3.375 \(\int \frac{\sqrt{a+b x^3}}{x^{10}} \, dx\)

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

[Out]

-Sqrt[a + b*x^3]/(9*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))

________________________________________________________________________________________

Rubi [A]  time = 0.0532572, antiderivative size = 95, normalized size of antiderivative = 1., 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 = {266, 47, 51, 63, 208} \[ \frac{b^2 \sqrt{a+b x^3}}{24 a^2 x^3}-\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{24 a^{5/2}}-\frac{b \sqrt{a+b x^3}}{36 a x^6}-\frac{\sqrt{a+b x^3}}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3]/x^10,x]

[Out]

-Sqrt[a + b*x^3]/(9*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 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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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{\sqrt{a+b x^3}}{x^{10}} \, dx &=\frac{1}{3} \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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*}

Mathematica [C]  time = 0.0088009, size = 39, normalized size = 0.41 \[ \frac{2 b^3 \left (a+b x^3\right )^{3/2} \, _2F_1\left (\frac{3}{2},4;\frac{5}{2};\frac{b x^3}{a}+1\right )}{9 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^3]/x^10,x]

[Out]

(2*b^3*(a + b*x^3)^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, 1 + (b*x^3)/a])/(9*a^4)

________________________________________________________________________________________

Maple [A]  time = 0.02, size = 76, normalized size = 0.8 \begin{align*} -{\frac{{b}^{3}}{24}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}}-{\frac{1}{9\,{x}^{9}}\sqrt{b{x}^{3}+a}}-{\frac{b}{36\,{x}^{6}a}\sqrt{b{x}^{3}+a}}+{\frac{{b}^{2}}{24\,{x}^{3}{a}^{2}}\sqrt{b{x}^{3}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)/x^3/a^2

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.49517, size = 370, normalized size = 3.89 \begin{align*} \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{align*}

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)]

________________________________________________________________________________________

Sympy [A]  time = 8.30894, size = 129, normalized size = 1.36 \begin{align*} - \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{align*}

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))

________________________________________________________________________________________

Giac [A]  time = 1.13218, size = 108, normalized size = 1.14 \begin{align*} \frac{1}{72} \, b^{3}{\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{5}{2}} - 8 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} a - 3 \, \sqrt{b x^{3} + a} a^{2}}{a^{2} b^{3} x^{9}}\right )} \end{align*}

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