\(\int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx\) [96]

Optimal result

Integrand size = 17, antiderivative size = 74 \[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=-\frac {2 \sqrt {a x+b x^4}}{15 a x^8}+\frac {8 b \sqrt {a x+b x^4}}{45 a^2 x^5}-\frac {16 b^2 \sqrt {a x+b x^4}}{45 a^3 x^2} \]

[Out]

-2/15*(b*x^4+a*x)^(1/2)/a/x^8+8/45*b*(b*x^4+a*x)^(1/2)/a^2/x^5-16/45*b^2*(b*x^4+a*x)^(1/2)/a^3/x^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2041, 2039} \[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=-\frac {16 b^2 \sqrt {a x+b x^4}}{45 a^3 x^2}+\frac {8 b \sqrt {a x+b x^4}}{45 a^2 x^5}-\frac {2 \sqrt {a x+b x^4}}{15 a x^8} \]

[In]

Int[1/(x^8*Sqrt[a*x + b*x^4]),x]

[Out]

(-2*Sqrt[a*x + b*x^4])/(15*a*x^8) + (8*b*Sqrt[a*x + b*x^4])/(45*a^2*x^5) - (16*b^2*Sqrt[a*x + b*x^4])/(45*a^3*
x^2)

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {a x+b x^4}}{15 a x^8}-\frac {(4 b) \int \frac {1}{x^5 \sqrt {a x+b x^4}} \, dx}{5 a} \\ & = -\frac {2 \sqrt {a x+b x^4}}{15 a x^8}+\frac {8 b \sqrt {a x+b x^4}}{45 a^2 x^5}+\frac {\left (8 b^2\right ) \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx}{15 a^2} \\ & = -\frac {2 \sqrt {a x+b x^4}}{15 a x^8}+\frac {8 b \sqrt {a x+b x^4}}{45 a^2 x^5}-\frac {16 b^2 \sqrt {a x+b x^4}}{45 a^3 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.75 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=-\frac {2 \sqrt {x \left (a+b x^3\right )} \left (3 a^2-4 a b x^3+8 b^2 x^6\right )}{45 a^3 x^8} \]

[In]

Integrate[1/(x^8*Sqrt[a*x + b*x^4]),x]

[Out]

(-2*Sqrt[x*(a + b*x^3)]*(3*a^2 - 4*a*b*x^3 + 8*b^2*x^6))/(45*a^3*x^8)

Maple [A] (verified)

Time = 2.93 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55

method	result	size
trager	\(-\frac {2 \left (8 b^{2} x^{6}-4 a b \,x^{3}+3 a^{2}\right ) \sqrt {b \,x^{4}+a x}}{45 x^{8} a^{3}}\)	\(41\)
pseudoelliptic	\(-\frac {2 \left (8 b^{2} x^{6}-4 a b \,x^{3}+3 a^{2}\right ) \sqrt {x \left (b \,x^{3}+a \right )}}{45 x^{8} a^{3}}\)	\(41\)
gosper	\(-\frac {2 \left (b \,x^{3}+a \right ) \left (8 b^{2} x^{6}-4 a b \,x^{3}+3 a^{2}\right )}{45 x^{7} a^{3} \sqrt {b \,x^{4}+a x}}\)	\(48\)
risch	\(-\frac {2 \left (b \,x^{3}+a \right ) \left (8 b^{2} x^{6}-4 a b \,x^{3}+3 a^{2}\right )}{45 a^{3} x^{7} \sqrt {x \left (b \,x^{3}+a \right )}}\)	\(48\)
default	\(-\frac {2 \sqrt {b \,x^{4}+a x}}{15 a \,x^{8}}+\frac {8 b \sqrt {b \,x^{4}+a x}}{45 a^{2} x^{5}}-\frac {16 b^{2} \sqrt {b \,x^{4}+a x}}{45 a^{3} x^{2}}\)	\(63\)
elliptic	\(-\frac {2 \sqrt {b \,x^{4}+a x}}{15 a \,x^{8}}+\frac {8 b \sqrt {b \,x^{4}+a x}}{45 a^{2} x^{5}}-\frac {16 b^{2} \sqrt {b \,x^{4}+a x}}{45 a^{3} x^{2}}\)	\(63\)

[In]

int(1/x^8/(b*x^4+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/45*(8*b^2*x^6-4*a*b*x^3+3*a^2)/x^8/a^3*(b*x^4+a*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=-\frac {2 \, {\left (8 \, b^{2} x^{6} - 4 \, a b x^{3} + 3 \, a^{2}\right )} \sqrt {b x^{4} + a x}}{45 \, a^{3} x^{8}} \]

[In]

integrate(1/x^8/(b*x^4+a*x)^(1/2),x, algorithm="fricas")

[Out]

-2/45*(8*b^2*x^6 - 4*a*b*x^3 + 3*a^2)*sqrt(b*x^4 + a*x)/(a^3*x^8)

Sympy [F]

\[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=\int \frac {1}{x^{8} \sqrt {x \left (a + b x^{3}\right )}}\, dx \]

[In]

integrate(1/x**8/(b*x**4+a*x)**(1/2),x)

[Out]

Integral(1/(x**8*sqrt(x*(a + b*x**3))), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=-\frac {2 \, {\left (8 \, b^{3} x^{10} + 4 \, a b^{2} x^{7} - a^{2} b x^{4} + 3 \, a^{3} x\right )}}{45 \, \sqrt {b x^{3} + a} a^{3} x^{\frac {17}{2}}} \]

[In]

integrate(1/x^8/(b*x^4+a*x)^(1/2),x, algorithm="maxima")

[Out]

-2/45*(8*b^3*x^10 + 4*a*b^2*x^7 - a^2*b*x^4 + 3*a^3*x)/(sqrt(b*x^3 + a)*a^3*x^(17/2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=-\frac {2 \, \sqrt {b + \frac {a}{x^{3}}} b^{2}}{3 \, a^{3}} - \frac {2 \, {\left (3 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {5}{2}} - 10 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {3}{2}} b\right )}}{45 \, a^{3}} \]

[In]

integrate(1/x^8/(b*x^4+a*x)^(1/2),x, algorithm="giac")

[Out]

-2/3*sqrt(b + a/x^3)*b^2/a^3 - 2/45*(3*(b + a/x^3)^(5/2) - 10*(b + a/x^3)^(3/2)*b)/a^3

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^8 \sqrt {a x+b x^4}} \, dx=-\frac {2\,\sqrt {b\,x^4+a\,x}\,\left (3\,a^2-4\,a\,b\,x^3+8\,b^2\,x^6\right )}{45\,a^3\,x^8} \]

[In]

int(1/(x^8*(a*x + b*x^4)^(1/2)),x)

[Out]

-(2*(a*x + b*x^4)^(1/2)*(3*a^2 + 8*b^2*x^6 - 4*a*b*x^3))/(45*a^3*x^8)