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

Optimal. Leaf size=109 \[ a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{3} a^2 e x^3+\frac {2}{5} a b c x^5+\frac {1}{3} a b d x^6+\frac {2}{7} a b e x^7+\frac {f \left (a+b x^4\right )^3}{12 b}+\frac {1}{9} b^2 c x^9+\frac {1}{10} b^2 d x^{10}+\frac {1}{11} b^2 e x^{11} \]

[Out]

a^2*c*x+1/2*a^2*d*x^2+1/3*a^2*e*x^3+2/5*a*b*c*x^5+1/3*a*b*d*x^6+2/7*a*b*e*x^7+1/9*b^2*c*x^9+1/10*b^2*d*x^10+1/
11*b^2*e*x^11+1/12*f*(b*x^4+a)^3/b

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Rubi [A]  time = 0.07, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1582, 1657} \[ a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{3} a^2 e x^3+\frac {2}{5} a b c x^5+\frac {1}{3} a b d x^6+\frac {2}{7} a b e x^7+\frac {f \left (a+b x^4\right )^3}{12 b}+\frac {1}{9} b^2 c x^9+\frac {1}{10} b^2 d x^{10}+\frac {1}{11} b^2 e x^{11} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*c*x + (a^2*d*x^2)/2 + (a^2*e*x^3)/3 + (2*a*b*c*x^5)/5 + (a*b*d*x^6)/3 + (2*a*b*e*x^7)/7 + (b^2*c*x^9)/9 +
(b^2*d*x^10)/10 + (b^2*e*x^11)/11 + (f*(a + b*x^4)^3)/(12*b)

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 1657

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

Rubi steps

\begin {align*} \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^2 \, dx &=\frac {f \left (a+b x^4\right )^3}{12 b}+\int \left (c+d x+e x^2\right ) \left (a+b x^4\right )^2 \, dx\\ &=\frac {f \left (a+b x^4\right )^3}{12 b}+\int \left (a^2 c+a^2 d x+a^2 e x^2+2 a b c x^4+2 a b d x^5+2 a b e x^6+b^2 c x^8+b^2 d x^9+b^2 e x^{10}\right ) \, dx\\ &=a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{3} a^2 e x^3+\frac {2}{5} a b c x^5+\frac {1}{3} a b d x^6+\frac {2}{7} a b e x^7+\frac {1}{9} b^2 c x^9+\frac {1}{10} b^2 d x^{10}+\frac {1}{11} b^2 e x^{11}+\frac {f \left (a+b x^4\right )^3}{12 b}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 124, normalized size = 1.14 \[ a^2 c x+\frac {1}{2} a^2 d x^2+\frac {1}{3} a^2 e x^3+\frac {1}{4} a^2 f x^4+\frac {2}{5} a b c x^5+\frac {1}{3} a b d x^6+\frac {2}{7} a b e x^7+\frac {1}{4} a b f x^8+\frac {1}{9} b^2 c x^9+\frac {1}{10} b^2 d x^{10}+\frac {1}{11} b^2 e x^{11}+\frac {1}{12} b^2 f x^{12} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*c*x + (a^2*d*x^2)/2 + (a^2*e*x^3)/3 + (a^2*f*x^4)/4 + (2*a*b*c*x^5)/5 + (a*b*d*x^6)/3 + (2*a*b*e*x^7)/7 +
(a*b*f*x^8)/4 + (b^2*c*x^9)/9 + (b^2*d*x^10)/10 + (b^2*e*x^11)/11 + (b^2*f*x^12)/12

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fricas [A]  time = 0.36, size = 102, normalized size = 0.94 \[ \frac {1}{12} x^{12} f b^{2} + \frac {1}{11} x^{11} e b^{2} + \frac {1}{10} x^{10} d b^{2} + \frac {1}{9} x^{9} c b^{2} + \frac {1}{4} x^{8} f b a + \frac {2}{7} x^{7} e b a + \frac {1}{3} x^{6} d b a + \frac {2}{5} x^{5} c b a + \frac {1}{4} x^{4} f a^{2} + \frac {1}{3} x^{3} e a^{2} + \frac {1}{2} x^{2} d a^{2} + x c a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^12*f*b^2 + 1/11*x^11*e*b^2 + 1/10*x^10*d*b^2 + 1/9*x^9*c*b^2 + 1/4*x^8*f*b*a + 2/7*x^7*e*b*a + 1/3*x^6*
d*b*a + 2/5*x^5*c*b*a + 1/4*x^4*f*a^2 + 1/3*x^3*e*a^2 + 1/2*x^2*d*a^2 + x*c*a^2

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giac [A]  time = 0.16, size = 105, normalized size = 0.96 \[ \frac {1}{12} \, b^{2} f x^{12} + \frac {1}{11} \, b^{2} x^{11} e + \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{9} \, b^{2} c x^{9} + \frac {1}{4} \, a b f x^{8} + \frac {2}{7} \, a b x^{7} e + \frac {1}{3} \, a b d x^{6} + \frac {2}{5} \, a b c x^{5} + \frac {1}{4} \, a^{2} f x^{4} + \frac {1}{3} \, a^{2} x^{3} e + \frac {1}{2} \, a^{2} d x^{2} + a^{2} c x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b^2*f*x^12 + 1/11*b^2*x^11*e + 1/10*b^2*d*x^10 + 1/9*b^2*c*x^9 + 1/4*a*b*f*x^8 + 2/7*a*b*x^7*e + 1/3*a*b*
d*x^6 + 2/5*a*b*c*x^5 + 1/4*a^2*f*x^4 + 1/3*a^2*x^3*e + 1/2*a^2*d*x^2 + a^2*c*x

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maple [A]  time = 0.04, size = 103, normalized size = 0.94 \[ \frac {1}{12} b^{2} f \,x^{12}+\frac {1}{11} b^{2} e \,x^{11}+\frac {1}{10} b^{2} d \,x^{10}+\frac {1}{9} b^{2} c \,x^{9}+\frac {1}{4} a b f \,x^{8}+\frac {2}{7} a b e \,x^{7}+\frac {1}{3} a b d \,x^{6}+\frac {2}{5} a b c \,x^{5}+\frac {1}{4} a^{2} f \,x^{4}+\frac {1}{3} a^{2} e \,x^{3}+\frac {1}{2} a^{2} d \,x^{2}+a^{2} c x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b^2*f*x^12+1/11*b^2*e*x^11+1/10*b^2*d*x^10+1/9*b^2*c*x^9+1/4*a*b*f*x^8+2/7*a*b*e*x^7+1/3*a*b*d*x^6+2/5*a*
b*c*x^5+1/4*f*a^2*x^4+1/3*a^2*e*x^3+1/2*a^2*d*x^2+a^2*c*x

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maxima [A]  time = 1.36, size = 102, normalized size = 0.94 \[ \frac {1}{12} \, b^{2} f x^{12} + \frac {1}{11} \, b^{2} e x^{11} + \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{9} \, b^{2} c x^{9} + \frac {1}{4} \, a b f x^{8} + \frac {2}{7} \, a b e x^{7} + \frac {1}{3} \, a b d x^{6} + \frac {2}{5} \, a b c x^{5} + \frac {1}{4} \, a^{2} f x^{4} + \frac {1}{3} \, a^{2} e x^{3} + \frac {1}{2} \, a^{2} d x^{2} + a^{2} c x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b^2*f*x^12 + 1/11*b^2*e*x^11 + 1/10*b^2*d*x^10 + 1/9*b^2*c*x^9 + 1/4*a*b*f*x^8 + 2/7*a*b*e*x^7 + 1/3*a*b*
d*x^6 + 2/5*a*b*c*x^5 + 1/4*a^2*f*x^4 + 1/3*a^2*e*x^3 + 1/2*a^2*d*x^2 + a^2*c*x

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mupad [B]  time = 0.08, size = 102, normalized size = 0.94 \[ \frac {f\,a^2\,x^4}{4}+\frac {e\,a^2\,x^3}{3}+\frac {d\,a^2\,x^2}{2}+c\,a^2\,x+\frac {f\,a\,b\,x^8}{4}+\frac {2\,e\,a\,b\,x^7}{7}+\frac {d\,a\,b\,x^6}{3}+\frac {2\,c\,a\,b\,x^5}{5}+\frac {f\,b^2\,x^{12}}{12}+\frac {e\,b^2\,x^{11}}{11}+\frac {d\,b^2\,x^{10}}{10}+\frac {c\,b^2\,x^9}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*d*x^2)/2 + (b^2*c*x^9)/9 + (a^2*e*x^3)/3 + (b^2*d*x^10)/10 + (a^2*f*x^4)/4 + (b^2*e*x^11)/11 + (b^2*f*x^1
2)/12 + a^2*c*x + (2*a*b*c*x^5)/5 + (a*b*d*x^6)/3 + (2*a*b*e*x^7)/7 + (a*b*f*x^8)/4

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sympy [A]  time = 0.09, size = 121, normalized size = 1.11 \[ a^{2} c x + \frac {a^{2} d x^{2}}{2} + \frac {a^{2} e x^{3}}{3} + \frac {a^{2} f x^{4}}{4} + \frac {2 a b c x^{5}}{5} + \frac {a b d x^{6}}{3} + \frac {2 a b e x^{7}}{7} + \frac {a b f x^{8}}{4} + \frac {b^{2} c x^{9}}{9} + \frac {b^{2} d x^{10}}{10} + \frac {b^{2} e x^{11}}{11} + \frac {b^{2} f x^{12}}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**2,x)

[Out]

a**2*c*x + a**2*d*x**2/2 + a**2*e*x**3/3 + a**2*f*x**4/4 + 2*a*b*c*x**5/5 + a*b*d*x**6/3 + 2*a*b*e*x**7/7 + a*
b*f*x**8/4 + b**2*c*x**9/9 + b**2*d*x**10/10 + b**2*e*x**11/11 + b**2*f*x**12/12

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