3.4.75 \(\int \frac {\sqrt {a+b x^3}}{x^{10}} \, dx\) [375]

Optimal. Leaf size=95 \[ -\frac {\sqrt {a+b x^3}}{9 x^9}-\frac {b \sqrt {a+b x^3}}{36 a x^6}+\frac {b^2 \sqrt {a+b x^3}}{24 a^2 x^3}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{24 a^{5/2}} \]

[Out]

-1/24*b^3*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(5/2)-1/9*(b*x^3+a)^(1/2)/x^9-1/36*b*(b*x^3+a)^(1/2)/a/x^6+1/24*b
^2*(b*x^3+a)^(1/2)/a^2/x^3

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Rubi [A]
time = 0.04, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 44, 65, 214} \begin {gather*} -\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{24 a^{5/2}}+\frac {b^2 \sqrt {a+b x^3}}{24 a^2 x^3}-\frac {\sqrt {a+b x^3}}{9 x^9}-\frac {b \sqrt {a+b x^3}}{36 a x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3]/x^10,x]

[Out]

-1/9*Sqrt[a + b*x^3]/x^9 - (b*Sqrt[a + b*x^3])/(36*a*x^6) + (b^2*Sqrt[a + b*x^3])/(24*a^2*x^3) - (b^3*ArcTanh[
Sqrt[a + b*x^3]/Sqrt[a]])/(24*a^(5/2))

Rule 43

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

Rule 272

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 {\sqrt {a+b x^3}}{x^{10}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^4} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {a+b x^3}}{9 x^9}+\frac {1}{18} b \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {a+b x^3}}{9 x^9}-\frac {b \sqrt {a+b x^3}}{36 a x^6}-\frac {b^2 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^3\right )}{24 a}\\ &=-\frac {\sqrt {a+b x^3}}{9 x^9}-\frac {b \sqrt {a+b x^3}}{36 a x^6}+\frac {b^2 \sqrt {a+b x^3}}{24 a^2 x^3}+\frac {b^3 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )}{48 a^2}\\ &=-\frac {\sqrt {a+b x^3}}{9 x^9}-\frac {b \sqrt {a+b x^3}}{36 a x^6}+\frac {b^2 \sqrt {a+b x^3}}{24 a^2 x^3}+\frac {b^2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^3}\right )}{24 a^2}\\ &=-\frac {\sqrt {a+b x^3}}{9 x^9}-\frac {b \sqrt {a+b x^3}}{36 a x^6}+\frac {b^2 \sqrt {a+b x^3}}{24 a^2 x^3}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{24 a^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 73, normalized size = 0.77 \begin {gather*} \frac {\sqrt {a+b x^3} \left (-8 a^2-2 a b x^3+3 b^2 x^6\right )}{72 a^2 x^9}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{24 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^3]/x^10,x]

[Out]

(Sqrt[a + b*x^3]*(-8*a^2 - 2*a*b*x^3 + 3*b^2*x^6))/(72*a^2*x^9) - (b^3*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(24*a
^(5/2))

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Maple [A]
time = 0.15, size = 76, normalized size = 0.80

method	result	size
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (-3 b^{2} x^{6}+2 a b \,x^{3}+8 a^{2}\right )}{72 x^{9} a^{2}}-\frac {b^{3} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{24 a^{\frac {5}{2}}}\)	\(62\)
default	\(-\frac {b^{3} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{24 a^{\frac {5}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{9 x^{9}}-\frac {b \sqrt {b \,x^{3}+a}}{36 x^{6} a}+\frac {b^{2} \sqrt {b \,x^{3}+a}}{24 a^{2} x^{3}}\)	\(76\)
elliptic	\(-\frac {b^{3} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{24 a^{\frac {5}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{9 x^{9}}-\frac {b \sqrt {b \,x^{3}+a}}{36 x^{6} a}+\frac {b^{2} \sqrt {b \,x^{3}+a}}{24 a^{2} x^{3}}\)	\(76\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/2)/x^10,x,method=_RETURNVERBOSE)

[Out]

-1/24*b^3*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(5/2)-1/9*(b*x^3+a)^(1/2)/x^9-1/36*b*(b*x^3+a)^(1/2)/x^6/a+1/24*b
^2*(b*x^3+a)^(1/2)/a^2/x^3

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Maxima [A]
time = 0.50, size = 137, normalized size = 1.44 \begin {gather*} \frac {b^{3} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{48 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} b^{3} - 8 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {b x^{3} + a} a^{2} b^{3}}{72 \, {\left ({\left (b x^{3} + a\right )}^{3} a^{2} - 3 \, {\left (b x^{3} + a\right )}^{2} a^{3} + 3 \, {\left (b x^{3} + a\right )} a^{4} - a^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/2)/x^10,x, algorithm="maxima")

[Out]

1/48*b^3*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(5/2) + 1/72*(3*(b*x^3 + a)^(5/2)*b^3
- 8*(b*x^3 + a)^(3/2)*a*b^3 - 3*sqrt(b*x^3 + a)*a^2*b^3)/((b*x^3 + a)^3*a^2 - 3*(b*x^3 + a)^2*a^3 + 3*(b*x^3 +
a)*a^4 - a^5)

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Fricas [A]
time = 0.36, size = 159, normalized size = 1.67 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{3} x^{9} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (3 \, a b^{2} x^{6} - 2 \, a^{2} b x^{3} - 8 \, a^{3}\right )} \sqrt {b x^{3} + a}}{144 \, a^{3} x^{9}}, \frac {3 \, \sqrt {-a} b^{3} x^{9} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{2} x^{6} - 2 \, a^{2} b x^{3} - 8 \, a^{3}\right )} \sqrt {b x^{3} + a}}{72 \, a^{3} x^{9}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/2)/x^10,x, algorithm="fricas")

[Out]

[1/144*(3*sqrt(a)*b^3*x^9*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(3*a*b^2*x^6 - 2*a^2*b*x^3 -
8*a^3)*sqrt(b*x^3 + a))/(a^3*x^9), 1/72*(3*sqrt(-a)*b^3*x^9*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + (3*a*b^2*x^6
- 2*a^2*b*x^3 - 8*a^3)*sqrt(b*x^3 + a))/(a^3*x^9)]

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Sympy [A]
time = 5.11, size = 129, normalized size = 1.36 \begin {gather*} - \frac {a}{9 \sqrt {b} x^{\frac {21}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {5 \sqrt {b}}{36 x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b^{\frac {3}{2}}}{72 a x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b^{\frac {5}{2}}}{24 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{24 a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/2)/x**10,x)

[Out]

-a/(9*sqrt(b)*x**(21/2)*sqrt(a/(b*x**3) + 1)) - 5*sqrt(b)/(36*x**(15/2)*sqrt(a/(b*x**3) + 1)) + b**(3/2)/(72*a
*x**(9/2)*sqrt(a/(b*x**3) + 1)) + b**(5/2)/(24*a**2*x**(3/2)*sqrt(a/(b*x**3) + 1)) - b**3*asinh(sqrt(a)/(sqrt(
b)*x**(3/2)))/(24*a**(5/2))

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Giac [A]
time = 1.42, size = 92, normalized size = 0.97 \begin {gather*} \frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {5}{2}} b^{4} - 8 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x^{3} + a} a^{2} b^{4}}{a^{2} b^{3} x^{9}}}{72 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/2)/x^10,x, algorithm="giac")

[Out]

1/72*(3*b^4*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x^3 + a)^(5/2)*b^4 - 8*(b*x^3 + a)^(3/2)*a
*b^4 - 3*sqrt(b*x^3 + a)*a^2*b^4)/(a^2*b^3*x^9))/b

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Mupad [B]
time = 1.50, size = 96, normalized size = 1.01 \begin {gather*} \frac {b^3\,\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 )}{48\,a^{5/2}}-\frac {\sqrt {b\,x^3+a}}{9\,x^9}-\frac {b\,\sqrt {b\,x^3+a}}{36\,a\,x^6}+\frac {b^2\,\sqrt {b\,x^3+a}}{24\,a^2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/2)/x^10,x)

[Out]

(b^3*log((((a + b*x^3)^(1/2) - a^(1/2))^3*((a + b*x^3)^(1/2) + a^(1/2)))/x^6))/(48*a^(5/2)) - (a + b*x^3)^(1/2
)/(9*x^9) - (b*(a + b*x^3)^(1/2))/(36*a*x^6) + (b^2*(a + b*x^3)^(1/2))/(24*a^2*x^3)

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