\(\int (d+e x) (c d^2+2 c d e x+c e^2 x^2)^p \, dx\) [1102]

Optimal result

Integrand size = 28, antiderivative size = 39 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{1+p}}{2 c e (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {643} \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{p+1}}{2 c e (p+1)} \]

[In]

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

[Out]

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

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{1+p}}{2 c e (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x)^2 \left (c (d+e x)^2\right )^p}{e (2+2 p)} \]

[In]

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

[Out]

((d + e*x)^2*(c*(d + e*x)^2)^p)/(e*(2 + 2*p))

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.97

method	result	size
risch	\(\frac {\left (x^{2} e^{2}+2 d e x +d^{2}\right ) \left (c \left (e x +d \right )^{2}\right )^{p}}{2 e \left (1+p \right )}\)	\(38\)
gosper	\(\frac {\left (e x +d \right )^{2} \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{p}}{2 e \left (1+p \right )}\)	\(40\)
parallelrisch	\(\frac {x^{2} {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d \,e^{2}+2 x {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d^{2} e +{\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d^{3}}{2 d e \left (1+p \right )}\)	\(94\)
norman	\(\frac {d x \,{\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{1+p}+\frac {d^{2} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{2 e \left (1+p \right )}+\frac {e \,x^{2} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{2+2 p}\)	\(106\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, {\left (e p + e\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (32) = 64\).

Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 3.56 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\begin {cases} \frac {x}{c d} & \text {for}\: e = 0 \wedge p = -1 \\d x \left (c d^{2}\right )^{p} & \text {for}\: e = 0 \\\frac {\log {\left (\frac {d}{e} + x \right )}}{c e} & \text {for}\: p = -1 \\\frac {d^{2} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 2 e} + \frac {2 d e x \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 2 e} + \frac {e^{2} x^{2} \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + 2 e} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x/(c*d), Eq(e, 0) & Eq(p, -1)), (d*x*(c*d**2)**p, Eq(e, 0)), (log(d/e + x)/(c*e), Eq(p, -1)), (d**2
*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*e*p + 2*e) + 2*d*e*x*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*e*p +
2*e) + e**2*x**2*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*e*p + 2*e), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p + 1}}{2 \, c e {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p + 1}}{2 \, c e {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.87 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.46 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx={\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p\,\left (\frac {d^2}{2\,e\,\left (p+1\right )}+\frac {d\,x}{p+1}+\frac {e\,x^2}{2\,\left (p+1\right )}\right ) \]

[In]

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

[Out]

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