3.142 \(\int (b x+d x^3) (e+f x^4)^2 \, dx\)

Optimal. Leaf size=50 \[ \frac{1}{2} b e^2 x^2+\frac{1}{3} b e f x^6+\frac{1}{10} b f^2 x^{10}+\frac{d \left (e+f x^4\right )^3}{12 f} \]

[Out]

(b*e^2*x^2)/2 + (b*e*f*x^6)/3 + (b*f^2*x^10)/10 + (d*(e + f*x^4)^3)/(12*f)

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Rubi [A]  time = 0.0198986, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1582, 12, 270} \[ \frac{1}{2} b e^2 x^2+\frac{1}{3} b e f x^6+\frac{1}{10} b f^2 x^{10}+\frac{d \left (e+f x^4\right )^3}{12 f} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + d*x^3)*(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 + (d*(e + f*x^4)^3)/(12*f)

Rule 1582

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] && I
GtQ[p, 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] &&  !MatchQ[P
x, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ[{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Co
eff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]

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 \left (b x+d x^3\right ) \left (e+f x^4\right )^2 \, dx &=\frac{d \left (e+f x^4\right )^3}{12 f}+\int b x \left (e+f x^4\right )^2 \, dx\\ &=\frac{d \left (e+f x^4\right )^3}{12 f}+b \int x \left (e+f x^4\right )^2 \, dx\\ &=\frac{d \left (e+f x^4\right )^3}{12 f}+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}+\frac{d \left (e+f x^4\right )^3}{12 f}\\ \end{align*}

Mathematica [A]  time = 0.002989, size = 65, normalized size = 1.3 \[ \frac{1}{2} b e^2 x^2+\frac{1}{3} b e f x^6+\frac{1}{10} b f^2 x^{10}+\frac{1}{4} d e^2 x^4+\frac{1}{4} d e f x^8+\frac{1}{12} d f^2 x^{12} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + d*x^3)*(e + f*x^4)^2,x]

[Out]

(b*e^2*x^2)/2 + (d*e^2*x^4)/4 + (b*e*f*x^6)/3 + (d*e*f*x^8)/4 + (b*f^2*x^10)/10 + (d*f^2*x^12)/12

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Maple [A]  time = 0.042, size = 54, normalized size = 1.1 \begin{align*}{\frac{d{f}^{2}{x}^{12}}{12}}+{\frac{b{f}^{2}{x}^{10}}{10}}+{\frac{def{x}^{8}}{4}}+{\frac{bef{x}^{6}}{3}}+{\frac{d{e}^{2}{x}^{4}}{4}}+{\frac{b{e}^{2}{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+b*x)*(f*x^4+e)^2,x)

[Out]

1/12*d*f^2*x^12+1/10*b*f^2*x^10+1/4*d*e*f*x^8+1/3*b*e*f*x^6+1/4*d*e^2*x^4+1/2*b*e^2*x^2

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Maxima [A]  time = 1.00185, size = 72, normalized size = 1.44 \begin{align*} \frac{1}{12} \, d f^{2} x^{12} + \frac{1}{10} \, b f^{2} x^{10} + \frac{1}{4} \, d e f x^{8} + \frac{1}{3} \, b e f x^{6} + \frac{1}{4} \, d e^{2} x^{4} + \frac{1}{2} \, b e^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+b*x)*(f*x^4+e)^2,x, algorithm="maxima")

[Out]

1/12*d*f^2*x^12 + 1/10*b*f^2*x^10 + 1/4*d*e*f*x^8 + 1/3*b*e*f*x^6 + 1/4*d*e^2*x^4 + 1/2*b*e^2*x^2

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Fricas [A]  time = 1.02542, size = 134, normalized size = 2.68 \begin{align*} \frac{1}{12} x^{12} f^{2} d + \frac{1}{10} x^{10} f^{2} b + \frac{1}{4} x^{8} f e d + \frac{1}{3} x^{6} f e b + \frac{1}{4} x^{4} e^{2} d + \frac{1}{2} x^{2} e^{2} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+b*x)*(f*x^4+e)^2,x, algorithm="fricas")

[Out]

1/12*x^12*f^2*d + 1/10*x^10*f^2*b + 1/4*x^8*f*e*d + 1/3*x^6*f*e*b + 1/4*x^4*e^2*d + 1/2*x^2*e^2*b

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Sympy [A]  time = 0.065855, size = 60, normalized size = 1.2 \begin{align*} \frac{b e^{2} x^{2}}{2} + \frac{b e f x^{6}}{3} + \frac{b f^{2} x^{10}}{10} + \frac{d e^{2} x^{4}}{4} + \frac{d e f x^{8}}{4} + \frac{d f^{2} x^{12}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+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 + d*e**2*x**4/4 + d*e*f*x**8/4 + d*f**2*x**12/12

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Giac [A]  time = 1.05355, size = 72, normalized size = 1.44 \begin{align*} \frac{1}{12} \, d f^{2} x^{12} + \frac{1}{10} \, b f^{2} x^{10} + \frac{1}{4} \, d f x^{8} e + \frac{1}{3} \, b f x^{6} e + \frac{1}{4} \, d x^{4} e^{2} + \frac{1}{2} \, b x^{2} e^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+b*x)*(f*x^4+e)^2,x, algorithm="giac")

[Out]

1/12*d*f^2*x^12 + 1/10*b*f^2*x^10 + 1/4*d*f*x^8*e + 1/3*b*f*x^6*e + 1/4*d*x^4*e^2 + 1/2*b*x^2*e^2