\(\int (c+d x)^3 \, dx\) [1263]

Optimal result

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

[Out]

1/4*(d*x+c)^4/d

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} \[ \int (c+d x)^3 \, dx=\frac {(c+d x)^4}{4 d} \]

[In]

Int[(c + d*x)^3,x]

[Out]

(c + d*x)^4/(4*d)

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 {(c+d x)^4}{4 d} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[(c + d*x)^3,x]

[Out]

(c + d*x)^4/(4*d)

Maple [A] (verified)

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

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

[In]

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

[Out]

1/4*(d*x+c)^4/d

Fricas [B] (verification not implemented)

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

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

[In]

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

[Out]

1/4*d^3*x^4 + c*d^2*x^3 + 3/2*c^2*d*x^2 + c^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 \[ \int (c+d x)^3 \, dx=c^{3} x + \frac {3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac {d^{3} x^{4}}{4} \]

[In]

integrate((d*x+c)**3,x)

[Out]

c**3*x + 3*c**2*d*x**2/2 + c*d**2*x**3 + d**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 = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.21 \[ \int (c+d x)^3 \, dx=\frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, c^{2} d x^{2} + c^{3} x \]

[In]

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

[Out]

1/4*d^3*x^4 + c*d^2*x^3 + 3/2*c^2*d*x^2 + c^3*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int (c+d x)^3 \, dx=\frac {{\left (d x + c\right )}^{4}}{4 \, d} \]

[In]

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

[Out]

1/4*(d*x + c)^4/d

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.21 \[ \int (c+d x)^3 \, dx=c^3\,x+\frac {3\,c^2\,d\,x^2}{2}+c\,d^2\,x^3+\frac {d^3\,x^4}{4} \]

[In]

int((c + d*x)^3,x)

[Out]

c^3*x + (d^3*x^4)/4 + (3*c^2*d*x^2)/2 + c*d^2*x^3