\(\int (b x+c x^2)^3 \, dx\) [248]

Optimal result

Integrand size = 11, antiderivative size = 43 \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {b^3 x^4}{4}+\frac {3}{5} b^2 c x^5+\frac {1}{2} b c^2 x^6+\frac {c^3 x^7}{7} \]

[Out]

1/4*b^3*x^4+3/5*b^2*c*x^5+1/2*b*c^2*x^6+1/7*c^3*x^7

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, 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} \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {b^3 x^4}{4}+\frac {3}{5} b^2 c x^5+\frac {1}{2} b c^2 x^6+\frac {c^3 x^7}{7} \]

[In]

Int[(b*x + c*x^2)^3,x]

[Out]

(b^3*x^4)/4 + (3*b^2*c*x^5)/5 + (b*c^2*x^6)/2 + (c^3*x^7)/7

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^3 x^3+3 b^2 c x^4+3 b c^2 x^5+c^3 x^6\right ) \, dx \\ & = \frac {b^3 x^4}{4}+\frac {3}{5} b^2 c x^5+\frac {1}{2} b c^2 x^6+\frac {c^3 x^7}{7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {b^3 x^4}{4}+\frac {3}{5} b^2 c x^5+\frac {1}{2} b c^2 x^6+\frac {c^3 x^7}{7} \]

[In]

Integrate[(b*x + c*x^2)^3,x]

[Out]

(b^3*x^4)/4 + (3*b^2*c*x^5)/5 + (b*c^2*x^6)/2 + (c^3*x^7)/7

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {x^{4} \left (20 c^{3} x^{3}+70 b \,c^{2} x^{2}+84 b^{2} c x +35 b^{3}\right )}{140}\)	\(36\)
default	\(\frac {1}{4} b^{3} x^{4}+\frac {3}{5} b^{2} c \,x^{5}+\frac {1}{2} b \,c^{2} x^{6}+\frac {1}{7} c^{3} x^{7}\)	\(36\)
norman	\(\frac {1}{4} b^{3} x^{4}+\frac {3}{5} b^{2} c \,x^{5}+\frac {1}{2} b \,c^{2} x^{6}+\frac {1}{7} c^{3} x^{7}\)	\(36\)
risch	\(\frac {1}{4} b^{3} x^{4}+\frac {3}{5} b^{2} c \,x^{5}+\frac {1}{2} b \,c^{2} x^{6}+\frac {1}{7} c^{3} x^{7}\)	\(36\)
parallelrisch	\(\frac {1}{4} b^{3} x^{4}+\frac {3}{5} b^{2} c \,x^{5}+\frac {1}{2} b \,c^{2} x^{6}+\frac {1}{7} c^{3} x^{7}\)	\(36\)

[In]

int((c*x^2+b*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/140*x^4*(20*c^3*x^3+70*b*c^2*x^2+84*b^2*c*x+35*b^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {1}{7} \, c^{3} x^{7} + \frac {1}{2} \, b c^{2} x^{6} + \frac {3}{5} \, b^{2} c x^{5} + \frac {1}{4} \, b^{3} x^{4} \]

[In]

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

[Out]

1/7*c^3*x^7 + 1/2*b*c^2*x^6 + 3/5*b^2*c*x^5 + 1/4*b^3*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {b^{3} x^{4}}{4} + \frac {3 b^{2} c x^{5}}{5} + \frac {b c^{2} x^{6}}{2} + \frac {c^{3} x^{7}}{7} \]

[In]

integrate((c*x**2+b*x)**3,x)

[Out]

b**3*x**4/4 + 3*b**2*c*x**5/5 + b*c**2*x**6/2 + c**3*x**7/7

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {1}{7} \, c^{3} x^{7} + \frac {1}{2} \, b c^{2} x^{6} + \frac {3}{5} \, b^{2} c x^{5} + \frac {1}{4} \, b^{3} x^{4} \]

[In]

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

[Out]

1/7*c^3*x^7 + 1/2*b*c^2*x^6 + 3/5*b^2*c*x^5 + 1/4*b^3*x^4

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {1}{7} \, c^{3} x^{7} + \frac {1}{2} \, b c^{2} x^{6} + \frac {3}{5} \, b^{2} c x^{5} + \frac {1}{4} \, b^{3} x^{4} \]

[In]

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

[Out]

1/7*c^3*x^7 + 1/2*b*c^2*x^6 + 3/5*b^2*c*x^5 + 1/4*b^3*x^4

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \left (b x+c x^2\right )^3 \, dx=\frac {b^3\,x^4}{4}+\frac {3\,b^2\,c\,x^5}{5}+\frac {b\,c^2\,x^6}{2}+\frac {c^3\,x^7}{7} \]

[In]

int((b*x + c*x^2)^3,x)

[Out]

(b^3*x^4)/4 + (c^3*x^7)/7 + (3*b^2*c*x^5)/5 + (b*c^2*x^6)/2