\(\int (d+e x) (c d^2+2 c d e x+c e^2 x^2) \, dx\) [979]

Optimal result

Integrand size = 26, antiderivative size = 15 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=\frac {c (d+e x)^4}{4 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {27, 12, 32} \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=\frac {c (d+e x)^4}{4 e} \]

[In]

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

[Out]

(c*(d + e*x)^4)/(4*e)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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}& = \int c (d+e x)^3 \, dx \\ & = c \int (d+e x)^3 \, dx \\ & = \frac {c (d+e x)^4}{4 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=\frac {c (d+e x)^4}{4 e} \]

[In]

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

[Out]

(c*(d + e*x)^4)/(4*e)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(30\) vs. \(2(13)=26\).

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.07

method	result	size
default	\(\frac {\left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{2}}{4 c e}\)	\(31\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) c}{4}\)	\(34\)
norman	\(c \,d^{3} x +c d \,e^{2} x^{3}+\frac {1}{4} c \,x^{4} e^{3}+\frac {3}{2} c \,d^{2} e \,x^{2}\)	\(36\)
parallelrisch	\(c \,d^{3} x +c d \,e^{2} x^{3}+\frac {1}{4} c \,x^{4} e^{3}+\frac {3}{2} c \,d^{2} e \,x^{2}\)	\(36\)
risch	\(\frac {c \,x^{4} e^{3}}{4}+c d \,e^{2} x^{3}+\frac {3 c \,d^{2} e \,x^{2}}{2}+c \,d^{3} x +\frac {c \,d^{4}}{4 e}\)	\(45\)

[In]

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

[Out]

1/4*(c*e^2*x^2+2*c*d*e*x+c*d^2)^2/c/e

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (13) = 26\).

Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.33 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=\frac {1}{4} \, c e^{3} x^{4} + c d e^{2} x^{3} + \frac {3}{2} \, c d^{2} e x^{2} + c d^{3} x \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (10) = 20\).

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.60 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=c d^{3} x + \frac {3 c d^{2} e x^{2}}{2} + c d e^{2} x^{3} + \frac {c e^{3} x^{4}}{4} \]

[In]

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

[Out]

c*d**3*x + 3*c*d**2*e*x**2/2 + c*d*e**2*x**3 + c*e**3*x**4/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (13) = 26\).

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 2.00 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{2}}{4 \, c e} \]

[In]

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

[Out]

1/4*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^2/(c*e)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (13) = 26\).

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.20 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=\frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} c d^{2} + \frac {1}{4} \, {\left (e x^{2} + 2 \, d x\right )}^{2} c e \]

[In]

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

[Out]

1/2*(e*x^2 + 2*d*x)*c*d^2 + 1/4*(e*x^2 + 2*d*x)^2*c*e

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.33 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx=c\,d^3\,x+\frac {3\,c\,d^2\,e\,x^2}{2}+c\,d\,e^2\,x^3+\frac {c\,e^3\,x^4}{4} \]

[In]

int((d + e*x)*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x),x)

[Out]

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