\(\int x^{4/3} (a+b x)^3 \, dx\) [667]

Optimal result

Integrand size = 13, antiderivative size = 51 \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3}{7} a^3 x^{7/3}+\frac {9}{10} a^2 b x^{10/3}+\frac {9}{13} a b^2 x^{13/3}+\frac {3}{16} b^3 x^{16/3} \]

[Out]

3/7*a^3*x^(7/3)+9/10*a^2*b*x^(10/3)+9/13*a*b^2*x^(13/3)+3/16*b^3*x^(16/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3}{7} a^3 x^{7/3}+\frac {9}{10} a^2 b x^{10/3}+\frac {9}{13} a b^2 x^{13/3}+\frac {3}{16} b^3 x^{16/3} \]

[In]

Int[x^(4/3)*(a + b*x)^3,x]

[Out]

(3*a^3*x^(7/3))/7 + (9*a^2*b*x^(10/3))/10 + (9*a*b^2*x^(13/3))/13 + (3*b^3*x^(16/3))/16

Rule 45

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*} \text {integral}& = \int \left (a^3 x^{4/3}+3 a^2 b x^{7/3}+3 a b^2 x^{10/3}+b^3 x^{13/3}\right ) \, dx \\ & = \frac {3}{7} a^3 x^{7/3}+\frac {9}{10} a^2 b x^{10/3}+\frac {9}{13} a b^2 x^{13/3}+\frac {3}{16} b^3 x^{16/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.76 \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3 x^{7/3} \left (1040 a^3+2184 a^2 b x+1680 a b^2 x^2+455 b^3 x^3\right )}{7280} \]

[In]

Integrate[x^(4/3)*(a + b*x)^3,x]

[Out]

(3*x^(7/3)*(1040*a^3 + 2184*a^2*b*x + 1680*a*b^2*x^2 + 455*b^3*x^3))/7280

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
gosper	\(\frac {3 x^{\frac {7}{3}} \left (455 b^{3} x^{3}+1680 a \,b^{2} x^{2}+2184 a^{2} b x +1040 a^{3}\right )}{7280}\)	\(36\)
derivativedivides	\(\frac {3 a^{3} x^{\frac {7}{3}}}{7}+\frac {9 a^{2} b \,x^{\frac {10}{3}}}{10}+\frac {9 a \,b^{2} x^{\frac {13}{3}}}{13}+\frac {3 b^{3} x^{\frac {16}{3}}}{16}\)	\(36\)
default	\(\frac {3 a^{3} x^{\frac {7}{3}}}{7}+\frac {9 a^{2} b \,x^{\frac {10}{3}}}{10}+\frac {9 a \,b^{2} x^{\frac {13}{3}}}{13}+\frac {3 b^{3} x^{\frac {16}{3}}}{16}\)	\(36\)
trager	\(\frac {3 x^{\frac {7}{3}} \left (455 b^{3} x^{3}+1680 a \,b^{2} x^{2}+2184 a^{2} b x +1040 a^{3}\right )}{7280}\)	\(36\)
risch	\(\frac {3 x^{\frac {7}{3}} \left (455 b^{3} x^{3}+1680 a \,b^{2} x^{2}+2184 a^{2} b x +1040 a^{3}\right )}{7280}\)	\(36\)

[In]

int(x^(4/3)*(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

3/7280*x^(7/3)*(455*b^3*x^3+1680*a*b^2*x^2+2184*a^2*b*x+1040*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.78 \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3}{7280} \, {\left (455 \, b^{3} x^{5} + 1680 \, a b^{2} x^{4} + 2184 \, a^{2} b x^{3} + 1040 \, a^{3} x^{2}\right )} x^{\frac {1}{3}} \]

[In]

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

[Out]

3/7280*(455*b^3*x^5 + 1680*a*b^2*x^4 + 2184*a^2*b*x^3 + 1040*a^3*x^2)*x^(1/3)

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3 a^{3} x^{\frac {7}{3}}}{7} + \frac {9 a^{2} b x^{\frac {10}{3}}}{10} + \frac {9 a b^{2} x^{\frac {13}{3}}}{13} + \frac {3 b^{3} x^{\frac {16}{3}}}{16} \]

[In]

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

[Out]

3*a**3*x**(7/3)/7 + 9*a**2*b*x**(10/3)/10 + 9*a*b**2*x**(13/3)/13 + 3*b**3*x**(16/3)/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3}{16} \, b^{3} x^{\frac {16}{3}} + \frac {9}{13} \, a b^{2} x^{\frac {13}{3}} + \frac {9}{10} \, a^{2} b x^{\frac {10}{3}} + \frac {3}{7} \, a^{3} x^{\frac {7}{3}} \]

[In]

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

[Out]

3/16*b^3*x^(16/3) + 9/13*a*b^2*x^(13/3) + 9/10*a^2*b*x^(10/3) + 3/7*a^3*x^(7/3)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3}{16} \, b^{3} x^{\frac {16}{3}} + \frac {9}{13} \, a b^{2} x^{\frac {13}{3}} + \frac {9}{10} \, a^{2} b x^{\frac {10}{3}} + \frac {3}{7} \, a^{3} x^{\frac {7}{3}} \]

[In]

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

[Out]

3/16*b^3*x^(16/3) + 9/13*a*b^2*x^(13/3) + 9/10*a^2*b*x^(10/3) + 3/7*a^3*x^(7/3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int x^{4/3} (a+b x)^3 \, dx=\frac {3\,a^3\,x^{7/3}}{7}+\frac {3\,b^3\,x^{16/3}}{16}+\frac {9\,a^2\,b\,x^{10/3}}{10}+\frac {9\,a\,b^2\,x^{13/3}}{13} \]

[In]

int(x^(4/3)*(a + b*x)^3,x)

[Out]

(3*a^3*x^(7/3))/7 + (3*b^3*x^(16/3))/16 + (9*a^2*b*x^(10/3))/10 + (9*a*b^2*x^(13/3))/13