\(\int \frac {(a+b x^3)^5}{x} \, dx\) [264]

Optimal result

Integrand size = 13, antiderivative size = 65 \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=\frac {5}{3} a^4 b x^3+\frac {5}{3} a^3 b^2 x^6+\frac {10}{9} a^2 b^3 x^9+\frac {5}{12} a b^4 x^{12}+\frac {b^5 x^{15}}{15}+a^5 \log (x) \]

[Out]

5/3*a^4*b*x^3+5/3*a^3*b^2*x^6+10/9*a^2*b^3*x^9+5/12*a*b^4*x^12+1/15*b^5*x^15+a^5*ln(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=a^5 \log (x)+\frac {5}{3} a^4 b x^3+\frac {5}{3} a^3 b^2 x^6+\frac {10}{9} a^2 b^3 x^9+\frac {5}{12} a b^4 x^{12}+\frac {b^5 x^{15}}{15} \]

[In]

Int[(a + b*x^3)^5/x,x]

[Out]

(5*a^4*b*x^3)/3 + (5*a^3*b^2*x^6)/3 + (10*a^2*b^3*x^9)/9 + (5*a*b^4*x^12)/12 + (b^5*x^15)/15 + a^5*Log[x]

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])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {(a+b x)^5}{x} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (5 a^4 b+\frac {a^5}{x}+10 a^3 b^2 x+10 a^2 b^3 x^2+5 a b^4 x^3+b^5 x^4\right ) \, dx,x,x^3\right ) \\ & = \frac {5}{3} a^4 b x^3+\frac {5}{3} a^3 b^2 x^6+\frac {10}{9} a^2 b^3 x^9+\frac {5}{12} a b^4 x^{12}+\frac {b^5 x^{15}}{15}+a^5 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=\frac {5}{3} a^4 b x^3+\frac {5}{3} a^3 b^2 x^6+\frac {10}{9} a^2 b^3 x^9+\frac {5}{12} a b^4 x^{12}+\frac {b^5 x^{15}}{15}+a^5 \log (x) \]

[In]

Integrate[(a + b*x^3)^5/x,x]

[Out]

(5*a^4*b*x^3)/3 + (5*a^3*b^2*x^6)/3 + (10*a^2*b^3*x^9)/9 + (5*a*b^4*x^12)/12 + (b^5*x^15)/15 + a^5*Log[x]

Maple [A] (verified)

Time = 3.61 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {5 a^{4} b \,x^{3}}{3}+\frac {5 a^{3} b^{2} x^{6}}{3}+\frac {10 a^{2} b^{3} x^{9}}{9}+\frac {5 a \,b^{4} x^{12}}{12}+\frac {b^{5} x^{15}}{15}+a^{5} \ln \left (x \right )\)	\(56\)
norman	\(\frac {5 a^{4} b \,x^{3}}{3}+\frac {5 a^{3} b^{2} x^{6}}{3}+\frac {10 a^{2} b^{3} x^{9}}{9}+\frac {5 a \,b^{4} x^{12}}{12}+\frac {b^{5} x^{15}}{15}+a^{5} \ln \left (x \right )\)	\(56\)
risch	\(\frac {5 a^{4} b \,x^{3}}{3}+\frac {5 a^{3} b^{2} x^{6}}{3}+\frac {10 a^{2} b^{3} x^{9}}{9}+\frac {5 a \,b^{4} x^{12}}{12}+\frac {b^{5} x^{15}}{15}+a^{5} \ln \left (x \right )\)	\(56\)
parallelrisch	\(\frac {5 a^{4} b \,x^{3}}{3}+\frac {5 a^{3} b^{2} x^{6}}{3}+\frac {10 a^{2} b^{3} x^{9}}{9}+\frac {5 a \,b^{4} x^{12}}{12}+\frac {b^{5} x^{15}}{15}+a^{5} \ln \left (x \right )\)	\(56\)

[In]

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

[Out]

5/3*a^4*b*x^3+5/3*a^3*b^2*x^6+10/9*a^2*b^3*x^9+5/12*a*b^4*x^12+1/15*b^5*x^15+a^5*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=\frac {1}{15} \, b^{5} x^{15} + \frac {5}{12} \, a b^{4} x^{12} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {5}{3} \, a^{3} b^{2} x^{6} + \frac {5}{3} \, a^{4} b x^{3} + a^{5} \log \left (x\right ) \]

[In]

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

[Out]

1/15*b^5*x^15 + 5/12*a*b^4*x^12 + 10/9*a^2*b^3*x^9 + 5/3*a^3*b^2*x^6 + 5/3*a^4*b*x^3 + a^5*log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=a^{5} \log {\left (x \right )} + \frac {5 a^{4} b x^{3}}{3} + \frac {5 a^{3} b^{2} x^{6}}{3} + \frac {10 a^{2} b^{3} x^{9}}{9} + \frac {5 a b^{4} x^{12}}{12} + \frac {b^{5} x^{15}}{15} \]

[In]

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

[Out]

a**5*log(x) + 5*a**4*b*x**3/3 + 5*a**3*b**2*x**6/3 + 10*a**2*b**3*x**9/9 + 5*a*b**4*x**12/12 + b**5*x**15/15

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=\frac {1}{15} \, b^{5} x^{15} + \frac {5}{12} \, a b^{4} x^{12} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {5}{3} \, a^{3} b^{2} x^{6} + \frac {5}{3} \, a^{4} b x^{3} + \frac {1}{3} \, a^{5} \log \left (x^{3}\right ) \]

[In]

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

[Out]

1/15*b^5*x^15 + 5/12*a*b^4*x^12 + 10/9*a^2*b^3*x^9 + 5/3*a^3*b^2*x^6 + 5/3*a^4*b*x^3 + 1/3*a^5*log(x^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=\frac {1}{15} \, b^{5} x^{15} + \frac {5}{12} \, a b^{4} x^{12} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {5}{3} \, a^{3} b^{2} x^{6} + \frac {5}{3} \, a^{4} b x^{3} + a^{5} \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/15*b^5*x^15 + 5/12*a*b^4*x^12 + 10/9*a^2*b^3*x^9 + 5/3*a^3*b^2*x^6 + 5/3*a^4*b*x^3 + a^5*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^3\right )^5}{x} \, dx=a^5\,\ln \left (x\right )+\frac {b^5\,x^{15}}{15}+\frac {5\,a^4\,b\,x^3}{3}+\frac {5\,a\,b^4\,x^{12}}{12}+\frac {5\,a^3\,b^2\,x^6}{3}+\frac {10\,a^2\,b^3\,x^9}{9} \]

[In]

int((a + b*x^3)^5/x,x)

[Out]

a^5*log(x) + (b^5*x^15)/15 + (5*a^4*b*x^3)/3 + (5*a*b^4*x^12)/12 + (5*a^3*b^2*x^6)/3 + (10*a^2*b^3*x^9)/9