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

Optimal result

Integrand size = 17, antiderivative size = 55 \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=\frac {1}{2} b d^2 x^2+\frac {1}{3} d (c d+2 b e) x^3+\frac {1}{4} e (2 c d+b e) x^4+\frac {1}{5} c e^2 x^5 \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {712} \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=\frac {1}{4} e x^4 (b e+2 c d)+\frac {1}{3} d x^3 (2 b e+c d)+\frac {1}{2} b d^2 x^2+\frac {1}{5} c e^2 x^5 \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=\frac {1}{60} x^2 \left (30 b d^2+20 d (c d+2 b e) x+15 e (2 c d+b e) x^2+12 c e^2 x^3\right ) \]

[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

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 52, normalized size of antiderivative = 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	\(\frac {c \,e^{2} x^{5}}{5}+\frac {\left (e^{2} b +2 c d e \right ) x^{4}}{4}+\frac {\left (2 b d e +c \,d^{2}\right ) x^{3}}{3}+\frac {b \,d^{2} x^{2}}{2}\)	\(52\)
norman	\(\frac {c \,e^{2} x^{5}}{5}+\left (\frac {1}{4} e^{2} b +\frac {1}{2} c d e \right ) x^{4}+\left (\frac {2}{3} b d e +\frac {1}{3} c \,d^{2}\right ) x^{3}+\frac {b \,d^{2} x^{2}}{2}\)	\(52\)
risch	\(\frac {1}{5} c \,e^{2} x^{5}+\frac {1}{4} x^{4} e^{2} b +\frac {1}{2} x^{4} c d e +\frac {2}{3} x^{3} b d e +\frac {1}{3} c \,d^{2} x^{3}+\frac {1}{2} b \,d^{2} x^{2}\)	\(54\)
parallelrisch	\(\frac {1}{5} c \,e^{2} x^{5}+\frac {1}{4} x^{4} e^{2} b +\frac {1}{2} x^{4} c d e +\frac {2}{3} x^{3} b d e +\frac {1}{3} c \,d^{2} x^{3}+\frac {1}{2} b \,d^{2} x^{2}\)	\(54\)

[In]

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

[Out]

1/60*x^2*(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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=\frac {1}{5} \, c e^{2} x^{5} + \frac {1}{2} \, b d^{2} x^{2} + \frac {1}{4} \, {\left (2 \, c d e + b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (c d^{2} + 2 \, b d e\right )} x^{3} \]

[In]

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

[Out]

1/5*c*e^2*x^5 + 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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=\frac {b d^{2} x^{2}}{2} + \frac {c e^{2} x^{5}}{5} + x^{4} \left (\frac {b e^{2}}{4} + \frac {c d e}{2}\right ) + x^{3} \cdot \left (\frac {2 b d e}{3} + \frac {c d^{2}}{3}\right ) \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=\frac {1}{5} \, c e^{2} x^{5} + \frac {1}{2} \, b d^{2} x^{2} + \frac {1}{4} \, {\left (2 \, c d e + b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (c d^{2} + 2 \, b d e\right )} x^{3} \]

[In]

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

[Out]

1/5*c*e^2*x^5 + 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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=\frac {1}{5} \, c e^{2} x^{5} + \frac {1}{2} \, c d e x^{4} + \frac {1}{4} \, b e^{2} x^{4} + \frac {1}{3} \, c d^{2} x^{3} + \frac {2}{3} \, b d e x^{3} + \frac {1}{2} \, b d^{2} x^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (d+e x)^2 \left (b x+c x^2\right ) \, dx=x^3\,\left (\frac {c\,d^2}{3}+\frac {2\,b\,e\,d}{3}\right )+x^4\,\left (\frac {b\,e^2}{4}+\frac {c\,d\,e}{2}\right )+\frac {b\,d^2\,x^2}{2}+\frac {c\,e^2\,x^5}{5} \]

[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