3.1312 \(\int \frac {1}{x^{17/2} \sqrt {a+b x^5}} \, dx\)

Optimal. Leaf size=48 \[ \frac {4 b \sqrt {a+b x^5}}{15 a^2 x^{5/2}}-\frac {2 \sqrt {a+b x^5}}{15 a x^{15/2}} \]

[Out]

-2/15*(b*x^5+a)^(1/2)/a/x^(15/2)+4/15*b*(b*x^5+a)^(1/2)/a^2/x^(5/2)

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {271, 264} \[ \frac {4 b \sqrt {a+b x^5}}{15 a^2 x^{5/2}}-\frac {2 \sqrt {a+b x^5}}{15 a x^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(17/2)*Sqrt[a + b*x^5]),x]

[Out]

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

Rule 264

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

Rule 271

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

Rubi steps

\begin {align*} \int \frac {1}{x^{17/2} \sqrt {a+b x^5}} \, dx &=-\frac {2 \sqrt {a+b x^5}}{15 a x^{15/2}}-\frac {(2 b) \int \frac {1}{x^{7/2} \sqrt {a+b x^5}} \, dx}{3 a}\\ &=-\frac {2 \sqrt {a+b x^5}}{15 a x^{15/2}}+\frac {4 b \sqrt {a+b x^5}}{15 a^2 x^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.65 \[ -\frac {2 \left (a-2 b x^5\right ) \sqrt {a+b x^5}}{15 a^2 x^{15/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(17/2)*Sqrt[a + b*x^5]),x]

[Out]

(-2*(a - 2*b*x^5)*Sqrt[a + b*x^5])/(15*a^2*x^(15/2))

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fricas [A]  time = 1.07, size = 27, normalized size = 0.56 \[ \frac {2 \, {\left (2 \, b x^{5} - a\right )} \sqrt {b x^{5} + a}}{15 \, a^{2} x^{\frac {15}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(17/2)/(b*x^5+a)^(1/2),x, algorithm="fricas")

[Out]

2/15*(2*b*x^5 - a)*sqrt(b*x^5 + a)/(a^2*x^(15/2))

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giac [A]  time = 0.21, size = 38, normalized size = 0.79 \[ -\frac {2 \, {\left (b + \frac {a}{x^{5}}\right )}^{\frac {3}{2}}}{15 \, a^{2}} + \frac {2 \, \sqrt {b + \frac {a}{x^{5}}} b}{5 \, a^{2}} - \frac {4 \, b^{\frac {3}{2}}}{15 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(17/2)/(b*x^5+a)^(1/2),x, algorithm="giac")

[Out]

-2/15*(b + a/x^5)^(3/2)/a^2 + 2/5*sqrt(b + a/x^5)*b/a^2 - 4/15*b^(3/2)/a^2

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maple [A]  time = 0.00, size = 26, normalized size = 0.54 \[ -\frac {2 \sqrt {b \,x^{5}+a}\, \left (-2 b \,x^{5}+a \right )}{15 a^{2} x^{\frac {15}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(17/2)/(b*x^5+a)^(1/2),x)

[Out]

-2/15*(b*x^5+a)^(1/2)*(-2*b*x^5+a)/x^(15/2)/a^2

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maxima [A]  time = 1.10, size = 35, normalized size = 0.73 \[ \frac {2 \, {\left (\frac {3 \, \sqrt {b x^{5} + a} b}{x^{\frac {5}{2}}} - \frac {{\left (b x^{5} + a\right )}^{\frac {3}{2}}}{x^{\frac {15}{2}}}\right )}}{15 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(17/2)/(b*x^5+a)^(1/2),x, algorithm="maxima")

[Out]

2/15*(3*sqrt(b*x^5 + a)*b/x^(5/2) - (b*x^5 + a)^(3/2)/x^(15/2))/a^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{x^{17/2}\,\sqrt {b\,x^5+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(17/2)*(a + b*x^5)^(1/2)),x)

[Out]

int(1/(x^(17/2)*(a + b*x^5)^(1/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(17/2)/(b*x**5+a)**(1/2),x)

[Out]

Timed out

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