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]

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

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 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]

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]]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.85, size = 137, normalized size = 1.85 \[ \left [\frac {3 \, \sqrt {a} b^{2} x^{6} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (3 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a^{3} x^{6}}, \frac {3 \, \sqrt {-a} b^{2} x^{6} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x^{3} - 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a^{3} x^{6}}\right ] \]

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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giac [A]  time = 0.17, size = 75, normalized size = 1.01 \[ \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3} - 5 \, \sqrt {b x^{3} + a} a b^{3}}{a^{2} b^{2} x^{6}}}{12 \, b} \]

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

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maple [A]  time = 0.02, size = 59, normalized size = 0.80 \[ -\frac {b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {5}{2}}}+\frac {\sqrt {b \,x^{3}+a}\, b}{4 a^{2} x^{3}}-\frac {\sqrt {b \,x^{3}+a}}{6 a \,x^{6}} \]

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)/a^2/x^3

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maxima [A]  time = 3.01, size = 104, normalized size = 1.41 \[ \frac {b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {b x^{3} + a} a b^{2}}{12 \, {\left ({\left (b x^{3} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{3} + a\right )} a^{3} + a^{4}\right )}} \]

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]

1/8*b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(5/2) + 1/12*(3*(b*x^3 + a)^(3/2)*b^2 -
5*sqrt(b*x^3 + a)*a*b^2)/((b*x^3 + a)^2*a^2 - 2*(b*x^3 + a)*a^3 + a^4)

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mupad [B]  time = 1.32, size = 79, normalized size = 1.07 \[ \frac {b^2\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{8\,a^{5/2}}-\frac {\sqrt {b\,x^3+a}}{6\,a\,x^6}+\frac {b\,\sqrt {b\,x^3+a}}{4\,a^2\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a + b*x^3)^(1/2)),x)

[Out]

(b^2*log((((a + b*x^3)^(1/2) - a^(1/2))^3*((a + b*x^3)^(1/2) + a^(1/2)))/x^6))/(8*a^(5/2)) - (a + b*x^3)^(1/2)
/(6*a*x^6) + (b*(a + b*x^3)^(1/2))/(4*a^2*x^3)

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sympy [A]  time = 10.87, size = 104, normalized size = 1.41 \[ - \frac {1}{6 \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {\sqrt {b}}{12 a x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b^{\frac {3}{2}}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{4 a^{\frac {5}{2}}} \]

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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