3.288 \(\int \frac{1}{x^3 \sqrt{a x^2+b x^5}} \, dx\)

Optimal. Leaf size=59 \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^5}}\right )}{3 a^{3/2}}-\frac{\sqrt{a x^2+b x^5}}{3 a x^4} \]

[Out]

-Sqrt[a*x^2 + b*x^5]/(3*a*x^4) + (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^5]])/(3*a^(3/2))

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Rubi [A]  time = 0.0545933, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2025, 2008, 206} \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^5}}\right )}{3 a^{3/2}}-\frac{\sqrt{a x^2+b x^5}}{3 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a*x^2 + b*x^5]),x]

[Out]

-Sqrt[a*x^2 + b*x^5]/(3*a*x^4) + (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^5]])/(3*a^(3/2))

Rule 2025

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt{a x^2+b x^5}} \, dx &=-\frac{\sqrt{a x^2+b x^5}}{3 a x^4}-\frac{b \int \frac{1}{\sqrt{a x^2+b x^5}} \, dx}{2 a}\\ &=-\frac{\sqrt{a x^2+b x^5}}{3 a x^4}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{x}{\sqrt{a x^2+b x^5}}\right )}{3 a}\\ &=-\frac{\sqrt{a x^2+b x^5}}{3 a x^4}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b x^5}}\right )}{3 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.061394, size = 71, normalized size = 1.2 \[ \frac{2 b \sqrt{x^2 \left (a+b x^3\right )} \left (\frac{\tanh ^{-1}\left (\sqrt{\frac{b x^3}{a}+1}\right )}{2 \sqrt{\frac{b x^3}{a}+1}}-\frac{a}{2 b x^3}\right )}{3 a^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a*x^2 + b*x^5]),x]

[Out]

(2*b*Sqrt[x^2*(a + b*x^3)]*(-a/(2*b*x^3) + ArcTanh[Sqrt[1 + (b*x^3)/a]]/(2*Sqrt[1 + (b*x^3)/a])))/(3*a^2*x)

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Maple [A]  time = 0.006, size = 66, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,{x}^{2}}\sqrt{b{x}^{3}+a} \left ( -b{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) a{x}^{3}+\sqrt{b{x}^{3}+a}{a}^{{\frac{3}{2}}} \right ){\frac{1}{\sqrt{b{x}^{5}+a{x}^{2}}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{5} + a x^{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^5 + a*x^2)*x^3), x)

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Fricas [A]  time = 0.809663, size = 294, normalized size = 4.98 \begin{align*} \left [\frac{\sqrt{a} b x^{4} \log \left (\frac{b x^{4} + 2 \, a x + 2 \, \sqrt{b x^{5} + a x^{2}} \sqrt{a}}{x^{4}}\right ) - 2 \, \sqrt{b x^{5} + a x^{2}} a}{6 \, a^{2} x^{4}}, -\frac{\sqrt{-a} b x^{4} \arctan \left (\frac{\sqrt{b x^{5} + a x^{2}} \sqrt{-a}}{a x}\right ) + \sqrt{b x^{5} + a x^{2}} a}{3 \, a^{2} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(sqrt(a)*b*x^4*log((b*x^4 + 2*a*x + 2*sqrt(b*x^5 + a*x^2)*sqrt(a))/x^4) - 2*sqrt(b*x^5 + a*x^2)*a)/(a^2*x
^4), -1/3*(sqrt(-a)*b*x^4*arctan(sqrt(b*x^5 + a*x^2)*sqrt(-a)/(a*x)) + sqrt(b*x^5 + a*x^2)*a)/(a^2*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{x^{2} \left (a + b x^{3}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**5+a*x**2)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(x**2*(a + b*x**3))), x)

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Giac [A]  time = 1.16582, size = 77, normalized size = 1.31 \begin{align*} -\frac{b \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{3 \, \sqrt{-a} a \mathrm{sgn}\left (x\right )} - \frac{\sqrt{\frac{b}{x} + \frac{a}{x^{4}}}}{3 \, a x \mathrm{sgn}\left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")

[Out]

-1/3*b*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a*sgn(x)) - 1/3*sqrt(b/x + a/x^4)/(a*x*sgn(x))