\(\int (a+b x)^3 \, dx\) [68]

Optimal result

Integrand size = 7, antiderivative size = 14 \[ \int (a+b x)^3 \, dx=\frac {(a+b x)^4}{4 b} \]

[Out]

1/4*(b*x+a)^4/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {32} \[ \int (a+b x)^3 \, dx=\frac {(a+b x)^4}{4 b} \]

[In]

Int[(a + b*x)^3,x]

[Out]

(a + b*x)^4/(4*b)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x)^4}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int (a+b x)^3 \, dx=\frac {(a+b x)^4}{4 b} \]

[In]

Integrate[(a + b*x)^3,x]

[Out]

(a + b*x)^4/(4*b)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\left (b x +a \right )^{4}}{4 b}\)	\(13\)
gosper	\(\frac {1}{4} b^{3} x^{4}+a \,b^{2} x^{3}+\frac {3}{2} a^{2} b \,x^{2}+a^{3} x\)	\(32\)
norman	\(\frac {1}{4} b^{3} x^{4}+a \,b^{2} x^{3}+\frac {3}{2} a^{2} b \,x^{2}+a^{3} x\)	\(32\)
parallelrisch	\(\frac {1}{4} b^{3} x^{4}+a \,b^{2} x^{3}+\frac {3}{2} a^{2} b \,x^{2}+a^{3} x\)	\(32\)
risch	\(\frac {b^{3} x^{4}}{4}+a \,b^{2} x^{3}+\frac {3 a^{2} b \,x^{2}}{2}+a^{3} x +\frac {a^{4}}{4 b}\)	\(40\)

[In]

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

[Out]

1/4*(b*x+a)^4/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (12) = 24\).

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.21 \[ \int (a+b x)^3 \, dx=\frac {1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} + a^{3} x \]

[In]

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

[Out]

1/4*b^3*x^4 + a*b^2*x^3 + 3/2*a^2*b*x^2 + a^3*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (8) = 16\).

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.29 \[ \int (a+b x)^3 \, dx=a^{3} x + \frac {3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac {b^{3} x^{4}}{4} \]

[In]

integrate((b*x+a)**3,x)

[Out]

a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (12) = 24\).

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.21 \[ \int (a+b x)^3 \, dx=\frac {1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} + a^{3} x \]

[In]

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

[Out]

1/4*b^3*x^4 + a*b^2*x^3 + 3/2*a^2*b*x^2 + a^3*x

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int (a+b x)^3 \, dx=\frac {{\left (b x + a\right )}^{4}}{4 \, b} \]

[In]

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

[Out]

1/4*(b*x + a)^4/b

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.21 \[ \int (a+b x)^3 \, dx=a^3\,x+\frac {3\,a^2\,b\,x^2}{2}+a\,b^2\,x^3+\frac {b^3\,x^4}{4} \]

[In]

int((a + b*x)^3,x)

[Out]

a^3*x + (b^3*x^4)/4 + (3*a^2*b*x^2)/2 + a*b^2*x^3