3.2024 \(\int \frac {x^7}{\sqrt {a+\frac {b}{x^3}}} \, dx\)
Optimal. Leaf size=294 \[ \frac {91 b^2 x^2 \sqrt {a+\frac {b}{x^3}}}{320 a^3}-\frac {13 b x^5 \sqrt {a+\frac {b}{x^3}}}{80 a^2}+\frac {91 \sqrt {2+\sqrt {3}} b^{8/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{320 \sqrt [4]{3} a^3 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}+\frac {x^8 \sqrt {a+\frac {b}{x^3}}}{8 a} \]
[Out]
91/320*b^2*x^2*(a+b/x^3)^(1/2)/a^3-13/80*b*x^5*(a+b/x^3)^(1/2)/a^2+1/8*x^8*(a+b/x^3)^(1/2)/a+91/960*b^(8/3)*(a
^(1/3)+b^(1/3)/x)*EllipticF((b^(1/3)/x+a^(1/3)*(1-3^(1/2)))/(b^(1/3)/x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/
2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)+b^(2/3)/x^2-a^(1/3)*b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/
4)/a^3/(a+b/x^3)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)
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Rubi [A] time = 0.15, antiderivative size = 294, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{335, 325, 218} \[ \frac {91 b^2 x^2 \sqrt {a+\frac {b}{x^3}}}{320 a^3}+\frac {91 \sqrt {2+\sqrt {3}} b^{8/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{320 \sqrt [4]{3} a^3 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {13 b x^5 \sqrt {a+\frac {b}{x^3}}}{80 a^2}+\frac {x^8 \sqrt {a+\frac {b}{x^3}}}{8 a} \]
Antiderivative was successfully verified.
[In]
Int[x^7/Sqrt[a + b/x^3],x]
[Out]
(91*b^2*Sqrt[a + b/x^3]*x^2)/(320*a^3) - (13*b*Sqrt[a + b/x^3]*x^5)/(80*a^2) + (Sqrt[a + b/x^3]*x^8)/(8*a) + (
91*Sqrt[2 + Sqrt[3]]*b^(8/3)*(a^(1/3) + b^(1/3)/x)*Sqrt[(a^(2/3) + b^(2/3)/x^2 - (a^(1/3)*b^(1/3))/x)/((1 + Sq
rt[3])*a^(1/3) + b^(1/3)/x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b
^(1/3)/x)], -7 - 4*Sqrt[3]])/(320*3^(1/4)*a^3*Sqrt[a + b/x^3]*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)/x))/((1 + Sqrt[
3])*a^(1/3) + b^(1/3)/x)^2])
Rule 218
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]
Rule 325
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 335
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]
Rubi steps
\begin {align*} \int \frac {x^7}{\sqrt {a+\frac {b}{x^3}}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^9 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\sqrt {a+\frac {b}{x^3}} x^8}{8 a}+\frac {(13 b) \operatorname {Subst}\left (\int \frac {1}{x^6 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{16 a}\\ &=-\frac {13 b \sqrt {a+\frac {b}{x^3}} x^5}{80 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^8}{8 a}-\frac {\left (91 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{160 a^2}\\ &=\frac {91 b^2 \sqrt {a+\frac {b}{x^3}} x^2}{320 a^3}-\frac {13 b \sqrt {a+\frac {b}{x^3}} x^5}{80 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^8}{8 a}+\frac {\left (91 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{640 a^3}\\ &=\frac {91 b^2 \sqrt {a+\frac {b}{x^3}} x^2}{320 a^3}-\frac {13 b \sqrt {a+\frac {b}{x^3}} x^5}{80 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^8}{8 a}+\frac {91 \sqrt {2+\sqrt {3}} b^{8/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{320 \sqrt [4]{3} a^3 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}\\ \end {align*}
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Mathematica [C] time = 0.03, size = 91, normalized size = 0.31 \[ \frac {40 a^3 x^9-12 a^2 b x^6-91 b^3 \sqrt {\frac {a x^3}{b}+1} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\frac {a x^3}{b}\right )+39 a b^2 x^3+91 b^3}{320 a^3 x \sqrt {a+\frac {b}{x^3}}} \]
Antiderivative was successfully verified.
[In]
Integrate[x^7/Sqrt[a + b/x^3],x]
[Out]
(91*b^3 + 39*a*b^2*x^3 - 12*a^2*b*x^6 + 40*a^3*x^9 - 91*b^3*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/6, 1/2, 7/
6, -((a*x^3)/b)])/(320*a^3*Sqrt[a + b/x^3]*x)
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fricas [F] time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{10} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{a x^{3} + b}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7/(a+b/x^3)^(1/2),x, algorithm="fricas")
[Out]
integral(x^10*sqrt((a*x^3 + b)/x^3)/(a*x^3 + b), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{7}}{\sqrt {a + \frac {b}{x^{3}}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7/(a+b/x^3)^(1/2),x, algorithm="giac")
[Out]
integrate(x^7/sqrt(a + b/x^3), x)
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maple [B] time = 0.03, size = 2233, normalized size = 7.60 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^7/(a+b/x^3)^(1/2),x)
[Out]
1/320/((a*x^3+b)/x^3)^(1/2)/x*(a*x^3+b)/(-a^2*b)^(1/3)/a^4*(40*I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a
*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*(-a^
2*b)^(1/3)*3^(1/2)*x^6*a^3+182*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1
/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3
)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a
^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(1/2)*x^2*a^2*b^3-1
20*(a*x^4+b*x)^(1/2)*(-a^2*b)^(1/3)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2
*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*a^3*x^6-364*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(
-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3))
)^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF
((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*
3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*3^(1/2)*x*a*b^3-52*I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/
2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*(-a^2*b)^(1/3)
*3^(1/2)*x^3*a^2*b+182*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^
2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*
b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1
/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*3^(1/2)*b^3-18
2*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/
3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)
/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(
1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b^3+364*(-a^2*b)^(1/3)*(-(I*3^(1/2)-3)/(I*3^
(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x
+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))
^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(
1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x*a*b^3+156*(a*x^4+b*x)^(1/2)*(-a^2*b)^(1/3)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+
I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*a^2*b*x
^3-182*(-a^2*b)^(2/3)*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b
)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^
(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)
)*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*b^3+91*I*(a*x^4+b*x)^(1/2)*((-a*
x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1
/3))/a^2*x)^(1/2)*(-a^2*b)^(1/3)*3^(1/2)*a*b^2-273*(a*x^4+b*x)^(1/2)*(-a^2*b)^(1/3)*((-a*x+(-a^2*b)^(1/3))*(2*
a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*a*b
^2)/((a*x^3+b)*x)^(1/2)/(I*3^(1/2)-3)/((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(
-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{7}}{\sqrt {a + \frac {b}{x^{3}}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^7/(a+b/x^3)^(1/2),x, algorithm="maxima")
[Out]
integrate(x^7/sqrt(a + b/x^3), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^7}{\sqrt {a+\frac {b}{x^3}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^7/(a + b/x^3)^(1/2),x)
[Out]
int(x^7/(a + b/x^3)^(1/2), x)
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sympy [A] time = 1.46, size = 46, normalized size = 0.16 \[ - \frac {x^{8} \Gamma \left (- \frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{3}, \frac {1}{2} \\ - \frac {5}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt {a} \Gamma \left (- \frac {5}{3}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**7/(a+b/x**3)**(1/2),x)
[Out]
-x**8*gamma(-8/3)*hyper((-8/3, 1/2), (-5/3,), b*exp_polar(I*pi)/(a*x**3))/(3*sqrt(a)*gamma(-5/3))
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