\(\int \frac {x^9}{(a+b x^5)^2} \, dx\) [1280]

Optimal result

Integrand size = 13, antiderivative size = 33 \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=\frac {a}{5 b^2 \left (a+b x^5\right )}+\frac {\log \left (a+b x^5\right )}{5 b^2} \]

[Out]

1/5*a/b^2/(b*x^5+a)+1/5*ln(b*x^5+a)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, 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} \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=\frac {a}{5 b^2 \left (a+b x^5\right )}+\frac {\log \left (a+b x^5\right )}{5 b^2} \]

[In]

Int[x^9/(a + b*x^5)^2,x]

[Out]

a/(5*b^2*(a + b*x^5)) + Log[a + b*x^5]/(5*b^2)

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}{5} \text {Subst}\left (\int \frac {x}{(a+b x)^2} \, dx,x,x^5\right ) \\ & = \frac {1}{5} \text {Subst}\left (\int \left (-\frac {a}{b (a+b x)^2}+\frac {1}{b (a+b x)}\right ) \, dx,x,x^5\right ) \\ & = \frac {a}{5 b^2 \left (a+b x^5\right )}+\frac {\log \left (a+b x^5\right )}{5 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=\frac {\frac {a}{a+b x^5}+\log \left (a+b x^5\right )}{5 b^2} \]

[In]

Integrate[x^9/(a + b*x^5)^2,x]

[Out]

(a/(a + b*x^5) + Log[a + b*x^5])/(5*b^2)

Maple [A] (verified)

Time = 4.53 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {a}{5 b^{2} \left (b \,x^{5}+a \right )}+\frac {\ln \left (b \,x^{5}+a \right )}{5 b^{2}}\)	\(30\)
norman	\(\frac {a}{5 b^{2} \left (b \,x^{5}+a \right )}+\frac {\ln \left (b \,x^{5}+a \right )}{5 b^{2}}\)	\(30\)
risch	\(\frac {a}{5 b^{2} \left (b \,x^{5}+a \right )}+\frac {\ln \left (b \,x^{5}+a \right )}{5 b^{2}}\)	\(30\)
parallelrisch	\(\frac {b \ln \left (b \,x^{5}+a \right ) x^{5}+a \ln \left (b \,x^{5}+a \right )+a}{5 b^{2} \left (b \,x^{5}+a \right )}\)	\(40\)

[In]

int(x^9/(b*x^5+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/5*a/b^2/(b*x^5+a)+1/5*ln(b*x^5+a)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=\frac {{\left (b x^{5} + a\right )} \log \left (b x^{5} + a\right ) + a}{5 \, {\left (b^{3} x^{5} + a b^{2}\right )}} \]

[In]

integrate(x^9/(b*x^5+a)^2,x, algorithm="fricas")

[Out]

1/5*((b*x^5 + a)*log(b*x^5 + a) + a)/(b^3*x^5 + a*b^2)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=\frac {a}{5 a b^{2} + 5 b^{3} x^{5}} + \frac {\log {\left (a + b x^{5} \right )}}{5 b^{2}} \]

[In]

integrate(x**9/(b*x**5+a)**2,x)

[Out]

a/(5*a*b**2 + 5*b**3*x**5) + log(a + b*x**5)/(5*b**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=\frac {a}{5 \, {\left (b^{3} x^{5} + a b^{2}\right )}} + \frac {\log \left (b x^{5} + a\right )}{5 \, b^{2}} \]

[In]

integrate(x^9/(b*x^5+a)^2,x, algorithm="maxima")

[Out]

1/5*a/(b^3*x^5 + a*b^2) + 1/5*log(b*x^5 + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=-\frac {\frac {\log \left (\frac {{\left | b x^{5} + a \right |}}{{\left (b x^{5} + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x^{5} + a\right )} b}}{5 \, b} \]

[In]

integrate(x^9/(b*x^5+a)^2,x, algorithm="giac")

[Out]

-1/5*(log(abs(b*x^5 + a)/((b*x^5 + a)^2*abs(b)))/b - a/((b*x^5 + a)*b))/b

Mupad [B] (verification not implemented)

Time = 5.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {x^9}{\left (a+b x^5\right )^2} \, dx=\frac {\ln \left (b\,x^5+a\right )}{5\,b^2}+\frac {a}{5\,b^2\,\left (b\,x^5+a\right )} \]

[In]

int(x^9/(a + b*x^5)^2,x)

[Out]

log(a + b*x^5)/(5*b^2) + a/(5*b^2*(a + b*x^5))