3.410 \(\int \frac{1}{x^7 \sqrt{a+b x^3}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{5/2}}+\frac{b \sqrt{a+b x^3}}{4 a^2 x^3}-\frac{\sqrt{a+b x^3}}{6 a x^6} \]

[Out]

-Sqrt[a + b*x^3]/(6*a*x^6) + (b*Sqrt[a + b*x^3])/(4*a^2*x^3) - (b^2*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a^(5/
2))

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Rubi [A]  time = 0.0381319, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{5/2}}+\frac{b \sqrt{a+b x^3}}{4 a^2 x^3}-\frac{\sqrt{a+b x^3}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*Sqrt[a + b*x^3]),x]

[Out]

-Sqrt[a + b*x^3]/(6*a*x^6) + (b*Sqrt[a + b*x^3])/(4*a^2*x^3) - (b^2*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a^(5/
2))

Rule 266

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]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{x^7 \sqrt{a+b x^3}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,x^3\right )\\ &=-\frac{\sqrt{a+b x^3}}{6 a x^6}-\frac{b \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^3\right )}{4 a}\\ &=-\frac{\sqrt{a+b x^3}}{6 a x^6}+\frac{b \sqrt{a+b x^3}}{4 a^2 x^3}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )}{8 a^2}\\ &=-\frac{\sqrt{a+b x^3}}{6 a x^6}+\frac{b \sqrt{a+b x^3}}{4 a^2 x^3}+\frac{b \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^3}\right )}{4 a^2}\\ &=-\frac{\sqrt{a+b x^3}}{6 a x^6}+\frac{b \sqrt{a+b x^3}}{4 a^2 x^3}-\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0077304, size = 39, normalized size = 0.53 \[ -\frac{2 b^2 \sqrt{a+b x^3} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{b x^3}{a}+1\right )}{3 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*Sqrt[a + b*x^3]),x]

[Out]

(-2*b^2*Sqrt[a + b*x^3]*Hypergeometric2F1[1/2, 3, 3/2, 1 + (b*x^3)/a])/(3*a^3)

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Maple [A]  time = 0.016, size = 59, normalized size = 0.8 \begin{align*} -{\frac{{b}^{2}}{4}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}}-{\frac{1}{6\,{x}^{6}a}\sqrt{b{x}^{3}+a}}+{\frac{b}{4\,{x}^{3}{a}^{2}}\sqrt{b{x}^{3}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(b*x^3+a)^(1/2),x)

[Out]

-1/4*b^2*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(5/2)-1/6*(b*x^3+a)^(1/2)/x^6/a+1/4*b*(b*x^3+a)^(1/2)/x^3/a^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56045, size = 325, normalized size = 4.39 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} x^{6} \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) + 2 \,{\left (3 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt{b x^{3} + a}}{24 \, a^{3} x^{6}}, \frac{3 \, \sqrt{-a} b^{2} x^{6} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt{b x^{3} + a}}{12 \, a^{3} x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[1/24*(3*sqrt(a)*b^2*x^6*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(3*a*b*x^3 - 2*a^2)*sqrt(b*x^3
+ a))/(a^3*x^6), 1/12*(3*sqrt(-a)*b^2*x^6*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + (3*a*b*x^3 - 2*a^2)*sqrt(b*x^3
+ a))/(a^3*x^6)]

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Sympy [A]  time = 4.53451, size = 104, normalized size = 1.41 \begin{align*} - \frac{1}{6 \sqrt{b} x^{\frac{15}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{\sqrt{b}}{12 a x^{\frac{9}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{b^{\frac{3}{2}}}{4 a^{2} x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} - \frac{b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )}}{4 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(b*x**3+a)**(1/2),x)

[Out]

-1/(6*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) + sqrt(b)/(12*a*x**(9/2)*sqrt(a/(b*x**3) + 1)) + b**(3/2)/(4*a**
2*x**(3/2)*sqrt(a/(b*x**3) + 1)) - b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(4*a**(5/2))

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Giac [A]  time = 1.09947, size = 89, normalized size = 1.2 \begin{align*} \frac{1}{12} \, b^{2}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} - 5 \, \sqrt{b x^{3} + a} a}{a^{2} b^{2} x^{6}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

1/12*b^2*(3*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x^3 + a)^(3/2) - 5*sqrt(b*x^3 + a)*a)/(a^2
*b^2*x^6))