3.1.65 \(\int x^3 (a+b x)^3 \, dx\)

Optimal. Leaf size=43 \[ \frac {a^3 x^4}{4}+\frac {3}{5} a^2 b x^5+\frac {1}{2} a b^2 x^6+\frac {b^3 x^7}{7} \]

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Rubi [A]  time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {43} \begin {gather*} \frac {3}{5} a^2 b x^5+\frac {a^3 x^4}{4}+\frac {1}{2} a b^2 x^6+\frac {b^3 x^7}{7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x)^3,x]

[Out]

(a^3*x^4)/4 + (3*a^2*b*x^5)/5 + (a*b^2*x^6)/2 + (b^3*x^7)/7

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int x^3 (a+b x)^3 \, dx &=\int \left (a^3 x^3+3 a^2 b x^4+3 a b^2 x^5+b^3 x^6\right ) \, dx\\ &=\frac {a^3 x^4}{4}+\frac {3}{5} a^2 b x^5+\frac {1}{2} a b^2 x^6+\frac {b^3 x^7}{7}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 43, normalized size = 1.00 \begin {gather*} \frac {a^3 x^4}{4}+\frac {3}{5} a^2 b x^5+\frac {1}{2} a b^2 x^6+\frac {b^3 x^7}{7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x)^3,x]

[Out]

(a^3*x^4)/4 + (3*a^2*b*x^5)/5 + (a*b^2*x^6)/2 + (b^3*x^7)/7

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^3 (a+b x)^3 \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^3*(a + b*x)^3,x]

[Out]

IntegrateAlgebraic[x^3*(a + b*x)^3, x]

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fricas [A]  time = 0.69, size = 35, normalized size = 0.81 \begin {gather*} \frac {1}{7} x^{7} b^{3} + \frac {1}{2} x^{6} b^{2} a + \frac {3}{5} x^{5} b a^{2} + \frac {1}{4} x^{4} a^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*x^7*b^3 + 1/2*x^6*b^2*a + 3/5*x^5*b*a^2 + 1/4*x^4*a^3

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giac [A]  time = 1.27, size = 35, normalized size = 0.81 \begin {gather*} \frac {1}{7} \, b^{3} x^{7} + \frac {1}{2} \, a b^{2} x^{6} + \frac {3}{5} \, a^{2} b x^{5} + \frac {1}{4} \, a^{3} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*x^7 + 1/2*a*b^2*x^6 + 3/5*a^2*b*x^5 + 1/4*a^3*x^4

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maple [A]  time = 0.00, size = 36, normalized size = 0.84 \begin {gather*} \frac {1}{7} b^{3} x^{7}+\frac {1}{2} a \,b^{2} x^{6}+\frac {3}{5} a^{2} b \,x^{5}+\frac {1}{4} a^{3} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^3*x^4+3/5*a^2*b*x^5+1/2*a*b^2*x^6+1/7*b^3*x^7

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maxima [A]  time = 1.35, size = 35, normalized size = 0.81 \begin {gather*} \frac {1}{7} \, b^{3} x^{7} + \frac {1}{2} \, a b^{2} x^{6} + \frac {3}{5} \, a^{2} b x^{5} + \frac {1}{4} \, a^{3} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*x^7 + 1/2*a*b^2*x^6 + 3/5*a^2*b*x^5 + 1/4*a^3*x^4

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mupad [B]  time = 0.04, size = 35, normalized size = 0.81 \begin {gather*} \frac {a^3\,x^4}{4}+\frac {3\,a^2\,b\,x^5}{5}+\frac {a\,b^2\,x^6}{2}+\frac {b^3\,x^7}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x)^3,x)

[Out]

(a^3*x^4)/4 + (b^3*x^7)/7 + (3*a^2*b*x^5)/5 + (a*b^2*x^6)/2

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sympy [A]  time = 0.07, size = 37, normalized size = 0.86 \begin {gather*} \frac {a^{3} x^{4}}{4} + \frac {3 a^{2} b x^{5}}{5} + \frac {a b^{2} x^{6}}{2} + \frac {b^{3} x^{7}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*x**4/4 + 3*a**2*b*x**5/5 + a*b**2*x**6/2 + b**3*x**7/7

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