\(\int (b+2 c x) (b x+c x^2)^p \, dx\) [172]

Optimal result

Integrand size = 18, antiderivative size = 19 \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\frac {\left (b x+c x^2\right )^{1+p}}{1+p} \]

[Out]

(c*x^2+b*x)^(p+1)/(p+1)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {643} \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\frac {\left (b x+c x^2\right )^{p+1}}{p+1} \]

[In]

Int[(b + 2*c*x)*(b*x + c*x^2)^p,x]

[Out]

(b*x + c*x^2)^(1 + p)/(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 (b x+c x^2\right )^{1+p}}{1+p} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\frac {(x (b+c x))^{1+p}}{1+p} \]

[In]

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

[Out]

(x*(b + c*x))^(1 + p)/(1 + p)

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\left (c \,x^{2}+b x \right )^{1+p}}{1+p}\)	\(20\)
default	\(\frac {\left (c \,x^{2}+b x \right )^{1+p}}{1+p}\)	\(20\)
risch	\(\frac {x \left (c x +b \right ) \left (x \left (c x +b \right )\right )^{p}}{1+p}\)	\(22\)
gosper	\(\frac {x \left (c x +b \right ) \left (c \,x^{2}+b x \right )^{p}}{1+p}\)	\(24\)
parallelrisch	\(\frac {x^{2} \left (x \left (c x +b \right )\right )^{p} b c +x \left (x \left (c x +b \right )\right )^{p} b^{2}}{b \left (1+p \right )}\)	\(40\)
norman	\(\frac {b x \,{\mathrm e}^{p \ln \left (c \,x^{2}+b x \right )}}{1+p}+\frac {c \,x^{2} {\mathrm e}^{p \ln \left (c \,x^{2}+b x \right )}}{1+p}\)	\(46\)

[In]

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

[Out]

(c*x^2+b*x)^(1+p)/(1+p)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\frac {{\left (c x^{2} + b x\right )} {\left (c x^{2} + b x\right )}^{p}}{p + 1} \]

[In]

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

[Out]

(c*x^2 + b*x)*(c*x^2 + b*x)^p/(p + 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.42 \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\begin {cases} \frac {b x \left (b x + c x^{2}\right )^{p}}{p + 1} + \frac {c x^{2} \left (b x + c x^{2}\right )^{p}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (x \right )} + \log {\left (\frac {b}{c} + x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((2*c*x+b)*(c*x**2+b*x)**p,x)

[Out]

Piecewise((b*x*(b*x + c*x**2)**p/(p + 1) + c*x**2*(b*x + c*x**2)**p/(p + 1), Ne(p, -1)), (log(x) + log(b/c + x
), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\frac {{\left (c x^{2} + b x\right )}^{p + 1}}{p + 1} \]

[In]

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

[Out]

(c*x^2 + b*x)^(p + 1)/(p + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\frac {{\left (c x^{2} + b x\right )}^{p + 1}}{p + 1} \]

[In]

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

[Out]

(c*x^2 + b*x)^(p + 1)/(p + 1)

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int (b+2 c x) \left (b x+c x^2\right )^p \, dx=\frac {x\,{\left (c\,x^2+b\,x\right )}^p\,\left (b+c\,x\right )}{p+1} \]

[In]

int((b*x + c*x^2)^p*(b + 2*c*x),x)

[Out]

(x*(b*x + c*x^2)^p*(b + c*x))/(p + 1)