3.2.37 \(\int (b+2 c x) (b x+c x^2)^p \, dx\) [137]

Optimal. Leaf size=19 \[ \frac {\left (b x+c x^2\right )^{1+p}}{1+p} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {643} \begin {gather*} \frac {\left (b x+c x^2\right )^{p+1}}{p+1} \end {gather*}

Antiderivative was successfully verified.

[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*} \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}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 17, normalized size = 0.89 \begin {gather*} \frac {(x (b+c x))^{1+p}}{1+p} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b + c*x))^(1 + p)/(1 + p)

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Maple [A]
time = 0.17, size = 20, normalized size = 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	\(\frac {x \left (c x +b \right ) \left (c \,x^{2}+b x \right )^{p}}{1+p}\)	\(24\)
norman	\(\frac {b x \,{\mathrm e}^{p \ln \left (c \,x^{2}+b x \right )}}{1+p}+\frac {c \,x^{2} {\mathrm e}^{p \ln \left (c \,x^{2}+b x \right )}}{1+p}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]
time = 0.28, size = 19, normalized size = 1.00 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{p + 1}}{p + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.35, size = 26, normalized size = 1.37 \begin {gather*} \frac {{\left (c x^{2} + b x\right )} {\left (c x^{2} + b x\right )}^{p}}{p + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (14) = 28\).
time = 0.25, size = 46, normalized size = 2.42 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]
time = 4.97, size = 19, normalized size = 1.00 \begin {gather*} \frac {{\left (c x^{2} + b x\right )}^{p + 1}}{p + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 2.03, size = 23, normalized size = 1.21 \begin {gather*} \frac {x\,{\left (c\,x^2+b\,x\right )}^p\,\left (b+c\,x\right )}{p+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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