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

Optimal. Leaf size=77 \[ a e^2 x+\frac{2}{5} a e f x^5+\frac{1}{9} a f^2 x^9+\frac{1}{2} b e^2 x^2+\frac{1}{3} b e f x^6+\frac{1}{10} b f^2 x^{10}+\frac{d \left (e+f x^4\right )^3}{12 f} \]

[Out]

a*e^2*x + (b*e^2*x^2)/2 + (2*a*e*f*x^5)/5 + (b*e*f*x^6)/3 + (a*f^2*x^9)/9 + (b*f^2*x^10)/10 + (d*(e + f*x^4)^3
)/(12*f)

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Rubi [A]  time = 0.0511904, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1582, 1850} \[ a e^2 x+\frac{2}{5} a e f x^5+\frac{1}{9} a f^2 x^9+\frac{1}{2} b e^2 x^2+\frac{1}{3} b e f x^6+\frac{1}{10} b f^2 x^{10}+\frac{d \left (e+f x^4\right )^3}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*e^2*x + (b*e^2*x^2)/2 + (2*a*e*f*x^5)/5 + (b*e*f*x^6)/3 + (a*f^2*x^9)/9 + (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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin{align*} \int \left (a+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 (a+b x) \left (e+f x^4\right )^2 \, dx\\ &=\frac{d \left (e+f x^4\right )^3}{12 f}+\int \left (a e^2+b e^2 x+2 a e f x^4+2 b e f x^5+a f^2 x^8+b f^2 x^9\right ) \, dx\\ &=a e^2 x+\frac{1}{2} b e^2 x^2+\frac{2}{5} a e f x^5+\frac{1}{3} b e f x^6+\frac{1}{9} a f^2 x^9+\frac{1}{10} b f^2 x^{10}+\frac{d \left (e+f x^4\right )^3}{12 f}\\ \end{align*}

Mathematica [A]  time = 0.0033786, size = 92, normalized size = 1.19 \[ a e^2 x+\frac{2}{5} a e f x^5+\frac{1}{9} a f^2 x^9+\frac{1}{2} b e^2 x^2+\frac{1}{3} b e f x^6+\frac{1}{10} b f^2 x^{10}+\frac{1}{4} d e^2 x^4+\frac{1}{4} d e f x^8+\frac{1}{12} d f^2 x^{12} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*e^2*x + (b*e^2*x^2)/2 + (d*e^2*x^4)/4 + (2*a*e*f*x^5)/5 + (b*e*f*x^6)/3 + (d*e*f*x^8)/4 + (a*f^2*x^9)/9 + (b
*f^2*x^10)/10 + (d*f^2*x^12)/12

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Maple [A]  time = 0.04, size = 77, normalized size = 1. \begin{align*}{\frac{d{f}^{2}{x}^{12}}{12}}+{\frac{b{f}^{2}{x}^{10}}{10}}+{\frac{a{f}^{2}{x}^{9}}{9}}+{\frac{def{x}^{8}}{4}}+{\frac{bef{x}^{6}}{3}}+{\frac{2\,aef{x}^{5}}{5}}+{\frac{d{e}^{2}{x}^{4}}{4}}+{\frac{b{e}^{2}{x}^{2}}{2}}+a{e}^{2}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*d*f^2*x^12+1/10*b*f^2*x^10+1/9*a*f^2*x^9+1/4*d*e*f*x^8+1/3*b*e*f*x^6+2/5*a*e*f*x^5+1/4*d*e^2*x^4+1/2*b*e^
2*x^2+a*e^2*x

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Maxima [A]  time = 1.00375, size = 103, normalized size = 1.34 \begin{align*} \frac{1}{12} \, d f^{2} x^{12} + \frac{1}{10} \, b f^{2} x^{10} + \frac{1}{9} \, a f^{2} x^{9} + \frac{1}{4} \, d e f x^{8} + \frac{1}{3} \, b e f x^{6} + \frac{2}{5} \, a e f x^{5} + \frac{1}{4} \, d e^{2} x^{4} + \frac{1}{2} \, b e^{2} x^{2} + a e^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*d*f^2*x^12 + 1/10*b*f^2*x^10 + 1/9*a*f^2*x^9 + 1/4*d*e*f*x^8 + 1/3*b*e*f*x^6 + 2/5*a*e*f*x^5 + 1/4*d*e^2*
x^4 + 1/2*b*e^2*x^2 + a*e^2*x

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Fricas [A]  time = 1.09603, size = 190, normalized size = 2.47 \begin{align*} \frac{1}{12} x^{12} f^{2} d + \frac{1}{10} x^{10} f^{2} b + \frac{1}{9} x^{9} f^{2} a + \frac{1}{4} x^{8} f e d + \frac{1}{3} x^{6} f e b + \frac{2}{5} x^{5} f e a + \frac{1}{4} x^{4} e^{2} d + \frac{1}{2} x^{2} e^{2} b + x e^{2} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^12*f^2*d + 1/10*x^10*f^2*b + 1/9*x^9*f^2*a + 1/4*x^8*f*e*d + 1/3*x^6*f*e*b + 2/5*x^5*f*e*a + 1/4*x^4*e^
2*d + 1/2*x^2*e^2*b + x*e^2*a

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Sympy [A]  time = 0.068377, size = 88, normalized size = 1.14 \begin{align*} a e^{2} x + \frac{2 a e f x^{5}}{5} + \frac{a f^{2} x^{9}}{9} + \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+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 + 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 = 1.05752, size = 103, normalized size = 1.34 \begin{align*} \frac{1}{12} \, d f^{2} x^{12} + \frac{1}{10} \, b f^{2} x^{10} + \frac{1}{9} \, a f^{2} x^{9} + \frac{1}{4} \, d f x^{8} e + \frac{1}{3} \, b f x^{6} e + \frac{2}{5} \, a f x^{5} e + \frac{1}{4} \, d x^{4} e^{2} + \frac{1}{2} \, b x^{2} e^{2} + a x e^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*d*f^2*x^12 + 1/10*b*f^2*x^10 + 1/9*a*f^2*x^9 + 1/4*d*f*x^8*e + 1/3*b*f*x^6*e + 2/5*a*f*x^5*e + 1/4*d*x^4*
e^2 + 1/2*b*x^2*e^2 + a*x*e^2