\(\int (c d^2+2 c d e x+c e^2 x^2)^p \, dx\) [1103]

Optimal result

Integrand size = 22, antiderivative size = 38 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e (1+2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {623} \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e (2 p+1)} \]

[In]

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

[Out]

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

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x) \left (c (d+e x)^2\right )^p}{e (1+2 p)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\left (e x +d \right ) \left (c \left (e x +d \right )^{2}\right )^{p}}{e \left (1+2 p \right )}\)	\(28\)
gosper	\(\frac {\left (e x +d \right ) \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{p}}{e \left (1+2 p \right )}\)	\(39\)
parallelrisch	\(\frac {x {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d e +{\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d^{2}}{\left (1+2 p \right ) d e}\)	\(64\)
norman	\(\frac {x \,{\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{1+2 p}+\frac {d \,{\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{e \left (1+2 p \right )}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (e x + d\right )} {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, e p + e} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (34) = 68\).

Time = 0.55 (sec) , antiderivative size = 116, normalized size of antiderivative = 3.05 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\begin {cases} \frac {x}{\sqrt {c d^{2}}} & \text {for}\: e = 0 \wedge p = - \frac {1}{2} \\x \left (c d^{2}\right )^{p} & \text {for}\: e = 0 \\\frac {\left (\frac {d}{e} + x\right ) \log {\left (\frac {d}{e} + x \right )}}{\sqrt {c e^{2} \left (\frac {d}{e} + x\right )^{2}}} & \text {for}\: p = - \frac {1}{2} \\\frac {d \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + e} + \frac {e x \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 e p + e} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (c^{p} e x + c^{p} d\right )} {\left (e x + d\right )}^{2 \, p}}{e {\left (2 \, p + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.55 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p} e x + {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p} d}{2 \, e p + e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.89 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.18 \[ \int \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\left (\frac {x}{2\,p+1}+\frac {d}{e\,\left (2\,p+1\right )}\right )\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p \]

[In]

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

[Out]

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