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

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Rubi [A]  time = 0.0116512, antiderivative size = 48, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0094371, 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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Maple [A]  time = 0.005, size = 26, normalized size = 0.5 \begin{align*} -{\frac{-4\,b{x}^{5}+2\,a}{15\,{a}^{2}}\sqrt{b{x}^{5}+a}{x}^{-{\frac{15}{2}}}} \end{align*}

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 = 0.993716, size = 47, normalized size = 0.98 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 1.74947, size = 69, normalized size = 1.44 \begin{align*} \frac{2 \,{\left (2 \, b x^{5} - a\right )} \sqrt{b x^{5} + a}}{15 \, a^{2} x^{\frac{15}{2}}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.22213, size = 49, normalized size = 1.02 \begin{align*} -\frac{4 \, b^{\frac{3}{2}}}{15 \, a^{2}} - \frac{2 \,{\left ({\left (b + \frac{a}{x^{5}}\right )}^{\frac{3}{2}} - 3 \, \sqrt{b + \frac{a}{x^{5}}} b\right )}}{15 \, a^{2}} \end{align*}

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]

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