3.7.59 \(\int x^{4/3} (a+b x)^2 \, dx\)

Optimal. Leaf size=36 \[ \frac {3}{7} a^2 x^{7/3}+\frac {3}{5} a b x^{10/3}+\frac {3}{13} b^2 x^{13/3} \]

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {43} \begin {gather*} \frac {3}{7} a^2 x^{7/3}+\frac {3}{5} a b x^{10/3}+\frac {3}{13} b^2 x^{13/3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(4/3)*(a + b*x)^2,x]

[Out]

(3*a^2*x^(7/3))/7 + (3*a*b*x^(10/3))/5 + (3*b^2*x^(13/3))/13

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^{4/3} (a+b x)^2 \, dx &=\int \left (a^2 x^{4/3}+2 a b x^{7/3}+b^2 x^{10/3}\right ) \, dx\\ &=\frac {3}{7} a^2 x^{7/3}+\frac {3}{5} a b x^{10/3}+\frac {3}{13} b^2 x^{13/3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.78 \begin {gather*} \frac {3}{455} x^{7/3} \left (65 a^2+91 a b x+35 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(4/3)*(a + b*x)^2,x]

[Out]

(3*x^(7/3)*(65*a^2 + 91*a*b*x + 35*b^2*x^2))/455

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IntegrateAlgebraic [A]  time = 0.01, size = 28, normalized size = 0.78 \begin {gather*} \frac {3}{455} x^{7/3} \left (65 a^2+91 a b x+35 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(4/3)*(a + b*x)^2,x]

[Out]

(3*x^(7/3)*(65*a^2 + 91*a*b*x + 35*b^2*x^2))/455

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fricas [A]  time = 1.02, size = 29, normalized size = 0.81 \begin {gather*} \frac {3}{455} \, {\left (35 \, b^{2} x^{4} + 91 \, a b x^{3} + 65 \, a^{2} x^{2}\right )} x^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(4/3)*(b*x+a)^2,x, algorithm="fricas")

[Out]

3/455*(35*b^2*x^4 + 91*a*b*x^3 + 65*a^2*x^2)*x^(1/3)

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giac [A]  time = 1.06, size = 24, normalized size = 0.67 \begin {gather*} \frac {3}{13} \, b^{2} x^{\frac {13}{3}} + \frac {3}{5} \, a b x^{\frac {10}{3}} + \frac {3}{7} \, a^{2} x^{\frac {7}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(4/3)*(b*x+a)^2,x, algorithm="giac")

[Out]

3/13*b^2*x^(13/3) + 3/5*a*b*x^(10/3) + 3/7*a^2*x^(7/3)

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maple [A]  time = 0.00, size = 25, normalized size = 0.69 \begin {gather*} \frac {3 \left (35 b^{2} x^{2}+91 a b x +65 a^{2}\right ) x^{\frac {7}{3}}}{455} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(4/3)*(b*x+a)^2,x)

[Out]

3/455*x^(7/3)*(35*b^2*x^2+91*a*b*x+65*a^2)

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maxima [A]  time = 1.32, size = 24, normalized size = 0.67 \begin {gather*} \frac {3}{13} \, b^{2} x^{\frac {13}{3}} + \frac {3}{5} \, a b x^{\frac {10}{3}} + \frac {3}{7} \, a^{2} x^{\frac {7}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(4/3)*(b*x+a)^2,x, algorithm="maxima")

[Out]

3/13*b^2*x^(13/3) + 3/5*a*b*x^(10/3) + 3/7*a^2*x^(7/3)

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mupad [B]  time = 0.04, size = 24, normalized size = 0.67 \begin {gather*} \frac {3\,x^{7/3}\,\left (65\,a^2+91\,a\,b\,x+35\,b^2\,x^2\right )}{455} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(4/3)*(a + b*x)^2,x)

[Out]

(3*x^(7/3)*(65*a^2 + 35*b^2*x^2 + 91*a*b*x))/455

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sympy [A]  time = 2.65, size = 34, normalized size = 0.94 \begin {gather*} \frac {3 a^{2} x^{\frac {7}{3}}}{7} + \frac {3 a b x^{\frac {10}{3}}}{5} + \frac {3 b^{2} x^{\frac {13}{3}}}{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(4/3)*(b*x+a)**2,x)

[Out]

3*a**2*x**(7/3)/7 + 3*a*b*x**(10/3)/5 + 3*b**2*x**(13/3)/13

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