3.250 \(\int x^{1+p} (2 b+3 c x) (b x+c x^2)^p \, dx\)

Optimal. Leaf size=24 \[ \frac{x^{p+1} \left (b x+c x^2\right )^{p+1}}{p+1} \]

[Out]

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

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Rubi [A]  time = 0.011545, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {763} \[ \frac{x^{p+1} \left (b x+c x^2\right )^{p+1}}{p+1} \]

Antiderivative was successfully verified.

[In]

Int[x^(1 + p)*(2*b + 3*c*x)*(b*x + c*x^2)^p,x]

[Out]

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

Rule 763

Int[((e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(e*x)^m*(b*
x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] /; FreeQ[{b, c, e, f, g, m, p}, x] && EqQ[b*g*(m + p + 1) - c*f*(m +
2*p + 2), 0] && NeQ[m + 2*p + 2, 0]

Rubi steps

\begin{align*} \int x^{1+p} (2 b+3 c x) \left (b x+c x^2\right )^p \, dx &=\frac{x^{1+p} \left (b x+c x^2\right )^{1+p}}{1+p}\\ \end{align*}

Mathematica [A]  time = 0.0149723, size = 22, normalized size = 0.92 \[ \frac{x^{p+1} (x (b+c x))^{p+1}}{p+1} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(1 + p)*(2*b + 3*c*x)*(b*x + c*x^2)^p,x]

[Out]

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

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Maple [A]  time = 0.003, size = 28, normalized size = 1.2 \begin{align*}{\frac{{x}^{2+p} \left ( cx+b \right ) \left ( c{x}^{2}+bx \right ) ^{p}}{1+p}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1+p)*(3*c*x+2*b)*(c*x^2+b*x)^p,x)

[Out]

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

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Maxima [A]  time = 1.1926, size = 43, normalized size = 1.79 \begin{align*} \frac{{\left (c x^{3} + b x^{2}\right )} e^{\left (p \log \left (c x + b\right ) + 2 \, p \log \left (x\right )\right )}}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+p)*(3*c*x+2*b)*(c*x^2+b*x)^p,x, algorithm="maxima")

[Out]

(c*x^3 + b*x^2)*e^(p*log(c*x + b) + 2*p*log(x))/(p + 1)

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Fricas [A]  time = 1.97082, size = 66, normalized size = 2.75 \begin{align*} \frac{{\left (c x^{2} + b x\right )}{\left (c x^{2} + b x\right )}^{p} x^{p + 1}}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+p)*(3*c*x+2*b)*(c*x^2+b*x)^p,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 59.3413, size = 56, normalized size = 2.33 \begin{align*} \begin{cases} \frac{b x^{2} x^{p} \left (b x + c x^{2}\right )^{p}}{p + 1} + \frac{c x^{3} x^{p} \left (b x + c x^{2}\right )^{p}}{p + 1} & \text{for}\: p \neq -1 \\2 \log{\left (x \right )} + \log{\left (\frac{b}{c} + x \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1+p)*(3*c*x+2*b)*(c*x**2+b*x)**p,x)

[Out]

Piecewise((b*x**2*x**p*(b*x + c*x**2)**p/(p + 1) + c*x**3*x**p*(b*x + c*x**2)**p/(p + 1), Ne(p, -1)), (2*log(x
) + log(b/c + x), True))

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Giac [B]  time = 1.14773, size = 66, normalized size = 2.75 \begin{align*} \frac{c x^{2} e^{\left (p \log \left (c x + b\right ) + 2 \, p \log \left (x\right ) + \log \left (x\right )\right )} + b x e^{\left (p \log \left (c x + b\right ) + 2 \, p \log \left (x\right ) + \log \left (x\right )\right )}}{p + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1+p)*(3*c*x+2*b)*(c*x^2+b*x)^p,x, algorithm="giac")

[Out]

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