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

Optimal. Leaf size=50 \[ a c x+\frac {1}{2} a d x^2+\frac {1}{3} a e x^3+\frac {1}{4} b c x^4+\frac {1}{5} b d x^5+\frac {1}{6} b e x^6 \]

[Out]

a*c*x+1/2*a*d*x^2+1/3*a*e*x^3+1/4*b*c*x^4+1/5*b*d*x^5+1/6*b*e*x^6

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1657} \[ a c x+\frac {1}{2} a d x^2+\frac {1}{3} a e x^3+\frac {1}{4} b c x^4+\frac {1}{5} b d x^5+\frac {1}{6} b e x^6 \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x + e*x^2)*(a + b*x^3),x]

[Out]

a*c*x + (a*d*x^2)/2 + (a*e*x^3)/3 + (b*c*x^4)/4 + (b*d*x^5)/5 + (b*e*x^6)/6

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

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Mathematica [A]  time = 0.00, size = 50, normalized size = 1.00 \[ a c x+\frac {1}{2} a d x^2+\frac {1}{3} a e x^3+\frac {1}{4} b c x^4+\frac {1}{5} b d x^5+\frac {1}{6} b e x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*c*x + (a*d*x^2)/2 + (a*e*x^3)/3 + (b*c*x^4)/4 + (b*d*x^5)/5 + (b*e*x^6)/6

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fricas [A]  time = 0.46, size = 40, normalized size = 0.80 \[ \frac {1}{6} x^{6} e b + \frac {1}{5} x^{5} d b + \frac {1}{4} x^{4} c b + \frac {1}{3} x^{3} e a + \frac {1}{2} x^{2} d a + x c a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*e*b + 1/5*x^5*d*b + 1/4*x^4*c*b + 1/3*x^3*e*a + 1/2*x^2*d*a + x*c*a

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giac [A]  time = 0.15, size = 42, normalized size = 0.84 \[ \frac {1}{6} \, b x^{6} e + \frac {1}{5} \, b d x^{5} + \frac {1}{4} \, b c x^{4} + \frac {1}{3} \, a x^{3} e + \frac {1}{2} \, a d x^{2} + a c x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*x^6*e + 1/5*b*d*x^5 + 1/4*b*c*x^4 + 1/3*a*x^3*e + 1/2*a*d*x^2 + a*c*x

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maple [A]  time = 0.04, size = 41, normalized size = 0.82 \[ \frac {1}{6} b e \,x^{6}+\frac {1}{5} b d \,x^{5}+\frac {1}{4} b c \,x^{4}+\frac {1}{3} a e \,x^{3}+\frac {1}{2} a d \,x^{2}+a c x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c*x+1/2*a*d*x^2+1/3*a*e*x^3+1/4*b*c*x^4+1/5*b*d*x^5+1/6*b*e*x^6

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maxima [A]  time = 1.34, size = 40, normalized size = 0.80 \[ \frac {1}{6} \, b e x^{6} + \frac {1}{5} \, b d x^{5} + \frac {1}{4} \, b c x^{4} + \frac {1}{3} \, a e x^{3} + \frac {1}{2} \, a d x^{2} + a c x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*e*x^6 + 1/5*b*d*x^5 + 1/4*b*c*x^4 + 1/3*a*e*x^3 + 1/2*a*d*x^2 + a*c*x

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mupad [B]  time = 0.02, size = 40, normalized size = 0.80 \[ \frac {b\,e\,x^6}{6}+\frac {b\,d\,x^5}{5}+\frac {b\,c\,x^4}{4}+\frac {a\,e\,x^3}{3}+\frac {a\,d\,x^2}{2}+a\,c\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)*(c + d*x + e*x^2),x)

[Out]

a*c*x + (a*d*x^2)/2 + (b*c*x^4)/4 + (a*e*x^3)/3 + (b*d*x^5)/5 + (b*e*x^6)/6

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sympy [A]  time = 0.07, size = 46, normalized size = 0.92 \[ a c x + \frac {a d x^{2}}{2} + \frac {a e x^{3}}{3} + \frac {b c x^{4}}{4} + \frac {b d x^{5}}{5} + \frac {b e x^{6}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c*x + a*d*x**2/2 + a*e*x**3/3 + b*c*x**4/4 + b*d*x**5/5 + b*e*x**6/6

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