3.134 \(\int b x (e+f x^4)^2 \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{2} b e^2 x^2+\frac {1}{3} b e f x^6+\frac {1}{10} b f^2 x^{10} \]

[Out]

1/2*b*e^2*x^2+1/3*b*e*f*x^6+1/10*b*f^2*x^10

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 270} \[ \frac {1}{2} b e^2 x^2+\frac {1}{3} b e f x^6+\frac {1}{10} b f^2 x^{10} \]

Antiderivative was successfully verified.

[In]

Int[b*x*(e + f*x^4)^2,x]

[Out]

(b*e^2*x^2)/2 + (b*e*f*x^6)/3 + (b*f^2*x^10)/10

Rule 12

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

Rule 270

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

Rubi steps

\begin {align*} \int b x \left (e+f x^4\right )^2 \, dx &=b \int x \left (e+f x^4\right )^2 \, dx\\ &=b \int \left (e^2 x+2 e f x^5+f^2 x^9\right ) \, dx\\ &=\frac {1}{2} b e^2 x^2+\frac {1}{3} b e f x^6+\frac {1}{10} b f^2 x^{10}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 32, normalized size = 0.97 \[ b \left (\frac {e^2 x^2}{2}+\frac {1}{3} e f x^6+\frac {f^2 x^{10}}{10}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[b*x*(e + f*x^4)^2,x]

[Out]

b*((e^2*x^2)/2 + (e*f*x^6)/3 + (f^2*x^10)/10)

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fricas [A]  time = 0.49, size = 27, normalized size = 0.82 \[ \frac {1}{10} x^{10} f^{2} b + \frac {1}{3} x^{6} f e b + \frac {1}{2} x^{2} e^{2} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*x^10*f^2*b + 1/3*x^6*f*e*b + 1/2*x^2*e^2*b

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giac [A]  time = 0.17, size = 27, normalized size = 0.82 \[ \frac {1}{30} \, {\left (3 \, f^{2} x^{10} + 10 \, f x^{6} e + 15 \, x^{2} e^{2}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(3*f^2*x^10 + 10*f*x^6*e + 15*x^2*e^2)*b

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maple [A]  time = 0.04, size = 27, normalized size = 0.82 \[ \left (\frac {1}{10} f^{2} x^{10}+\frac {1}{3} e f \,x^{6}+\frac {1}{2} e^{2} x^{2}\right ) b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(b*x*(f*x^4+e)^2,x)

[Out]

b*(1/10*f^2*x^10+1/3*e*f*x^6+1/2*e^2*x^2)

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maxima [A]  time = 1.40, size = 27, normalized size = 0.82 \[ \frac {1}{30} \, {\left (3 \, f^{2} x^{10} + 10 \, e f x^{6} + 15 \, e^{2} x^{2}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(3*f^2*x^10 + 10*e*f*x^6 + 15*e^2*x^2)*b

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mupad [B]  time = 0.03, size = 27, normalized size = 0.82 \[ \frac {b\,x^2\,\left (15\,e^2+10\,e\,f\,x^4+3\,f^2\,x^8\right )}{30} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(b*x*(e + f*x^4)^2,x)

[Out]

(b*x^2*(15*e^2 + 3*f^2*x^8 + 10*e*f*x^4))/30

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sympy [A]  time = 0.07, size = 29, normalized size = 0.88 \[ \frac {b e^{2} x^{2}}{2} + \frac {b e f x^{6}}{3} + \frac {b f^{2} x^{10}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*e**2*x**2/2 + b*e*f*x**6/3 + b*f**2*x**10/10

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