3.2 \(\int (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3)^p \, dx\)

Optimal. Leaf size=30 \[ \frac{\left (\frac{a}{b}+x\right ) \left (b^3 \left (\frac{a}{b}+x\right )^3\right )^p}{3 p+1} \]

[Out]

((a/b + x)*(b^3*(a/b + x)^3)^p)/(1 + 3*p)

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Rubi [A]  time = 0.0177064, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2067, 15, 30} \[ \frac{\left (\frac{a}{b}+x\right ) \left (b^3 \left (\frac{a}{b}+x\right )^3\right )^p}{3 p+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a/b + x)*(b^3*(a/b + x)^3)^p)/(1 + 3*p)

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.054513, size = 23, normalized size = 0.77 \[ \frac{(a+b x) \left ((a+b x)^3\right )^p}{3 b p+b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*((a + b*x)^3)^p)/(b + 3*b*p)

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Maple [A]  time = 0.002, size = 46, normalized size = 1.5 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({b}^{3}{x}^{3}+3\,a{b}^{2}{x}^{2}+3\,x{a}^{2}b+{a}^{3} \right ) ^{p}}{b \left ( 1+3\,p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)/b/(1+3*p)*(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^p

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Maxima [A]  time = 1.17738, size = 34, normalized size = 1.13 \begin{align*} \frac{{\left (b x + a\right )}{\left (b x + a\right )}^{3 \, p}}{b{\left (3 \, p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)*(b*x + a)^(3*p)/(b*(3*p + 1))

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Fricas [A]  time = 1.25522, size = 90, normalized size = 3. \begin{align*} \frac{{\left (b x + a\right )}{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p}}{3 \, b p + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x + a)*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p/(3*b*p + b)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**3*x**3+3*a*b**2*x**2+3*a**2*b*x+a**3)**p,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.0984, size = 99, normalized size = 3.3 \begin{align*} \frac{{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p} b x +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p} a}{3 \, b p + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*b*x + (b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)^p*a)/(3*b*p + b)