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

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

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {763} \begin {gather*} \frac {x^{p+1} \left (b x+c x^2\right )^{p+1}}{p+1} \end {gather*}

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*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.92 \begin {gather*} \frac {x^{p+1} (x (b+c x))^{p+1}}{p+1} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.13, size = 0, normalized size = 0.00 \begin {gather*} \int x^{1+p} (2 b+3 c x) \left (b x+c x^2\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 31, normalized size = 1.29 \begin {gather*} \frac {{\left (c x^{2} + b x\right )} {\left (c x^{2} + b x\right )}^{p} x^{p + 1}}{p + 1} \end {gather*}

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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giac [B]  time = 0.27, size = 49, normalized size = 2.04 \begin {gather*} \frac {c x^{2} e^{\left (p \log \left (c x + b\right ) + 2 \, p \log \relax (x) + \log \relax (x)\right )} + b x e^{\left (p \log \left (c x + b\right ) + 2 \, p \log \relax (x) + \log \relax (x)\right )}}{p + 1} \end {gather*}

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)

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maple [A]  time = 0.04, size = 28, normalized size = 1.17 \begin {gather*} \frac {\left (c x +b \right ) x^{p +2} \left (c \,x^{2}+b x \right )^{p}}{p +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.80, size = 32, normalized size = 1.33 \begin {gather*} \frac {{\left (c x^{3} + b x^{2}\right )} e^{\left (p \log \left (c x + b\right ) + 2 \, p \log \relax (x)\right )}}{p + 1} \end {gather*}

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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mupad [B]  time = 1.17, size = 41, normalized size = 1.71 \begin {gather*} {\left (c\,x^2+b\,x\right )}^p\,\left (\frac {b\,x\,x^{p+1}}{p+1}+\frac {c\,x^{p+1}\,x^2}{p+1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 43.34, size = 56, normalized size = 2.33 \begin {gather*} \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 {\relax (x )} + \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(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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