\(\int (d+e x) (b x+c x^2)^2 \, dx\) [234]

Optimal result

Integrand size = 17, antiderivative size = 55 \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=\frac {1}{3} b^2 d x^3+\frac {1}{4} b (2 c d+b e) x^4+\frac {1}{5} c (c d+2 b e) x^5+\frac {1}{6} c^2 e x^6 \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (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 = {645} \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=\frac {1}{3} b^2 d x^3+\frac {1}{5} c x^5 (2 b e+c d)+\frac {1}{4} b x^4 (b e+2 c d)+\frac {1}{6} c^2 e x^6 \]

[In]

Int[(d + e*x)*(b*x + c*x^2)^2,x]

[Out]

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

Rule 645

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=\frac {1}{60} x^3 \left (5 b^2 (4 d+3 e x)+6 b c x (5 d+4 e x)+2 c^2 x^2 (6 d+5 e x)\right ) \]

[In]

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

[Out]

(x^3*(5*b^2*(4*d + 3*e*x) + 6*b*c*x*(5*d + 4*e*x) + 2*c^2*x^2*(6*d + 5*e*x)))/60

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {x^{3} \left (10 c^{2} e \,x^{3}+24 x^{2} b c e +12 c^{2} d \,x^{2}+15 b^{2} e x +30 b c d x +20 b^{2} d \right )}{60}\)	\(52\)
default	\(\frac {c^{2} e \,x^{6}}{6}+\frac {\left (2 b c e +c^{2} d \right ) x^{5}}{5}+\frac {\left (b^{2} e +2 b c d \right ) x^{4}}{4}+\frac {d \,x^{3} b^{2}}{3}\)	\(52\)
norman	\(\frac {c^{2} e \,x^{6}}{6}+\left (\frac {2}{5} b c e +\frac {1}{5} c^{2} d \right ) x^{5}+\left (\frac {1}{4} b^{2} e +\frac {1}{2} b c d \right ) x^{4}+\frac {d \,x^{3} b^{2}}{3}\)	\(52\)
risch	\(\frac {1}{6} c^{2} e \,x^{6}+\frac {2}{5} x^{5} b c e +\frac {1}{5} c^{2} d \,x^{5}+\frac {1}{4} b^{2} e \,x^{4}+\frac {1}{2} c b d \,x^{4}+\frac {1}{3} d \,x^{3} b^{2}\)	\(54\)
parallelrisch	\(\frac {1}{6} c^{2} e \,x^{6}+\frac {2}{5} x^{5} b c e +\frac {1}{5} c^{2} d \,x^{5}+\frac {1}{4} b^{2} e \,x^{4}+\frac {1}{2} c b d \,x^{4}+\frac {1}{3} d \,x^{3} b^{2}\)	\(54\)

[In]

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

[Out]

1/60*x^3*(10*c^2*e*x^3+24*b*c*e*x^2+12*c^2*d*x^2+15*b^2*e*x+30*b*c*d*x+20*b^2*d)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=\frac {1}{6} \, c^{2} e x^{6} + \frac {1}{3} \, b^{2} d x^{3} + \frac {1}{5} \, {\left (c^{2} d + 2 \, b c e\right )} x^{5} + \frac {1}{4} \, {\left (2 \, b c d + b^{2} e\right )} x^{4} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=\frac {b^{2} d x^{3}}{3} + \frac {c^{2} e x^{6}}{6} + x^{5} \cdot \left (\frac {2 b c e}{5} + \frac {c^{2} d}{5}\right ) + x^{4} \left (\frac {b^{2} e}{4} + \frac {b c d}{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=\frac {1}{6} \, c^{2} e x^{6} + \frac {1}{3} \, b^{2} d x^{3} + \frac {1}{5} \, {\left (c^{2} d + 2 \, b c e\right )} x^{5} + \frac {1}{4} \, {\left (2 \, b c d + b^{2} e\right )} x^{4} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=\frac {1}{6} \, c^{2} e x^{6} + \frac {1}{5} \, c^{2} d x^{5} + \frac {2}{5} \, b c e x^{5} + \frac {1}{2} \, b c d x^{4} + \frac {1}{4} \, b^{2} e x^{4} + \frac {1}{3} \, b^{2} d x^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int (d+e x) \left (b x+c x^2\right )^2 \, dx=x^4\,\left (\frac {e\,b^2}{4}+\frac {c\,d\,b}{2}\right )+x^5\,\left (\frac {d\,c^2}{5}+\frac {2\,b\,e\,c}{5}\right )+\frac {b^2\,d\,x^3}{3}+\frac {c^2\,e\,x^6}{6} \]

[In]

int((b*x + c*x^2)^2*(d + e*x),x)

[Out]

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