\(\int (b x+c x^2)^2 \, dx\) [235]

Optimal result

Integrand size = 11, antiderivative size = 30 \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {b^2 x^3}{3}+\frac {1}{2} b c x^4+\frac {c^2 x^5}{5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {625} \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {b^2 x^3}{3}+\frac {1}{2} b c x^4+\frac {c^2 x^5}{5} \]

[In]

Int[(b*x + c*x^2)^2,x]

[Out]

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

Rule 625

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && (EqQ[a, 0] ||  !PerfectSquareQ[b^2 - 4*a*c])

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {b^2 x^3}{3}+\frac {1}{2} b c x^4+\frac {c^2 x^5}{5} \]

[In]

Integrate[(b*x + c*x^2)^2,x]

[Out]

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

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {x^{3} \left (6 c^{2} x^{2}+15 b c x +10 b^{2}\right )}{30}\)	\(25\)
default	\(\frac {1}{3} b^{2} x^{3}+\frac {1}{2} b c \,x^{4}+\frac {1}{5} x^{5} c^{2}\)	\(25\)
norman	\(\frac {1}{3} b^{2} x^{3}+\frac {1}{2} b c \,x^{4}+\frac {1}{5} x^{5} c^{2}\)	\(25\)
risch	\(\frac {1}{3} b^{2} x^{3}+\frac {1}{2} b c \,x^{4}+\frac {1}{5} x^{5} c^{2}\)	\(25\)
parallelrisch	\(\frac {1}{3} b^{2} x^{3}+\frac {1}{2} b c \,x^{4}+\frac {1}{5} x^{5} c^{2}\)	\(25\)

[In]

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

[Out]

1/30*x^3*(6*c^2*x^2+15*b*c*x+10*b^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {1}{5} \, c^{2} x^{5} + \frac {1}{2} \, b c x^{4} + \frac {1}{3} \, b^{2} x^{3} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {b^{2} x^{3}}{3} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} \]

[In]

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

[Out]

b**2*x**3/3 + b*c*x**4/2 + c**2*x**5/5

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {1}{5} \, c^{2} x^{5} + \frac {1}{2} \, b c x^{4} + \frac {1}{3} \, b^{2} x^{3} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {1}{5} \, c^{2} x^{5} + \frac {1}{2} \, b c x^{4} + \frac {1}{3} \, b^{2} x^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x+c x^2\right )^2 \, dx=\frac {b^2\,x^3}{3}+\frac {b\,c\,x^4}{2}+\frac {c^2\,x^5}{5} \]

[In]

int((b*x + c*x^2)^2,x)

[Out]

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