3.3.23 \(\int (d+e x)^2 (b x+c x^2) \, dx\) [223]

Optimal. Leaf size=55 \[ \frac {1}{2} b d^2 x^2+\frac {1}{3} d (c d+2 b e) x^3+\frac {1}{4} e (2 c d+b e) x^4+\frac {1}{5} c e^2 x^5 \]

[Out]

1/2*b*d^2*x^2+1/3*d*(2*b*e+c*d)*x^3+1/4*e*(b*e+2*c*d)*x^4+1/5*c*e^2*x^5

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Rubi [A]
time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {712} \begin {gather*} \frac {1}{4} e x^4 (b e+2 c d)+\frac {1}{3} d x^3 (2 b e+c d)+\frac {1}{2} b d^2 x^2+\frac {1}{5} c e^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(b*x + c*x^2),x]

[Out]

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

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 49, normalized size = 0.89 \begin {gather*} \frac {1}{60} x^2 \left (30 b d^2+20 d (c d+2 b e) x+15 e (2 c d+b e) x^2+12 c e^2 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(30*b*d^2 + 20*d*(c*d + 2*b*e)*x + 15*e*(2*c*d + b*e)*x^2 + 12*c*e^2*x^3))/60

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Maple [A]
time = 0.39, size = 52, normalized size = 0.95

method	result	size
gosper	\(\frac {x^{2} \left (12 c \,e^{2} x^{3}+15 b \,e^{2} x^{2}+30 c d e \,x^{2}+40 b d e x +20 c \,d^{2} x +30 b \,d^{2}\right )}{60}\)	\(52\)
default	\(\frac {c \,e^{2} x^{5}}{5}+\frac {\left (b \,e^{2}+2 d e c \right ) x^{4}}{4}+\frac {\left (2 b d e +c \,d^{2}\right ) x^{3}}{3}+\frac {b \,d^{2} x^{2}}{2}\)	\(52\)
norman	\(\frac {c \,e^{2} x^{5}}{5}+\left (\frac {1}{4} b \,e^{2}+\frac {1}{2} d e c \right ) x^{4}+\left (\frac {2}{3} b d e +\frac {1}{3} c \,d^{2}\right ) x^{3}+\frac {b \,d^{2} x^{2}}{2}\)	\(52\)
risch	\(\frac {1}{5} c \,e^{2} x^{5}+\frac {1}{4} x^{4} b \,e^{2}+\frac {1}{2} x^{4} d e c +\frac {2}{3} x^{3} b d e +\frac {1}{3} x^{3} c \,d^{2}+\frac {1}{2} b \,d^{2} x^{2}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(c*x^2+b*x),x,method=_RETURNVERBOSE)

[Out]

1/5*c*e^2*x^5+1/4*(b*e^2+2*c*d*e)*x^4+1/3*(2*b*d*e+c*d^2)*x^3+1/2*b*d^2*x^2

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Maxima [A]
time = 0.27, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{5} \, c x^{5} e^{2} + \frac {1}{2} \, b d^{2} x^{2} + \frac {1}{4} \, {\left (2 \, c d e + b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (c d^{2} + 2 \, b d e\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*c*x^5*e^2 + 1/2*b*d^2*x^2 + 1/4*(2*c*d*e + b*e^2)*x^4 + 1/3*(c*d^2 + 2*b*d*e)*x^3

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Fricas [A]
time = 1.55, size = 55, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, c d^{2} x^{3} + \frac {1}{2} \, b d^{2} x^{2} + \frac {1}{20} \, {\left (4 \, c x^{5} + 5 \, b x^{4}\right )} e^{2} + \frac {1}{6} \, {\left (3 \, c d x^{4} + 4 \, b d x^{3}\right )} e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*c*d^2*x^3 + 1/2*b*d^2*x^2 + 1/20*(4*c*x^5 + 5*b*x^4)*e^2 + 1/6*(3*c*d*x^4 + 4*b*d*x^3)*e

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Sympy [A]
time = 0.01, size = 54, normalized size = 0.98 \begin {gather*} \frac {b d^{2} x^{2}}{2} + \frac {c e^{2} x^{5}}{5} + x^{4} \left (\frac {b e^{2}}{4} + \frac {c d e}{2}\right ) + x^{3} \cdot \left (\frac {2 b d e}{3} + \frac {c d^{2}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.04, size = 53, normalized size = 0.96 \begin {gather*} \frac {1}{5} \, c x^{5} e^{2} + \frac {1}{2} \, c d x^{4} e + \frac {1}{3} \, c d^{2} x^{3} + \frac {1}{4} \, b x^{4} e^{2} + \frac {2}{3} \, b d x^{3} e + \frac {1}{2} \, b d^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.05, size = 51, normalized size = 0.93 \begin {gather*} x^3\,\left (\frac {c\,d^2}{3}+\frac {2\,b\,e\,d}{3}\right )+x^4\,\left (\frac {b\,e^2}{4}+\frac {c\,d\,e}{2}\right )+\frac {b\,d^2\,x^2}{2}+\frac {c\,e^2\,x^5}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)*(d + e*x)^2,x)

[Out]

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

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