\(\int x (a+b x^3)^5 \, dx\) [277]

Optimal result

Integrand size = 11, antiderivative size = 66 \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {a^5 x^2}{2}+a^4 b x^5+\frac {5}{4} a^3 b^2 x^8+\frac {10}{11} a^2 b^3 x^{11}+\frac {5}{14} a b^4 x^{14}+\frac {b^5 x^{17}}{17} \]

[Out]

1/2*a^5*x^2+a^4*b*x^5+5/4*a^3*b^2*x^8+10/11*a^2*b^3*x^11+5/14*a*b^4*x^14+1/17*b^5*x^17

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {276} \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {a^5 x^2}{2}+a^4 b x^5+\frac {5}{4} a^3 b^2 x^8+\frac {10}{11} a^2 b^3 x^{11}+\frac {5}{14} a b^4 x^{14}+\frac {b^5 x^{17}}{17} \]

[In]

Int[x*(a + b*x^3)^5,x]

[Out]

(a^5*x^2)/2 + a^4*b*x^5 + (5*a^3*b^2*x^8)/4 + (10*a^2*b^3*x^11)/11 + (5*a*b^4*x^14)/14 + (b^5*x^17)/17

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^5 x+5 a^4 b x^4+10 a^3 b^2 x^7+10 a^2 b^3 x^{10}+5 a b^4 x^{13}+b^5 x^{16}\right ) \, dx \\ & = \frac {a^5 x^2}{2}+a^4 b x^5+\frac {5}{4} a^3 b^2 x^8+\frac {10}{11} a^2 b^3 x^{11}+\frac {5}{14} a b^4 x^{14}+\frac {b^5 x^{17}}{17} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00 \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {a^5 x^2}{2}+a^4 b x^5+\frac {5}{4} a^3 b^2 x^8+\frac {10}{11} a^2 b^3 x^{11}+\frac {5}{14} a b^4 x^{14}+\frac {b^5 x^{17}}{17} \]

[In]

Integrate[x*(a + b*x^3)^5,x]

[Out]

(a^5*x^2)/2 + a^4*b*x^5 + (5*a^3*b^2*x^8)/4 + (10*a^2*b^3*x^11)/11 + (5*a*b^4*x^14)/14 + (b^5*x^17)/17

Maple [A] (verified)

Time = 3.63 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86

method	result	size
gosper	\(\frac {1}{2} a^{5} x^{2}+a^{4} b \,x^{5}+\frac {5}{4} a^{3} b^{2} x^{8}+\frac {10}{11} a^{2} b^{3} x^{11}+\frac {5}{14} a \,b^{4} x^{14}+\frac {1}{17} b^{5} x^{17}\)	\(57\)
default	\(\frac {1}{2} a^{5} x^{2}+a^{4} b \,x^{5}+\frac {5}{4} a^{3} b^{2} x^{8}+\frac {10}{11} a^{2} b^{3} x^{11}+\frac {5}{14} a \,b^{4} x^{14}+\frac {1}{17} b^{5} x^{17}\)	\(57\)
norman	\(\frac {1}{2} a^{5} x^{2}+a^{4} b \,x^{5}+\frac {5}{4} a^{3} b^{2} x^{8}+\frac {10}{11} a^{2} b^{3} x^{11}+\frac {5}{14} a \,b^{4} x^{14}+\frac {1}{17} b^{5} x^{17}\)	\(57\)
risch	\(\frac {1}{2} a^{5} x^{2}+a^{4} b \,x^{5}+\frac {5}{4} a^{3} b^{2} x^{8}+\frac {10}{11} a^{2} b^{3} x^{11}+\frac {5}{14} a \,b^{4} x^{14}+\frac {1}{17} b^{5} x^{17}\)	\(57\)
parallelrisch	\(\frac {1}{2} a^{5} x^{2}+a^{4} b \,x^{5}+\frac {5}{4} a^{3} b^{2} x^{8}+\frac {10}{11} a^{2} b^{3} x^{11}+\frac {5}{14} a \,b^{4} x^{14}+\frac {1}{17} b^{5} x^{17}\)	\(57\)

[In]

int(x*(b*x^3+a)^5,x,method=_RETURNVERBOSE)

[Out]

1/2*a^5*x^2+a^4*b*x^5+5/4*a^3*b^2*x^8+10/11*a^2*b^3*x^11+5/14*a*b^4*x^14+1/17*b^5*x^17

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {1}{17} \, b^{5} x^{17} + \frac {5}{14} \, a b^{4} x^{14} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {5}{4} \, a^{3} b^{2} x^{8} + a^{4} b x^{5} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

integrate(x*(b*x^3+a)^5,x, algorithm="fricas")

[Out]

1/17*b^5*x^17 + 5/14*a*b^4*x^14 + 10/11*a^2*b^3*x^11 + 5/4*a^3*b^2*x^8 + a^4*b*x^5 + 1/2*a^5*x^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {a^{5} x^{2}}{2} + a^{4} b x^{5} + \frac {5 a^{3} b^{2} x^{8}}{4} + \frac {10 a^{2} b^{3} x^{11}}{11} + \frac {5 a b^{4} x^{14}}{14} + \frac {b^{5} x^{17}}{17} \]

[In]

integrate(x*(b*x**3+a)**5,x)

[Out]

a**5*x**2/2 + a**4*b*x**5 + 5*a**3*b**2*x**8/4 + 10*a**2*b**3*x**11/11 + 5*a*b**4*x**14/14 + b**5*x**17/17

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {1}{17} \, b^{5} x^{17} + \frac {5}{14} \, a b^{4} x^{14} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {5}{4} \, a^{3} b^{2} x^{8} + a^{4} b x^{5} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

integrate(x*(b*x^3+a)^5,x, algorithm="maxima")

[Out]

1/17*b^5*x^17 + 5/14*a*b^4*x^14 + 10/11*a^2*b^3*x^11 + 5/4*a^3*b^2*x^8 + a^4*b*x^5 + 1/2*a^5*x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {1}{17} \, b^{5} x^{17} + \frac {5}{14} \, a b^{4} x^{14} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {5}{4} \, a^{3} b^{2} x^{8} + a^{4} b x^{5} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

integrate(x*(b*x^3+a)^5,x, algorithm="giac")

[Out]

1/17*b^5*x^17 + 5/14*a*b^4*x^14 + 10/11*a^2*b^3*x^11 + 5/4*a^3*b^2*x^8 + a^4*b*x^5 + 1/2*a^5*x^2

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int x \left (a+b x^3\right )^5 \, dx=\frac {a^5\,x^2}{2}+a^4\,b\,x^5+\frac {5\,a^3\,b^2\,x^8}{4}+\frac {10\,a^2\,b^3\,x^{11}}{11}+\frac {5\,a\,b^4\,x^{14}}{14}+\frac {b^5\,x^{17}}{17} \]

[In]

int(x*(a + b*x^3)^5,x)

[Out]

(a^5*x^2)/2 + (b^5*x^17)/17 + a^4*b*x^5 + (5*a*b^4*x^14)/14 + (5*a^3*b^2*x^8)/4 + (10*a^2*b^3*x^11)/11