{8}}{4} + \frac {c^{2} d x^{9}}{9} + \frac {c^{2} e x^{10}}{10} + x^{6} \left (\frac {a c e}{3} + \frac {b^{2} e}{6}\right ) + x^{5} \cdot \left (\frac {2 a c d}{5} + \frac {b^{2} d}{5}\right ) \]

[In]

integrate((e*x+d)*(c*x**4+b*x**2+a)**2,x)

[Out]

a**2*d*x + a**2*e*x**2/2 + 2*a*b*d*x**3/3 + a*b*e*x**4/2 + 2*b*c*d*x**7/7 + b*c*e*x**8/4 + c**2*d*x**9/9 + c**
2*e*x**10/10 + x**6*(a*c*e/3 + b**2*e/6) + x**5*(2*a*c*d/5 + b**2*d/5)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=\frac {1}{10} \, c^{2} e x^{10} + \frac {1}{9} \, c^{2} d x^{9} + \frac {1}{4} \, b c e x^{8} + \frac {2}{7} \, b c d x^{7} + \frac {1}{6} \, {\left (b^{2} + 2 \, a c\right )} e x^{6} + \frac {1}{2} \, a b e x^{4} + \frac {1}{5} \, {\left (b^{2} + 2 \, a c\right )} d x^{5} + \frac {2}{3} \, a b d x^{3} + \frac {1}{2} \, a^{2} e x^{2} + a^{2} d x \]

[In]

integrate((e*x+d)*(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

1/10*c^2*e*x^10 + 1/9*c^2*d*x^9 + 1/4*b*c*e*x^8 + 2/7*b*c*d*x^7 + 1/6*(b^2 + 2*a*c)*e*x^6 + 1/2*a*b*e*x^4 + 1/
5*(b^2 + 2*a*c)*d*x^5 + 2/3*a*b*d*x^3 + 1/2*a^2*e*x^2 + a^2*d*x

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=\frac {1}{10} \, c^{2} e x^{10} + \frac {1}{9} \, c^{2} d x^{9} + \frac {1}{4} \, b c e x^{8} + \frac {2}{7} \, b c d x^{7} + \frac {1}{6} \, b^{2} e x^{6} + \frac {1}{3} \, a c e x^{6} + \frac {1}{5} \, b^{2} d x^{5} + \frac {2}{5} \, a c d x^{5} + \frac {1}{2} \, a b e x^{4} + \frac {2}{3} \, a b d x^{3} + \frac {1}{2} \, a^{2} e x^{2} + a^{2} d x \]

[In]

integrate((e*x+d)*(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

1/10*c^2*e*x^10 + 1/9*c^2*d*x^9 + 1/4*b*c*e*x^8 + 2/7*b*c*d*x^7 + 1/6*b^2*e*x^6 + 1/3*a*c*e*x^6 + 1/5*b^2*d*x^
5 + 2/5*a*c*d*x^5 + 1/2*a*b*e*x^4 + 2/3*a*b*d*x^3 + 1/2*a^2*e*x^2 + a^2*d*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int (d+e x) \left (a+b x^2+c x^4\right )^2 \, dx=\frac {a^2\,e\,x^2}{2}+\frac {c^2\,d\,x^9}{9}+\frac {c^2\,e\,x^{10}}{10}+\frac {d\,x^5\,\left (b^2+2\,a\,c\right )}{5}+\frac {e\,x^6\,\left (b^2+2\,a\,c\right )}{6}+a^2\,d\,x+\frac {2\,a\,b\,d\,x^3}{3}+\frac {a\,b\,e\,x^4}{2}+\frac {2\,b\,c\,d\,x^7}{7}+\frac {b\,c\,e\,x^8}{4} \]

[In]

int((d + e*x)*(a + b*x^2 + c*x^4)^2,x)

[Out]

(a^2*e*x^2)/2 + (c^2*d*x^9)/9 + (c^2*e*x^10)/10 + (d*x^5*(2*a*c + b^2))/5 + (e*x^6*(2*a*c + b^2))/6 + a^2*d*x
+ (2*a*b*d*x^3)/3 + (a*b*e*x^4)/2 + (2*b*c*d*x^7)/7 + (b*c*e*x^8)/4