3.2010 \(\int (a+\frac{b}{x^3})^{3/2} x^5 \, dx\)
Optimal. Leaf size=68 \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^3}}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{6} x^6 \left (a+\frac{b}{x^3}\right )^{3/2}+\frac{1}{4} b x^3 \sqrt{a+\frac{b}{x^3}} \]
[Out]
(b*Sqrt[a + b/x^3]*x^3)/4 + ((a + b/x^3)^(3/2)*x^6)/6 + (b^2*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(4*Sqrt[a])
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Rubi [A] time = 0.0364547, antiderivative size = 68, 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, 47, 63, 208} \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^3}}}{\sqrt{a}}\right )}{4 \sqrt{a}}+\frac{1}{6} x^6 \left (a+\frac{b}{x^3}\right )^{3/2}+\frac{1}{4} b x^3 \sqrt{a+\frac{b}{x^3}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b/x^3)^(3/2)*x^5,x]
[Out]
(b*Sqrt[a + b/x^3]*x^3)/4 + ((a + b/x^3)^(3/2)*x^6)/6 + (b^2*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(4*Sqrt[a])
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 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 \left (a+\frac{b}{x^3}\right )^{3/2} x^5 \, dx &=-\left (\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^3} \, dx,x,\frac{1}{x^3}\right )\right )\\ &=\frac{1}{6} \left (a+\frac{b}{x^3}\right )^{3/2} x^6-\frac{1}{4} b \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^2} \, dx,x,\frac{1}{x^3}\right )\\ &=\frac{1}{4} b \sqrt{a+\frac{b}{x^3}} x^3+\frac{1}{6} \left (a+\frac{b}{x^3}\right )^{3/2} x^6-\frac{1}{8} b^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^3}\right )\\ &=\frac{1}{4} b \sqrt{a+\frac{b}{x^3}} x^3+\frac{1}{6} \left (a+\frac{b}{x^3}\right )^{3/2} x^6-\frac{1}{4} b \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^3}}\right )\\ &=\frac{1}{4} b \sqrt{a+\frac{b}{x^3}} x^3+\frac{1}{6} \left (a+\frac{b}{x^3}\right )^{3/2} x^6+\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^3}}}{\sqrt{a}}\right )}{4 \sqrt{a}}\\ \end{align*}
Mathematica [A] time = 0.145227, size = 80, normalized size = 1.18 \[ \frac{1}{12} x^{3/2} \sqrt{a+\frac{b}{x^3}} \left (\frac{3 b^{3/2} \sinh ^{-1}\left (\frac{\sqrt{a} x^{3/2}}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{\frac{a x^3}{b}+1}}+x^{3/2} \left (2 a x^3+5 b\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b/x^3)^(3/2)*x^5,x]
[Out]
(Sqrt[a + b/x^3]*x^(3/2)*(x^(3/2)*(5*b + 2*a*x^3) + (3*b^(3/2)*ArcSinh[(Sqrt[a]*x^(3/2))/Sqrt[b]])/(Sqrt[a]*Sq
rt[1 + (a*x^3)/b])))/12
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Maple [C] time = 0.01, size = 3569, normalized size = 52.5 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b/x^3)^(3/2)*x^5,x)
[Out]
1/12*((a*x^3+b)/x^3)^(3/2)*x^5/a^2*(-36*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^2+18*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^2-18*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^2+2*I*3^(1/2)*(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)*x^4*a^3+5*I*3^(1/2)
*(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)*x*a^2*b+18*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^2+18*(-(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^2-18*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^2-
18*(-(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^2-6*x^4*
(a*x^4+b*x)^(1/2)*a^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)-36*(-b*a^2)^(1/3)*(-(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*a*b^2+36*(-b*a^2)^(1/3)*(-(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*a*b^2+36*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^2+18*(-b*a^2)^(2/3)*(-(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^2-18*(-b*a^2)^(2/3)*
(-(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^2-15*b*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))/(a*x^3+b)/(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)
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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((a+b/x^3)^(3/2)*x^5,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 2.46139, size = 404, normalized size = 5.94 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} \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 ) + 4 \,{\left (2 \, a^{2} x^{6} + 5 \, a b x^{3}\right )} \sqrt{\frac{a x^{3} + b}{x^{3}}}}{48 \, a}, -\frac{3 \, \sqrt{-a} b^{2} \arctan \left (\frac{2 \, \sqrt{-a} x^{3} \sqrt{\frac{a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) - 2 \,{\left (2 \, a^{2} x^{6} + 5 \, a b x^{3}\right )} \sqrt{\frac{a x^{3} + b}{x^{3}}}}{24 \, a}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b/x^3)^(3/2)*x^5,x, algorithm="fricas")
[Out]
[1/48*(3*sqrt(a)*b^2*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)) + 4
*(2*a^2*x^6 + 5*a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a, -1/24*(3*sqrt(-a)*b^2*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3 + b
)/x^3)/(2*a*x^3 + b)) - 2*(2*a^2*x^6 + 5*a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a]
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Sympy [A] time = 4.17313, size = 76, normalized size = 1.12 \begin{align*} \frac{a \sqrt{b} x^{\frac{9}{2}} \sqrt{\frac{a x^{3}}{b} + 1}}{6} + \frac{5 b^{\frac{3}{2}} x^{\frac{3}{2}} \sqrt{\frac{a x^{3}}{b} + 1}}{12} + \frac{b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} x^{\frac{3}{2}}}{\sqrt{b}} \right )}}{4 \sqrt{a}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b/x**3)**(3/2)*x**5,x)
[Out]
a*sqrt(b)*x**(9/2)*sqrt(a*x**3/b + 1)/6 + 5*b**(3/2)*x**(3/2)*sqrt(a*x**3/b + 1)/12 + b**2*asinh(sqrt(a)*x**(3
/2)/sqrt(b))/(4*sqrt(a))
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Giac [A] time = 1.27808, size = 69, normalized size = 1.01 \begin{align*} \frac{1}{12} \, \sqrt{a x^{4} + b x}{\left (2 \, a x^{3} + 5 \, b\right )} x - \frac{b^{2} \arctan \left (\frac{\sqrt{a + \frac{b}{x^{3}}}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b/x^3)^(3/2)*x^5,x, algorithm="giac")
[Out]
1/12*sqrt(a*x^4 + b*x)*(2*a*x^3 + 5*b)*x - 1/4*b^2*arctan(sqrt(a + b/x^3)/sqrt(-a))/sqrt(-a)