3.2018 \(\int \frac{x^2}{\sqrt{a+\frac{b}{x^3}}} \, dx\)
Optimal. Leaf size=50 \[ \frac{x^3 \sqrt{a+\frac{b}{x^3}}}{3 a}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^3}}}{\sqrt{a}}\right )}{3 a^{3/2}} \]
[Out]
(Sqrt[a + b/x^3]*x^3)/(3*a) - (b*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(3*a^(3/2))
________________________________________________________________________________________
Rubi [A] time = 0.0257762, antiderivative size = 50, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used =
{266, 51, 63, 208} \[ \frac{x^3 \sqrt{a+\frac{b}{x^3}}}{3 a}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^3}}}{\sqrt{a}}\right )}{3 a^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[x^2/Sqrt[a + b/x^3],x]
[Out]
(Sqrt[a + b/x^3]*x^3)/(3*a) - (b*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(3*a^(3/2))
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{x^2}{\sqrt{a+\frac{b}{x^3}}} \, dx &=-\left (\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x^3}\right )\right )\\ &=\frac{\sqrt{a+\frac{b}{x^3}} x^3}{3 a}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^3}\right )}{6 a}\\ &=\frac{\sqrt{a+\frac{b}{x^3}} x^3}{3 a}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^3}}\right )}{3 a}\\ &=\frac{\sqrt{a+\frac{b}{x^3}} x^3}{3 a}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^3}}}{\sqrt{a}}\right )}{3 a^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.0259459, size = 81, normalized size = 1.62 \[ \frac{\sqrt{a} x^{3/2} \left (a x^3+b\right )-b \sqrt{a x^3+b} \tanh ^{-1}\left (\frac{\sqrt{a} x^{3/2}}{\sqrt{a x^3+b}}\right )}{3 a^{3/2} x^{3/2} \sqrt{a+\frac{b}{x^3}}} \]
Antiderivative was successfully verified.
[In]
Integrate[x^2/Sqrt[a + b/x^3],x]
[Out]
(Sqrt[a]*x^(3/2)*(b + a*x^3) - b*Sqrt[b + a*x^3]*ArcTanh[(Sqrt[a]*x^(3/2))/Sqrt[b + a*x^3]])/(3*a^(3/2)*Sqrt[a
+ b/x^3]*x^(3/2))
________________________________________________________________________________________
Maple [C] time = 0.013, size = 3347, normalized size = 66.9 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2/(a+b/x^3)^(1/2),x)
[Out]
1/3/((a*x^3+b)/x^3)^(1/2)/x*(a*x^3+b)/a^3*(6*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)
*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a
^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+
I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(1
/2)*x^2*a^2*b-6*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2
*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3
))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1
/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(
1/2)*x^2*a^2*b-12*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)
+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1
/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(
1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(1/3)*3^(1/2)*x*a*b+12
*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(
1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2
))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),(-1
+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(1/3)*3^(
1/2)*x*a*b+6*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*
x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/
(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3))
)^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(2/3)*3^(1/2)*b-6*(-(I*3^(1
/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3
^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b
*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(
-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b-6*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a
^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*
((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticPi((-(I*3
^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^
(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(2/3)*3^(1/2)*b+6*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x
+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^
(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticPi(
(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-
1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b+12*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2
)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((
I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1
/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-
3))^(1/2))*(-b*a^2)^(1/3)*x*a*b-12*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)
*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*
a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/
(-a*x+(-b*a^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/
2)-3))^(1/2))*(-b*a^2)^(1/3)*x*a*b+I*(a*x^4+b*x)^(1/2)*(1/a^2*x*(-a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3
)+2*a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3)))^(1/2)*3^(1/2)*x*a^2-6*(-(I*3^(1/2)-3)
*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2)
)/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^
(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3
^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(2/3)*b+6*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^
2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*(
(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticPi((-(I*3^
(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(
1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(2/3)*b-3*x*(a*x^4+b*x)^(1/2)*a^2*(1/a^2*x*(-a*x+(-b*a^2)^(
1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3)))^(1/2))/
(x*(a*x^3+b))^(1/2)/(I*3^(1/2)-3)/(1/a^2*x*(-a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3
))*(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3)))^(1/2)
________________________________________________________________________________________
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(x^2/(a+b/x^3)^(1/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [A] time = 2.40123, size = 347, normalized size = 6.94 \begin{align*} \left [\frac{4 \, a x^{3} \sqrt{\frac{a x^{3} + b}{x^{3}}} + \sqrt{a} b \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} + 4 \,{\left (2 \, a x^{6} + b x^{3}\right )} \sqrt{a} \sqrt{\frac{a x^{3} + b}{x^{3}}}\right )}{12 \, a^{2}}, \frac{2 \, a x^{3} \sqrt{\frac{a x^{3} + b}{x^{3}}} + \sqrt{-a} b \arctan \left (\frac{2 \, \sqrt{-a} x^{3} \sqrt{\frac{a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right )}{6 \, a^{2}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(a+b/x^3)^(1/2),x, algorithm="fricas")
[Out]
[1/12*(4*a*x^3*sqrt((a*x^3 + b)/x^3) + sqrt(a)*b*log(-8*a^2*x^6 - 8*a*b*x^3 - b^2 + 4*(2*a*x^6 + b*x^3)*sqrt(a
)*sqrt((a*x^3 + b)/x^3)))/a^2, 1/6*(2*a*x^3*sqrt((a*x^3 + b)/x^3) + sqrt(-a)*b*arctan(2*sqrt(-a)*x^3*sqrt((a*x
^3 + b)/x^3)/(2*a*x^3 + b)))/a^2]
________________________________________________________________________________________
Sympy [A] time = 2.2916, size = 49, normalized size = 0.98 \begin{align*} \frac{\sqrt{b} x^{\frac{3}{2}} \sqrt{\frac{a x^{3}}{b} + 1}}{3 a} - \frac{b \operatorname{asinh}{\left (\frac{\sqrt{a} x^{\frac{3}{2}}}{\sqrt{b}} \right )}}{3 a^{\frac{3}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2/(a+b/x**3)**(1/2),x)
[Out]
sqrt(b)*x**(3/2)*sqrt(a*x**3/b + 1)/(3*a) - b*asinh(sqrt(a)*x**(3/2)/sqrt(b))/(3*a**(3/2))
________________________________________________________________________________________
Giac [A] time = 1.3076, size = 90, normalized size = 1.8 \begin{align*} \frac{1}{3} \, b{\left (\frac{\arctan \left (\frac{\sqrt{\frac{a x^{3} + b}{x^{3}}}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{\sqrt{\frac{a x^{3} + b}{x^{3}}}}{{\left (a - \frac{a x^{3} + b}{x^{3}}\right )} a}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/(a+b/x^3)^(1/2),x, algorithm="giac")
[Out]
1/3*b*(arctan(sqrt((a*x^3 + b)/x^3)/sqrt(-a))/(sqrt(-a)*a) - sqrt((a*x^3 + b)/x^3)/((a - (a*x^3 + b)/x^3)*a))