\(\int a (e+f x^4)^2 \, dx\) [133]

Optimal result

Integrand size = 11, antiderivative size = 28 \[ \int a \left (e+f x^4\right )^2 \, dx=a e^2 x+\frac {2}{5} a e f x^5+\frac {1}{9} a f^2 x^9 \]

[Out]

a*e^2*x+2/5*a*e*f*x^5+1/9*a*f^2*x^9

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 200} \[ \int a \left (e+f x^4\right )^2 \, dx=a e^2 x+\frac {2}{5} a e f x^5+\frac {1}{9} a f^2 x^9 \]

[In]

Int[a*(e + f*x^4)^2,x]

[Out]

a*e^2*x + (2*a*e*f*x^5)/5 + (a*f^2*x^9)/9

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int a \left (e+f x^4\right )^2 \, dx=a \left (e^2 x+\frac {2}{5} e f x^5+\frac {f^2 x^9}{9}\right ) \]

[In]

Integrate[a*(e + f*x^4)^2,x]

[Out]

a*(e^2*x + (2*e*f*x^5)/5 + (f^2*x^9)/9)

Maple [A] (verified)

Time = 1.43 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
default	\(\left (\frac {1}{9} f^{2} x^{9}+\frac {2}{5} e f \,x^{5}+e^{2} x \right ) a\)	\(24\)
parallelrisch	\(\left (\frac {1}{9} f^{2} x^{9}+\frac {2}{5} e f \,x^{5}+e^{2} x \right ) a\)	\(24\)
norman	\(a \,e^{2} x +\frac {2}{5} a e f \,x^{5}+\frac {1}{9} a \,f^{2} x^{9}\)	\(25\)
risch	\(a \,e^{2} x +\frac {2}{5} a e f \,x^{5}+\frac {1}{9} a \,f^{2} x^{9}\)	\(25\)
gosper	\(\frac {x \left (5 f^{2} x^{8}+18 e f \,x^{4}+45 e^{2}\right ) a}{45}\)	\(26\)

[In]

int(a*(f*x^4+e)^2,x,method=_RETURNVERBOSE)

[Out]

(1/9*f^2*x^9+2/5*e*f*x^5+e^2*x)*a

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int a \left (e+f x^4\right )^2 \, dx=\frac {1}{9} \, a f^{2} x^{9} + \frac {2}{5} \, a e f x^{5} + a e^{2} x \]

[In]

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

[Out]

1/9*a*f^2*x^9 + 2/5*a*e*f*x^5 + a*e^2*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int a \left (e+f x^4\right )^2 \, dx=a e^{2} x + \frac {2 a e f x^{5}}{5} + \frac {a f^{2} x^{9}}{9} \]

[In]

integrate(a*(f*x**4+e)**2,x)

[Out]

a*e**2*x + 2*a*e*f*x**5/5 + a*f**2*x**9/9

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int a \left (e+f x^4\right )^2 \, dx=\frac {1}{45} \, {\left (5 \, f^{2} x^{9} + 18 \, e f x^{5} + 45 \, e^{2} x\right )} a \]

[In]

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

[Out]

1/45*(5*f^2*x^9 + 18*e*f*x^5 + 45*e^2*x)*a

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int a \left (e+f x^4\right )^2 \, dx=\frac {1}{45} \, {\left (5 \, f^{2} x^{9} + 18 \, e f x^{5} + 45 \, e^{2} x\right )} a \]

[In]

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

[Out]

1/45*(5*f^2*x^9 + 18*e*f*x^5 + 45*e^2*x)*a

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int a \left (e+f x^4\right )^2 \, dx=\frac {a\,x\,\left (45\,e^2+18\,e\,f\,x^4+5\,f^2\,x^8\right )}{45} \]

[In]

int(a*(e + f*x^4)^2,x)

[Out]

(a*x*(45*e^2 + 5*f^2*x^8 + 18*e*f*x^4))/45