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

Optimal result

Integrand size = 29, antiderivative size = 30 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {\left (\frac {a}{b}+x\right ) \left (b^3 \left (\frac {a}{b}+x\right )^3\right )^p}{1+3 p} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2092, 15, 30} \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {\left (\frac {a}{b}+x\right ) \left (b^3 \left (\frac {a}{b}+x\right )^3\right )^p}{3 p+1} \]

[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 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]

Rule 2092

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]

Rubi steps \begin{align*} \text {integral}& = \text {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 ) \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {(a+b x) \left ((a+b x)^3\right )^p}{b (1+3 p)} \]

[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*(1 + 3*p))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {\left (b x +a \right ) \left (\left (b x +a \right )^{3}\right )^{p}}{b \left (1+3 p \right )}\)	\(26\)
gosper	\(\frac {\left (b x +a \right ) \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p}}{b \left (1+3 p \right )}\)	\(46\)
parallelrisch	\(\frac {x \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p} a b +\left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )^{p} a^{2}}{\left (1+3 p \right ) a b}\)	\(82\)
norman	\(\frac {x \,{\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )}}{1+3 p}+\frac {a \,{\mathrm e}^{p \ln \left (b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}\right )}}{b \left (1+3 p \right )}\)	\(85\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p}}{3 \, b p + b} \]

[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)

Sympy [F]

\[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\begin {cases} \frac {x}{\sqrt [3]{a^{3}}} & \text {for}\: b = 0 \wedge p = - \frac {1}{3} \\x \left (a^{3}\right )^{p} & \text {for}\: b = 0 \\\int \frac {1}{\sqrt [3]{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}}\, dx & \text {for}\: p = - \frac {1}{3} \\\frac {a \left (a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}\right )^{p}}{3 b p + b} + \frac {b x \left (a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}\right )^{p}}{3 b p + b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x/(a**3)**(1/3), Eq(b, 0) & Eq(p, -1/3)), (x*(a**3)**p, Eq(b, 0)), (Integral((a**3 + 3*a**2*b*x + 3
*a*b**2*x**2 + b**3*x**3)**(-1/3), x), Eq(p, -1/3)), (a*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p/(3*
b*p + b) + b*x*(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)**p/(3*b*p + b), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b x + a\right )}^{3 \, p}}{b {\left (3 \, p + 1\right )}} \]

[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))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (30) = 60\).

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.43 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\frac {{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p} b x + {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}^{p} a}{3 \, b p + b} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \left (a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3\right )^p \, dx=\left (\frac {x}{3\,p+1}+\frac {a}{b\,\left (3\,p+1\right )}\right )\,{\left (a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3\right )}^p \]

[In]

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

[Out]

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